CHAPTER ONE

THE MOLE CONCEPT

The Mole (Mol)

A mole of a substance is the amount of substance which contains as many particles as there are
carbon particles in 12 grams of carbon -12.

Avogadro Number or Avogadro constant, NA

The number of particles in one mole of any substance is 6.02 x 1023. This number is called
Avogadro number or Avogadro constant, L.
The particles being referred to here can be atoms, molecules, ions, electrons, protons etc.
E.g
1 mole of hydrogen atoms contains 6.02 x 1023 atoms
1 mole of hydrogen molecules contains 6.02 x 1023 molecules
1 mole of hydrogen ions contains 6.02 x 1023 ions.

The following hints are useful when working with moles and Avogadro number.
Number of particles in a substance = Number of moles x Avogadro’s constant
Number of particles
Or Number of moles in a substance =
Avogadronumber

Worked examples;
1. Calculate (i) the number of particles in 0.5 moles of sodium hydroxide.
(ii) the number of moles of particles in 1.8 x 1024.

Solution
(i) Number of particles in a substance = Number of moles x Avogadro’s constant
= 0.5 x 6.02 x1023
= 3.01 x 1023 particles.
(ii) 1 mole of a substance contains 6.02 x 1023.
24 1.8 x 1024
Moles in 1.8 x 10 particles =
6.02 x 1023
= 2.99 x 100
= 3 moles

1
The terms used:
The following are the common terms used in the study of the mole concept.
 Relative Atomic Mass (RAM).
 Relative Molecular Mass (RAM).
 Molar Mass (MM).

(a) Relative Atomic Mass (RAM)

The relative atomic mass of an element is the mass of one atom of the element divide by
the mass of one-twelfth of an atom of Carbon – 12.

Mass of one atom of an element

I.e. Relative Atomic Mass =
1
x Mass of one atom of Carbon 12
12
(b) Relative Molecular Mass (RMM)
The Relative molecular mass of an element or compound is the mass of one molecule
divided by the mass of one-twelfth of an atom of Carbon-12.

Mass of one atom of an element

Relative Molecular Mass =
1
x Mass of one atom of Carbon 12
12
Example
1. Given, H=1.008, calculate
(i) RAM
(ii) RMM of hydrogen

Solution
Mass of one atom of an element
(i) Relative Atomic Mass =
1
x Mass of one atom of Carbon 12
12
Mass of one atom of hydrogen
=
1
x Mass of one atom of Carbon 12
12
1.008
=
1
x 12
12
= 1.008
=1

2
 Mass of one molecule of a subs tan ce
(ii) Relative Molecular Mass =
1
x Mass of one atom of Carbon 12
12
Mass of one molecule of hydrogen
=
1
x Mass of one atom of Carbon 12
12
2 x 1.008
=
1
x 12
12
= 2 x 1.008
= 2.016
=2

(c) Molar Mass (MM)

The molar mass of a compound is the mass of one mole of a compound.
It is got by getting the sum of the relative atomic masses of the respective elements
in the compound.

The SI unit is - Kg/mol or Kgmol-1

The common unit used is - g/mol or gmol-1

Examples
1. Calculate the molar masses of the following compounds.
(i) H2SO4
(ii) Na2CO3
(iii) (NH4)2SO4 (H = 1, S = 32, O = 16, N = 14, C = 12, Na = 23)

Solution
(i) The molar mass of sulphuric acid, H2SO4 = (2 x 1) + 32 + (4 x 16)
= 2 + 32 + 64
= 98 g/mol
(ii) The molar mass of sodium carbonate, Na2CO3,
= (2 x 23) + 12 + (3 x 16)
= 46 + 12 + 48
= 106 g/mol
3
(iii) The molar mass of Ammonium sulphate = (NH4)2SO4
= 2[14 + (1 x 4)] + 32 + (4 x 16)
= (2 x 18) + 32 + 64
= 36 + 32 + 64
= 132 g/mol

Relative Atomic Masses of the common Elements

Name Symbol RAM

Aluminum Al 27

Calcium Ca 40

Carbon C 12

Chlorine Cl 35.5

Copper Cu 63.5

Hydrogen H 1

Iodine I 127

Iron Fe 56

Lead Pb 207

Magnesium Mg 24

Nitrogen N 14

Oxygen O 16

Silver Ag 108

Sodium Na 23

Sulphur S 32

Zinc Zn 65

4
Calculations from Formulae and Equations
(a) To Calculate the Formula Mass
 Write the chemical formula of the compound.
 Get the sum of the relative atomic masses of the element present in the compound.

Examples
1. Calculate the formula masses of the following compounds
(a) Sulphuric acid.
(b) Calcium Nitrate.
(c) Ammonium Nitrate
(H = 1, S = 32, O = 16, N = 14, Ca = 40)

Solution
(a) Formula of Sulphuric acid = H2SO4
= (2 x 1) + 32 + (4 x 16)
= 2 + 32 + 64
= 98 g

(b) Formula of Calcium Nitrate = Ca (NO3)2

= 40 + 2(14 + 3 x 16)
= 40 + 2 x 62
= 40 + 124
= 164 g

(c) Formula of Ammonium Sulphate = (NH4)2SO4

= 2(14 + 4 x 1) + 32 + (4 x 16)
= 2(14 + 4) + 32 + 64
= (2 x 18) + 96
= 36 + 96
= 132 g

5
(b) To Calculate the Percentage by mass of an Element in a compound

The percentage by mass of an element in a compound can be calculated by using the

formula:
Mass of Element in a Compound
Percentage by Mass of an element = x 100
Formula Mass

Examples
1. Calculate the percentage by mass of oxygen in Calcium oxide.
(Ca = 40, O = 16)
Solution
Formula of Calcium oxide = CaO
Mass of oxygen in CaO = 16
Formula mass of CaO = 40 + 16
= 56

Mass of Element in a Compound

Percentage by mass of an element = x 100
Formula Mass
Mass of oxygen
= x 100
Formula mass
16
= x 100
56
= 28.57 %

(c) To Calculate the Percentage Composition by mass of a compound

Examples
1. Calculate the percentage composition by mass of sodium hydroxide.
(Na = 23, O = 16, H = 1)
The formula mass of NaOH = 23 + 16 + 1
= 40 g
Mass of sodium
Percentage by mass of sodium = x 100
Formula mass
23
= x 100
40
= 57.5%

6
 Mass of oxygen
Percentage by mass of oxygen = x 100
Formula mass
16
= x 100
40
= 40 %

Mass of hydrogen
Percentage by mass of hydrogen = x 100
Formula mass
1
= x 100
40
= 2.5%

Or Percentage by mass of hydrogen = 100% – (57.5 + 40) %

= (100 - 97.5) %
= 2.5%

2. Calculate the percentage composition by mass of the following compounds.

(a) Calcium oxide.
(b) Sodium sulphate.
(c) Ammonium sulphate.
(Ca = 40, O = 16, Na = 23, S = 32, N = 14, H = 1)

Solution
(a) Formula of Calcium oxide = CaO
Mass of oxygen in CaO = 16
Formula mass of CaO = 40 + 16
= 56

Mass of oxygen
Percentage by mass of oxygen = x 100
Formula mass
16
= x 100
56
= 28.57 %

7
 Mass of calcium
Percentage by mass of calcium = x 100
Formula mass
40
= x 100
56
= 71.43 %

Or Percentage by mass of calcium = (100 – 28.57) %

= 71.43 %

(b) Formula of Sodium sulphate = Na2SO4

Formula mass of Sodium sulphate = (2 x 23) + 32 + (4 x 16)
= 46 + 32 + 64
= 142

Mass of sodium in Na2SO4 = 2 x 23

= 46

Mass of sodium
Percentage by mass of sodium = x 100
Formula mass
46
= x 100
142
= 32.39 %

Mass of sulphur
Percentage by mass of sulphur = x 100
Formula mass
32
= x 100
142
= 22.54 %

Mass of oxygen
Percentage by mass of oxygen = x 100
Formula mass
64
= x 100
142
= 45.07 %

Or Percentage by mass of oxygen = 100% - (32.39% + 22.54%)

= 100% - 54.93%
= 45.07 %
8
(d) To calculate the Number of moles for a given mass of an element or a compound

Mass of element
(i) Number of moles of an element =
RAM

Mass of compound
(ii) Number of moles of a compound =
RFM

Examples
1. Calculate the number of moles in 2.3 g of sodium (Na = 23).

Solution
Mass of element
Number of moles of an element =
RAM
2 .3
=
23
= 0.1 moles

2. Calculate the number of moles in 50.3g of sodium carbonate.

(Na = 23, C = 12, O = 16)

Solution
Mass of sodium carbonate = 50.3 g,
RFM of sodium carbonate = (2 x 23) + 12 + (3 x 16)
= 46 + 12 + 48
= 106 g

Mass of compound
Number of moles of a compound =
RFM
50.3
=
106
= 0.5 moles

(e) To Calculate Empirical formula and Molecular formula

(i) Empirical formula

Empirical formula of a compound is the formula which shows the simplest
ratio of the different atoms in the compound.

Or Empirical formula of a compound is the formula which expresses its

composition by mass and the ratio of the numbers of the different atoms
present in the compound.

9
 (ii) Molecular formula
Molecular formula of a compound is the formula which expresses the actual
number of each kind of atoms present in one molecule of the compound.

Note: In some compounds, the empirical formula is the same as the molecular formula.
While in others they are different as shown in the table below.

Name of compound Molecular formula Empirical formula

Methane CH4 CH4

Ethane C2H6 CH3

Ethene C2H4 CH2

Propene C3H8 C3H8

Glucose C6H12O6 CH2O

Benzene C6H6 CH

Butane C4H10 C2H5

Water H2O H2O

Examples
 To calculate empirical formula
1. A compound contains 40% carbon, 6.7% hydrogen the rest being oxygen. Calculate
the empirical formula of the compound. (C = 12, H = 1, O = 16)
Percentage of oxygen = [100 – (40 + 6.7)] % = 53.3%

Elements C H O
Percentages by mass 40 6.7 53.33
40 6.7 53.3
Moles
12 1 16
0.33 6.7 3.33
3.33 6.7 3.33
Mole ratios
3.33 3.33 3.33
1 2 1

Hence the empirical formula of the compound is CH2O

10
2. (a) A hydrated salt has the following percentage composition: iron
20.14%; sulphur 11.53%; oxygen 23.01% and water 45.32%.
Find the simplest formula.
(Fe = 56, S = 32, O = 16, RFM of water = 18

Elements Fe S O Water

Percentages 20.14 11.53 23.01 45.32

20.14 11.53 23.01 45.32

Moles
56 32 16 18

0.3596 0.3603 1.4381 2.5178

0.3596 0.3609 1.4381 2.5178

Mole ratios
0.3596 0.3596 0.3596 0.3596

1 1.0019 3.9991 7.0222

1 1 4 7

Therefore the simplest formula = FeSO4.7H2O

3. A compound W contains 3.6 g of P and 1.4 g of Q. Calculate the empirical formula

of W. (P = 24, Q = 14)

Solution
Elements P Q
Mass 3.6 1.4
3 .6 1 .4
Moles
24 14
0.15 0.1
0.15 0 .1
Mole ratios
0.1 0 .1
1.5 1
2 x 1.5 2x1
3 2
The empirical formula of W is P3Q2

11
  To calculate molecular formula of a compound

The molecular formula of a compound is calculated from:

(i) (Empirical formula)n = Relative Molecular Mass
(ii) (Empirical formula)n = 2 x Vapour density

Where:  n = a whole number 1, 2, 3, 4, ….

 Vapour density is the ratio of the mass of certain
volume of the gas or vapour to the mass of the same volume of
hydrogen at the same temperature and pressure.

Example
1. A gaseous hydrog carbon contains 80% carbon by mass. If the molecular mass of
the gas is 30 g, find the molecular formula.

Solution:
Percentage of H = (100 – 80) % = 20%
Elements C H
Percentages 80 20
80 20
Moles
12 1
6.7 20
6.7 20
Mole ratios
6.7 6.7

1 3

The empirical formula is CH3

(Empirical formula)n= Relative Molecular Mass

(CH3)n = 30
(12 + 1 x 3 )n = 30
15n
= 30
15
30
n =
15
=2
(CH3)2 = C2H6

12
2. A compound X contains 85.7% carbon and 14.3% hydrogen. If the vapour density
of the gas is 14, calculate:
(a) The simplest formula of X.
(b) (i) Determine the molecular formula of X.
(ii) Write the name of X.
(iii) Write the structural formula of X.

Solution
(a) Elements C H
Percentages 85.7 14.3

85.7 14.3
Moles
12 1
7.14 14.3

7.14 14.3
Mole ratios
7.14 7.14

1 2

The empirical formula is CH2

(b) (i) (Empirical formula) n = 2 x Vapour density

(CH2) n = 2 x 14
[12 + (1 x 2)] n = 28
14n
= 28
14
28
n =
14
=2
(CH2)2 = C2H4

The molecular formula is C2H4

(ii) X is ethene.

(iii) CH2 = CH2

13
(d) Calculations from Chemical Equations
 Write a balanced equation for the reaction or use the reaction equation written
on the question paper.
 Take the numbers in front of each formula or symbol to represent moles.
 Convert the moles to masses if necessary.
 Compare the masses of the compounds concerned.

(i) To calculate mass of a reactant or product.

1. Excess hydrogen gas was passed over 20 g of heated copper (II) oxide.
Calculate: (a) the mass of copper formed.
(b) the mass of hydrogen used up in the reaction.

Solution:
CuO (s) + H2 (g)  Cu (s) + H2O (g)
1 mol 1 mol 1 mol 1 mol
80g 2g 64 g

(a) 80 g of CuO forms 64 g of copper.

64
1 g of CuO forms g of copper.
80
64
20 g of CuO forms x 20 = 16 g of copper.
80

(b) 80 g of CuO requires 2 g of hydrogen.

2
1 g of CuO requires g of hydrogen.
80
2
20 g of CuO requires x 20 = 0.5 g of hydrogen
80

2. 5.00 g of zinc carbonate was heated to a constant mass.

(a) State what was observed.
(b) Write an ionic equation for the reaction that took place.
(c) Calculate the apparent loss in mass.

14
 Solution:
(a) The carbonate decomposed to give:
 a colourless gas that turns lime water milky and
 a solid residue which is yellow when hot and turned to white on cooling.

(b) CO32- (s)  O2- (s) + CO2 (g)

(c) RFM of ZnCO3 = 65 + 12 + (3 x 16)

= 65 + 12 + 48
= 125 g
125 g of zinc carbonate gives a loss in mass of 44 g of carbon dioxide.
44
1 g of zinc carbonate gives a loss in mass of g of carbon.
125
44
5 g of zinc carbonate gives a loss in mass of x 5 = 1.76 g
125
3. 5.00 g of calcium carbonate was heated until there was no change in mass.
(a) (i) Write equation for the reaction that took place.
(ii) Calculate the mass of the solid left.
(b) The residue was shaken with water and the product was tested with blue
litmus paper. State what was observed.
(Ca = 40, C = 12, O = 16)

Solution
(a) (i) CaCO3 (s)  CaO (s) + CO2 (g)

(ii) RFM of CaCO3 = 40 + 12 + (3 x 16)

= 40 + 12 + 48
= 100 g

RFM of CaO = 40 + 16
= 56 g

100 g of calcium carbonate produces 56 g of calcium oxide.

56
1 g of calcium carbonate produces g of calcium oxide.
100
56
5.00 g of calcium carbonate produces x 5 = 2.8 g
100
(c) No observable change (The blue litmus paper remained blue because the
resultant solution is basic).

15
 (e) Calculations involving Gas volumes
To calculate volume of a reactant or a product (Calculations involving Gas
Volumes)

Summary of data on gas volumes

1. The volume of all gases varies with temperature and pressure.
2. Therefore, always the standards of temperature and pressure must be stated.
The standard chosen are:
stp - Standard temperature (0 C or 273 K) and pressure
760 mmHg or 1 atmosphere.
rt - room temperature.
3. Molar gas Volume (MGV).
The volume occupied by one mole of any gas at:
stp = 22.4 dm3 or 22,400 cm3.
rt = 24 dm3 or 24,00 cm3.
3. Equal volume of all gases at the same temperature and pressure contain the
same number of molecules (Avogadro’s law).
The following statement is also true:
Equal numbers of molecules of different substances in the gaseous state
occupy the same volume, provided all volume measurement take place under
the same temperature and pressure.
Volume of gas
4. Number of moles of a gas =
Molar Gas Volume

EXAMPLES
1. When 0.107 g of ammonium chloride was heated with calcium hydroxide, a gas was
evolved. (a) Write equation for the reaction.
(b) Calculate the volume of the gas that was evolved at room
temperature. (1 mole of gas occupies 24 dm3 at rt)
Solution
(a) Ca(OH)2 (s) + 2NH4Cl (s)  CaCl2 (s) + 2 H2O (l) + 2 NH3 (g)
(b) Mass of NH4Cl that reacted = 2[14 + (1 x 4) + 35.5]
= 2(14 + 4 + 35.5)
= 2 x 53.5
= 107 g
107 g of ammonium chloride liberates (2 x 24) dm3 of ammonia.
2 x 24
1 g of ammonium chloride liberates dm3 of ammonia.
107
2 x 24
0.107 g of ammonium chloride will liberate x 0.107 = 0.048 dm3.
107
16
2. 6.0 g of carbon are burnt completely in air at s.t.p. Calculate the volume of carbon dioxide
evolved. [C = 12, Molar volume = 22400cm3 at s.t.p.]
C (s) + O2 (g)  CO2 (g)

12 g of carbon liberates 22400 cm3 of carbon dioxide.

2 2400
1 g of carbon liberates cm3 of carbon dioxide.
12
2 2400
6.0 g of carbon will liberate x 6.0 = 11200 cm3
12
3. Zinc nitrate decomposes on heating according to the equation
2Zn(NO3)2(s)  2ZnO(s) + 4NO2(g) + O2(g).
Calculate the maximum volume of oxygen evolved in this reaction at s.t.p. when
7.56 g of zinc nitrate is heated.
[Zn = 65, N = 14, O = 16, molar gas volume at s.t.p. = 22.4 dm3]
Mass of Zn(NO3)2 that reacted = 2[65 + 2(14 + 3 x 16)]
= 2 [65 + 2(14 + 48)]
= 2[65 + (2 x 62)]
= 2(65 + 124)
= 2 x 189
= 378 g

378 g of zinc nitrate liberates 22.4 dm3 of oxygen.

2 2 .4
1 g of zinc nitrate liberates dm3 of oxygen.
378
2 2 .4
7.56 g of zinc nitrate will liberate x 7.56 = 0.448 dm3
378
4. Ammonia is oxidised by copper (II) oxide according to the equation
2NH3 (g) + 3CuO (s)  3H2O (l) + N2 (g) + 3Cu (s).
Calculate the volume of ammonia that will be oxidised by 6.0 g of copper (II) oxide at
s.t.p.? (1 mole of a gas at s.t.p. occupies 22400 cm3)
Mass of CuO that oxidised ammonia = 3(64 + 16)
= 3 x 80
= 240 g
240 g of CuO oxidised 2 x 22.4 dm3 of ammonia.
2 x 22.4
1 g of CuO oxidises dm3 of ammonia.
240
2 x 22.4
6.0 g of CuO will oxidise x 6 = 1.12 dm3
240

17
 GAS LAWS
Points to know:
Gas laws deal with the influence of physical factors such as:
 Temperature and
 Pressure - on the behaviour of gases.

Recall that:
The gas molecules are spread far apart compared to the molecules of liquids and solids.
The gas molecules are always in a continuous random motion (Brownian motion)

There are three gas laws, namely:

Boyle’s law,
Charles law and
Pressure law.

(a) Boyle’s law.

Boyle’s law states that:
The volume of a fixed mass of gas is inversely proportional to the pressure at constant
temperature.

Boyle’s law implies that at constant temperature as the pressure of a given mass of a gas
increases, the volume decreases.

Mathematically, Boyle’s law is expressed as;

1
p  or PV = Constant
V
That is when pressure of a gas changes from initial value P1 to final value P2, its volume
changes from initial value V1 to final value V2.

In general, this is written as; P1V1 = P2V2

Example:
The volume of a fixed mass of gas at constant temperature is 20 cm3 when the pressure is
75cmHg. Find the new pressure when the volume increases to 50 cm3.
Solution: P1 = 75cmHg, P2 =?, V1 = 20 cm3, V2 = 50 cm3
P1V1 = P2V2
75 x 20 = P2 x 50
75 x 20
P2 =
50
 P2 = 30 cmHg

18
(b) Charles’s law
Charles’s law states that:
The volume of a fixed mass of gas at constant pressure is directly proportional to the
absolute temperature.

Where the absolute temperature is the temperature at which a gas occupies zero volume.

Charles law implies that the volume of a fixed mass of a gas increases or decreases in the
same proportion as the absolute temperature provided the pressure is kept constant.

Mathematically, Charles law is expressed as:

V
V  T or = Constant
T
That is at a constant pressure, when the volume of a gas changes from V1 to V2, the
temperature changes from T1 to T2.

V1 V
In general, this is written as;  2
T1 T2

Example
The volume of a fixed mass of gas at constant pressure is 50 cm3 at a temperature of
27 oC. What will the volume be at 42 oC.

Solution: T1 = 27 C = 273 + 27 = 300 K, T2 = 42 C = 273 + 42 = 315 K

V1 = 50 cm3, V2 = ?
V1 V 50 V2 50  315
 2    V2 =
T1 T2 300 315 300
 V2 = 52.5 cm3
(c) Pressure Law:
Pressure Law states that:
The pressure of a fixed mass of a gas is directly proportional to its absolute temperature if
the volume is kept constant.

Mathematically, pressure law is expressed as:

P
P  T or = Constant
T
That is at a constant volume, when the pressure of a gas changes from initial value P1 to
final value P2, the temperature changes from initial value T1 to final value T2.

19
 P1 P2
In general, this is written as; 
T1 T2

Example:
The pressure of a given mass of gas at constant volume at a temperature of 100 oC is 300
mmHg. What will the pressure be if the temperature falls to 27 oC.

Solution: T1 = 273 + 100 = 373K, T2 = 273+ 27 = 300K

P1 = 300 mm Hg, P2 =?
P1 P 300 P2
 2  
T1 T2 373 300
300x300
P2 =
373
P2 = 241.3 mmHg

Equation of state or Ideal gas equation

The three equations can be combined into one equation.
If a fixed mass of a gas of volume V1 exerts a pressure P1 at absolute temperature T1, the
P1V1
expression may be written as: = Constant. ………………….. 1
T1
Now suppose the same mass of a gas has a volume V2 and exerts a pressure P2 at absolute
temperature T2, then new expression becomes:
P2V2
= Constant. ………………….. 2
T2
Equating equation (1) to equation (2), we have the equation of state expressed as:
P1V1 P2V2
=
T1 T2
 This equation holds for ideal gases under all conditions.

Examples
1. Air in a 2.5 litre vessel at 127 C exerts a pressure of 3 atm. Calculate the pressure that the
same mass of air would exert if contained in a 4 litre vessel at – 43 C.

Solution: T1 = 273 + 127 = 400K, T2 = 273 – 43 = 230K

P1 =3 atm, P2 =? atm
V1 = 2.5 dm3 V2 = 4 dm3

20
 P1V1 P2V2 3 x 2.5 P2 x 4
  
T1 T2 400 230

3 x 2.5 x 230
P2 
400 x 4
P2 = 1.078 atm

2. The volume of a gas is 90 cm3 at 25 oC and 780 mmHg pressure. Calculate the volume of
the gas at s.t.p.
Solution: T1 = 250 oC = 25 + 273 = 298K, T2 = 0 oC = 273K (at stp)
P1 = 780 mmHg, P2 = 760 mmHg (at stp)
V1 = 90 cm3, V2 = ?
P1V1 P2V2 780 x 90 760 x V2
=  =
T1 T2 298 273
780 x 90 x 273
V2 =
298 x 760
V2 = 84.6 cm3

Graham’s law of diffusion of gases

Graham’s law of diffusion of gases states that:
Under the same conditions of temperature and pressure, the rate of diffusion of a gas is
inversely proportional to the square root of its density.

The following formulae are used when using Graham’s law.

Rate of diffusion of gas A Molecular mass of gas B

1. =
Rate of diffusion of gas B Molecular mass of gas A

Time taken by gas A to diffuse Molecular mass of gas B

2. =
Time taken by gas B to diffuse Molecular mass of gas A

Where: A and B are gases of equal volumes.

Examples
1. Equal volumes of carbon monoxide and carbon dioxide are allowed to diffuse
through the same medium. Calculate the relative rate of diffusion of carbon
monoxide. (C = 12, O = 16).
Solution: Molecular mass of CO = 12 + 16
= 28 g
Molecular mass of CO2 = 12 + (2 x 16)
= 44 g
21
 Rate of diffusion of CO Molecular mass of CO2
=
Rate of diffusion of CO2 Molecular mass of CO
44
=
28
= 1.254
 Carbon monoxide diffuses 1.254 times faster than carbon dioxide.

2. If it takes 20 s for 200 cm3 of oxygen gas to diffuse across a porous material. How
long will it take an equal volume of sulphur dioxide to diffuse across the same
porous material? (S =32, O = 16)
Solution:
Molecular mass of O2 = 2 x 16 = 32 g, Time taken by O2 = 20 s
Molecular mass of SO2 = 32 + (2 x 16) = 64 g, Time taken by SO2 = ?
Time taken by O2 to diffuse Molecular mass of O2
=
Time taken by SO2 to diffuse Molecular mass of SO2
20 32
=
Time taken by SO2 to diffuse 64
1
=
2
1
=
2
1
=
2
Time taken by SO2 to diffuse = 20 2
 Time taken by SO2 to diffuse = 28.3 s

3. Determine the molecular mass of gas Y when oxygen diffuses 1½ times faster than
gas Y. (O = 16)
Solution: Rate of diffusion of O2 = 1½, Molecular mass of O2 = 32 g
Rate of diffusion of Y =1 Molecular mass of Y = ?
Rate of diffusion of gas Y Molecular mass of O2
=
Rate of diffusion of gas O2 Molecular mass of Y
1 32
=
1½ Molecular mass of Y
12 32
=
1.5 2 Y
Y= 1.52 x 32
= 2.25 x 32
The molecular mass of Y = 72 g
22
4. If it takes 30 seconds for 100 cm3 of carbon dioxide to diffuse across a porous plate.
How long will it take 150 cm3 of nitrogen dioxide to diffuse across the same plate
under similar conditions? (C = 12, N = 14, O = 16)

Solution:
Volume of CO2 = 100 cm3, Volume of NO2 = 150 cm3
Molecular mass of CO2 = 44 g, Molecular mass of NO2 = 46 g
Time taken by CO2 = 30 s, Time taken by NO2 = ?
v 100
Rate of diffusion of CO2 = = = 3.33 cm3s-1
t 30
Rate of diffusion of CO 2 Molecular mass of NO2
=
Rate of diffusion of NO2 Molecular mass of CO2

3.33 46
=
Rate of diffusion of NO2 44

3.33
= 1.0225
Rate of diffusion of NO2
3.33
Rate of diffusion of NO2 =
1.0225
= 3.26
 Rate of diffusion of NO2 = 3.26 cm3s-1

Now the time taken by NO2 to diffuse is calculated from the formula:
Volume of gas
Rate of diffusion of NO2 =
Time taken
150 cm 3
3.26 cm3s-1 =
Time taken
150 cm 3
Time taken = = 46 s
3.26 cm 3 s 1

Points to recall:
 Diffusion is the “spreading” of molecules from a region of high concentration to a
region of low concentration through a medium.
 Diffusion occurs faster in gases than in liquids.
 Gases with low densities diffuse faster than those with higher densities.

23
 Examples
1. (1990 Q.6)
(a) Smoke was put in a glass-cell and viewed under a microscope.
(i) State what was observed.
(ii) Explain the observation in (i).
(b) One piece of cotton wool was soaked in concentrated ammonia solution and another
in concentrated hydrochloric acid. The two pieces of cotton wool were placed in a
glass tube as shown in figure below.

(i) Write the formula of the substance that formed the white ring.
(ii) Explain why the white ring is formed in position A and not in the middle of
the tube.
Solution:
(a) (i) Smoke particles were seen moving randomly in the glass cell.
(ii) The smoke particles collided with air molecules, hence the random motion.
(b) (i) NH4Cl
(ii) Ammonia is less dense than hydrogen chloride gas and therefore diffuses
faster than hydrogen chloride hence the formation of the white ring near the
cotton wool soaked in concentrated hydrochloric acid.
2. (1998 Q.3)
A crystal of potassium manganese (VII) was placed at the corner in a trough of water as
shown in figure 1 below and the experiment was allowed to stand for about 30 minutes.

(a) State what was observed after 30 minutes.

(b) Name the process that occurred.
(c) State the purpose of the experiment.
Solution
(a) Purple colour of potassium manganate (VII) spread through out the liquid.
(b) Diffusion.
(c) To show that water molecules are in a constant random motion. And so collide with
the particles of potassium manganate (VII).

24
 The Mole Concept and Volumetric Analysis

Stoichiometry:
- Deals with ratios in which substances combine with one another when chemical reactions
occur.
- I t also deals with composition of chemical substances i.e. ratios by mass in which elements
are combined in a given compound.
- The fundamental unit of Stoichiometry is the mole.

1. Volumetric Analysis:
Volumetric analysis is the analytical process to determine quantities of compounds
(usually acids and alkalis) which involves measurement of volumes of solutions, using
pipette, burette and ( for approximate measurements) measuring cylinder.

(a) Uses of Volumetric Analysis:

The object of volumetric analysis is to:
(i) Measure concentration of a given compound either in g/l or mol/l.
(ii) Determine the relative atomic mass of an element in a compound.
(iii) Determine the number of moles of water of crystallization in a hydrated
compound.
(iv) Determine the percentage purity or impurity of a compound in an impure
sample.
(v) Determine the basicity of an acid.
(vi) Determine the reaction ratio.

(b) Terms Used.

(i) Concentration:
Concentration is a measure of the amount of a substance dissolved in a
known volume of a solution.

Units: It is measured in:

- mol/dm3 (moles per cubic decimeter)
- mol/l (moles per litre)
- g/dm3 (grams per cubic decimeter)
- g/l (grams per litre)
- g/cm3 (grams per cubic centimeter)
Note: 1 dm3 = 1 litre
1 dm3 = 1,000 cm3
1l = 1,000 cm3
25
 (ii) Standard Solution:
Standard solution is a solution of known concentration. Or - is one for which the
concentration is known.

(iii) Molar Solution:

Molar Solution is one, which contains one mole of a substance in a litre or a cubic
decimeter.

(iv) Molarity:
Molarity is the number of moles per litre.
Or Molarity is the number of moles per cubic decimeter.

(v) Molar Mass or Relative Formula Mass (RFM)

Molar mass is the mass of one mole of a compound.
RFM = the sum of the relative atomic masses of the elements present in the
compound.

The Table below shows the RFMs of the common compounds used in Volumetric analysis.
(H = 1, Na = 23, Cl = 35.5, O = 16, C = 12)

Compound Name of Compound Chemical formula Molar Mass(RFM)/g

Hydrochloric Acid HCl 36.5
Sulphuric Acid H2SO4 98
Acid Anhydrous Oxalic Acid H2C2O4 90
Hydrated Oxalic Acid (COOH)2 126
H2C2O4.2H2O
(COOH)2.2H2O
Sodium Hydroxide NaOH 40
Base Anhydrous sodium carbonate Na2CO3 106
Hydrated Sodium Carbonate Na2CO3.10H2O 286

(b) Relationship Between Molarity, Concentration and RFM

Concentration in g / l
Molarity =
RFM

Concentration in g / dm3
Or Molarity =
RFM

26
2. (a) Titration:
Titration is the estimation of concentration of an acid or alkali in solution by
reacting it with a standard alkali or acid respectively.

- In the analysis, two aqueous solutions are used.

- The concentration of one solution is always known and the other unknown.

(b) Titration Process:

- The titration process involves running one solution from burette on to the other
solution of a fixed volume (usually 20 or 25 cm3 ) in which 2-3 drops of
indicator are added in a conical flask until the two have just reacted completely.

- This is noticed when the colour of the indicator just changes.. This shows that
the neutral point (end point) of the titration has been reached and the volume of
the solution used from the burette is accurately noted.

- The volume used from the burette is calculated by using the formula:
Volume used = (Final burette reading) - (Initial burette reading)
- The value is always written in two places of decimal.
- The procedure is repeated 3 - 4 times and the values are entered in a table
usually drawn in advance on the question paper.

3. Indicators:
An indicator is substance that has different colour in acid and alkali solutions.

The common indicators used are:

- Phenolphthalein indicator.
- Methyl Orange.
The table below shows the colours of the indicators in the laboratory in different media.

Colour in
Indicator Neutral Acid Alkali
Phenolphthalein Colourless Colourless Red
Methyl Orange Orange Red Yellow

27
4. CALCULATIONS:

Steps followed in the calculations:

The steps followed in the calculations involve the following:

(i) Write the volume of the pipette used in two places of decimal in the space usually
provided on the question paper.
(ii) Find the average volume, using the consistent results, of the solution used from the
burette.
(iii) Write a balanced reaction equation or ionic equation for the reaction that took place.
Or use the written equation of the reaction on the question paper.
However, sometimes he equations of the reaction is written on the qustion paper. In
such a case, refer to the equation.

(iv) From the balanced equation, extract the mole ratio and express it in full.
e.g. 1 mole of acid : 2 mole of the base or vice versa.
(v) Always start the calculation with the solution of known concentration and volume.
(vi) Calculate the number of moles of the other solution using the mole ratio.

NB: Do NOT use the "titration equation".

M aV a M V M aV a n
= b b Or = a
na nb M bV b nb

Where: M a and Mb = the molarities of the acid and the base respectively.
Va and Vb = the volumes of the acid and the base respectively.
na and nb = the number of moles of the acid and the base respectively.

28
 EXAMPLES
1. Determination of Molarity of a Solution (Concentration mol/l)
You are provided with BA1 and BA2 solutions.
BA1 is a solution of Hydrochloric acid.
BA2 is a solution of anhydrous sodium carbonate that contains 7.2g/dm3.
(Na = 23, C = 12, O = 16)

You are required to determine the molarity of BA1.

Procedure:
Pipette 25 cm3 or 20 cm3 of BA2 into a conical flask. Add two drops of the indicator and
titrate it with BA1 from burette. Repeat the procedure until you get consistent results.

Enter the results in the table below.

Sample results
Volume of pipette used 20.00. cm3
Burette Readings 1 2 3 4
Final Burette Reading (cm3) 33.00 31.80 32.30 30.80
Initial Burette Reading (cm3) 1.50 2.00 2.50 1.00
Volume of BA1 used (cm3) 31.50 29.80 29.80 29.80

Titre values used to find the average volume = 29.80, 29.80, 29.80
29.80  29.80  29.80
Average volume BA1 used =
3
89.4
=
3
= 29.80 cm3
Questions:
(a) Write the equation for the reaction.
2HCl (aq) + Na2CO3 (aq)  2NaCl (aq) + CO2 (g) + H2O (l)
Mole ratio: 2 mol of acid : 1 mol of base
(b) Calculate:
(i) the molarity of BA2.
Concentration = 7.2 g/l, RFM of Na2CO3 = (2 x 23) + 12 + (3 x 16)
=106 g

29
 Concentration in g / l
Molarity =
RFM
7 .2
=
106
= 0.07M

(ii) The number of moles of BA2 that reacted with BA1.

Solution: Molarity = 0.07M, Vol = 20 cm3, Number of moles = ?

0.07 moles of BA2 are contained in 1000 cm3 of solution.

The number of moles contained in 20 cm3 of BA2 solution
0.07 x 20
=
1000
= 0.0014 moles

(iii) The number of moles of BA1 that reacted with BA2.

From the mole ratio, 2 moles of acid : 1 mole of base, the number of moles of BA1
that reacted with BA2 = 2 x the number of moles of BA2
= 2 x 0.0014
= 0.0028 moles

(iv) The molarity of BA1.

Solution: Molarity = ?, Number of moles = 0.0028, Vol = 29.80 cm3

0.0028 moles of BA1 are contained in 29.80 cm3 of solution.

The number of moles contained in 1000 cm3 of BA1 solution
0.0028 x 1000
=
29.80
= 0.09
= 0.1M

30
2. Determination of Atomic mass of an Element in a compound

You are provided with BA1 and BA2 solutions.

BA1 is a solution 0.1M Hydrochloric acid.
BA2 is a solution obtained by dissolving 5.3 g of a metal carbonate, M2CO3, per litre.
(C = 12, O = 16)
You are required to determine the atomic mass of the element M in the carbonate.

Procedure:
Pipette 25 cm3 or 20 cm3 of BA2 into a conical flask. Add two drops of methyl orange
indicator and titrate it with BA1 from burette. Repeat the procedure until you get consistent
results.
Enter the results in the table below.

Sample results
Volume of pipette used 20.00. cm3

Burette Readings 1 2 3 4
Final Burette Reading (cm3) 21.70 30.20 21.90 31.30
Initial Burette Reading (cm3) 1.50 21.70 2.50 21.90
Volume of BA1 used (cm3) 20.20 19.50 19.40 19.40

Titre Values used to find the average volume= 19.50, 19.40, 19.40
19.50  19.40  19.40
Average volume of BA1 used =
3
58.3
=
3
= 19.40 cm3
Questions:
(a) Write the equation for the reaction.
2HCl (aq) + M2CO3 (aq)  2MCl (aq) + CO2 (g) + H2O (l)

31
(b) Calculate:
(i) The number of moles of BA1 that reacted with BA2.
Volume of BA1 (acid) = average volume
= 19.40 cm3
Molarity of BA1 = 0.1M, Number of moles = ?
0.1 moles of acid are contained in 1000 cm3 of solution.
The number of moles contained 19.40 cm3 of soltuion
19.40 x 0.1
=
1000
= 0.00194
(ii) The number of moles of BA2 that reacted with BA1.
From the mole ratio: 2 moles of acid : 1 mole of base
The number of moles of M2CO3 that reacted with BA1
= ½ of number of moles of the acid
= ½ x 0.00194
= 0.00097
= 0.001
(iii) The molarity of BA2.
Volume of BA2 (Base) = 20.00 cm3, Number of moles = 0.001,
Molarity of BA2 = ?,

0.001 moles of BA2 are contained in 20 cm3 of solution.

The number of moles contained 1000 cm3 of soltuion
0.001 x 1000
=
20
= 0.05 M
(c) Determine:
(i) The formula mass of BA2.
Solution: Molarity of BA1 = 0.05M, Concentration = 5.3 g/l, RFM = ?
Concentration in g / l
Molarity =
RFM
Concentration in g / l
RFM =
Molarity
5.3
=
0.05
= 106

32
 (ii) The atomic mass of M.

2M + 12 + 3 x 16 = 106
2M + 12 + 48 = 106
2M + 60 = 106
2M = 106 - 60
2M = 46
46
M =
2
M = 23

3. Determination of the number of moles of water of crystallization in a hydrated

compound

Hydrated compounds are compounds that contain water of crystallization.

The hydrated compounds used for this purpose are:
- Na2CO3.10H2O and
- H2C2O4.2H2O

In the calculations, the figures 2 and 10 and the symbols Na and C2O4 in the above
compounds are replaced by a letters. The common letters used for the figures are x and n.

Let us consider the following examples.

I. You are provided with BA1 and BA2 solutions.

BA1 is a solution of ethanoic acid, H2X.nH2O made by dissolving 12.6 g of the acid
per litre.
BA2 is 0.2M solution of sodium hydroxide solution.

You are required to determine the value of n in the formula of the acid.

Procedure:
Pipette 25cm3 or 20 cm3 of BA1 into a conical flask. Add two drops of
phenolphthalein indicator and titrate it with BA2 from burette. Repeat the procedure
until you get consistent results.
Enter the results in the table below.

33
 Sample results
Volume of pipette used 20.00. cm3

Burette Readings 1 2 3 4
Final Burette Reading (cm3) 22.00 42.10 22.50 42.50
Initial Burette Reading (cm3) 1.50 22.00 2.50 22.50
Volume of BA2 used (cm3) 20.20 19.10 19.00 18.90

Titre values used to find the average volume= 19.10, 19.00, 18.90 cm3
19.10  19.00  18.90
Average volume of BA2 used =
3
57
=
3
= 19.00 cm3
Questions:
(a) Write the equation for the reaction.

H2X (aq) + 2NaOH (aq)  2Na2X (aq) + 2H2O (l)

(b) Calculate:
(i) The number of moles of BA2 that reacted with BA1.

Volume of BA2 (base) = average volume = 19.00 cm3,

Molarity of BA2 = 0.2M, Number of moles = ?

0.2 moles of acid are contained in 1000 cm3 of solution.

The number of moles of BA2 contained 19.00 cm3 of solution
19 x 0.2
=
1000
= 0.0038
(ii) The number of moles of BA1 that reacted with BA2.
From the mole ratio: 2 moles of base : 1 mole of acid
0.004 moles react with = ½ of the number of moles of the BA1
= ½ x 0.0038
= 0.0019

34
(c) Determine:

(i) The molarity of BA1.

Volume of BA1 (acid) = 20.00 cm3, Molarity of BA1 =?,

Number of moles = 0.0019.

20.00 cm3 of BA1 solution contains 0.0019 moles,

0.0019 x 1000
1000 cm3 of solution contains = 0.095
20
= 0.1 moles
Therefore, the molarity of BA1 = 0.1M.

(ii) The RFM of BA1.

Concentration in g / l
Molarity =
RFM
Concentration in g / l
RFM =
Molarity
12.6
=
0 .1
 RFM = 126

(iii) The value of n (Given that the mass of H2X = 90)

90 + 18n = 126
18n = 126 – 90
18n = 36
36
n =
18
n = 2

35
II. You are provided with BA1 and BA2 solutions.

BA1 is a solution of 0.2M solution of Hydrochloric acid.

BA2 is a solution of a metal hydrated carbonate M2CO3.nH2O, made by dissolving

4g of the hydrated carbonate in 250 cm3 of solution.

You are required to determine the value of n in the formula of the hydrated
compound.
Procedure:
Pipette 25cm3 or 20 cm3 of BA2 into a conical flask. Add two drops of methyl
orange indicator and titrate it with BA1 from burette. Repeat the procedure until
you get consistent results.
Enter the results in the table below.
Sample results
Volume of pipette used 20.00. cm3

Burette Readings 1 2 3 4
Final Burette Reading (cm3) 13.50 24.70 35.90 13.10
Initial Burette Reading (cm3) 1.00 13.50 24.70 2.00
Volume of BA1 used (cm3) 12.50 11.20 11.20 11.10

Titre values used to find the average volume = 11.20, 11.20, 11.10 cm3.
11.20  11.20  11.20
Average volume of BA1 used =
3
33.60
=
3
= 11.20 cm3
Questions:
(a) Write the equation for the reaction.
2HCl (aq) + M2CO3 (aq)  2MCl (aq) + CO2 (g) + H2O (l)
(b) Calculate:
(i) The number of moles of BA1 that reacted with BA2.
Volume of BA1 (acid) = average volume = 11.20 cm3
Molarity of BA1 = 0.2M, Number of moles = ?
0.2 moles of acid are contained in 1000 cm3 of solution.
11.20 x 0.2
Then 11.20 cm3 of BA1 solution contains
1000
= 0.00224 moles

36
 (ii) The number of moles of BA2 that reacted with BA1.

From the mole ratio: 2 moles of BA1 (acid): 1 BA2 mole of (base).
0.00224 moles of BA2 require ½ the number of moles of the BA1.
= ½ x 0.00224
= 0.00112 moles
(c) Determine:
(i) The molarity of BA2.

Volume of BA2 (base) = 20.00 cm3, Number of moles = 0.00112,

Molarity of BA2 =?,

20.00 cm3 of BA2 solution contains 0.00112 moles.

0.00112 x 1000
1000 cm3 of solution contains = 0.056
20
= 0.056 moles
Therefore, the molarity of BA2 = 0.056M.

(ii) The RFM of BA2.

250 cm3 of solution contains 4 g, then the mass contained in 1000 cm3
4 x 1000
=
250
= 16 g/l

Concentration in g / l
From Molarity =
RFM
Concentration in g / l
RFM =
Molarity
16
=
0.056
= 285.7
 RFM = 286 g

(iii) The value of n (Given that the mass of M2CO3 = 106)

106 + 18n = 286
18n = 286 – 106
18n = 180
180
n =
18
n = 10
37
 III. A solution of ethanedoic acid was made by dissolving 10g of acid crystals,
H2C2O4.xH2O in water and the solution made up to 500 cm3. 25.0cm3 of the acid
solution required 15.90 cm3 of 0.5M sodium hydroxide solution for complete
reaction.

(a) Write the equation for the reaction.

H2C2O4.xH2O (aq) + 2 NaOH (aq)  C2O4Na2 (aq) + 2 H2O (l) + xH2O (l)
Or (COOH)2.xH2O (aq) + 2 NaOH (aq)  (COONa)2 (aq) + 2 H2O (l) + xH2O (l)

 COOH   COONa 
Or Ι  .xH2O (aq) + 2NaOH (aq)   Ι  (aq) + 2H2O (l) + xH2O (l)
 COOH   COONa 

Mole ratio is 1 mole of acid: 2 moles of Sodium Hydroxide.

(b) Calculate:
(i) the concentration of acid in mol dm-3.

Acid: Volume = 25.0cm3, Conc. = 10g/500cm3 of water,

Molarity = ? No. of moles = ?

Base: Vol = 15.90cm3, Molarity = 0.5M; No. of moles =?

Moles of Sodium hydroxide:

1000 cm3 of sodium hydroxide solution contains 0.5 moles.
0 .5
1 cm3 of sodium hydroxide solution contains moles.
1000
0.5 x 15.90
15.90 cm3 of sodium hydroxide solution contains
1000

= 0.00795 moles

Moles of acid:
From the mole ratio, 1 mole of acid: 2 moles of Sodium Hydroxide,
the number of moles of the acid is equal to half the number of moles of sodium
hydroxide.
= ½ x 0.00795
= 0.003975 moles

38
 Molarity of the acid
25.0 cm3 of the acid solution contains 0.003975 moles.
0.003975
1 cm3 of acid solution contains moles
25
0.003975 x 1000
1000 cm3 of the acid solution contains = 0.159 mol dm-3
25

(ii) The number of molecules of water of crystallization in the acid.

Data: Molarity of acid = 0.159M, Conc. = 10g/500cm3(20g/l), RFM = ?

Relative Formula Mass of the Acid

Concentration in g / l
From Molarity =
RFM
Concentration in g / l
RFM =
Molarity
20
=
0.159
= 125.786
= 126 g

The value of x:

H2C2O4.xH2O = 126
(1 x 2) + (2 x 12) + (4 x 16) + 18x = 126
2 + 24 + 64 + 18x = 126
90 + 18x = 126
18x = 126 – 90
18x 36
=
18 18
x =2

39
4. Determination of the percentage purity of a compound in an impure sample

You are provided with BA1 and BA2 solutions.

BA1 is a solution of 0.1M solution of Hydrochloric acid.

BA2 is a solution of an impure sample of anhydrous sodium carbonate, Na2CO3, made by

dissolving 6.2g of the carbonate per litre.

You are required to determine the percentage purity of the compound.

Procedure:
Pipette 25cm3 or 20 cm3 of BA2 into a conical flask. Add two drops of methyl orange
indicator and titrate it with BA1 from burette. Repeat the procedure until you get consistent
results.

Enter the results in the table below.

Sample results
Volume of pipette used 20.00. cm3

Burette Readings 1 2 3 4
Final Burette Reading (cm3) 13.50 24.70 35.90 13.10
Initial Burette Reading (cm3) 1.00 13.50 24.70 2.00
Volume of BA1 used (cm3) 20.40 19.00 19.00 19.00

Titre values used to find the average volume = 19.00, 19.00, 19.00 cm3
19.00  19.00  19.00
Average volume of BA2 used =
3
57
=
3
= 19.00 cm3

Questions:
(a) Write the equation for the reaction.

2HCl (aq) + Na2CO3 (aq)  2NaCl (aq) + CO2 (g) + H2O (l)

40
(b) Calculate:
(i) The number of moles of BA1 that reacted with BA2.

Molarity = 0.1M, Vol. = 19.00 cm3, Number of moles = ?

0.1 moles of BA1 are contained in 1000 cm3 of solution.

The number of moles of BA1 contained 19.00 cm3 of solution
19 x 0.1
=
1000
= 0.0019
(ii) The number of moles of BA2 that reacted with BA1.

From the mole ratio, 2 mol of acid: 1 mol of base,

the number of BA2 hat reacted with BA1
= ½ the number of moles of BA1
= ½ x 0.0019
= 0.00095 moles
(c) Determine:
(i) The molarity of BA2.

Solution:
Molarity = ?, Number of moles = 0.0095, Vol = 20.00 cm3

20 cm3 of BA2 solution contains 0.0095 moles.

0.00095 x 1000
1000 cm3 of BA2 contains = 0.0475
20
= 0.05M

(ii) The concentration of BA2 in g/l. (Na = 23, C =12, O = 16)

Molarity = 0.05M, RFM = 106, Concentration in g/l = ?

Concentration in g / l
Molarity =
RFM
Concentration
0.05 =
106

Concentration = 0.05 x 106

 Concentration = 5.3 g/l
41
 (iii) The percentage purity.
Concentration in g / l of pure salt
Percentage purity = x 100
Concentration in g / l of impure salt

5.3
=
6.2

 Percentage purity = 85.48%

5. Determination of basicity of an acid

You are provided with BA1 and BA2 solutions.

BA1 is a solution of sodium hydroxide obtained by dissolving 8 g of a metal carbonate
M2CO3 per litre. ( Na = 23, C = 12 O = 16 )
BA2 is a solution 0.1M acid of formula HnX.

You are required to determine the basicity of the acid.

Procedure:
Pipette 25cm3 or 20 cm3 of BA2 into a conical flask. Add two drops of phenolpthaein
indicator and titrate it with BA1 from burette. Repeat the procedure until you get consistent
results. Enter the results in the table below.
Sample results
Volume of pipette used 20.00. cm3

Burette Readings 1 2 3 4
Final Burette Reading (cm3) 22.00 42.10 22.50 42.50
Initial Burette Reading (cm3) 1.50 22.00 2.50 22.50
Volume of BA1 used (cm3) 21.50 20.10 20.00 19.90

Titre Values used to find the average volume= 20.10, 20.00, 19.90 cm3
20.10  20.00  19.90
Average volume of BA1 used =
3
60.00
=
3
= 20.00 cm3

42
Questions:
(a) Write the equation for the reaction.
HnX (aq) + nNaOH (aq)  NaX (aq) + nH2O (l)

Mole ratio = 1 mol of acid: n mol of base

(b) Calculate:
(i) Molarity of sodium hydroxide.
Molarity of BA1 = ?, Concentration = 8 g/l, RFM = 40
Concentration in g / l
Molarity =
RFM
8
=
40
= 0.2 M

(ii) The number of moles of BA1 that reacted with BA2.

Volume of BA1 (acid) = average volume

= 20.00 cm3

Molarity of BA1 = 0.2M, Number of moles = ?

0.2 moles of acid are contained in 1000 cm3 of solution.
The number of moles of BA1 contained 20.00 cm3 of soltuion
20 x 0.2
=
1000
= 0.004

(iii) The number of moles of BA2 that reacted with BA1.

Volume of BA2 = 20.00 cm3. Molarity of BA1 = 0.1M, Number of moles =
?
0.1 moles of acid are contained in 1000 cm3 of solution.
The number of moles of BA1 contained 20.00 cm3 of solution
20 x 0.1
=
1000
= 0.002
43
 (c) Determine the basicity of the acid.
From the reaction equation, 1 mole of BA2 requires n moles of BA1.
i.e 1 mole of BA2 : n moles of BA1.
0.002 moles : 0.004 moles
0.002 n = 0.004
0.004
n =
0.002
n =2
Therefore, the basicity of the acid = 2

SELF-CHECK 1.1
1. (2005 Q.26)
The concentration, in grams per litre, of a 0.05 M sodium carbonate solution is [N=23;
O=16; C=12]
A. 0.05 x 83 B. 0.05 x 106
106 83
C. D.
0.05 0.05

2. (2004 Q.6)
Magnesium reacts with hydrochloric acid according to the following equation:
Mg(s) + 2HCl (aq)  MgCl2 (aq) + H2 (g)
The volume of hydrogen formed at s.t.p. when 2.32 g of magnesium reacts completely with
dilute hydrochloric acid is
[Molar gas volume at s.t.p. is 22.4 dm3, mg = 24]
22.4 x 2.32 3 22.4 x 24
A. dm B. dm3
24 2.32
2.32 x 24
C. dm3 D. 2.32 x 24 x 22.4dm3
22.4

3. (2004 Q.10)
Under a certain temperature and pressure, hydrogen reacted with nitrogen according to the
equation below: 3H2 (g) + N2 (g)  2NH3 (g)
The volume of nitrogen required to react with 150cm3 of hydrogen under the same
temperature and pressure is
A. 15.0 cm3 B. 50.0 cm3
3
C. 300.0 cm D. 450.0 cm3

44
4. (2004 Q.29)
What mass in grams, of sodium carbonate -10- water (Na2CO3 •10H2O) is contained in
50cm3 of 0.1 m solution? [H = 1, C = 12, O = 16, Na = 23]
106 x 0.1 x 100 106 x 0.1 x 50
A. B.
50 1000
286 x 0.1 x 1000 286 x 0.1 x 50
C. D.
50 1000

5. (2003 Q.9)
Hydrochloric acid reacts with sodium hydroxide according to the equation: HCl (aq)
+ NaOH (aq)  NaCl (aq) + H2O(l)
25.0 cm3 of 0.1M hydrochloric acid reacted completely with 20 cm3 of sodium hydroxide.
What is the molarity of sodium hydroxide?
25 x 0.1 20 x 0.1
A. B.
20 25
25 x 0.1
C. 20 x 0.1 x 25 D.
20

6. (2003 Q.11)
100 cm3 of nitrogen were reacted with 300 cm3 of hydrogen at s.t.p. What was the volume
of ammonia produced?
A. 100 cm3 B. 200 cm3
C. 300 cm3 D. 400 cm3

7. (2003 Q.18)
The volume of carbon dioxide evolved when 6.0 g of carbon are burnt completely in air at
s.t.p. is [C = 12, Molar volume = 22400cm3 at s.t.p.]
12 x 22400 12 x 6
A. cm3. B. cm3.
6 22400
6 x 22400
C. cm3. D. 6 x 12 x 22400 cm3.
12
8. (2003 Q.21)
20.0 cm3 of 0.1 M sodium hydroxide solution reacted with 0.1M of solution Y. The volume
of solution Y that reacted completely with the alkali is [Mole ratio NaOH: Y = 2:1]
3
A. 40 cm . B. 30 cm3.
3
C. 20 cm . D. 10 cm3.
9. (2003 Q.26)
20 cm3 of an acid HX neutralized 25 cm3 of 0.05 M sodium carbonate solution. The
molarity of the acid is
25 x 0.05 2 x 25 x 0.05
A. B.
20 20
2 x 20 x 0.05 188
C. D.
25 1.88 x 80

45
10. (2002 Q.19)
Calculate the relative molecular mass of gas P if 8.4 dm3 of the gas has a mass of
0.93 g. (1 mole of gas occupies 22.4 dm3 at s.t.p.)
0.93 x 22.4 22.4 x 8.4
A. B.
8.4 8 .4

0.93 x 8.4 0.93

C. D.
22.4 22.4 x 8.4
11. (2002 Q.25)
The mass of silver nitrate, AgNO3 in 0.2 M solution of the salt is (Ag = 108, O = 16, N = 14)
A. 17.0 B. 34.0
C. 85.0 D. 170.0

12. (2002 Q.35)

The mass in g of OH- ions in 0.25M NaOH solution is (H = 1, O = 16)
0.25
A. B. 17 x 0.25
17
0.25 x 4 17 x 0.25
C. D.
17 4
13. (2002 Q.36)
The minimum volume of 1M HCl (aq) required to produce 0.25 g of hydrogen with excess
magnesium is
A. 25 cm3. B. 100 cm3.
3
C. 250 cm . D. 1000 cm3.

14. (2001 Q.26)

25 cm3 of a 0.25M acid required 25 cm3 of 0.5M sodium hydroxide solution for
neutralisation. The basicity of the acid is
A. 1 B. 2
C. 3 D. 4

15. (2000 Q.19)

What is the mass of sulphuric acid (Mr = 98) in 5 cm3 of a 0.2 M solution of the acid?
98 x 5 98 x 0.2 x5
A. B.
0.2 x 1000 1000
98 x 0.2 98 x 5 x 1000
C. D.
5 x 1000 0 .2
16. (2000 Q.21)
25 cm3 of M sodium carbonate required 22.70 cm3 of hydrochloric acid for complete
neutralisation. The molarity of the acid id given by
0.00125 x 000 0.00125 x 1000
A. B.
2 x 22.7 2 x 25
0.00125 x 2 0.00125 x 2 x 1000
C. D.
22.7 25

46
17. (1999 Q.13)
Calcium carbonate reacts with hydrochloric acid according to the following equation:
CaCO3(s) + 2HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l)
The mass of CaCO3 that will react completely with 50 c.c of 2M hydrochloric acid is
(CaCO3 = 100)
2 x 50 x 100 2 x 1000
A. B.
2 x 1000 50 x 2 x 100
50 x 100 100 x 50 x 2
C. D.
200 x 1000 1000

18. (1999 Q.30)

What is the molarity of sodium hydroxide solution if 30 cm3 of 0.2 M hydrochloric acid
just neutralises 20 cm3 of the alkali?
20 20 x 0.2
A. B.
0.2 x 30 30
30 30 x 0.2
C. D.
0.2 x 20 20
19. (1998 Q.4)
20 cm3 of 0.2 M HCl reacts with 25 cm3 of sodium hydroxide solution. The molarity of the
hydroxide is
25 x 0.2 20 x 0.2
A. B.
20 25
25 20
C. D.
20 x 0.2 25 x 0.2
20. (1997 Q.10)
0.02 moles of calcium chloride (CaCl2) is dissolved to make 200 cm3 of solution. What is
the concentration of chloride ions in moles per litre, in this solution?
A. 0.05 M. B. 0.1 M.
C. 0.2 M. D. 0.3 M.

SELF-CHECK 1.2
1. A gaseous hydrocarbon X contains 20% hydrogen by mass. 7.5 g of X occupies
5.6 dm3 at stp.
(a) Calculate: (i) The empirical formula of X.
(ii) The molar mass of X.
(iii) The molecular formula of X.
(b) Write: (i) The name of X.
(ii) The structural formula of X.
(C = 12, H = 1)

47
 CHAPTER TWO
EFFECT OF ELECTRICITY ON SUBSTANCES
(ELECTROLYSIS)

Electrolysis is divided into two parts, namely:

- Part I. Qualitative Electrolysis
- Part II. Quantitative Electrolysis

QUALITATIVE ELECTROLYSIS

Experiment:
To investigate the conduction of Electricity by a variety of substances

Apparatus:
(a) (i) Solids:
Small pieces of: -
- Metals e.g. Magnesium, Copper, Lead, etc.
- Non-metallic substances: e.g. Rubber, Plastic, etc.
- Compounds e.g. salts, sugar etc.
(ii) Liquids:- Water, alcohol (ethanol), urea, urine
(iii) Aqueous solutions of:
- Salts: E.g. Sodium chloride, Copper (II) sulphate, Silver nitrate etc.
- Strong acids (Mineral Acids).
E.g. Hychrochloric acid, Sulphuric acid,
Nitric acid and Phosphoric acid.
- Weak acids: E.g. Ethanoic acid, Oxalic acid etc.
- Strong alkalis: E.g. Sodium hydroxide, Potassium hydroxide, Aluminium
hydroxide etc.
- Weak alkalis: E.g. Ammonia Solution.

(b) A beaker, 2 graphite rods, 2 Crocodile clips, 6V D.C. supply, 1 6V lamp bulb and 1 switch.
Procedure:
 Arrange the apparatus as shown in the diagram below, in which a 6V D.C. supply is
connected in series with a 6V lamp to a pair of crocodile clips.

48
 Fig. 1
 Test the conductivity of the substances in the list above in turn and in each case,
observe the bulb.

Notes:
If : (i) the bulb lights or glows, it indicates that the circuit is
complete, which means that the substance under
investigation is a good conductor of electricity.
(ii) the bulb does not light or glow, it indicates that the circuit is
incomplete, which means that the substance under investigation is a
bad conductor of electricity.

NB: For the solid substances simply attach the crocodile clips to the ends of the
spacemen while for the liquids/solutions attach the crocodile clips to the
graphite rods and then insert them into the spacemen under investigation.

Observations:
(i) The bulb glows, brightly, for some solids especially all metals and some aqueous
solutions and dimly, for some solutions.
(ii) While for some solids e.g. rubber and some liquids the bulb did not glow at all.

Results:
The results showed that:
(i) All metals are good conductors of electricity.
(ii) Ionic compounds in a solid state e.g. salts do not conduct electricity.
(iii) Ionic compounds in solution or molten state are good conductors of electricity.
(iv) Covalent compounds are bad conductors of electricity.
(v) Weak acids and alkaline solutions and water conduct electricity weakly.

49
 Terms Used:
1. Conductors and Insulators(Non-conductors)
(a) Conductors:
A conductor is a substance, which allows electric current to pass
through it.
E.g. - All metals and carbon in form of graphite.

(b) Insulators (Non-conductors):

These are substances, which do not allow electricity to pass through. Examples of
insulators include: rubber, plastic, wood etc.

2. Electrolytes and Non-electrolytes:

(a) Electrolytes:
An electrolyte is a compound which, when in solution or molten, conducts an
electric current and is decomposed by it.
Examples of electrolytes include:
- Molten ionic compounds.
- Aqueous solutions of ionic compounds (salts, acids and alkalis).

NB:
 All electrolytes are ionic compounds. I.e. are composed of ions.
 The above compounds only conduct electricity in molten (fused) and aqueous state.
They DO NOT conduct electricity in solid state.

Reason:
- In solid state, the ions are rigidly held in regular positions by strong forces
as such are unable to move freely.
- In molten and aqueous state, the ions move freely hence conduct electricity.

 The passage of electricity through the electrolyte is accompanied by a chemical

decomposition.
 The decomposition of an electrolyte, when electric current passes through it is called
electrolysis.

(a) Non-electrolyte:
A non-electrolyte is a compound which, when in aqueous or molten does not
conduct electricity.
Examples include: - all covalent compounds. E.g. alcohol, sugar etc.

50
 Differences between Electrolyte and Non-electrolyte

Electrolyte Non-electrolyte

- Conducts electricity in molten or - Do not conduct electricity in molten

solution form. or solution form.

- Consists of free moving oppositely - Consists of covalent molecules and

charged ions. therefore no free moving ions.

(c) Types of Electrolytes:

There are two types of electrolytes, namely:-
- Strong Electrolytes and
- Weak Electrolytes.

(i) Strong Electrolyte:

A strong electrolyte is a compound, which ionizes or dissociates, completely
in dilute solution or in the molten state into ions.

Examples of strong electrolytes are shown in the table below.

Compound Name of Compound State Dissociation equation

Lead bromide Molten PbBr2 (s)  Pb2+ (l) + 2Br-(l)

Salts Sodium Chloride Aqueous NaCl (aq)  Na+(aq) + Cl-(aq)

Copper (II) Sulphate Aqueous CuSO (aq)

4  Cu2+ (aq) + SO42- (aq)

Acids Hydrochloric acid HCl (aq)  H+ (aq) + Cl-(aq)

Sulphuric acid Aqueous H SO (aq)

2 4  2H+(aq) + SO42- (aq)

Phospharic acid H3PO4 (aq)  3H+(aq) + PO43- (aq)

Alkali Sodium Hydroxide Aqueous NaOH (aq)  Na+(aq) + OH-(aq)

51
 (ii) Weak Electrolyte:
A weak electrolyte is a compound, which ionizes or dissociates, slightly in dilute
solution or in the molten state.
Examples of weak electrolytes are shown in the table below.

Compound Name of Compound State Dissociation equation

Alkali Ammonium Hydroxide Aqueous NH4OH(aq) NH4+(aq) + OH-(aq)

Acid Ethanoic acid Aqueous CH3COOH(aq) H+(aq) + CH3COO-(aq)

Water Hydrogen Oxide Liquid H2O(l) H+(aq) + OH-(aq)

(d) Difference between Strong Electrolyte and Weak Electrolyte

Strong Electrolyte Weak Electrolyte

- Ionizes completely in aqueous solutions - Ionizes slightly in aqueous solutions

- Contains ions and no molecules - Contains ions and molecules

- Produces a higher current thus causing - Produces small current thus causing
a bulb to glow brightly. a bulb to glow dimly.

- Causes large deflection in an ammeter - Causes small deflection in an ammeter

3. Electrolysis:
Electrolysis is the decomposition of a substance (an electrolyte) by passing an electric
current through it.

NB: Electrolysis is accompanied by chemical changes, which occur at the electrodes.

4. (a) Electrodes:
Electrodes are the two poles (conductors) through which electric current enters
and leaves the electrolyte.

NB: (i) The electrodes are connected to the terminal of the D.C supply.
(ii) Before the switch is closed, the electrodes have no charges.
(iii) They acquire charges, on the closer of the switch, hence giving the types of
electrodes.

52
 (b) Types of electrodes:
There are two types of electrodes, namely:-
- anode and
- cathode.

(i) Anode:
Anode is the positively charged electrode through which electric current
enters the electrolyte.
- It is electron deficient.
- It is connected to the positive terminal of D.C supply.
- It is usually tall in size, in the circuit symbol.

(ii) Cathode:
Cathode is the negatively charged electrode at which the electric current
leaves the electrolyte.

- It is reach with electrons.

- It is connected to the negative terminal D.C supply
- It is usually short in size, in the circuit symbol.

The Ionic Theory (Put forward by Arrhenius about 1880)

States that:
An electrolyte consists of free positively and negatively charged particles called ions,
which are responsible for the transmission of electricity through the electrolytes.

(a) Types of ions:

There are two types of ions, namely:-
- Cations - positively charged ions and
- Anions - negatively charged ions.

Examples of ions are shown in the table below.

Cations Anions
++ 2+ 2+ 3+ 2+
(i) Metallic ions: K , Na , Ca , Mg , Al , Zn Non-metallic ions: Cl-, Br-, I-

Fe2+, Fe3+, Pb2+, (H+), Cu 2+, Ag+

(ii) Radicals: NH4+ Radicals: SO4 2-, NO3-, CO32- , OH-

53
(b) Experiment:
To investigate the movement of ions during electrolysis

(i) Apparatus:
 20 V D.C supply,
 a microscope slide,
 2 crocodile clips.
 A strip of filter paper,
 4 pieces of connecting wire and
 a crystal of coloured salt e.g. Potassium permanganate or Copper (II) sulphate.

(ii) Procedure:
 Cut a piece of filter paper approximately the size of microscope slide.
 Attach the piece of the filter paper on top of the microscope slide using
crocodile clips.
 Moisten the filter paper with water.
 Place a small crystal of Potassium permanganate mid-way between the points of
attachment of the clips.
 Connect the slide and its contents in a circuit with a 20V D.C. supply as
shown in figure 2 and observe the paper over a period of 10 - 15 minutes.

(iii) Observation:
- A purple colour is seen moving towards the anode.
- Since in aqueous state the Potassium permanganate ionizes according to
the following equation:
KMnO4 (aq)  K+(aq) + MnO4- (aq)

The permanganate ions are attracted by the anode and therefore move towards the anode.
This purple colour of permanganate is due to the MnO4- ions.

54
 (c) Explanation of Electrolysis using the ionic theory:
Consider the electrolytic cell diagram in fıg.3 below.

Fig. 3

- When the circuit is complete, the process of electrolysis begins.

- During the process, the law of electricity, which states that: like charges repel and
unlike charges attract is obeyed and the ions migrate to their respective electrodes.
- That is the attraction between the electrodes and the ions begins.
- The cathode attracts cations (netatively charged ions) while the anode attracts
anions (positively charged ions).
- Thus resulting to ion migration to the respective electrode.

Memory Aid:
You can remember which ion goes to which electrode by the statement:
"Cat at Cat" and "An at An"

(d) The process of discharge:

When ions arrive at their respective electrodes, they become discharged i.e. they loose
their charges and form products.
The anions loose electrons to the anode and the cations gain electrons from the cathode.

(i) Discharge at the Cathode:

When cations arrive at the cathode, which is electron rich, they 'strike' the cathode
and pick electrons from the cathode equal to their charges.
The electrons neutralize the positive charge on the ion and the ion turns back to
atom(s).
Metal cations become metal atoms and are deposited on the cathode. For hydrogen,
two atoms combine to form a molecule of hydrogen gas leading to the formation of
colourless bubbles on the cathode.

55
 (ii) Discharge at the Anode:
When anions arrive at the anode, which is electron deficient, they loose their excess
electron(s) to the anode equal to their valences. Thus they become discharged and
revert to atom(s).
Insoluble gaseous atoms form molecules and are evolved and bubbles of colourless
gas are seen on the lower part of the anode.

(e) Selective/Preferential Discharge of Ions at the Electrodes

When two or more ions of the same charge arrive at an electrode, only one is selected
preferentially discharged depending on the factors of preferential discharge.
The Factors which Determine Preferential Discharge of Ions at the Electrodes

1. Position of the ion in the Electro-Chemical Series:

When two or more ions of a similar charge are present in a dilute electrolyte, the one
lower in the series is discharged in preference to that above it.

The electrochemical series

Highly electropositive and highly electronegative ions are not easily discharged unless
there are no other less electropositive cations or less electronegative anions.

2. Concentration of the ion:

When ions of similar charge are present, the one in greater concentration is
discharged preferentially even if it is higher in the E.C.S.
3. Nature of the electrode.
Different electrodes for a given electrolyte may cause the formation of different
products despite the above factors.

56
For examples:
(i) Electrolysis of Sodium Chloride.
When Sodium chloride solution is electrolyzed:
 using platinum cathode, H+ ions are discharged instead of Na+ ions
and
 when mercury is used as a cathode, Na+ ions are discharged. The
product formed is called Sodium amalgam.

(ii) Electrolysis of Copper (II) Sulphate solution

When Copper (II) Sulphate solution is electrolyzed using copper anode,
neither OH- nor SO42- ions are discharged, but rather electrode ionization
occurs according to the equation:

Cu (s)  Cu2+(aq) + 2e-

The copper ions formed are released into the electrolyte and the electrons are
immediately passed to the cathode.

Note that:
 A passive/un-reactive electrode does not affect the process of electrolysis.
 For this reason platinum or carbon (in form of graphite) which are passive are
commonly used as electrodes.
 However, Platinum (Pt) is not used where chlorine is produced. This is
because chlorine attacks platinum.

57
 EXAMPLES OF ELECTROLYSIS

1. Electrolysis of Salts:
(a) Electrolysis of Molten Lead bromide using graphite electrodes.

Ions present: PbBr2(s)  Pb2+(l) + 2Br-(l)

The ions formed migrate to the electrode of opposite charge, i.e. Pb2+ ions move to
the cathode (negative electrode) and Br- ions move to anode (positive electrode). On
their arrival at the respective electrode, they become discharged.

Equations for the reactions at the electrodes

(i) At the anode:

The negatively charged bromide ions (Br-) become discharged by loss of electrons
to the anode according to the equation:

2Br-(l) - 2e  Br2 (g) or

2Br-(l)  Br2 (g) + 2e (Oxidation)

Observations:
 Bubbles of reddish brown gas with choking and irritating smell are formed.
 The gas is slightly soluble in water, forming a reddish brown solution.
 The bulb glows brightly.

58
(ii) At the cathode:
The positively charged lead ions (Pb2+) become discharged by gain of electrons
from the cathode one lead ion gains two electrons to become lead atom according to
the equation:

Pb2+(l) + 2e  Pb (l) (Reduction)

Observations:
A grey solid formed melts and sinks to the bottom of the crucible.

NB: The electrolysis of molten Lead (II) bromide is comparatively easy to understand
because only one type of cation and one type of anion are present in the electrolyte.
In aqueous solutions, there are two or more cations and anions except for dilute
sulphuric acid where there is only one cation (H+ ion).

(b) Electrolysis of Sodium chloride solution using graphite or platinum cathode

and graphite anode (to resist attack by chlorine gas).

Ions present: NaCl (aq)  Na+(aq) + Cl-(aq)

H2O (l) H+(aq) + OH-(aq)

The ions formed migrate to the electrode of opposite charge, i.e. Na+ and H+ ions move to
the cathode (negative electrode) and Cl - and OH - ions move to anode (positive electrode)
and then a preferential discharge occurs.

Equations for the reactions at the electrodes:

(i) At anode:
The negatively charged Cl - ions are preferentially discharged by loss of electrons to
the anode according to the equation:
2Cl-(aq) - 2e  Cl2 (g) (Oxidation)

Observations:
 Bubbles of greenish-yellow gas with a choking, unpleasant and irritating smell are
seen.
 The bulb glows brightly.

(ii) At the cathode:

The positively charged H+ ions are preferentially discharged by gain of electrons
from the cathode according to the equation:
2H+(aq) + 2e-  H2 (g) (Reduction)

59
 Observations:
 Bubbles of colourless gas that burns with a pop sound are formed.

NB:
 Discharge of H+ ions disturbs the ionic equilibrium of water and causes more water
molecule to ionize to restore it.
 Excess OH-ions so produced, with the presence of Na+ ions in the solution, is
equivalent to presence of sodium hydroxide.
 The ratio of the products indicates equal volumes. However, in practice the volume of
chlorine produced is less than the expected because chlorine formed reacts with OH-
ions according to the equation:
Cl2 (g) + 2OH - (aq)  Cl - (aq) + OCl -(aq) + H2O (l)
 At some concentration, both chlorine and oxygen are produced at the anode.

(d) Electrolysis of Copper (II) sulphate solution using graphite electrode or

platinum anode
Ions present: CuSO4(aq)  Cu2+(aq) + SO42-(aq)

H2O(l) H+(aq) + OH-(aq)

The ions formed migrate to the electrode of opposite charge, i.e. Cu2+ and H+ ions
move to the cathode (negative electrode) and SO42- and OH- ions move to anode
(positive electrode) and then a preferential discharge occurs.

Equations for the reactions at the electrodes

(i) At the anode:
The negatively charged OH - ions are preferentially discharged by loss of electrons to the
anode according to the equation:
4OH-(aq) - 4e-  2H2O (l) + O2 (g) (Oxidation)

Observations:
 Bubbles of colourless gas that rekindles a glowing splint are formed.
 The bulb glows brightly.

(ii) At the cathode:

The positively charged Cu2+ ions are preferentially discharged by gain of electrons from the
cathode according to the equation:
Cu2+(aq) + 2e-  Cu (s) (Reduction)
Observations:
 A brown solid formed at the lower part of cathode.
 The blue colour of the electrolyte fades. This is because the copper (II) ions which
are responsible for the blue colour are being converted to copper.
If the electrolysis is left for a long period of time, the solution becomes colourless.

60
(e) Electrolysis of Copper (II) sulphate solution using graphite cathode and copper anode

Ions present: CuSO4(aq)  Cu2+(aq) + SO42-(aq)

H2O(l) H+(aq) + OH-(aq)

The ions formed migrate or move towards the electrode of opposite charge, i.e. Cu2+ and H+
ions move to the cathode (negative electrode) and SO42- and OH- ions move to anode
(positive electrode) and then a preferential discharge occurs.
Equations for the reactions at the electrodes
(i) At the anode:
Although both OH- and SO42- ions arrive at the anode, none of them is discharged.
Instead, the copper anode dissolves into the solution according to the equation:
Cu(s)  Cu2+(aq) + 2e (Oxidation)

Observations:
 The copper anode dissolves into the solution thus reducing in size.

(ii) At the cathode:

The positively charged Cu2+ ions are preferentially discharged by gain of electrons
from the cathode according to the equation:
Cu2+(aq) + 2e  Cu (s) (Reduction)

Observations:
 A brown solid formed at the lower part of cathode.
 The blue colour of the electrolyte remains.
 This is because the copper ions discharged are being replaced by the ones
formed from the anode. That is copper ions are being effectively transferred
from the anode to the cathode. Hence, the concentration of the copper (II)
ions remains constant.

61
2. Electrolysis of Acids
(a) Electrolysis of Sulphuric acid using platinum electrodes.
Ions present: H2SO4 (aq)  2H+(aq) + SO42-(aq)
H2O (l) H+(aq) + OH-(aq)
The ions formed migrate to the electrode of opposite charge, i.e. H+ ions the only cations
move to the cathode (negative electrode) and SO42- and OH- ions move to anode (positive
electrode) and then a preferential discharge occurs.

Equations for the reactions at the electrodes

(i) At the anode:
The negatively charged OH- ions are preferentially discharged by loss of electrons
to the anode according to the equation:
4OH-(aq) - 4e  2H2O (l) + O2 (g) (Oxidation)

Observations:
 Bubbles of colourless gas that relights a glowing splint are formed.

(ii) At the cathode:

The only positively charged H+ ions are discharged by gain of two electrons from
the cathode according to the equation:
2H+(aq) + 2e  H2 (g) (Reduction)
Observations:
 Bubbles of colourless gas that burns with a pop sound are formed.
NB: Total acidity remains constant.
Explanation:
The discharge of H+ ions at the cathode is equivalent to decrease of the concentration of the
acid. But the discharge of OH- ions disturbs the ionic equilibrium of water causing more
water molecules to ionize to restore it according to the equation:
H2O (l) H+(aq) + OH-(aq)
Excess H+ ions so produced plus the SO42- ions is equivalent to increased concentration of
sulphuric acid. Therefore, at the cathode the acidity decreases while at the anode it
increases, hence the total acidity remains constant.

The following points about electrolysis of solutions should be remembered:

(i) Metals, if produced, are discharged at the cathode.
(ii) Hydrogen is produced at the cathode only.
(iii) Non-metals, apart from hydrogen are produced at the anode.
(iv) Reactive metals are not formed at the cathode during electrolysis of aqueous
solutions. An exception is during electrolysis of sodium chloride using a mercury
cathode.
(v) The products produced depend upon the concentration of the electrolyte in the
solution. For example, electrolysis of concentrated sodium chloride produces
chlorine at the anode but electrolysis of dilute sodium chloride can produce oxygen
at the anode.
62
 APPLICATIONS OF ELECTROLYSIS
1. Purification of Copper.
In the purification process, the anode is impure copper rod or plate and the cathode is pure
copper plate or copper rod.
The electrolyte used must contain copper ions e.g. copper (II) sulphate solution.
The diagram showing the purification of copper

Explanation
During the electrolysis, copper dissolves from the anode, according to the equation.
Cu(s)  Cu2+(aq) + 2e- (Oxidation)

The positively charged Cu2+ ions so formed move to the cathode and are preferentially
discharged by gain of two electrons from the cathode according to the equation:
Cu2+(aq) + 2e-  Cu (s) (Reduction)

NB: At the bottom of the cell, “anode slime” collects. The slime is rich in precious metals, e.g.
Silver, Gold, and is formed from the impurities in the original impure copper.

2. Electroplating
Electroplating is the coating of the surfaces of objects/metal surface with a thin layer of
another metal.
It is done to: (i) Decorate ornaments and other goods.
(ii) Prevent corrosion.
The process is the same as for the purification of copper.
Points to note:
(i) The metal/object to be plated or coated is made the cathode and its surface must be
clean and grease-free.
(ii) The metal to be used for coating is made the anode.
(iii) The electrolyte used should contain the cations of the metal used for the coating.

Observations
 The cathode surface is coated with a metal layer.
 The anode dissolves and goes into solution as ions.

63
 Example UNEB 1992 Q.3 (b)
Draw a labeled diagram of an apparatus that can be used in the laboratory to copper plate an object.
Solution

3. Extraction of reactive metals

E.g. Potassium, Sodium, Calcium, Magnesium and Aluminium.

Extraction of reactive metals e.g Sodium from Sodium Chloride

Sodium is extracted by the electrolysis of molten sodium chloride in the Downs Cell.
The Downs Cell consists of a cylindrical steel cathode and graphite anode.
Sodium chloride is heated and melts at 801 ºC. Calcium chloride is added as an impurity to
lower the melting point to about 600 ºC.

The Diagram of Downs Cell showing the Extraction of Sodium from Molten Sodium chloride

Explanation:
Ions present: NaCl (s)  Na+ (l) + Cl- (l)
CaCl2 (s)  Ca2+ (l) + 2Cl- (l)
   2Cl-(l)  Cl2 (g) + 2e-

64
 The products Sodium and Chlorine produced are kept apart to prevent them from reacting
and reforming sodium chloride.
The liquid Sodium is collected under dry nitrogen gas. This prevents the metal from
reacting with the atmosphere.

4. Manufacture of Sodium Hydroxide.

Sodium Hydroxide is produced industrially by the electrolysis of sodium chloride solution
(brine) using either the:
- Diaphragm cell or
- Kellner Solvay Cell.
In both cases the products are sodium hydroxide, hydrogen gas and chlorine gas.

In the diaphragm cell the electrolysis of nearly saturated solution of sodium chloride is
carried out using a graphite anode and a steel cathode.
The cell is made of an asbestos substitute. It allows the solution to pass through but
prevents the products chlorine and sodium hydroxide solution from coming into contact.

The Diagram of a Diaphragm Cell showing the manufacture of Sodium hydroxide

QUANTITATIVE ELECTROLYSIS

(a) The Laws of Electrolysis

The laws expressing the quantitative results of electrolysis were first stated by Faraday.
The laws assert that:
The amount (expressed in moles) of an element liberated during electrolysis depends upon:
(i) The magnitude of the steady current passed
(ii) The time taken for the current to pass through the electrolyte.
(iii) The charge on the ion of the element.

65
(b) The Quantity of Electricity
The quantity of electricity, Q, (measured in units called coulombs) is given by the product
of the current (measured in amperes) and the time (measured in seconds).

Thus: Quantity of electricity = Current x Time

(Coulombs) (Amps) x (Seconds)

i.e Q = It

(c) Faraday’s Laws of Electrolysis:

Law I
Faraday’s first law of electrolysis states that:
The mass of a given element liberated at (or dissolved from) an electrode during
electrolysis is proportional to the quantity of electricity consumed or passed through the
electrolyte.

Law II
The second law of electrolysis states that:
When the same quantity of electricity is passed through different electrolytes, the relative
numbers of moles of the elements deposited are inversely proportional to the charges on
the ions of each of the elements respectively.
Or
The masses of the different elements liberated by the same quantity of electricity form
simple whole number ratios when divided by their relative atomic masses.

Faraday’s Constant:
The least quantity of electricity required to liberate one mole of a univalent element is the
Faraday. Its approximate value is 96500C/mol. (96487C/mol).

Let the equation below represent chemical reaction at the cathode for a univalent metal.

Since M+(aq) + e-  M (s)

1 mol 1 mol 1 mol
From the equation, it follows that for one mole of uni-positive ion to be discharged, the
charge carried by one mole of electrons will be required.
Hence,
One Faraday = One mole of electrons
i.e. 1F = 1 mol of e-

66
 This information may be used as follows:
(i) Univalent cation
Ag+(aq) + e-  Ag (s)
1 mol 1 mol 1 mol (108g)
(1F = 96500c)

(ii) Divalent cation

Cu2+ (aq) + 2e-  Cu(s)
1 mol 2 mol 1 mol (64g)
(2F = 2 x 96500C)

(iii) Trivalent cation

Al3+(aq) + 3e-  Al (s)
1 mol 3 mol 1 mol (27g)
(3F = 3 x 96500 C)

(iv) For a diatomic gas (except oxygen) e.g. Hydrogen or Chlorine.

2H+ (aq) + 2e-  H2 (s)
1 mol 2 mol 1 mol (2g)
(2F = 2 x 96500 C)

(v) For oxygen:

4OH-(aq) - 4e-  2H2O (l) + O2 (g)
4mol 4mol 2mol 1mol (32g) or
(4F = 4 x 96500 C) (22400cm3 at stp or 24000cm3 at rt)

From the above, it can be seen that the higher the charge on the ion of an element, the
more the quantity of electricity required to liberate one mole of the element.

EXAMPLES
1. A steady current of 2.50 A was passed through Copper (II) sulphate solution for 40
minutes and 50 seconds.
Calculate (i) the quantity of electricity passed.
(ii) the mass of copper deposited at the cathode.
(iii) the volume of oxygen gas liberated at the anode in cm3
( 1F = 96500C, Cu = 64, 1 mole of gas at rt = 24l )

Solution: I = 2.50 A, t = 40 x 60 + 50 = 2450 s, Q = ?

(i) Q = It
= 2.50 x 2450
= 6125 C

67
 (ii) Equation at cathode.
Cu2+ (aq) + 2e-  Cu (s)
1 mol 2 mol 1 mol (64g)
(2F = 2 x 96500C)

If (2 x 96500) C liberates 64g of copper,

64
1 coulomb liberates g of copper.
2 x 96500
64
6125C liberates x 6125 = 2.031g
2 x 96500
= 2.03g

(iii) Equation at anode:

4OH-(aq) - 4e-  2H2O (l) + O2 (g)
4mol 4mol 2mol 1mol (32g) or
(4F = 4 x 96500C) (24000cm3 at rt)

If 4 x 96500 C liberates 24000 cm3 of oxygen gas at room temperature,

24000
1 Coulomb of electricity liberates of oxygen gas.
4 x 96500
24000
6125C will liberates x 6125 =380.975
4 x 96500
= 380.98 cm3
2. A steady current of 5.36 A was passed through an electrolyte of Silver nitrate solution for
30 minutes.
Calculate (i) the quantity of electricity passed.
(ii) the mass of silver deposited at the cathode.
(iii) the volume of oxygen gas liberated at the anode at stp.
( 1F = 96,500C, Ag = 108, 1 mole of gas at stp = 22,400cm3 )

Solution: I = 5.36 A, t = 30 x 60 = 1800s, Q = ?

(i) Q = It
= 5.36 x 1800
= 9648 C

(ii) Equation at cathode.

Ag+ (aq) + e-  Ag (s)
1 mol 1 mol 1 mol (108g)
(1F = 96500C)
68
 If 96500C liberates 108g of silver,
108
1 coulomb liberates g of silver.
96500
108
9648 C liberates x 9648 = 10.8g
96500

(iii) Equation at anode:

4OH-(aq) - 4e-  2H2O (l) + O2 (g)
4mol 4mol 2mol 1mol (32g) or
(4F = 4 x 96500C) (24,000cm3 at rt)

If (4 x 96500) C liberates 22,400 cm3 of oxygen gas at stp,

22400
1 coulomb liberates cm3 of oxygen gas.
4 x 96500
22400
9648 C liberates x 9648 = 559.88 cm3
4 x 96500

SELF-CHECK 2.1
1. 1987 Q.10
A circuit was connected as shown in the diagram below and a steady current of 0.20
amperes was passed for 20 minutes. (1F = 96500Cmol-1; Cu = 64)

(a) Write equation for the reaction that took place at the cathode.
(b) Calculate:
(i) The number of coulombs of electricity used.
(ii) The number of moles of coulombs of electricity.
(iii) The mass of the substance formed at the cathode.

2. 1988 Q.8
A steady current of 0.65 A was passed for 35 minutes through acidified water to electrolyse
it using carbon electrodes.
(a) State the electrode at which oxygen was liberated.
(b) Calculate the mass of oxygen liberated. (1F = 96500 coulombs)
69
3. 1990 Q.8
The circuit shown in the diagram in figure 3 was used in an experiment to study the effect
of electricity on lead (II) bromide.

(a) State what was observed.

(i) Before lead (II) bromide had melted.
(ii) After lead (II) bromide had completely melted.
(b) Explain your answer in (a).
(c) Write equation for the reaction that took place at:
(i) P
(ii) Q

4. 1992 Q.5
When a steady current was passed through the circuit shown in the diagram
below, 0.02 moles of a substance was deposited at P.

(a) (i) Name the electrodes: P, Q, R and S.

(ii) State what was observed at R.
(b) Write equation(s) for the reaction(s) that took place at:
(i) P.
(ii) S
(c) Calculate the number of moles of the gaseous substance formed at S.

70
5. 1993 Q.4
Copper (II) sulphate was electrolysed using carbon electrodes.
(a) State what was observed at the:
(i) Anode.
(ii) Cathode.
(b) Write equation(s) for the reaction(s) that took place at the anode.

6. 1994 Q.4

(a) State what was observed when:

(i) The switch was turned on.
(ii) Lead (II) bromide was melted and the switch turned on.
(b) Explain your observations in (a) (i) and (ii).

7. 1997 Q.3
(a) Molten lead (II) bromide was electrolyzed between carbon electrodes.
(i) State what was observed at the:
Anode.
Cathode
(ii) Write an equation for the reaction that took place at each electrode.
(b) Calculate the mass of the product formed at the cathode when a current of 2 amps
is passed for 1 hour and 30 minutes.

8. 2000 Q.2
The diagram below shows an arrangement of the apparatus used for the purification of copper.

(a) Name the substance used as:

(i) anode.
(ii) cathode.

71
 (b) Name the electrolyte.
(c) Write equation for the reaction that took place at:
(i) anode.
(ii) cathode.
9. 2003 Q.7
Molten lead (II) bromide was electrolyzed between two carbon electrodes.
(a) Explain why lead (II) bromide was electrolyzed in the molten state and not in the solid state.
(b) State what was observed at the:
(i) anode.
(ii) cathode.
(c) Write equation for the reaction that took place at the anode.

10. 2003 Q.13

(a) The diagram below shows an electrolytic cell in which electrolysis of
dilute sulphuric acid occurs.

(i) Name the gases X and Y that are evolved during electrolysis.
(ii) Give equation for the reaction occurring at the anode.
(iii) Indicate the direction of electron flow in the circuit.
(iv) Calculate the volume of gas X produced when a current of one ampere flows for
10 minutes through the electrolyte.
[1F = 9.6 x 104 coulombs, 1 mole of a gas occupies 2.4 x 104 cm3 at room
temperature and pressure]
(v) State two industrial applications of electrolysis other than the manufacture of
sodium hydroxide.
(c) Sodium hydroxide can be manufactured using mercury cell. How would this
manufacturing process affect the environment?

72
 ELECTROCHEMICAL CELL
(a) Definition:
Electrochemical cell is a cell which changes chemical energy into electrical energy.
(b) Structure of a Simple cell
It consists of two metal strips dipped into an electrolyte and connected by pieces of
conducting wires through a lamp bulb or voltmeter as shown in the diagrams below.

(a) (b)
NB: In this cell, the metal higher in the electrochemical series (zinc) is the negative
electrode but acts as the anode while metal lower in the electrochemical series
(copper) is the positive plate of the cell and acts as the cathode.

(c) Mechanism of the Cell

The metal (zinc) which is higher in the electrochemical series dissolves leaving electrons
on the strip and releasing Zn2+ ions into the electrolyte according to the equation:
Zn (s)  Zn2+ (aq) + 2e- (Oxidation)
The electrons travel though the external circuit and arrive at the metal lower in the
electrochemical series (copper) where they are picked by hydrogen ions to form molecules
of hydrogen gas according to the equation:
2H+ (aq) + 2e-  H2 (g) (Reduction)

Observations:
In diagram (a);
‫ ־‬Bubbles of a colourless gas which burns with a pop sound are produced.
‫ ־‬The bulb lights (glows) brightly and then gradually dims until eventually it dies off.
In diagram (b);
‫ ־‬Bulbs of a colourless gas which burns with a pop sound are produced.
‫ ־‬The pointer of the voltmeter deflects through a large angle and then the angle of
deflection gradually decreases until eventual it becomes zero.
Explanation:
The cell deteriorates due to polarization. Polarization is the building up of hydrogen
bubbles on the copper strip. This prevents the hydrogen ions from being discharged
quickly. The effect is that the quantity of electric current flowing is reduced. Shaking the
electrode or cleaning its surface can overcome this effect.
NB: The problem of polarization is best solved by dipping each metal in its own solution and
the electrical contact between the electrolytes is completed by using a salt bridge, which is
an inverted U-tube containing saturated potassium chloride, as seen in the Daniel Cell.

73
(d) Convention for current flow
The flow of electrons in the external circuit is from the element higher (i.e zinc) in the
electrochemical series towards the one lower (i.e copper). However, since electrons are
negative, current is conventionally considered to flow in the opposite direction to the
direction of the flow of electrons. Current therefore flows from the metal lower in the ECS
(cathode) to the metal higher in the ECS (anode) in the external circuit and vice versa in
the salt bridge.

The Daniel Cell

- A Daniel Cell is a modified simple (voltaic) cell.
- It comprises electrodes immersed in solutions of their own ions and connected with a
salt bridge or a porous partition.

Voltaic Cell: A Chemical Battery

The zinc and copper strips act as electrodes, and the salt bridge (in this case potassium
chloride) allows electrons to flow between the beakers without allowing the solutions to
mix. When the circuit joining the two systems is completed (as shown on the right), the
reaction generates electric current. Note that the zinc strip is used up (oxidation), and the
strip appears eaten away. The copper strip is built-up as additional electrons react with the
copper sulphate solution to produce additional metal (reduction).
NB: Replacing the light bulb with a battery would reverse the reaction, creating an
electrolytic cell.
(b) Cell Representation:
Zn(s)/Zn2+(aq) // Cu(s)/Cu2+(aq)
NB: - The metal higher in the ECS is always placed to the left hand side and the
one lower in the ECS is placed to the right hand side.
- // represents the salt bridge.

(c) Half Reaction Equations at the Electrodes:

At the anode [i.e the metal higher in the E.C.S (Zinc)]
Zn (s)  Zn 2+(aq) + 2e- (Oxidation)

74
 At the cathode: [i.e the metal lower in the E.C.S (Copper)]
Cu 2+(aq) + 2e-  Cu (s) (Reduction)

(d) The Cell Reaction/Overall Reaction:

The overall reaction equation is obtained by putting together the reactants and the
products of the half reaction equations and the electrons which appear on both
sides canceled.
Zn (s) + Cu 2+ (aq) + 2e-  Zn 2+ (aq) + Cu(s) + 2e-
Zn (s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)

(e) Position of the electrodes in the E.C.S.

If zinc/zinc sulphate solution is replaced by magnesium/magnesium sulphate
solution, a greater voltage is obtained. This is because the difference between the
electrode potentials is greater for Mg/Mg2+ and Cu/Cu2+ than for Zn/Zn2+ and
Cu/Cu2+.

Electrode potential
Electrode potential of an element is the potential difference between the element and
a solution of its own ions. The electrode potentials are negative for metals above
hydrogen and positive for metals below hydrogen in the electrochemical series.

Energy Supplied by the Cell (Not so important at this level)

The energy supplied by the cell is given by the equation:
Energy = Voltage x Charge
(Joules) (Volts) (Coulombs)
But Coulombs = nF
Where; n = Number of electrons and
F = Faraday (96500C).
If voltage = E, then:
Energy = EnF

Example
A voltmeter in a Daniel cell reads 1.20V for a Zn/Cu couple. Find the electrical energy
supplied by the cell. (1F = 96500C)

Solution: E = 1.20 V, n = 2 (Valence of Zn and Cu = 2)

From: Energy = EnF
= 1.20 x 2 x 96500
= 231.6 KJ
Electromotive Force (e.m.f) generated by the Cell
The e.m.f generated by the cell is given by the formula:
E.m.f = Eθright - Eθleft

75
 Example:
The following are half reactions for metals Y and Z.
Z (s)  Z3+(aq) + 3e Eθ = - 1.66 V
Y(s)  Y2+ (aq) + 2e Eθ = + 0.34V

(i) Draw a chemical cell for the metals Z and Y.

(ii) Calculate the e.m.f of the cell.
Recall that:
 Reactive metals above H have negative electrode potential and are always placed on
the left in the cell representation.
 Less reactive metals below H have positive electrode potential and are always placed
on the right hand side of the cell.
Solution: (i) Z (s) /Z3+(aq) // Y(s) /Y2+(aq)
(ii) E.m.f = Eθright - Eθleft
= + 0.34 – -1.66
= 0.34 + 1.66
= 2.00 V
SELF-CHECK 2.2

1. The diagram below shows a Daniel cell.

(a) Indicate with an arrow on the diagram the direction of flow of electrons.
(b) What is the purpose of the salt bridge?
(c) Write an equation for the reaction taking place at:
(i) Zinc rod
(ii) Copper rod.
(d) The voltmeter connected across the cell showed a reading of 0.90 V. Calculate the
energy supplied by the cell. (1F = 96500C)
(e) If the Zn/Zn2+ in the left hand side beaker is replaced by Mg/Mg2+, would the
reading on the voltmeter be less than, equal to or greater than 0.90 volts?
(f) Give a reason for your answer in (e).

76
2. UNEB 1991 Q.4
Figure 1 shows a diagram of an electrochemical cell.

(a) (i) Write an equation for the overall reaction.

(ii) State what would be observed if the reaction is allowed to continue for a
long time.
(b) The reading on the voltmeter, V, was 1.10 V. Calculate the energy in KJ produced.

3. UNEB 1997 Q.14

(a) Draw a diagram of Daniel cell consisting of a zinc rod dipped in zinc sulphate
solution and a copper rod dipped in copper (II) sulphate solution; the solutions
separated by a porous wall and the rods are connected by conducting wire.
(b) Indicate: (i) the charges on each electrode.
(ii) the direction of electron movement in the wire.
(c) Write: (i) equations at each electrode.
(ii) an equation for the overall reaction.

4. (1998 Q.5)
Figure 2 below shows a simple voltaic cell.

(a) (i) Write equation for the reaction taking place at:
 the cathode
 the anode
(ii) Write the overall equation of the cell reaction.
(b) Draw an arrow on the diagram to show the direction of flow of electrons.

77
5. (2002 Q.4).
The cell conversion for an electrochemical cell is shown below.
Zn (s) / Zn2+(aq) // Pb2+(aq) / Pb(s)

(a) Name two substances that could be used as electrolytes.

(b) State which one of the electrodes is the anode.
(c) Write equation for the reaction at.
(i) The anode.
(ii) The cathode.
(d) Write equation for the overall cell reaction.

78
 CHAPTER THREE
RATE OF REACTION
Rate Reaction of reaction - is the amount of reactant(s) used up per unit time.
Or - is the amount of product(s) formed per unit time.

Amount
i.e. Rate =
Time

The units of rate are: - mol l-1s-1

- g l-1s-1

Determination of rate of reaction

Consider the reaction between calcium carbonate and dilute hydrochloric acid.
The determination of rate of reaction can be done by using one of the following methods:
(i) Measuring the volume of a gas evolved with time.
(ii) Measuring the decrease in mass of the reactants with time.
(iii) Measuring the rate at which a colourless substance changes to coloured or the rate
at which a coloured substance changes to colourless.

Consider the reactions below.

(i) Reaction between Calcium carbonate and Hydrochloric acid
The equation for the reaction is:

CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g)

Facts about the reaction:

 If the reaction takes place in an open container e.g. a beaker and the reaction vessel
is place on a weighing machine, the total mass of reaction decreases with time. This
is due to the loss of carbon dioxide to the atmosphere.
 If the reaction takes place in a closed system where the gas collects in a syringe, the
volume in the syringe is seen increasing with time.
If the results are entered in a table and then plotted against time, the graph obtained is called
the Rate Curve.

79
 The Rate Curves
(a) When the mass of reactant is plotted against time, the shape of the graph below is obtained.
The graph of mass of calcium carbonate against time

Interpretation of the shape of the graph

At time t = 0, the mass of the reactants is high. As time increases, the calcium carbonate is used up and the mass of the reactants
drops very fast at first and the drop reduces as time increases. When all the calcium carbonate is used up, the mass remains
constant.

(b) When the volume of gas evolved is plotted against time, the shape of the graph below is
obtained.

The graph of volume of carbon dioxide against time

Interpretation of the shape of the graph

A rate curve shows that the rate of chemical reaction is high at the beginning, then
decreases with time and finally becomes zero.

Explanation
At the beginning of the reaction, the concentration of the reactants is high. As a result the
frequency of collision of the reacting particles is high. Hence the reaction is fast. But as the
reactants get used up the rate of reaction decreases with time (I.e. slows down) until finally
it becomes zero. The reaction stops when one of the reactants is used up, hence the volume
remains constant

80
Reading of volume of gas from the graph
(a) Reading volume of a gas at a given time
To find the volume of a gas produced at a given time, say, t1 from the graph, draw a
perpendicular line to the time axis at the time given, to meet the curve.
Read off the volume, v1, where the perpendicular line meets the rate curve.

The graph of volume of carbon dioxide against time

(b) To determine the rate at a given time

To obtain the rate of reaction at time, t2 draw a tangent to the curve at the time, t2,
given. Complete the tangent to obtain a right angle triangle.
y
Find the gradient of the tangent, , gives the rate of the reaction at time, t2,
x
given.

(c) To determine the average rate between two points

To find the average rate between points M and N, divide the change in volume by
v
the change in time i.e. . Units are cm3s-1.
t

81
(d) End of the reaction - is the point at which the graph starts to level off. i.e point X
in the graph below,

NB:
 The other substances that may be used instead of Calcium carbonate and dilute
hydrochloric acid are:
 Magnesium and Hydrochloric acid or dilute sulphuric acid.
 Zinc and dilute Hydrochloric or dilute sulphuric acid.
 Hydrogen peroxide and Manganese (IV) oxide.

Factors affecting rates of reaction

The factors that affect the rate of Chemical reactions include the following:
(i) Concentration of reactants.
(ii) Temperature.
(iii) Pressure.
(iv) Catalyst.
(v) Surface area of the reactants.
(vi) Light.

The study of rates of reactions is known as Chemical Kinetics. The explanations for many
of the effects of the above factors are based on the Kinetic Theory.

82
1. The effect of concentration of reactants on the rate of reaction
Experiment 1
To investigate the effect of concentration on the rate of reaction
Sodium thiosulphate reacts with hydrochloric acid to form sulphur which appears as yellow
colouration according to the following equation:

S2O32- (aq) + 2H+ (aq)  H2O (l) + SO2 (g) + S (s)

Since this reaction produces a precipitate from two colourless solution, the intensity of the
precipitate at any given moment in time represents the extent of the reaction.

Procedure
(a) Place 50 cm3 of sodium thiosulphate solution (containing 40 g dm-3 of the
compound) in 100 cm3 beaker.
(b) Add 5 cm3 2M hydrochloric acid and at the same time start a stop-clock, or note the
time on watch reading in seconds.
(c) Swirl the beaker carefully a couple of times to mix it and place it on a white piece
of paper with a cross drawn on it.
(d) Observe the cross by looking down through the solution from above the beaker and
stop the clock (or record the time) at the moment the cross is invisible.
(e) Take 40, 30, 20 and 10 cm3 of the original sodium thiosulphate solution and make
the total volume up to 50 cm3 by adding distilled water in different beakers. (This is
to ensure the same depth of solution in the beaker for each reaction, so that the time
of disappearance of the cross is made at the same stage in each of the reactions.)
(f) For each, solution, repeat the procedure (b) to (d).
(g) Enter your results in the table below.

Volume of original
Sodium thiosulphate 10 20 30 40 50
solution (cm3)
Time (s)

NOTE: You should notice from your table of results that as the concentration of
thiosulphate solution increases the time taken for the disappearance of the
cross decreases.

Conclusion: The rate of reaction increases with increase in concentration.

83
 Explanation
The rate of reaction therefore depends upon the frequency with which the particles collide,
which intern depends on their concentration.
With increase in concentration, there are more reacting particles so the chances of
collision and hence the rate of reaction is high. I.e. the more crowded the particles are the
more often they pump into one another.

2. The effect of temperature on the rate of reaction

Experiment 2
To investigate the rate of reaction by using sodium
thiosulphate

You are provided with the following:

BA1, which is a sodium thiosulphate solution.
BA2, which is a dilute hydrochloric acid.

Sodium thiosulphate reacts with hydrochloric acid to form sulphur which appears as yellow
colouration according to the following equation:
S2O32- (aq) + 2H+ (aq)  H2O (l) + SO2 (g) + S (s)
The rate of the reaction at a particular temperature can be followed by noting the time taken
for the yellow coloration to appear at that temperature.

You are required to investigate how the rate of the reaction varies with the
temperature for the reaction.

Procedure:
(a) Mark a small cross (X) with a blue/black pen on a sheet of white paper provided
and place it on the table.
(b) Place a 250 cm3 conical flask right onto the cross.
(c) Using a measuring cylinder, transfer 50 cm3 of BA1 into the conical flask which is
on the cross.
(d) Using another measuring cylinder, measure 5.0 cm3 of BA2 and add at once to the
solution of BA1 in the conical flask, and at the same time start the stop clock
(watch). Shake to mix and place the flask over the cross.
Note and record the temperature of the mixture. View the cross from above
through the mixture. Note and record the time taken for the yellow coloration to
just make the cross invisible. (This is the time, t (in seconds) for the reaction to
occur at room temperature.)
(e) Transfer a fresh 50 cm3 of BA1 into the conical flask, and heat the solution to 30 C.

84
 (f) Add 5.0 cm3 of BA2 to the hot solution and the same time start the Stop-clock
(watch).
(g) Shake to mix and place the flask over the cross.
(h) Look at the cross from above through the mixture.
(i) Note and record the time taken for the yellow coloration to just make the cross
invisible. (This is the time, t (in seconds) for the reaction to occur at 30 C
temperature.)
(j) Repeat procedures (e) to (i) for the reaction at 40, 50 and 60 C respectively.
Record your results in the table below.
Spacemen results:
Temperature (C) Room temperature 30 40 50 60
25
Time, t, for yellow coloration 46 45 25 16 10
to cover the cross (s)
1 -1 0.02 0.02 0.02 0.06 0.1
(s )
t

Questions:
1
(k) Calculate the reciptical of the time ( ) for each reaction temperature and record the
t
values in the table above.
1
(l) Plot the graph of (along the vertical axis) against temperature (along the
t
horizontal axis).
1
The sketch of the Graph of against temperature
t

(m) State how the rate of reaction varies with temperature.

The rate of reaction increase with temperature.

85
 Explanation
When the temperature of a reaction is increased, heat energy is supplied to the reacting
particles involved in the reaction. As the particles acquire K.E they move faster and thus
collide with one another at more frequent intervals, hence the reaction proceeds at a faster
rate.
Or
As temperature increases, the total energy of molecules increases so the number of
molecules having the activation energy required for the reaction also increases.

3. The effect of Surface area of a reactant on the rate of reaction

For Heterogeneous reaction (reaction where the reactants are at the different states), particle
size influences the rate of reaction. For example the reaction between marble chips and
dilute hydrochloric acid shows that, the smaller the pieces of marble the greater the rate of
reaction. This is attributed at the greater surface area of the marble for attack by the acid in
the case of the smaller pieces.
In other words, a powder has a larger total surface area for reaction than granules. So
more particles react at any given time.

4. The effect of Catalyst on the rate of reaction

A catalyst is a substance which alters the rate of a chemical reaction and remains
chemically unchanged at the end of the reaction.

A catalyst usually increases the rate of reaction by lowering the activation energy so,
although energy remains the same, the energy of a greater number of molecules now
exceeds the new activation. This is called positive catalysis.

5. The effect of light on the rate of reaction

Light is source of energy and can influence the rate of some chemical reactions by
energizing some of the molecules involved.
For example the reaction between chlorine and hydrogen at ordinary pressure is negligible
in darkness. Slow in diffuse light but explosive in sun light (at room temperature).
Light is also vital factor in photosynthesis.
One of the commonest examples of the effect of light on the rate of chemical reaction is in
photography.

6. The effect of Pressure on the rate of reaction

Pressure influences the reactions in which the reactants are gases.
High pressure brings gas molecules closer so they collide more often, leading to faster
chemical reaction.

86
Typical results of graphs obtained by taking the volume of gas produced against time for different
factor which determine the rate of reaction.

87
REACTION RATES
SELF-CHECK 3.1
1. (2001 Q.15). A catalyst is a substance which
A. controls the concentration of reactants.
B. stops a reaction from becoming violent.
C. increases the rate of a chemical reaction.
D. neutralises a base.
2. (2000 Q.14). The rate of the chemical reaction between calcium carbonate and
hydrochloric acid can be determined by the
A. concentration of carbon dioxide produced.
B. temperature of carbon dioxide produced.
C. volume of carbon dioxide produced.
D. pressure of carbon dioxide produced.
3. (2000 Q.28). Increasing concentration increases the rate of a reaction because the particles
A. move faster. B. collide more often.
C. have more energy. D. collide with more force.
4. (1999 Q.27). The graph in figure 1 shows the variation in the volume of hydrogen evolved
with time when Zinc reacts with dilute sulphuric acid using Copper (II) sulphate as a
catalyst.

The best explanation for the shape of the graph between X and P is
A. the zinc is used up. B. the products stopped the reaction.
C. sulphuric acid is used up. D. the catalyst is used up.
5. (1996 Q.21). The graph X below shows the variation in mass with time when 50 g of
calcium carbonate powder was reacted with excess 1M HCl at 25oC.

To obtain graph Y for the same reaction, one would keep all other conditions of the
reaction the same but
A. use 2M hydrochloric acid. B. use 50 g of calcium carbonate lumps.
C. reduce the temperature to 12,5oC. D. use 25 g of calcium carbonate powder.

88
 SELF-CHECK 3.2
1. (a) Hydrogen peroxide, H2O2, decomposes in aqueous solution to give
oxygen gas. The equation for the reaction is:
2H2O2 (aq)  2H2O (l) + O2 (g)
The reaction is catalyzed by manganese (IV) oxide. In two separate experiments 100 cm3 of
solution of hydrogen peroxide was stirred with the same quantity of manganese (IV) oxide
at 20C. In the first experiment coarse particles of the catalyst were used, while in the
second experiment a finely powdered form of the catalyst was used. The volume of oxygen
produced was measured at 1 minute intervals. The results obtained are shown of graphs
labeled A and B below.

120
100 A
80 B C
60
40
20
2 4 6 8 10 12 14 16

(i) State which graph corresponds to the coarse particles of catalyst. Give reason for
your answer.
(ii) In both experiments, the volume of oxygen evolved per minute decreased with time.
Give an explanation to this.
(iii) If experiment B were carried out at 10 C instead of 20C, sketch on the graph the
type of curve you would expect o obtain (label it C) and give an explanation to
support it.
(iv) Why do all the graphs level at the same value (100 cm3)?

89
 EXTRA STUDY QUESTIONS
(For better results plot the required graphs)
2. (2003 Q.11).
(a) What is meant by rate of a chemical reaction?
(b) State how the following factors affect the rate of a chemical reaction:
(i) Temperature.
(ii) Surface area of the reactants.
(c) The table below shows the volume of hydrogen collected at various time intervals
when magnesium was reached with a 2M hydrochloric acid.

Time (s) 0 1 2 3 4 5 6 7

Volume of hydrogen collected (cm3) 0 25 45 60 70 75 77 77

(i) Plot a graph of volume of hydrogen versus time.

(ii) Determine the rate of the reaction at 3 seconds.
(iii) Determine the volume of hydrogen evolved at 3.5 seconds.
(d) State how the rate of the reaction at 3 seconds would be affected if a 1M
hydrochloric acid was used.
3. (1999 Q. 2).
(a) State the factors that can affect the rate of a chemical reaction.
(b) A mixture of a known mass of magnesium and a certain volume of 2M hydrochloric
acid was put in a conical flask and the mass of the mixture was recorded at various
intervals. The result of the experiment is shown in the graph below.

On the same axes, draw a graph that would be obtained when same mass of
magnesium was reacted with the same volume of 1 M hydrochloric acid.
(c) 5.0 g of calcium carbonate was reacted with 20 cm3 of 2 M hydrochloric acid.
(i) Write the equation for the reaction between hydrochloric acid and calcium
carbonate.
(ii) Calculate the mass of calcium carbonate that was left.
(Ca = 40; O = 16; C = 12)

90
4. (1999 Q.11).
(a) (i) What is the rate of reaction?
(ii) How does particle size affect rate of reaction?
Explain your answer.
(b) The table of results below shows the time taken for Sulphur to form when various
concentrations of sodium thiosulphate were used.

Concentration of S2O32- (M) 0.2 0.6 0.8 1.2 1.6

Time for sulphur to form (Sec) 60 20 15 10 7.5

1/t (Sec-1) 0.017 0.05 0.07 0.10 0.13

Plot the graph of 1/t (sec-1) vertical against concentration of thiosulphate.

(c) (i) Explain the relationship between a rate of the reaction and 1/t.
(ii) Deduce from the graph, how the rate of reaction varies with the
concentration of thiosulphate.
(d) Name one reagent that you would use to test for sulphur dioxide and state what
would be observed if the reagent was used.

5. (1999 Q.12).
In an experiment to determine the rate of reaction between zinc and sulphuric acid, dilute
sulphuric acid was reacted with zinc granules to which some copper (II) sulphate solution
was added. The volumes of hydrogen gas evolved at various times were measured. The
results are shown in the table below.

Time in minutes 0 5 10 15 20 25 30
Volume of gas in cm3 0 10 20 25.5 29.5 32 32

(a) (i) What is the role of copper (II) sulphate solution?

(ii) Write an ionic equation for the solution in the reaction above.
(iii) Explain what would happen to the reaction if zinc granules were replaced
with zinc powder.
(b) (i) Plot the graph of volume of hydrogen evolved (vertical) against time.
(ii) Describe how you would determine the rate of the reaction at 12 minutes.
(iii) Compare the rate of reaction at 12 minutes with that of 20 minutes. Give a
reason for your answer.
(iv) What happens to the shape of the graph after 25 minutes? Explain your answer.

91
 CHAPTER FOUR
ORGANIC CHEMISTRY

Organic chemistry is a branch of chemistry which deals with the study of carbon and its
compounds.

The compounds referred to here exclude:

- the oxides of carbon, Carbon dioxide (CO2) and carbon monoxide, (CO),
- the metallic carbonates and hydrogen carbonates(bicarbonates).

Modern chemists consider organic compounds to be those containing carbon and one or
more other elements, most often: hydrogen, oxygen, nitrogen, sulphur, or the
halogens, but sometimes others as well.

Uniqueness of Carbon
Carbon is a unique element. Its atoms have the ability to join and form straight chains,
branched chains and rings. The unique properties of carbon are manifest in the simplest
class of organic compounds, the aliphatic, called hydrocarbons.

Hydrocarbons
Hydrocarbons are organic compounds that contain carbon and hydrogen only.
The general formula of hydrocarbons is - CxHy
Where x and y are whole numbers.

Classes of Hyrocarbons
Hydrocarbons are classified into:
- Alkanes,
- Alkenes and
- Alkynes.

Alkanes
Alkanes are saturated hydrocarbons with the general formula CnH2n + 2.
Where n = 1, 2, 3 . . . for successive members of the group.

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 The table below shows the molecular formula and names of the first ten members
of the alkanes class

Value of n Molecular formula Name

1 CH4 Methane
2 C2H6 Ethane
3 C3H8 Propane
4 C4H10 Butane
5 C5H12 Pentane
6 C6H14 Hexane
7 C7H16 Heptane
8 C8H18 Octane
9 C9H20 Nonane
10 C10H22 Decane

Structure of Alkanes
A saturated compound is a compound in which all the valency electrons are used in
bonding.

In alkanes, carbon atom(s) exercise a valency of four (4). I.e. all the valency electrons
are shared with other atoms to form covalent bonds. Therefore each carbon atom is
surrounded by four covalent bonds, thus making alkanes saturated compounds.

Molecular formula, Structural formula and Displayed formula

(i) Molecular formula

The molecular formula of a compound is the formula that indicates the number
of each kind of atom in a molecule of that substance. E.g. The molecular
formula of butane ( C4H10.), shows that, there are four carbon atoms chemically
bonded to 10 atoms of hydrogen.

(ii) Structural formula

The structural formula of a compound is the formula that shows the sequence
and arrangement of the atoms in a molecule of the compound.

Name Molecular formula Structural formula

93
 Ethane C2H6 CH3CH3

Propane C3H8 CH3CH2CH3

Butane C4H10 CH3CH2CH2CH3

(iii) Displayed (Graphic) formula

The displayed formula of a compound is the formula that shows the sequence
and arrangement of the atoms and their bonds in a plane.

Name Molecular formula Structural formula

H
Methane CH4 H-C-H
H
H H
Ethane C2H6 H-C–C-H
H H
H H H
Propane C3H8 H-C–C–C-H
H H H
H H H H
Butane C4H10 H-C–C–C–C-H
H H H H

Homologous Series
Homologous Series Is the family of related carbon compounds.
Compounds of the same homologous series are referred to as members.

Properties of the Members of Homologous series

(i) They have the same general formula.
(ii) They have the same name ending.
E.g. - ane – for alkanes,
- ene – for alkenes and
- yne – for alkynes.
(iii) They differ from each other by a methlene group (- CH2-).

94
 (iv) They show a gradation of physical properties due to sucessive increase
in molecular mass.
(v) They can be prepared in the laboratory by the same general method.
(vi) They have similar chemical properties.

Isomers and Isomerism

(i) Isomers
Isomers are compounds having the same molecular formulae but different
structural formulae.

(ii) Isomerism
Isomerism is the occurrence of two or more compounds with the same
molecular formula but different structural formulae.

The table below shows some isomers and their formulae

Molecular
Name Structural formula Displayed formula
formula
H H H H
Butane C4H10 CH3CH2CH2CH3
H - C – C – C – C -H
H H H H

H H H
CH3CHCH3
2-methylpropane C4H10
H-C–C– C-H
CH3
HH-C-H H
H

Properties of Alkanes
(a) Physical Properties
(i) The - lower (first four) members are gases.
- next twelve (C5H12 to C16H34) are liquids.
- rest are solid e.g. paraffin wax.
(ii) All are insoluble in water but soluble in organic solvents e.g. ethanol,
benzene, toluene etc.
(iii) All have low densities, less than the density of water and the
densities rise gradually with relative molecular mass.

95
(b) Chemical properties
(i) Combustion.
Combustion is divided into complete and incomplete combustion.
 Complete Combustion:
Alkanes burn in sufficient (plenty) air forming carbon dioxide and water.
CH4 (g) + 2O2 (g)  CO2 (g) + 2 H2O (g)

 Incomplete Combustion:
In insufficient (limited) air, alkanes burn forming carbon monoxide or
carbon and water.
2 CH4 (g) + 3 O2 (g)  2 CO (g) + 4 H2O (g)

CH4 (g) + O2 (g)  C (s) + 2 H2O (g)

(ii) Substitution Reaction

Since alkanes are saturated compounds, they undergo substitution
reaction with halogens. The halogen atoms such as chlorine and bromine
substitute one or more of the hydrogen atoms.

E.g. Reaction with chlorine in diffused sunlight.

CH4 (g) + Cl2 (g)  CH3Cl (g) + HCl (g)
CH3Cl (g) + Cl2 (g)  CH2Cl2 (g) + HCl (g)
CH2Cl2 (g) + Cl2 (g)  CHCl3(g) + HCl (g)
CHCl3 (g) + Cl2 (g)  CCl4 (g) + HCl (g)

Uses of Alkanes
They are used : (i) As fuel for cooking and lighting.
(ii) For making Lubricants.
(iii) For making Printer’s ink.
(iv) For making Paints.
(v) For making Carbon paper.

96
 Alkenes
Alkenes are unsaturated hydrocarbons with the general formula CnH2n .
Where n = 2, 3, 4 … for successive members.
This family of hydrocarbons is characterized by one or more double bonds between two
carbon atoms (C=C). Due to the double bond, alkenes are reactive. This is because the
double bond is readily converted to a single bond by addition of other atoms.

Structure of Alkenes
The table showing the molecular formula and names of the first three members of
alkenes class

Value of n Name Molecular Structural Displayed

formula formula formula

H H
2 Ethene C2H4 CH2=CH2 C=C
H H
H H H
3 Propene C3H6 CH3CH=CH2 H-C-C= C
H H

But-1-ene CH3CH2CH=CH2 H HH H
H-C- C-C=C
H H H
4 C4H8

But-2-ene CH3CH=CHCH3 H H H H
H-C- C=C-C-H
H H

NB: 1. But-1-ene and But-2-ene are isomers.

2. The most important alkene is ethene.

97
Ethene
Laboratory Preparation of Ethene
Ethene is prepared in the laboratory by dehydration of ethanol. A mixture of concentrated
sulphuric acid and ethanol is heated to a temperature of 170 C. The gas is passed through
solution of caustic soda to remove sulphur dioxide and is collected over water.

The Diagram Showing Lab Preparation of Ethene

C2H5OH (l)  C2H4 (g) + H2O (l)

Properties of Ethene
(a) Physical properties
(i) It is a colourless gas.
(ii) It is slightly soluble in water.
(iii) It has a faint smell.
(iv) Its density is about the same as that of air.
(v) It undergoes polymerization.

(b) Chemical Properties

(i) Combustion
Combustion is divided into complete and incomplete combustion.

98
  Complete Combustion:
Ethene burns in sufficient (plenty) air forming carbon dioxide and water
according to the equation.
C2H4 (g) + 3 O2 (g)  2 CO2 (g) + 2 H2O (g)

 Incomplete Combustion:
In insufficient (limited) air, ethene burns forming carbon monoxide or carbon
and water.
C2H4 (g) + 2 O2 (g)  2 CO (g) + 2 H2O (g)

(ii) Addition reaction of ethene

Due to the double bond in ethene, it undergoes addition reaction with hydrogen and
halogens.

 Hydrogenation:
Ethene reacts with hydrogen at 200C in presence of nickel catalyst to form ethane.
C2H4 (g) + H2 (g)  C2H6 (g)

 Bromination:
When ethene is bubbled into bromine water, which is brown, the colour of bromine
becomes decolourised.
C2H4 (g) + Br2 (aq)  C2H4Br2 (aq)
Or C2H4 (g) + Br2 (aq)  CH2Br CH2Br (aq)

Uses of ethene
Ethene is used in the manufacture of:
- plastics
- ethanol
- electrical insulators
- bottles and parts of some apparatus.

Test for Alkenes

(a) Add two drops of a solution of bromine in trichloroethane (CHCl3) to an alkene.
Observation: The brown colour becomes colourless.

(b) Add a few drops of acidified potassium manganate (VII) to an alkene.

Observation: The pink colour of the solution becomes colourless.

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 Alkynes
Alkynes are unsaturated hydrocarbons with the general formula CnH2n - 2.
Where n = 2, 3, 4, …
The functional group of the alkyne series is a triple bond between two carbon atoms (CC).
Alkynes are highly unsaturated compared to alkenes as a result they are more reactive than
alkenes.
Like alkenes, they undergo similar chemical reactions.

E.g. Combustion and addition reactions.

The most important member of alkynes is ethyne (acetylene), HCCH.

Laboratory Preparation of Ethyne (acetylene)

Ethyne is prepared in the laboratory by dropping water on to calcium dicarbide at room
temperature and the gas is collected over water.

Diagram showing laboratory preparation of ethyne

CaC2 (s) + 2 H2O (l)  C2H2 (g) + Ca(OH)2 (aq)

Properties of Ethyne

(a) Physical properties

(i) It is colourless gas.
(ii) It has a sweet smell.
(iii) It is insoluble in water.

100
(b) Chemical Properties
(i) Complete combustion.
Ethyne burns in sufficient air like any other hydrocarbon to form carbon dioxide
and water.
2 C2H2 (g) + 5 O2 (g)  4 CO2 (g) + 2H2O(g)

(ii) Incomplete combustion

In limited supply of air, ethyne burns to form carbon monoxide and water.
2 C2H2 (g) + 3 O2 (g)  4 CO (g) + 2H2O(g)

(iii) Addition reaction with hydrogen

Hydrogen combines in presence of nickel to form ethane.
C2H2 (g) + 2H2 (g)  C2H6 (g)

(iv) Addition reaction with bromine

When ethyne is bubbled through bromine water, the reddish-brown colour of
bromine water or liquid bromine becomes colourless.
C2H2 (g) + 2 Br2 (aq)  C2H2Br4 (aq)

(v) Addition reaction with chlorine

It reacts explosively with chlorine, forming carbon.
C2H2 (g) + Cl2 (aq)  2HCl (g) + 2 C (s)

NB: Under special conditions, chlorine and ethyne combine to form tetrachloroethane.
C2H2 (g) + 2 Cl2 (aq)  C2H2Cl4 (aq)

Uses of Ethyne
Ethyne is used in:
(i) The industrial manufacture of compounds like adhesive and plastics.
(ii) Oxy-acetylene flame, which is used in welding and metal cutting.

101
 Petroleum (Crude Oil)
The word “petroleum” means ‘rock oil’. It is found in a layer of porous rock such as
sandstone or between two layers of non-porous rock. It is a thick green or dark brown
liquid.

Above the oil is a natural gas. It is a mixture of many gases but contains mainly methane,
CH4 (between 80 – 99%).

Isolation of Alkanes from petroleum

Petroleum contains many different alkanes. The alkanes are separated by a process called
fractional distillation into a number of different mixtures called fractions.
The products (or fractions) are not pure compounds. They are simpler mixtures, each
mixture containing several alkanes with different range of boiling points.
A fractionating tower for fractional distillation of petroleum

NB: Petroleum refineries or distilleries are located in isolated areas near seaports rather than far
inland. E.g. Mombassa – Kenya, Dar es Salaam – Tanzania.

This is to avoid: - heavy transport costs of the bulky oil.

- the risks of fire out break.

102
The table below shows the fractions and their boiling point temperature range

Fraction Distilling
Temperature/C
Gas Below 40

Petrol (Naphtha) 40 - 175

Paraffin 175 - 250

Gas oil 250 - 300

Diesel 300 - 350

Lubricating oil 350 - 400

Waxes Above 400

Bitumen

NB: Note that:

As the boiling points of the fractions increase, the viscocity and intensity of colour
also increases while inflammability decreases.

Uses of the fractions

Fraction Uses

Gas Fuel - for heating and lighting in gas cookers

Petrol (Naphtha) Fuel - in motor aviation using petrol engine.

Paraffin Fuel - for heating and lighting.

Gas oil For making petrol, house warming.

Diesel Fuel - in diesel engines

Lubricating oil Lubrication of machine parts for smooth running.

Waxes For making Vaseline, grease, candles.

Bitumen For tar making roads and runways.

Polymers
A polymer is a complex molecule of high molecular mass formed when several molecules of
simple compounds called monomers join up by a process called polymerization.

103
 Polymerization
Polymerization is the joining together of simple molecules called monomers to form a
complex molecule called polymer.

Types of polymerization
There are two types of polymerization, namely:
(i) Addition polymerization
(ii) Condensation polymerization

Addition polymerization
Addition polymerization is the formation of a polymer by the combination of monomers, of
the same kind, without loss or gain of other atoms.

For example, under suitable conditions,

(i) Ethene molecules (monomers) join to form polythene (polymer).

Equation showing addition polymerization

Where: n is in thousands.

(ii) Propene molecules join to form polypropene.

104
 Condensation polymerization
Condensation polymerization is the formation of a polymer by the combination of
monomers with elimination of water molecules.

E.g. - Starch is a polymer formed by condensation polymerization of glucose

molecules with the elimination of water molecules.
- Nylon 6 and nylon 66 are condensation polymers
Types of polymers
There are two types of polymers, namely:
- Natural polymers.
- Synthetic polymers.

(a) Natural polymers:

Natural polymers are polymers that exist in nature. I.e. made by God.
They are formed by condensation polymerization.
Examples of natural polymers are
- Starch, proteins, Fats, Cellulose, Glycogen, Rubber, Wool, Silk, Etc.

(b) Synthetic Polymers

Synthetic Polymers are man made polymers.
Examples of synthetic polymers include:
- Polythene, Polypropene, Perspex, Nylon, Bakelite, Synthetic rubber, Poly Vinyl
Chloride (PVC).

Al the above polymers are plastic in nature, hence can be molded into plastics.
(b) Synthetic Rubber
It is elastic and not strong enough for use. However, its physical properties can be
improved (modified) by heating it with sulphur. The process by which synthetic
rubber is heated with sulphur is known as vulcanization of rubber.
Vulcanised rubber is strong, tough and durable.
105
 Cracking
Cracking is the process by which long chain alkanes are broken down to produce shorter
chain hydrocarbons.
Types of Cracking
There are two types of cracking, namely:
- Thermal cracking
- Catalytic cracking.

(a) Thermal Cracking

This is simply carried out by heating. E.g. cracking of paraffin in the lab.

C10H22 (l)  C7H16 (l) + C3H6 (g)

(b) Catalytic Cracking

Catalytic cracking is carried out using a catalyst.

Plastics
A plastic is a substance which when soft can be made into different shapes.

Classification of Plastics
Plastics are classified into two classes according to their behaviour when heated.
These are: - Thermoplastics (Thermo softening plastics)
- Thermosetting plastics.

106
 (a) Thermoplastics (Thermosoftenıng plastics)
Thermoplastics are plastics which soften on heating and harden on cooling.

Softening and hardening are reversible. As a result they can be molded or remolded
under hot conditions.

E.g. Polythene, polypropene, Perspex, polyvinylchloroethene (PVC). They have

low melting points. This is because they are made of molecules with long chains as
a result the forces between the chains are weak.

(b) Thermo sets (Thermosetting plastics)

These are plastics, which soften or melt on heating during manufacture and take the
shape of the mold in which they are processed on cooling.
They cannot be remolded by heating after manufacture. At high temperatures, they
decompose.
E.g. Bakelite (plastic used for making electric switches and some types of plates).
The molecules in thermosetting plastics have cross linkage. As a result the forces
between the molecules are strong and their structures are three-dimensional
networks.

Uses of Plastics
They are used for making:
- house hold utensils e.g. jags, plates, cups, jericans, combs, shoes etc.

Some uses of specific plastics

Type of Plastic Use

Polythene For making bags, plates, cups, electric insulators, rain coats

Poly Vinyl Chloride (PVC) For making suitcases, radio cases, television sets, computers.

Nylon For making ropes, clothes, fishing nets.

Perspex For making wind screens for cars and aircrafts.

Rubber For making tyres, soles of shoes and tubings

Bakelite For making electric switches, plugs and sockets.

107
Advantages and Disadvantages of Plastics
(a) Advantages
(i) They are light and potable.
(ii) They are good insulators.
(iii) They are good thermal radiators.
(iv) They are resistant to corrosion, hence are used to store acids and alkalis.
(v) Thermosoftening plastics are easy to manufacture and recylcle.
(vi) They can easily be made coloured and therefore become attractive.

(b) Disadvantages:
(i) They are poor heat resistant hence melt and burn easily.
(ii) They are poor weather resistant. As a result they deteriorate on long
exposure to high temperature; become brittle and break easily.
(iii) When burnt, they produce pollutant gases, such as carbon monoxide and
carbon dioxide.
(iv) They are non-biodegradable (i.e. do not rot) hence litter the environment
thus making it dirty.

Alcohols
Alcohols are saturated organic compounds which contain the hydroxyl group (OH) as a
functional group.

They form a homologous series of general formula CnH2n + 1OH or CnH2n + 2O.
Since the hydroxyl (OH) is characteristic of the alcohols, the first formula is usually preferred.
The names of alcohols end with - ol.
The first two members of this series are:
- methanol - CH3OH
- ethanol - C2H5OH
Both exist as liquids at room temperature.

108
 Manufacture of Ethanol by fermentation method
Starch is pressure cooked to release the starch granules and then treated with malt for an hour
at 60C. The starch is hydrolysed by an organic catalyst called diastase to sugar (maltose).
2 C6H10O5 (aq) + H2O (l) Diastase C12H22O11 (aq)
Yeast is added and the mixture is left at room temperature. The maltose is hydrolyzed to
glucose by an enzyme called maltase.
C12H22O11 (aq) + H2O (l) Maltase 2 C6H12O6 (aq)
Another enzyme of yeast, zymase, catalyses the decomposition of glucose to ethanol and
carbon dioxide.
C6H12O6 (aq) Zymase 2 C2H5OH (l) + 2CO2 (g)

NB: The main steps are summarized as follows:

Starch Hydrolysis Sugar Fermentation Alcohol

Concentrating Ethanol Solution:

The “wash” containing less than 11% of ethanol is fractionally distilled at 78 °C – 82°C to a
liquid containing about 95% of ethanol and 5% water. The water can be absorbed by absorption
using quicklime (CaO).

Local Brewing (Production) of Ethanol in Uganda

Maize, cassava and millet flour are mixed with water. The product is soaked in a tin or pot or
buried in the ground for about a week to allow starch to be converted into sugar.
The mixture is then removed, roasted and dried. The product is then allowed to ferment for two
to three days in suitable container. During this process, the sugar formed is converted to crude
ethanol.
The formation of the crude ethanol from the carbohydrates (starch) involves the participation of
enzymes.
The relevant equations for the main reactions involved are given below.

2 C6H10O5 (aq) + H2O (l) Diastase C12H22O11 (aq)

C12H22O11 (aq) + H2O (l) Maltase 2 C6H12O6 (aq)

C6H12O6 (aq) Zymase 2 C2H5OH (l) + CO2 (g)

109
Properties of Ethanol

(a) Physical Properties

(i) It is a colourless liquid.

(ii) It has a boiling point of 78.5C.
(iii) It is highly soluble in water.
(iv) It has a characteristic smell.

(b) Chemical Properties.

 Combustion.
Ethanol burns with a blue flame to form carbon dioxide and water.
C2H5OH (l) + 3 O2 (g)  2 CO2 (g) + 3 H2O (g)
 Reducing properties of ethanol

(i) With acidified potassium dichromate.

When acidified potassium dichromate, orange in colour, is warmed with
ethanol, it turns green.

(ii) With acidified potassium manganate (VII).

When ethanol is added to a solution of acidified potassium permanganate,
purple in colour, the purple colour becomes decolourised.

NB: In each case, ethanol is oxidized to ethanoic acid.

 Dehydration of ethanol
When excess concentrated Sulphuric acid is added to ethanol and the mixture is
heated to 180 C, ethene is formed. This reaction is called dehydration.

C2H5OH (l)  C2H4 (g) + H2O2 (g)

NB: At temperatures below 140 C, ether is predominantly formed.
Uses of Ethanol
Ethanol is used as:
(i) beverage.
(ii) fuel.
(iii) thermometric liquid in thermometers.
(iv) A solvent for organic compounds.

110
 Soaps and Detergents
(a) Soaps:
A soap is a sodium salt of a long chain carboxylic acid with general formula
CnH2n+1COONa. Where n  8.

The common soap has n = 17, (C17H35COONa) and is called sodium stearate in short
written as (NaSt).
Soap is manufactured by a process called saponification.
Manufacture of Soap (Saponification)
Soap is manufactured by boiling a vegetable oil with concentrated sodium hydroxide solution.

Oil or Fat + Sodium hydroxide  Soap + Glycerol

When the process of saponofocation is complete, brine (concentrated sodium chloride)

is added to the mixture to precipitate (to solidify) the soap. Perfumes and colour are also
added. The soap, which floats above the liquid, is removed and pressed into bars.

Fats and Oils are esters of long chain carboxylic acid and glycerol.
Oils occur naturally in plants while fats occur naturally in animals only.
Oils are got from plants such as Sunflower, Groundnut, simsim, corn etc.
Examples of fats include: lard in pigs, butter from milk.
(b) Detergents.
A detergent is a sodium salt of sulphuric acid or hydrogen sulphate.

Preparation of detergents
In the laboratory a detergent can be prepared by boiling a vegetable oil with
concentrated sulphuric acid. The product, hydrogen sulphate, is then neutralized by
adding sodium hydroxide solution. A precipitate forms. The mixture is evaporated
on a water bath to leave a white solid detergent.

Advantages and disadvantages of soaps and detergents

(a) Soaps
Advantages
(i) Soaps are cheap (inexpensive).
(ii) They are friendly to environment.
(iii) They are used for bathing and washing.
111
 Disadvantages
(i) Soap does not form lather easily with hard water. This leads to waste of it.
(ii) They react with calcium and magnesium ions in hard water to form scum
according to the equation:

2 C17H35COONa (aq) + Ca2+ (aq)  (C17H35COO)2Ca (s) + 2 Na+ (aq)

If the cloth is not rinsed properly, the scum leaves dirty stains on it.

Detergents
Advantages
(i) They dissolve in water and form lather easily.
(ii) They do not form scum in hard water.
(iii) They do not leave stains on clothes.

Disadvantages
(i) They are expensive.
(ii) The foam from detergents interferes with the bacterial breakdown of
sewage. The sewage not only pollutes the rivers and lakes but also kills the
aquatic life. E.g Fish.
(iii) Detergent contains fertilizer materials which nourishes algae and other
microorganisms. This reduces oxygen supply in the water. The result is that
fish and other aquatic animals die.

SELF-CHECH 4.1
1. (2005 Q.2).
Which one of the following formulae represents an alkane?
A. C2H4. B. C3H4. C. C4H8. D. C4H10.

2. (2005 Q.18).
The reaction between ethanol and concentrated sulphuric acid to form ethene is called
A. hydrogenation. B. catalysis. C. dehydration. D. hydration.

3. (2004 Q.17).
What name is given to the reaction leading to the formation of soap from oil?
A. Hydrogenation. B. Polymerisation.
C. Saponification. D. Degradation.

4. (2003 Q.14)
Which one of the following substances decolourises bromine water?
A. Ammonia. B. Ethene. C. Methane. D. Chlorine.

112
5. (2003 Q.28).
Which one of the following polymers is a synthetic?
A. Wool. B. Cotton. C. Sisal. D. Nylon.

6. (2003 Q.29).
Which one of the following is not produced during the fermentation of sugar solution?
A. Water vapour. B. Carbon dioxide. C. Methane. D. heat.

7. (2003 Q.32).
In which one of the following processes does ethene form a plastic?
A. Polymerisation. B. Precipitation.
C. Neutralisation. D. Electrolysis.

8. (2000 Q.5).
Which one of the following is not used as a fossil fuel?
A. Hydrogen. B. Charcoal.
C. Coal. D. Ethanol.

9. (2000 Q.12).
Concentrated sulphuric acid reacts with ethanol to form ethene. This shows that
A. the acid is an oxidising agent. B. ethene can be converted to ethanol.
C. ethene has a higher affinity for the acid. D. the acid is dehydrating agent.

10. (2000 Q.37).

An example of a non-biodegradable substance is
A. silk. B. wool.
C. polythene. D. paper.

11. (1999 Q.7).

The process by which property of rubber is improved by heating it with sulphur is called
A. polymerisation. B. vulcanisation.
C. catalysis. D. dehydration.

12. (1999 Q.38).

A gas that when bubbled through bromine water changes the colour of bromine water from
reddish-brown to colourless is
A. methane. B. ethene. C. ethane. D. hydrogen.

13. (Q.40).
Which one of the following sets contains natural fibres only?
A. Nylon, wool, cotton. B. Cotton, nylon, silk.
C. Silk, nylon, wool. D. Cotton, wool, silk.

113
14. (1998 Q.25).
One of the disadvantages of using detergents for washing is, they
A. are precipated in hard water to form scum.
B. sometimes cause stains on clothes.
C. cause water pollution in rivers and lakes.
D. wash away the colour of clothes.

15. (1998 Q.26).

The colourless gas produced durıng fermentatıon of sugar ıs
A. ammonia. B. carbondioxide. C. hydrogen. D. oxygen.
16. (1998 Q.28).
Which one of the following hydrocarbon contains multiple bonds?
A. CH4. B. C2H2. C. C2H4. D. C3H4.

17. (1998 Q.34).

A colourless gas which decolourises bromine water is
A. chlorine. B. ethene.
C. sulphur dioxide. D. ethane.

18. (1998 Q.40).

Which one of the following can be used to test for ethene?
A. Lime water. B. Bromine water.
C. Glowing water. D. Potassium dischromate paper.

19. (1997 Q.21).

Which one of the following is a synthetic polymer?
A. Wool. B. Cotton. C. Silk. D. Nylon.

20. (1996 Q.2).

Which one of the following is a monomer of protein?
A. Glucose. B. Amino. C. Isoprene. D. Ethene.

Each of the questions 21 to 24 consists of an assertion (statement) on the left-hand side and a
reason on the right-hand side.
Select:
A. if both the assertion and the reason are true statements and the reason is a correct
explanation of the assertion.
B. if both the assertion and the reason are true statements but the reason is not a correct
explanation of the assertion.
C. if the assertion is true but the reason is not a correct statement.
D. if the assertion is not correct but the reason is a correct statement.

114
 INSTRUCTIONS SUMMARISED:
Assertion Reason
A. True True (reason is a correct explanation.)
B. True True (Reason is not a correct explanation.)
C. True Incorrect.
D. Incorrect True.

21. (1995 Q.43)

Soap can remove both dirt and
because soap is made from cooking oil.
oil from cloth
22. (1995 Q.46)
When ethene burns in air, it produces a of the presence of unburnt
because
smoky flame carbon.
23. (1996 Q.44)
Ethene readily decolourises bromine
because it is saturated.
water
24. (1995 Q.34)
Crude petroleum is refined by fractional its fractions have different
because
crystallisation boiling points.

In each of the questions 25 to 26 one or more of the answers given may be correct. Read each
question carefully and then indicate the correct answer according to the following:
A. If 1, 2 and 3 only are correct. B. If 1 and 3 only are correct.
C. If 2 and 4 only are correct. D. If 4 only is correct.
25. (1997 Q.32)
How does soap differ from a detergent?
1. A detergent is made from vegetable oil but soap from animal oil.
2. A detergent is not efficient with hard water.
3. A detergent is in powder form only.
4. Soap is not efficient with hard water.

26. (1997 Q.41).

When a mixture of ethanol and concentrated sulphuric acid is heated, a gas is liberated.
Which of the following properties is/are shown by the gas?
1. It is soluble in water. 3. It decolourises potassium manganate (VII).
2. It decolourises bromine water. 4. It forms a white precipitate with limewater.
SELF-CHECK 4.2
1. (2004 Q.9) .
(a) The molecular formula of ethene is C2H4. Write the structural formula of ethene.
(b) Bromine water is one of the reagents that can be used to test for the presence of
ethene. (i) State what would be observed if ethene is tested with bromine
water and write an equation for the reaction.
(ii) Name one other reagent that can be used to test for the presence of
ethene.
(c) Name one compound from which ethene can be prepared.

115
2. (2003 Q.3)
During the manufacture of soap, sodium hydroxide was boiled with substance X.
(a) Identify substance X.
(b) What name is given to the process leading to the formation of soap?
(c) Name a substance that can be used to precipitate the soap from the solution.
(d) State what would be observed if soap solution was reacted with aqueous calcium
hydrogen carbonate.
3. (2001 Q.13 )
(a) Name the raw materials used in your locality to make an alcoholic drink.
(b) Briefly describe how ethanol can be obtained from the materials you have named in
(a).
(c) State how ethanol prepared in (b) can be concentrated and suggest one way of
determining whether the ethanol is pure or not.
(d) Ethene can be formed from ethanol.
(ı) Write equation and state the conditions for the reaction leading to the
formation of ethene.
(ıı) Name two uses of ethanol apart from the preparation of ethene.
4. (1998 Q. 12)
(a) (i) State the difference between fats and oils.
(ii) Give one example of each.
(b) Briefly describe how soap can be prepared.
(c) State what would be observed if soap solution was shaken with a solution
containing magnesium hydrogen carbonate.
(d) Explain your answer in (c) above.
(e) State what would be observed if a soap less detergent was used in (c) instead of
soap solution.
(f) Give one disadvantage of soap less detergents.
5. (1994 Q.7)
(a) Ethanol can be prepared by fermentation of sugars.
Write an equation to show how ethanol can be prepared from glucose, C6H12O6.
(b) When ethanol is heated with concentrated sulpuric acid at 170°C, a substance W is
formed.
(i) Name W.
(ii) Write the structural formula of W.
(iii) Name one reagent that can be used to identify W.

116
6. (1988 Q.2)
Glucose, C6H12O6, can be converted to ethanol by a catalytic reaction caused by an enzyme
produced from yeast.
(a) Name; (i) the reaction in which yeast converts glucose into alcohol.
(ii) The enzyme produced by yeast during the reaction.
(b) Write the equation for that leads to the formation of ethanol.
(c) Briefly describe how the ethanol produced can be concentrated.

7. (1988 Q.14 )
(a) (i) Explain what is meant by the term polymerization.
(ii) Name two naturally occurring polymers and one synthetic polymer.
(b) The structure of a polymer is shown below.

Write down the structural formula of the monomer of the polymer.

(c) Distinguish between a thermoplastic and a thermosetting plastic.
(d) Explain the term cracking.
(e) Draw a fully labeled diagram of the apparatus that can be used to crack liquid
paraffin in the laboratory.

8. (1987 Q. 3).
The structure of an organic substance A is shown below.

(a) Name A.
(b) A reacts with excess concentrated sulphuric acid at 170˚ C to form an organic product
B.
(i) Name B.
(ii) Write the structure of B.
(iii) Name one reagent that could be used to detect the presence of B.
(iv) State what would be observed if the reagent named in (iii) was used.

117
 CHAPTER FIVE

NITROGEN AND ITS COMPOUNDS

Nitrogen
Nitrogen, symbol N, with atomic number 7 and mass number 14, is a non-metal. It belongs
to group V and period II in the Periodic Table.

Occurrence:
It occurs as uncombined gas in air forming 78.9% by volume of air.
It also exists combined in: - minerals such as Nitrates and Nitrites,
- oxides e.g. NO, NO2.

Preparation of Nitrogen from air

A sample of air is first filtered, dried by passing it through silcagel and then passed through
a sodium hydroxide solution to remove carbon dioxide according to the equation: 2NaOH
(aq) + CO2 (g)  Na2CO3(aq) + H2O (l)

The remaining air is passed over heated copper turnings in a glass tube to remove oxygen
according to the equation:
2Cu (s) + O2 (g)  2 CuO (s)
The nitrogen left is then collected over water as shown in the diagram below.

Observation:
The brown copper turnings turn black due to the formation of copper (II) oxide (CuO).
NB: - The nitrogen prepared is not pure. This is because it contains noble gases
as impurities.
- Pure nitrogen gas is manufactured by Industrial distillation of liquid air.

118
 Industrial manufacture of nitrogen
Pure nitrogen is manufactured by fractional distillation of liquid air. A sample of air is first
filtered, dried by passing it through silica gel and then passed through a solution of sodium
hydroxide to remove carbon dioxide according to the equation:
2NaOH (aq) + CO2 (g)  Na2CO3 (aq) + H2O (l)

The resulting dry air is liquefied by successive compressing and cooling.

The liquid is then fractionally distilled. Nitrogen distills off at a boiling point of
–196C and is collected as pure nitrogen gas.

Properties of Nitrogen
(a) Physical Properties
(i) Nitrogen gas is colourless.
(ii) It is odorless.
(iii) It is insoluble in water.
(iv) It does not support combustion.

(b) Chemical Properties

Nitrogen is generally inert. This is because of the triple bond between the two
nitrogen (N  N) atoms in the molecule of nitrogen. To break the triple bond
requires a lot of energy. This makes Nitrogen stable. Therefore Nitrogen reacts
under special conditions such as:
- Very high temperature,
- Pressure and
- Presence of a catalyst.

(i) Formation of metallic Nitrides

Nitrogen reacts with some metals at very high temperatures forming nitrides.
E.g. Calcium and Magnesium
3 Ca(s) + N2 (g)  Ca3N2 (s)
3 Mg(s) + N2 (g)  Mg3N2 (s)
The Magnesium nitride and Calcium nitrides are white solids. They dissolve in
water forming hydroxides and ammonia gas according to the equations below.
Ca3N2 (s) + 6 H2O (l)  2 NH3 (g) + 3 Ca(OH)2 (aq)
Mg3N2 (s) + 6 H2O (l)  2 NH3 (g) + 3 Mg(OH)2 (aq)
The gas is identified by its characteristic smell and action on red
litmus paper.
Lithium also reacts with nitrogen similarly forming lithium nitride.
6Li (s) + N2 (g) 2Li3N (s)
119
 (ii) Formation of Hydride (Reaction with Hydrogen to form ammonia)
Nitrogen reacts with hydrogen in presence of a catalyst of finely divided iron heated at 450C and a pressure of about 200 atmospheres to form
ammonia according to the equation:

N2 (g) + 3H2 (g) 2NH3 (g)

Ammonia Gas (NH3)

Laboratory Preparation of Ammonia
Ammonia is prepared in the laboratory by heating a mixture of ammonium salt and an
alkali. Usually, a mixture of ammonium chloride and calcium hydroxide. The gas is
collected by upward delivery method since it is lighter (less dense) than air.

The diagram showing the laboratory preparation of dry Ammonia gas

Ca(OH)2 (s) + 2NH4Cl (s)  CaCl2 (s) + 2 H2O (l) + 2 NH3 (g)

Properties of Ammonia
(a) Physical Properties
(i) Ammonia is a colourless gas.
(ii) It has a characteristics pungent smell.
(iii) It is lighter than air.
(iv) It is the most soluble of all gases (1cm3 in water dissolves about 800cm3 of
ammonia at room temperature forming ammonia solution).
(v) It liquefies at –33 C.
(vi) It extinguishes a burning splint.
(vii) It does not burn in air.

120
(b) Chemical Properties
(i) Combustion.
 Aided combustion of ammonia.
Ammonia burns in oxygen or air enriched with oxygen giving a yellowish green
flame forming nitrogen and water.
4 NH3 (g) + 3 O2(g)  2 N2 (g) + 6 H2O (g)

The diagram showing burning of ammonia in enriched air

 Catalytic oxidation of ammonia

In presence of a catalyst platinum or copper, ammonia burns in oxygen forming
nitrogen monoxide and water.

4 NH3 (g) + 5O2 (g)  4 NO (g) + 6 H2O (g)

The oxygen oxidizes the nitrogen monoxide to nitrogen dioxide.

2NO (g) + O2 (g)  2 NO2 (g)

Later the gases turn white due to the formation of ammonium nitrate.

4 NO2 (g) + O2 (g) + 2H2O (g)  4HNO3 (g)

4 NH3 (g) + HNO3 (g)  NH4NO3 (s)
(White)
Observation: Reddish-brown fumes and white solid are formed as
explained in the above equations.

(ii) Reaction with Chlorine.

Ammonia burns in chlorine forming a mist of hydrogen chloride gas.
2 NH3 (g) + 3 Cl2 (g)  N2 (g) + 6 HCl (g)

(iii) Reaction with Hydrogen Chloride

Ammonia reacts with hydrogen chloride gas forming dense white fumes of
ammonium chloride according to the equation:
NH3 (g) + HCl (g)  NH4Cl (s)
121
(iv) Reduction of Copper (II) oxide or Lead (II) oxide
Ammonia reduces heated oxide of Copper (II) oxide and Lead (II) oxide to
their metals and it self is oxidized to nitrogen and water.
2 NH3 (g) + 3 CuO (s)  3 Cu (s) + 3 H2O (g) + N2 (g)
Black Brown

Observation:
 The black solid turned brown and
 A colourless liquid that turns white anhydrous copper (II) sulphate blue.

With lead (II) oxide, it reacts according to the equation:

2 NH3 (g) + 3 PbO (s)  3 Pb (s) + 3 H2O (l) + N2 (g)

(v) Solubility
It is the most soluble of all gases. 1cm3 of water dissolves about
800 cm3 of ammonia at room temperature forming ammonia solution.

NH3 (g) + H2O (l)  NH3 (aq)

Or NH3 (g) + H2O (l)  NH4OH (aq)

The high solubility of ammonia is seen in the fountain experiment.

Fountain Experiment
 Fill a large, round-bottomed flask with ammonia and cork it up with a rubber
stopper carrying two tubes fitted with clips.
 Invert and clamp firmly the flask carrying the tubes such that its mouth is under
water coloured with red litmus in a trough as shown in the diagram below.

122
  Open clip B for a moment, close it and allow the few drops of water that entered
to run down into the round part of the flask.
 Then replace the flask and the tubes under the water and open clip A.

Observation: - Water rises rapidly up the tube and it enters the flask as a
fountain and continues until the flask is full.
- The red litmus solution turns blue.

Explanation:
The few drops of water, which entered when tube B was opened, dissolved so much of
ammonia that there was a partial vacuum in the flask.
When clip A was opened, atmospheric pressure forced water rapidly up the tube, and the
water entering dissolves more ammonia thus creating more vacuum hence a fountain is
seen playing until the flask is full of water.
The turning of the red litmus solution to blue shows the alkaline nature of ammonia.

Manufacture of Ammonia (Haber process)

A mixture of dry nitrogen and hydrogen mixed in volume proportions of 1 : 3 respectively
is compressed in a compressing chamber at a relatively high pressure usually between 200
and 500 atmospheres and are then passed over finely divided iron impregnated or mixed
with alumina, aluminium oxide (a catalyst) at a temperature of 450C in a catalytic
chamber. (The aluminium oxide improves the performance of iron by making it porous)
The reaction is reversible and exothermic according to the equation:
N2 (g) + 3 H2 (g) 2NH3 (g) + Heat
The ammonia produced or formed is either liquefied by cooling it in a cooling chamber or
dissolved in water and tapped off. The uncombined nitrogen and hydrogen are passed back
to the compressing chamber for further recycling.

123
 Flow chart for the manufacture of ammonia by Haber process

Test of Ammonia
Ammonia turns moist red litmus paper blue.

NB: Ammonia is the only alkaline gas that turns red litmus blue.

Uses of Ammonia

(i) Manufacture of fertilizers.

(ii) Making of explosives.
(iii) In laundry, ammonia is used to soften water.
(iv) Ammonia is used to prevent fainting and dizziness.
(v) Liquid ammonia is a raw material for the manufacture of hydrogen.
(vi) Manufacture of nitric acid.

124
 Nitric Acid
Laboratory Preparation of Nitric Acid

Nitric acid can be prepared in the laboratory by heating a mixture of metallic nitrate such as
potassium nitrate and concentrated sulphuric acid as shown in the diagram.

The diagram showing the laboratory preparation of nitric acid

KNO3 (s) + H2SO4(l)  KHSO4 (aq) + HNO3 (l)

The nitric acid distils and collects in the condenser as a yellow liquid. The yellow
colouration is due to brown fumes of nitrogen dioxide formed by slight decomposition of
the nitric acid by heat and dissolving in it.

4HNO3 (l) 2 H2O (l) + 4 NO2 (g) + O2 (g)

The yellow colour may be removed by bubbling air or oxygen into the acid.

Manufacture of Nitric Acid

There are three main stages in the manufacture of nitric acid. These include:

1. Oxidation of Ammonia
Ammonia from Haber process is mixed with excess air and the mixture is passed
over platinum catalyst heated at about 800C and 8 atmospheres pressure in the
catalytic chamber. It is oxidized to nitrogen monoxide. The reaction is exothermic
and once, started, maintains the temperature of the catalyst.
4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g) + Heat

125
2. Oxidation of nitrogen monoxide to nitrogen dioxide.
The nitrogen monoxide formed is rapidly cooled and combines with the oxygen
from excess air to form nitrogen dioxide.
2 NO (g) + O2 (g)  2 NO2 (g)

3. Oxidation of nitrogen dioxide and absorption.

The nitrogen dioxide is oxidized by more air and then absorbed in hot water to form
nitric acid.
4 NO2 (g) + O2(g) + 2 H2O (g)  4 HNO3 (g)

The pure ordinary concentrated nitric acid as sold contains about 70% by mass of the
pure acid and 30% of water.

Properties of Nitric Acid

(a) Physical properties
(i) Nitric acid is a colourless, fuming liquid.
(ii) It has a density of 1.5 gcm-3
(iii) It has a boiling point of 85 C at ordinary atmospheric pressure.
(iv) The acid is corrosive and destroys organic matter very readily.

(b) Chemical properties

The chemical properties of nitric acid are divided under two main headings.
 Where nitric acid behaves as an acid (when dilute).
 Where nitric acid behaves as an oxidizing agent (when concentrated).

1. Nitric acid as an acid.
(i) Reaction with bases
It reacts with bases to form salt and water.
E.g 2 HNO3 (aq) + PbO (s)  Pb(NO3)2 (aq) + H2O (l)

126
 (ii) Reaction with carbonates
It reacts with carbonates to form carbon dioxide, metallic nitrate and water
according to the equation:.

2 HNO3 (aq) + CuCO3 (s)  Cu(NO3)2 (aq) + CO2 (g) + H2O (l)

(iii) Reaction with metals

Unlike other dilute acids, dilute nitric acid does not release hydrogen when heated
with metals, except with magnesium.

E.g 2 HNO3 (aq) + Mg (s)  Mg(NO3)2 (aq) + H2 (g)

2. Nitric acid as an oxidizing agent (When concentrated)

(i) Reaction with Copper
When copper is dropped into concentrated nitric acid, it dissolves with
effervescence of reddish brown fumes of nitrogen dioxide and a green
solution containing copper (II) nitrate is formed.
4HNO3 (aq) + Cu (s) Cu(NO3)2 + 2 H2O (l) + 2 NO2 (g)

On dilution with water, the solution turns from green to blue.

(ii) Reaction with sulphur

When sulphur is dropped into concentrated nitric acid and the mixture is
warmed gently in presence of bromine as a catalyst, effervescence of reddish
brown fumes of nitrogen dioxide are evolved according to the equation:
6 HNO3 (aq) + S (s)  H2SO4 (aq) + 2 H2O (l) + 6 NO2 (g)

NB: Whenever nitric acid is reduced, brown fumes of nitrogen dioxide are
evolved.

Uses of Nitric Acid

Nitric acid is used in the Manufacture of:
(i) fertilizers
(ii) explosives
(iii) dyes
(iv) drugs

127
 Nitrates
(a) Nitrates

Nitrates are salts of nitric acid.

The properties of nitrates vary according to the position of the metal in the
reactivity series. The lower the metal is in the series, the more readily and
completely the nitrate of that metal decomposes.

(i) Metallic nitrates that decompose to form two products

Potassium nitrate (KNO3) and Sodium nitrate (NaNO3)

Potassium nitrate and Sodium nitrate melt to form a colourless liquid. On further
heating, they slowly decompose to form a pale-yellow metallic nitrite and oxygen as
shown in the following equations:

Potassium nitrate  Potassium nitrite + Oxygen

2KNO3(s)  2 KNO2(s) + O2 (g)

Sodium nitrate  Sodium nitrite + Oxygen

2 NaNO3(s)  2 NaNO2(s) + O2 (g)

Ammonium nitrate (NH4NO3)

Ammonium nitrate, NH4NO3, decomposes on heating to form di-nitrogen oxide

(N2O) and water.

NH4NO3 (s)  N2O (g) + 2H2O (g)

NB: Samples of ammonium nitrate can be explosive and it is best done in Situ by
using a mixture of ammonium chloride and potassium nitrate.

(ii) Metallic nitrates that decompose to form three products

The metallic nitrates from calcium to copper decompose to form the oxide of the
metal, brown fumes of nitrogen dioxide and oxygen.

Metallic nitrate  Metal oxide + Nitrogen dioxide + Oxygen

128
Example:
 Calcium nitrate  Calcium oxide + Nitrogen dioxide + Oxygen
2Ca(NO3)2(s)  2CaO(s) + 4NO2(g) + O2 (g)

 Magnesium nitrate  Magnesium oxide + Nitrogen dioxide + Oxygen

2Mg(NO3)2 (s)  2MgO (s) + 4NO2 (g) + O2 (g)

 Zinc nitrate  Zinc oxide + Nitrogen dioxide + Oxygen

2Zn(NO3)2 (s)  2ZnO (s) + 4NO2 (g) + O2 (g)

NB: The zinc oxide is yellow when hot and white when cold.

 Iron (II) nitrate  Iron (II) oxide + Nitrogen dioxide + Oxygen

2Fe(NO3)2 (s)  2FeO (s) + 4NO2 (g) + O2 (g)

 Iron (III) nitrate  Iron (III) oxide + Nitrogen dioxide + Oxygen

4Fe(NO3)3 (s)  2Fe2O3 (s) + 12NO2 (g) + 3O2 (g)

 Lead (II) nitrate

When lead (II) nitrate is heated, it decrepitates. That is it decomposes with a
crackling sound giving off brown fumes of nitrogen dioxide and leaving a reddish
brown solid when hot and yellow when cold.

Lead (II) nitrate  Lead (II) oxide + Nitrogen dioxide + Oxygen

2Pb(NO3)2 (s)  2PbO (s) + 4NO2 (g) + O2 (g)

NB: The crackling sound is due to formation of gas molecules that form inside
the crystals and splits them.

 Copper (II) nitrate  Copper (II) oxide + Nitrogen dioxide + Oxygen

2Cu(NO3)2 (s)  2CuO (s) + 4NO2 (g) + O2 (g)

(iii) Metallic nitrates that decompose to form three (3) products, metal, nitrogen dioxide
and oxygen are:
- Silver nitrate and
- Mercury
Why do these nitrates form metals when heated?
The nitrates of these metals decompose to form metal oxides, nitrogen dioxide and
oxygen. But the oxides are unstable to heat and therefore they decompose to the
metal and oxygen.
129
 Silver nitrate  Silver + Nitrogen dioxide + Oxygen
2AgNO3(s)  2Ag (s) + 2NO2 (g) + O2 (g)

Mercury nitrate  Mercury + Nitrogen dioxide + Oxygen

2HgNO3(s)  2Hg (l) + 2NO2 (g) + O2 (g)

The summary of the effect of heat on nitrates

Metal Effect of heat Solubility

K
Nitrates of these metals are decomposed by
Na
heat to give two products:
- The nitrite and oxygen.
Ca
Mg Nitrates of these metals are decomposed on All
Al heating to give three products: Nitrates
Zn - The oxide of the metal, are
Fe - Nitrogen dioxide and soluble
- Oxygen in
Pb
Cu water

Hg Nitrates of these metals are decomposed on heating to give

Ag three products:
- Metal, nitrogen dioxide and oxygen.

130
 NITROGEN AND ITS COMPOUNDS)
SELF-CHECK 5.1
1. (2005Q.14)
Which one of the following nitrates will produce nitrogen dioxide when strongly
heated?
A. Potassium nitrate. B. Sodium nitrate.
C. Zinc nitrate. D. Ammonium nitrate.

2. (2005 Q.17)
The substance which is most suitable for drying ammonia is
A. concentrated sulphuric acid. B. calcium chloride.
C. phosphorus (V) oxide. D. calcium oxide.

3. (2005 Q.25)
The gas that turns brown when exposed in air from the following list is
A. sulphur dioxide. B. hydrogen chloride.
C. hydrogen sulphide. D. nitrogen monoxide.

4. (2005 Q.31)
The catalyst used during the manufacture of nitric acid by oxidation of
ammonia is
A. platinum. B. iron.
C. nickel. D. copper.

5. (2004Q.10)
Under a certain temperature and pressure, hydrogen reacted with nitrogen
according to the equation below: 3H2 (g) + N2 (g) 2NH3 (g)
The volume of nitrogen required to react with 150 cm3 of hydrogen under the
same temperature and pressure is
A. 15.0 cm3 B. 50.0 cm3
C. 300.0 cm3 D. 450.0 cm3
6. (2004 Q.24)
Which one of the following is not a property of ammonia? It is
A. an alkaline gas. B. a reducing agent.
C. soluble in water. D. is denser than air.

7. (2004 Q.28)
Which one of the following oxides can be reduced by dry ammonia?
A. Copper (II) oxide. B. Magnesium oxide.
C. Zinc oxide. D. Calcium oxide.

8. (2004 Q.34)
Ammonia burns in oxygen to yield
A. nitrogen and water. B. nitric acid.
C. nitrogen and hydrogen. D. nitric acid, nitrogen and water

131
9. (2004 Q.36)
Fuming nitric acid was heated and the gas evolved was collected over water.
This gas was
A. nitrogen dioxide. B. oxygen.
C. nitrogen monoxide. D. hydrogen.

10. (2004Q.39)
The process which does not require a catalyst is the manufacture of
A. nitric acid. B. ammonia.
C. sodium hydroxide. D. Sulphuric acid.
11. (2003Q.6)
Which one of the following metals reacts with cold dilute nitric acid?
A. Calcium. B. Copper.
C. Silver. D. Lead.

12. (2003 Q.10)

Which one of the following substances is used as a catalyst in the
manufacture of nitric acid?
A. Vanadian (V) oxide. B. Manganese (IV) oxide.
C. Platinised asbestos. D. Finely divided iron.

13. (2003 Q.35)

Which one of the following hydroxides is normally used in the laboratory to prepare
ammonia from ammonium chloride?
A. Sodium hydroxide. B. Calcium hydroxide.
C. Copper (II) hydroxide. D. Iron (II) hydroxide.

14. (2003 Q.38)

Which one of the following gases forms white fumes with ammonia?
A. Chlorine. B. Carbon dioxide.
C. Sulphur dioxide. D. Hydrogen chloride.

15. (2002Q.39)
When ammonium nitrate is heated, it produces
A. nitrogen dioxide. B. ammonia.
C. dinitrogen oxide. D. nitrogen monoxide

16. (2001Q.19)
Dilute nitric acid reacts with copper to produce
A. copper nitrate, water and nitrogen dioxide. C. copper nitrate, water and ammonia.
B. copper nitrate, water and nitrogen monoxide. D. copper nitrate, water and hydrogen.

17. (2001 Q.24)

The product formed when silver nitrate is heated until there is no further
change is
A. silver oxide, nitrogen dioxide and oxygen. C. silver metal and nitrogen dioxide
B. silver metal, nitrogen dioxide and oxygen. D. silver oxide and nitrogen dioxide.

132
18. (2000 Q.25). Which one of the following metals reacts with nitrogen when heated?
A. Calcium. B. Silver.
C. Copper. D. Sodium.

19. (1999 Q.9).

The catalyst used in the manufacture of nitric acid is
A. iron. B. platinum.
C. iron (III) oxide. D. vanadium (V) oxide

20. (1999 Q.23).

Which one of the following products are formed when concentrated nitric acid reacts
with copper?
A. Copper oxide, water and nitrogen dioxide.
B. Copper (II) nitrate, water and nitrogen monoxide.
C. Copper oxide, water and nitrogen monoxide.
D. Copper (II) nitrate, water and nitrogen dioxide.

21. (1999 Q.37).

Which one of the following nitrates does not decompose on heating?
A. Copper nitrate. B. Lead nitrate.
C. Sodium nitrate. D. Silver nitrate.

22. (1998 Q.3).

Which one of the following nitrates does not produce nitrogen oxide when heated
strongly?
A. KNO3. B. Ca(NO3)2.
C. NH4NO3. D. Cu(NO3)2.

23. (1998 Q.7).

Which one of the following gases is dried using calcium oxide?
A. Sulphur dioxide. B. Hydrogen chloride.
C. Ammonia. D. Hydrogen.

24. (1997Q.24).
The fountain experiment can be demonstrated with ammonia because
ammonia
A. reacts readily with water. C. is lighter than air.
B. is a very soluble gas in water. D. is denser than air.

25. (1997 Q.45).

Which of the following does not occur when sodium nitrate is strongly heated?
A. It melts. C. It liberates oxygen.
B. It liberates nitrogen dioxide D. It loses weight.

133
 SELF-CHECH 5.2
NITROGEN
1. In an experiment, ammonia was prepared by heating an ammonium salt with an
alkali. After drying, 240 cm3 of ammonia gas were collected at room temperature and
pressure. All the ammonia was then reacted completely with 250 cm3 solution of
Hydrochloric acid.
(a) What is meant by the term alkali?
(b) Explain, using the physical properties of the gas, why ammonia is not
collected: (i) Over water
(ii) By downward delivery method.
(c) Ammonia turns wet red litmus paper blue. Which ion in ammonia solution is
responsible for this reaction?
(d) Calculate the number of moles of ammonia gas that were collected in the
above experiment given that one mole of gas occupies a volume of 24,000
cm3 at room temperature.
(e) The equation below shows the reaction between ammonia and sulphuric acid.
2NH3 (g) + H2SO4 (aq)  (NH4)2SO4 (aq)
(i) Explain how crystals of ammonium sulphate could be obtained in this
experiment.
(ii) Calculate the maximum mass of ammonium chloride that could be
obtained in this experiment. (N = 14, H = 1, Cl = 35.5, S = 32, O = 16)

SELF-CHECK 5.3
2. (a) The diagram below shows a set-up that can be used to obtain nitrogen gas in
an experiment.

(i) Name gas A and liquid X.

(ii) What observation would be made in combustion tube after heating for
some time?
(iii) Write an equation for the reaction that took place in the combustion.

134
 (iv) If 320 cm3 of ammonia gas reacted completely with the copper?
Calculate:
(I) the volume of nitrogen gas produced.
(II) the mass of copper(II) oxide that reacted. (Cu = 63.5, O = 16.0,
1 mole of gas occupies 24 dm3 at rt and pressure).
(III) At the end of the experiment the pH of the water in the beaker
was found to be about 10. Explain.

(b) In another experiment a gas jar containing ammonia was inverted over a
burning splint. State what observation was made?
(c) Why is it advisable to obtain nitrogen from air instead ammonia?

3. (2005 Q.11).
(a) Describe how a dry sample of ammonia can be prepared in the laboratory.
(Diagram not required.)
(b) Name a reagent that can be used to test for ammonia and state what would be
observed if ammonia is tested with the reagent.
(c) (i) Draw a labelled diagram of the set up of the apparatus that can be used
to show that ammonia can burn in oxygen.
(ii) Write an equation for the combustion of ammonia in oxygen.
(d) Dry ammonia was passed over heated copper (II) oxide.
(i) State what was observed.
(ii) Write an equation for the reaction.

4. (2002 Q.13).
(a) (i) Draw a labelled diagram to show how a dry sample of ammonia can be
prepared from ammonium chloride in the laboratory.
(ii) Write equation for the reaction leading to the formation of ammonia.
(b) Dry ammonia gas was passed over-heated lead (II) oxide.
(i) State what was observed.
(ii) Write equation for the reaction that takes place.
(c) Describe how ammonia can be converted to nitric acid. Use equation to
illustrate your answer.

135
5. (2000 Q.11).
(a) (i) Name the raw materials used for the manufacture of ammonia.
(ii) Write equation for the reaction leading to the formation of ammonia.
(b) Explain how formation of ammonia is affected by:
(i) Pressure.
(ii) Temperature.
(c) State another factor that affects the formation of ammonia.
(d) Dry ammonia was passed over heated copper (II) oxide until there was no
further change. State what was observed and explain your answer.

6. (1993 Q.12).
(a) Draw a labelled diagram of the apparatus that can be used to prepare ammonia
in the laboratory.
(b) Describe an experiment that can be used to show that ammonia is a soluble
alkaline gas.
(c) A copper coil was heated strongly and held over a concentrated solution of
ammonia in a beaker. Oxygen was then bubbled into the ammonia solution.
(i) State what was observed.
(ii) Explain the observations in (i).
(d) Ammonia reacts with lead (II) oxide according to the equation.
2NH3(g) + 3PbO(s)  N2(g) + 3Pb(s) + 3H2O(l).
Calculate the volume of ammonia at room temperature that would be required
to completely react with 2.5 g of lead (II) oxide.
(N = 14, H= 1, Pb = 207, O= 16)
7. (1995 Q.4).
The diagram in the figure shows the apparatus which can be used to prepare nitrogen
in the laboratory.

(a) Name gas A.

(b) (i) State what would be observed in the combustion tube during the
reaction.
(ii) Write an equation for the reaction.
(c) Name (i) substance X.
(ii) One reagent that can be used to identify X.
136
8. (1991 Q.7).
Substances A and B were obtained from a reaction between ammonia gas and copper
oxide using the apparatus shown in the diagram below.

(a) Name substance

(i) A
(ii) B
(iii) X
(b) Write an equation for the reaction that takes place in the combustion tube.
(c) State why it is not possible to collect excess ammonia in the gas jar.
(d) Name one other oxide that can be used instead of copper (II) oxide.

9. (1992 Q.14).
(a) Describe the industrial preparation of nitric acid from ammonia (Diagram of
the plant is not required). Your description should include equations for the
reactions that occur.
(b) Explain what happens when concentrated nitric acid is added to copper.
(c) Describe one chemical test that can be used to confirm the presence of nitrate.

EXTRA STUDY QUESTIONS

1. (2005 Q.1).
Although nitrogen is generally un-reactive, it readily reacts with a burning
magnesium ribbon.
(a) Give a reason why nitrogen is generally inert.
(b) Burning magnesium reacts with nitrogen.
(i) Give a reason for the reaction.
(ii) Write an equation for the reaction.
(c) Water was added to the product in (b).
Write an equation for the reaction.

2. (1994 Q.8). (a)

(a) A piece of burning magnesium was introduced into a jar of nitrogen.
(i) State what was observed.
(ii) Write an equation for the reaction that took place.
(b) Water was added to product of the reaction in (a) and the resultant mixture
tested with litmus. State what was observed.
(c) Name one other metal that reacts with nitrogen in a similar way to
magnesium.
137
 CHAPTER SIX
SULPHUR AND ITS COMPOUNDS
Sulphur, symbol S, is a yellow non-metallic solid. It belongs to group VI and period III in
the Periodic Table.

Occurrence:
Sulphur occurs as a free element underground. It also occurs in combined form as
sulphides.

E.g. Iron sulphide - FeS

Copper pyrite - CuFeS2
Metallic sulphates and metallic sulphites.
Petroleum oil and natural gas often contain sulphur.

Extraction of Sulphur
Several methods are used to extract free sulphur from the earth. But the most
common method used is the Frasch method, invented in 1891 by the American
chemist Herman Frasch.

The Frasch process for Extraction of Sulphur

In this method, a device called sulphur pump which consists of three concentric pipes
(the largest being about 20 cm (8 in) in diameter) is sunk down to the sulphur deposit.
Through the outermost pipe,steam of water that is super heated at 170C and a
pressure of 10-15 atmospheres is directed underground into the sulphur deposit to
melt the sulphur.
When a sufficient quantity of sulphur has been melted, a blast of hot compressed air
at about 15 atmospheres is forced down the inmost pipe i.e the narrowest pipe and the
mixture of the molten sulphur, water and air is forced up to the surface through the
middle pipe. The sulphur is run into wooden bins where it solidifies on cooling as roll
sulphur, a product that is, about 99.5 percent pure.

138
 A diagram showing Frasch process for Extraction of Sulphur

Allotropes of Sulphur
Sulphur as an allotropic element exists in variety of allotropic crystalline forms at
room temperature and also amorphous (non-crystalline). The most common allotropes
The most common allotropes are:
- Rhombic sulphur and Monoclinic sulphur.

Rhombic Sulphur
Rhombic sulphur is the most stable allotrope of sulphur. It has the following
properties:
(i) It is a yellow, crystalline solid.
(ii) It has a density of 2.06 g/cm3 at 20° C.
(iii) It melts at 119C and
(iv) Boils at 444C.
(v) It is insoluble in water and slightly soluble in alcohol and ether,
moderately soluble in oils and extremely soluble in carbon disulphide.
(vi) When kept at temperatures above 96° C, but below 120° C, the rhombic
form changes to monoclinic sulphur.

The temperature at which rhombic and monoclinic sulphur are in equilibrium, 96

C, is known as the transition temperature.
Monoclinic sulphur
(i) It consists of elongated, transparent, needlelike structures
(ii) It has a density of 1.96 g/cm3 at 20° C.

139
 The difference between rhombic sulphur and monoclinic sulphur

Rhombic Sulphur Mono-clinic Sulphur

Stable below 96C Stable above 96C

Octahedral crystal Needle-shaped crystal
Bright yellow Pale yellow
Melting point 114C Melting point 119C
Density higher (2.08 g/cm3) at 20C Density lower (1.98 g/cm3) at 20C

Two facts prove that the rhombic and monoclinic sulphur are allotropes of sulphur.
1. One gram of monoclinic sulphur slowly changes at room temperature into one
gram of rhombic sulphur.
2. One gram of either form burns in oxygen to yield the same mass (2g) of
sulphur dioxide and nothing else.
S (s) + O2 (g) → SO2 (g)

Uses of Sulphur
(i) Medicinally, it is used to manufacture drugs and skin ointments.
(ii) Industrially, Sulphur is employed in the production of:
- Matches,
- Vulcanized rubber,
- Dyes, and
- Explosives e.g Gunpowder
(iii) In agriculture, finely divided sulphur, mixed with lime, is used as a
fungicide on plants.
(iv) In Photography, sulphur is used to make a salt called hypo (sodium
thiosulphate, Na2 S2O3·5H2O) which is used for “fixing” negatives and
prints.
(v) In education, the most important use of sulphur is in the manufacture of
sulphur compounds, such as:
- Salts e.g - Metallic Sulphites, Sulphates and suphides,
- Acids e.g - Sulphorous acid and Sulphuric acid
- Gases e.g - Sulphur dioxide and Sulphur trioxide.

140
 Sulphur Dioxide

Laboratory preparation of Sulphur dioxide

Sulphur dioxide is prepared in the laboratory by heating a mixture of dilute

hydrochloric acid and sodium sulphite in a round bottom flask and the gas is collected
by downward delivery method. If the gas is required dry, it is passed through
concentrated sulphuric acid.

The diagram showing Laboratory Preparation of dry Sulphur dioxide

Na2SO3 (s) + 2 HCl (aq)  2 NaCl (aq) + H2O (l) + SO2 (g)

Properties of Sulphur dioxide

(a) Physical Properties

Sulphur dioxide has the following properties:
(i) It is a colourless gas.
(ii) It has an irritating smell.
(iii) It is denser than air.
(iv) It is very soluble in water forming sulphorous acid.
(v) It turns moist (damp) blue litmus paper red.

141
(b) Chemical Properties
In its chemical nature sulphur dioxide is an oxidizing agent. It oxidizes
hydrogen sulphide and magnesium and is itself reduced to sulphur, as a result,
a yellow solid is observed.

(i) Reaction with hydrogen sulphide

It reacts with moist hydrogen sulphide forming yellow particles of sulphur.
SO2 (g) + 2 H2S (g)  2 H2O (l) + 3 S (s)

(ii) Reaction with magnesium metal

It reacts with burning magnesium forming yellow particles of sulphur.

SO2 (g) + 2 Mg (s)  2 MgO (s) + S (s)

(iii) Bleaching action

It bleaches flowers and brown paper.

Chemical Test for Sulphur dioxide

(i) A filter paper soaked in acidified solution of potassium dichromate is placed
in a steam of sulphur dioxide.

Observation: The solution turns from orange to green.

(ii) Sulphur dioxide is bubbled through acidified potassium permanganate
solution.

Observation:
The purple colour of the solution becomes colourless.

Uses of Sulphur dioxide

Sulphur dioxide is used in the following areas:
(i) Bleaching of pulp in the paper factory during the manufacture of
paper.
(ii) Preservation of tinned food and drinks.
(iii) Refrigeration as a refrigerant.
(iv) Manufacture of sulphuric acid by contact process.

142
 Sulphuric acid (H2SO4)
Manufacture of sulphuric acid by the Contact Process

The Contact Process

This process uses the following raw materials.
- Sulphur or
- Sulphur dioxide
- Air,
- Concentrated sulphuric acid and
- Water.
There are four main stages involved. They include:-
1. Oxidation of sulphur to sulphur dioxide

Sulphur is burned in air to form sulphur dioxide.

S (s) + O2 (g)  SO2 (g)

2. Oxidation of sulphur dioxide to sulphur trioxide

The sulphur dioxide is mixed with air. The mixture is dried and passed over
vanadium (V) oxide catalyst heated at 450C – 500C and pressure of 1
atmosphere. The sulphur dioxide is oxidized to sulphur trioxide.
2 SO2 (g) + O2 (g)  2SO3 (g)

3. Absorption of Sulphur trioxide

The sulphur trioxide formed is absorbed into concentrated sulphuric acid
forming a fuming liquid called ‘oleum’.
SO3 (g) + H2SO4 (l)  H2S2O7 (l)
(oleum)
4. Absorption, cooling and Dilution
The oleum formed is absorbed, cooled and carefully diluted with the correct
amount of water to form ordinary, sulphuric acid which is 98%.
H2S2O7 (l) + H2O (l)  2 H2SO4 (aq)

143
 The flow diagram for the Contact Process

NOTE:
(i) Vanadium (V) oxide is preferred to platinum as a catalyst in the Contact
process because of the following reasons:
- It is inexpensive i.e. it is cheaper than platinum.
- Unlike platinum it is not easily poisoned by impurities.
(ii) Sulphur trioxide is not dissolved in water directly but in Concentrated
sulphuric acid because:
- There is a violent reaction with the water resulting to a mist of fine
drops of sulphuric acid which fills the factory. This would damage
equipment and harm the plant operators.

Properties of Sulphuric acid

The properties of sulphuric acid are divided into three categories. These are where the
acid acts as: - an acid when dilute,
- an oxidizing agent and
- a dehydrating agent, when concentrated.
(a) Sulphuric acid as an acid when dilute
Dilute sulphuric acid dissociates completely into H+ ions and SO42- ions.
H2SO4 (aq)  2 H+ (aq) + SO42- (aq)
The dilute acid reacts with the reactive metals above hydrogen in the
electrochemical series, bases, carbonates in the usual way.
E.g. (i) With magnesium metal
Mg (s) + H2SO4 (aq)  MgSO4 (aq) + H2 (g)
(ii) With copper (II) oxide
CuO (s) + H2SO4 (aq)  CuSO4 (aq) + H2O (l)
144
 (iii) With sodium carbonate
Na2CO3 (s) + H2SO4 (aq)  Na2SO4 (aq) + H2O (l) + CO2 (g)
(b) Sulphuric acid as an oxidising agent when concentrated
Reaction with copper metal
When a mixture of concentrated sulphuric acid and copper is heated, the copper
dissolves with effervescence, giving off sulphur dioxide and leaving a blue solution
of copper (II) sulphate.
Cu (s) + 2 H2SO4 (aq)  CuSO4 (aq) + 2 H2O (l) + SO2 (g)
NB: This reaction is one of the reactions for the laboratory preparation of sulphur
dioxide.

(c) Sulphuric acid as a dehydrating agent when concentrated

Concentrated sulphuric acid is hygroscopic. That is it has a great affinity for water. It
removes the elements of water from substances.

The dehydrating properties of concentrated sulphuric acid are seen the following reactions:
(i) Dehydration of hydrated copper (II) sulphate
When concentrated sulphuric acid is added to blue crystals of copper (II)
sulphate, it removes the water molecules leaving white crystals of anhydrous
copper (II) sulphate crystals.
CuSO4.5H2O (s)  CuSO4 (s) + 5 H2O (l)
Blue White
Observation:
The blue crystals of copper (II) sulphate turn to white powder.
(ii) Dehydration of hydrated sugar crystals
When concentrated sulphuric acid is added to sugar crystals, the white crystals turn
dark, then swell up forming a black porous mass of carbon. A lot of heat is given out.
This shows that the reaction is exothermic.
C12H22O11 (s)  12 C (s) + 11 H2O (g)
(Black)
(iii) Dehydration of ethanol to form ethene
When a mixture of excess concentrated sulphuric acid and ethanol is heated at a
temparature of 180C, ethene is formed according to the equation:

C2H5OH (l)  C2H4 (g) + H2O (l)

Uses of Sulphuric acid
Sulphuric acid is used in the manufacture of:
(i) fertilizers
(ii) drugs
(iii) dyes
(iv) detergents
(v) plastics
145
 Hydrogen Sulphide
Laboratory Preparation of Hydrogen Sulphide
Hydrogen Sulphide is prepared in the laboratory by the action of dilute hydrochloric
acid on iron (II) sulphide. And the gas is collected by downward delivery method or
over warm water as shown ın the diagram below.

FeS (s) + 2HCl (aq)  FeCl2 (aq) + H2S (g)

If the gas is required dry, it is passed through calcium chloride packed in a u-tube and
is collected by down ward delivery method.

Properties of Hydrogen Sulphide

(a) Physical Properties

(i) It is colourless gas.
(ii) It is poisonous gas.
(iii) It has a foul smell of rotten egg.
(ıv) It is denser than air.
(b) Chemical Properties
It is a good reducing agent, being oxidized to sulphur, a yellow solid.
This is seen in the following reactions.

(i) Reaction with Sulphur dioxide

SO2 (g) + 2H2S (g)  3S (s) + H2O (g)

(ii) Reaction with Chlorine

H2S(g) + Cl2 (g)  S (s) + 2HCl (g)

146
 Pollution of Sulphur Compounds
Sulphur is a major impurity in petroleum and coal. The combustion of these fuels
releases hydrogen sulphide, sulphur dioxide, carbon monoxide and carbon dioxide
into the atmosphere. These gases are air pollutants. They lead to the degradation of
the environment.
Explanation
Sulphur dioxide and hydrogen sulphide are poisonous gases.
(i) Sulphur dioxide
Sulphur dioxide readily dissolves in rain water forming sulphorous acid. This
sulphorous acid is oxidized to sulphuric acid by aerial oxygen.

Effect of the acid rain

The acid rain increases soil acidity, which dissolves away minerals leading to
low yield of agricultural produce. The acid rain also dissolves away minerals
in stone works of buildings thus weakening the buildings.

(ii) Hydrogen sulphide

Hydrogen sulphide also leads to formation of acid rain, but more directly
darkens buildings, which have lead paints. The darkening is due to formation
of black lead (II) sulphide.

SELF-CHECK 6.1
1. (2005 Q.3).
Which one of the following compounds is formed when excess sulphur dioxide is
passed through sodium hydroxide solution?
A. Sodium sulphate. B. Sodium hydrogen sulphite.
C. Sodium sulphite. D. Sodium hydrogen sulphate.
2. (2004Q. 26)
The false statement about Sulphur dioxide is that sulphur dioxide
A. is a colourless gas. B. dissolves readily in water.
C. turns moist red litmus to blue. D. behaves as a bleaching agent.
3. (2002 Q.15) Sulphur dioxide is used in the following ways except in the
A. bleaching of materials such as wool and silk.
B. large-scale production of sulphuric acid.
C. treating of wool pulp when manufacturing paper.
D. manufacture of vulcanised rubber.
4. (2002 Q.29) Which one of the following processes adds sulphur dioxide into the
atmosphere?
A. Burning coal and oil. B. Decaying organic matter.
C. Heating limestone in a kiln. D. Fractional distillation of liquid air.

147
5. (2002 Q.38) Which one of the following statements about the reaction between iron and
sulphur is false? The reaction
A. is an oxidation-reduction one. B. produces iron (II) sulphate.
C. proceeds with evolution of heat. D. is induced by heat.
6. (2001 Q.12) Which one of the following compounds is used as catalyst in the manufacture
of sulphur trioxide?
A. Alumina. B. Vanadium (V) oxide.
C. Manganese (IV) oxide. D. Iron powder.
7. (2000 Q.13) Which one of the following reagents is used to test for sulphur dioxide?
A. Chlorine water. B. Acidified potassium permanganate.
C. Cobalt chloride. D. An hydrous copper sulphate.
8. (1997 Q.5) The products of the reaction between concentrated sulphuric acid and copper
are
A. water, sulphur dioxide, and copper (II) sulphate.
B. water and copper (II) sulphate.
C. water, copper (II) sulphate and hydrogen.
D. sulphur dioxide and copper (II) sulphate.
9. (1997 Q.24) Sulphur dioxide reacts with oxygen to form sulphur trioxide according to the
equation: SO2(g) + O2(g) 2SO3(g) + heat.
Which one of the following conditions favours the formation of sulphur trioxide?
A. Low pressure and low temperature. B. High pressure and high temperature.
C. Low pressure and high temperature. D. High pressure and low temperature.
10. (1996 Q.7) Which one of the following gases turns potassium dichromate solution green?
A. NH3. B. NO2. C. Cl2. D. SO2.
SELF-CHECK 6.2
SULPHUR AND ITS COMPOUNDS
1. (1991 Q.11).
(a) Describe how sulphur is extracted by the Frasch process.
(b) Write equations to show how fuming sulphuric acid can be obtained from
sulphur.
(c) State what would be observed if concentrated sulphuric acid is added to sugar.
(d) State any two uses of sulphur.

2. (2003 Q.5).
(a) Write an ionic equation to show how sulphur dioxide can be formed from
sodium sulphite and hydrochloric acid.
(b) (i) Name one reagent that can be used to test for sulphur dioxide.
(ii) State what would be observed if sulphur dioxide was reacted with the
reagent you have named in (b) (i).
(c) Sulphur dioxide was passed into a beaker containing a red flower and water.
(i) State what was observed.
(ii) Give a reason for your answer in (c) (i).

148
3. (2002 Q.11).
(a) (i) Name one substance that is reacted with hydrochloric acid to
produce sulphur dioxide in the laboratory.
(ii) State the conditions for the reaction.
(iii) Name a substance that can be used to dry the sulphur dioxide.
(iv) Write equation for the reaction leading to the formation of sulphur
dioxide.
(b) State what would be observed and explain what would happen if sulphur
dioxide is passed through a solution containing
(i) Acidified potassium dichromate.
(ii) A dye.
(c) Briefly describe how sulphur dioxide can be converted to sulphuric acid.
Your answer should include equations and conditions for the reaction(s).

4. (1999 Q.7).
(a) State what would be observed if concentrated sulphuric acid is added to sugar.
Explain your answer.
(b) Dilute sulphuric acid was added to zinc carbonate. State what was observed
and write equation for the reaction.
(c) State the conditions under which sulphuric acid reacts with copper and write
equation for the reaction.
(d) Describe a test that can be carried out to identify the sulphate ion in sulphuric
acid.
(e) State one use of sulphuric acid.

5. (1996 Q.13).
The following flow chart shows the steps in the manufacture of sulphuric acid by the
Contact process.

(a) Write an equation for the reaction that takes place in step I.
(b) Why is step II necessary?
(c) Name:
(i) The drying agent in step III.
(ii) The catalyst in step IV.
149
 (d) Describe the process that takes place in step V.
(e) Sulphur dioxide combines with air to form sulphur trioxide according to the
equation:
2SO2 (g) + O2(g)  2SO3(g) ∆H = -200 kJ mol-1
(i) Give three conditions for maximum yield of sulphur trioxide.
(ii) Calculate the volume of sulphur trioxide that would be formed when
20 cm3 of sulphur dioxide was reacted with 100 cm3 of oxygen.
(f) Give one use of sulphuric acid.
(g) Concentrated sulphuric acid is 18 M. Calculate the volume of concentrated
sulphuric acid required to make 2 liltre of 2 M sulphuric acid.

6. (1992 Q.11).
(a) Draw a labelled diagram to show how a dry sample of sulphur dioxide can be
prepared in the laboratory.
(b) Write equation for the reaction that takes place in (a).
(c) Describe a test that can be carried out to confirm the presence of sulphur
dioxide.
(d) Excess sulphur dioxide was bubbled through a solution of sodium hydroxide.
Write equation for the reaction that took place.
(e) 25.0 cm3 of 0.1 M sodium hydrogen carbonate solution reacted completely
with 27.8 cm3 of sulphuric acid. Calculate the concentration of sulphuric acid
in moles per litre.

150
 CHAPTER SEVEN
CHLORINE AND ITS COMPOUNDS
Chlorine
Chlorine, symbol Cl, with atomic number 17 and mass number 35.5, is one of the halogens.
It belongs to group VII and period III in the Periodic Table.

Occurrence:
Chlorine does not occur as a free element. It occurs in combined form as ionic and covalent
chloride compounds.

Preparation of Chlorine gas

In the laboratory, chlorine is prepared from concentrated hydrochloric acid by oxidation with:
 Potassium permanganate and
 Manganese (IV) oxide.

With potassium permanganate, the reaction takes place in the cold and with manganese (IV) oxide, the
mixture in heated. The gas produced in both reactions is passed through water to remove any hydrogen
chloride gas and then through concentrated sulphuric acid to dry the gas.

Laboratory Preparation of dry Chlorine using Conc. Hydrochloric acid and

potassium permanganate
Chlorine is prepared in the laboratory by the action of concentrated hydrochloric acid on
potassium permanganate crystals in a flat-bottomed flask. Effervescence occurs in the cold
evolving chlorine gas. And the gas is passed through water to absorb/remove any hydrogen
chloride gas and then through concentrated sulphuric acid to dry the gas. It is then collected
by downward delivery method since it is denser than air as shown in the diagram below.
The diagram showing the Laboratory Preparation of Dry Chlorine

2KMnO4(s) + 16HCl(aq)  2KCl(aq) + 2MnCl2(aq) + 8H2O(l) + 5Cl2(g)

151
 Laboratory Preparation of dry Chlorine using Conc. Hydrochloric acid and
manganese (IV) oxide
Chlorine is prepared in the laboratory by heating a mixture of concentrated hydrochloric acid
and manganese (IV) oxide crystals in a round-bottomed flask. Effervescence occurs,
evolving chlorine gas. And the gas is passed through water to absorb/remove any hydrogen
chloride gas and then through concentrated sulphuric acid to dry the gas. It is then collected
by downward delivery method since it is denser than air as shown in the diagram below.
The diagram showing the Laboratory Preparation of Dry Chlorine

MnO2 (s) + 4HCl (aq)  MnCl2 (aq) + 2H2O (l) + Cl2 (g)

Properties of Chlorine
(a) Physical Properties
(i) Chlorine is a yellowish green gas.
(ii) It has a choking, unpleasant, irritating smell.
(iii) It is denser than air (being about 2½ times as dense as air).
(iv) It is very poisonous gas.
(v) It is fairly soluble in water (which dissolves about 3 times of its volume of the
gas) forming yellowish water, contains hydrochloric acid and chloric (I) acid.
[ H2O (l) + Cl2 (g) HCl (aq) + HClO (aq) ]

(vi) It readily liquefies under pressure of 6.8 atmospheres, at 20°C.

(vii) It melts at -101° C.
(viii) It boils at -34.05° C, 1 atmosphere pressure, and
(ix) Has a specific gravity of 1.41 at -35°C.

152
 Chemical Properties
Chlorine is an active element. It reacts with many metals, non-metals, organic
compounds (hydrocarbons), alkalis, some metallic halogen compounds and with
water. These properties are seen in the following reactions.

(a) Reaction with metals (Sodium, Magnesium and Zinc)

When burning pieces of the above metals are plunged into gas gas jar of chlorine,
they continue to burn in chlorine forming white clouds, which settles as white solid.
2Na (s) + Cl2 (g)  2 NaCl (s)
Mg (s) + Cl2 (g)  MgCl2 (s)
Zn (s) + Cl2 (g)  ZnCl2 (s)

(b) Reaction with non-metals (Phosphorus, Sulphur and Hydrogen)

(i) Reaction with Phosphorus.
When a piece of white phosphorus is plunged into dry chlorine, it catches fire
and burns forming white fumes of phosphorus tri-chlorides and pent-chloride.
2P (s) + 3 Cl2 (g)  2PCl3 (s) (Phosphorus tri-chloride)
PCl3 (s) + Cl2 (g)  PCl5 (s) (Phosphorus pent-chloride)

(ii) Reaction with Sulphur

When dry chlorine is bubbled into molten sulphur, a red liquid, disulphur
dichloride is formed.

2 S (l) + Cl2 (g)  S2Cl2 (l)

(iii) Reaction with Hydrogen

The mixture of hydrogen and chlorine does not react in darkness, reacts
slowly in diffused sunlight and explodes if placed in bright sunlight.
Burning hydrogen continues to burn quietly in chlorine, forming hydrogen
chloride.

H2 (g) + Cl2 (g)  2HCl (g)

(c) Reaction with Hydrocarbons

Burning hydrocarbon continues to burn in chlorine forming clouds of black solid
particles (soot – carbon) and hydrogen chloride.
For example, when a filter paper or piece of cotton wool soaked in paraffin is plunged
into a gas jar of dry chlorine, it burns with a reddish flame forming clouds of soot and
hydrogen chloride.
C12H26 (l) + 13Cl2 (g)  12C (s) + 26HCl (g)

153
 (d) Reaction with alkalis
(i) Reaction with cold dilute alkaline solution
When chlorine is bubbled through cold dilute aqueous solution of sodium hydroxide
or potassium hydroxide, it is absorbed forming a pale yellow solution of the
hypochlorite and chloride of the metal.
Cl2 (g) + 2NaOH (aq)  NaOCl (aq) + NaCl (aq) + H2O (l)
Cl2 (g) + 2 KOH (aq)  KOCl (aq) + KCl (aq) + H2O (l)

Ionically the reaction equation is written as:

Cl2 (g) + 2 OH- (aq)  OCl- (aq) + Cl- (aq) + H2O (l)

(ii) Reaction with hot dilute alkaline solution.

When chlorine is bubbled through hot concentrated aqueous solution of sodium
hydroxide or potassium hydroxide, for some time, a mixture of the chloride and
chlorate of the metal is formed and the later can be obtained by crystallizing the
mixture where crystals of chlorate separate first. (These can be purified by
crystallization).
3Cl2 (g) + 6 NaOH(aq)  NaClO3 (aq) + 5NaCl (aq) +
3H2O (l)
3Cl2 (g) + 6KOH (aq)  KClO3 (aq) + 5KCl (aq) + 3H2O(l)

Ionically the reaction equation is written as:

3Cl2 (g) + 6OH- (aq)  ClO3- (aq) + 5Cl- (aq) + 3H2O (l)

(e) Reaction of Chlorine with the compounds of Halogens, bromides and iodides.
Chlorine displaces the halogens below it in the Periodic Table from their compounds.
(i) Reaction with Sodium bromide
When chlorine is bubbled into a solution of sodium bromide, the solution turns from
colourless to orange.

Cl2 (g) + 2NaBr (aq)  2NaCl (aq) + Br2 (l)

Or Cl2 (g) + 2Br- (aq)  2Cl- (aq) + Br2 (l)

(ii) Reaction with potassium iodide

When chlorine is bubbled into a solution of potassium iodide, the solution turns from
colourless to brown. And a black solid is also deposited. The brown-black colour is
due to the formation of solid iodine discharged.
Cl2 (g) + 2KI (aq)  2KCl (aq) + I2 (s)
Or Cl2 (g) + 2I- (aq)  2Cl- (aq) + I2 (s)
154
 The action of sunlight on Chlorine water
When chlorine water filled in a long glass tube is inverted in a beaker containing
some of the chlorine water exposed to bright sunlight, after some time, a colorless gas
that rekindles a glowing splint collects above the water in the U-tube.

2Cl2 (g) + 2H2O (aq)  4HCl (aq) + O2(g)

The above reaction probably occurs in two stages, as indicated in the equations below:
2Cl2 (g) + 2H2O (aq) HOCl (aq) + HCl (aq)
The unstable hypochlorous acid gives up its oxygen forming hydrochloric acid.
2HOCl (aq)  2HCl (aq) + O2 (g)

The Bleaching action of Chlorine

Chlorine dissolves in water and reacts with it forming hypochlorous acid and
hydrochloric acid.
Cl2 (g) + H2O (l)  HOCl (aq) + HCl (aq)
Chlorine dissolved in water is known as chlorine water. It is a bleaching agent.
The bleaching action of chlorine is due to the ion, OCl-, in the hypochlorous acid. The
hypochlorous acid is a very reactive compound and readily gives up its oxygen.
2HOCl (aq)  2HCl (aq) + O2 (g)
The oxygen given off reacts with the dye, to form a colourless compound.
2Dye + O2 (g)  2(Dye + O)
Colourless compound
Uses of Chlorine
Chlorine is used to:
(i) Sterilize (Treat) water for domestic and industrial and swimming baths.
(ii) Manufacture plastics.
(iii) Manufacture. insecticides
(iv) Manufacture antiseptics.
(v) Manufacture hydrochloric acid.
(vi) Bleach textiles and paper in industries.

155
 Hydrogen Chloride
Laboratory preparation of Hydrogen Chloride
Hydrogen Chloride is prepared in the laboratory by adding concentrated sulphuric acid on
sodium chloride in a flat bottom flask. Effervescence occurs and the gas is dried by passing it
through concentrated sulphuric acid and is collected by downward delivery method, as it is
denser than air.
Laboratory preparation of Hydrogen Chloride

H2SO4(aq) + NaCl (aq)  NaHSO4 (aq) + HCl (g)

Properties of Hydrogen Chloride
(a) Physical Properties
(i) It is a colourless gas.
(ii) It is denser than air.
(iii) It is very soluble in water, being the second to ammonia (1 cm3 of
water dissolves about 540 cm3 of the gas), as a result it can be used to
perform fountain experiment.
(iv) It is acidic and therefore turns a moist blue litmus paper red.

(b) Chemical Properties

(i) It forms white dense fumes with ammonia gas.
HCl (g) + NH3 (g)  NH4Cl (s)
(ii) It forms a white precipitate with silver nitrate solution.
HCl (g) + AgNO3 (aq)  AgCl (s) + HNO3 (aq)
The precipitate dissolves in excess ammonia solution forming a
colourless solution.
(iii) Dry hydrogen chloride reacts with heated iron to form iron (II)
chloride and hydrogen.
2HCl (g) + Fe (s)  FeCl2 (s) + H2 (g)
Chemical test for Hydrogen Chloride
The gas forms misty fumes in moist air, turns blue litmus red, forms white precipitate
with acidified silver nitrate solution.

156
 SELF-CHECK 7.1
1. (2003 Q.39)
Chlorine dissolves in cold aqueous solution of sodium hydroxide to produce the
following substance.
A. Sodium chlorate. B. Sodium chloride.
C. Sodium chlorite. D. Sodium hydrogen chloride

2. (2002 Q.40).
The reaction of iron and chlorine
A. occurs at ordinary temperature. B. requires platinum catalyse.
C. produces iron (II) chloride. D. produces iron (III) chloride

3. (2001Q.9).
The best reason for including water in the laboratory preparation of chlorine is to
A. remove chlorine gas. B. cool chlorine gas.
C. cool hydrogen chloride gas. D. remove hydrogen chloride gas.

4. (2000 Q.26)
A compound reacts with concentrated sulphuric acid to give a colourless gas which
fumes with ammonia. The compound contains a
A. chloride. B. carbonate. C. nitrate. D. sulphate

5. (1998 Q.20)
What is the correct equation for the reaction between chlorine and heated steel wool?
A. Cl2(g) + 2Fe(s)  2FeCl(s) B. Cl2(g) + Fe(s)  FeCl2(s)
C. 3Cl2(g) + 2Fe(s)  2FeCl3(s) D. 3Cl2(g) + 4Fe(s) 2Fe2Cl3(s)

6. (1997 Q.4)
The gas formed when chlorine water is exposed to sunlight is
A. oxygen. B. hydrogen.
C. chlorine. D. hydrogen chloride.

7. (1996 Q.13)
Which one of the following gases is an oxidising agent?
A. CO. B. H2S. C. Cl2. D. NH3.

8. (1996 Q.26) Chlorine can be prepared in the laboratory by reacting

A. sodium chloride and concentrated ethanoic acid.
B. potassium chlorate and concentrated sulphuric acid.
C. sodium chloride and concentrated sulphuric acid.
D. potassium permanganate and concentrated hydrochloric acid.

9. (1995 Q.11)
Which one of the methods below is used to prepare a sample of copper (II) chloride?
A. Heating copper in stream of dry chlorine.
B. Action of copper on dilute hydrochloric acid.
C. Reacting copper (II) oxide with hydrochloric acid.
D. Heating copper in a stream of dry hydrogen chloride

157
 SELF-CHECK 7.2
1. The diagram in Fig.1 shows a set up of the apparatus for the laboratory preparation of
dry chlorine from hydrochloric acid.

(a) (i) Name substances A, B and C.

(ii) State the role of substance B.
(b) State the conditions for the reaction.
(c) Write equation for the reaction.

2. (2002 Q.9).
(a) Chlorine can be prepared from concentrated hydrochloric acid.
(i) Name a substance that can react with hydrochloric acid to produce
chlorine.
(ii) Write equation for the reaction.

(b) Chlorine gas was passed through cold dilute sodium hydroxide solution.
(i) State what was observed.
(ii) Write equation for the reaction that took place.

3. (1999 Q.14). Chlorine

(a) (i) Describe with the aid of a well-labelled diagram how a dry sample of
chlorine can be prepared in the laboratory.
(ii) Write an equation for the reaction that took place.
(iii) State any three uses of chlorine.
(b) State with the aid of equations, what would be observed if chlorine was added
to: (i) Iron (II) chloride solution,
(ii) Potassium iodine solution.
(c) Burning sodium was plunged into a jar of chlorine.
(i) State and explain what was observed.
(ii) Write the equation for the reaction.
158
4. (1997 Q.2) Chlorine
During the preparation of chlorine in the laboratory, the gas may be passed through
water and concentrated sulphuric acid before collection.
(a) State the use of:
- water.
- concentrated sulphuric acid.
(b) Chlorine is a bleaching agent when in the presence of water.
(i) Write an equation for the reaction between chlorine and water.
(ii) Using equations explain the bleaching action of chlorine.
(c) (i) State what would be observed if chlorine was bubbled through a
solution of iron (II) sulphate solution.
(ii) Write an ionic equation for the reaction between chlorine and iron (II)
ions.

5. (a) When lead (IV) oxide was added to concentrated hydrochloric acid and
heated, the gas evolved bleached litmus.
(i) Identify the gas.
(ii) Write the equation for the reaction.
(b) When chlorine was passed into a solution of potassium bromide, a red
colouration developed owing to liberation of bromine
(i) Write the equation for the reaction.
(ii) Describe what will be observed when fluorine is passed into potassium
bromide solution, given that fluorine is just above chlorine in the group
(iii) Write the ionic equation for the reaction in b (ii) above.

6. (a) When concentrated hydrochloric acid was added to manganese(IV)oxide and

heated the gas, thus evolved, bleached litmus paper. Write the equation to
show what happened
(b) Chlorine dissolves in water to give chlorine water
(i) Write the equation for the reaction that occurs
(ii) When the chlorine water is exposed to sunlight a colourless gas which
gives a positive test with a glowing splint is given off. Explain.
(c) Chlorine can react with a cold dilute solution of potassium hydroxide to form
water and two other products.
(i) Write the equation for the reaction
(ii) Name the other two products
(iii) Give the use of the solution obtained above

159
 SELF-CHECK 7.3
HYDROGEN CHLORIDE
7. (2003 Q.12).
(a) Write equation to show how hydrogen chloride can be prepared from sodium
chloride.
(b) Draw a labelled diagram to show how aqueous hydrogen chloride can be
prepared in the laboratory.
(c) State what would be observed and write equation for the reaction that would
take place when aqueous hydrogen chloride is reacted with:
(i) Solid calcium carbonate.
(ii) Silver nitrate solution.
(iii) Magnesium.
(d) State why aqueous hydrogen chloride does not react with copper.
8. (1998 Q.11)
(a) Describe how a dry sample of hydrogen chloride can be prepared from a
named chloride. (Diagram not required). Your answer should include the
following: - Conditions for the reaction.
- Name of the drying agent.
- Method of collection.
- Equation for the reaction.
(b) Name the substance that is formed when hydrogen chloride is passed through water.
(c) (i) Name one reagent that can be used to test for the presence of chloride
ions in solution.
(ii) State what would be observed if the reagent was added to chloride solution.
(d) Write an ionic equation to show the reaction between aqueous hydrogen
chloride and calcium hydrogen carbonate solution.
(e) 25.0 cm3 of a 0.2 M lead (II) nitrate solution was shaken with excess aqueous
hydrogen chloride. Lead (II) ions react with chloride ions according to the
following equation: Pb2+(aq) + 2 Cl-(aq)  PbCl2(s).
Calculate the mass of lead (II) nitrate formed.
9. (1991 Q.14)
(a) Draw a well-labelled diagram to show how a sample of dry hydrogen chloride
can be prepared.
(b) Dry hydrogen chloride gas was passed over heated iron fillings. Write an
equation for the reaction that took place.
(c) The solid product in (b) was dissolved in water and aqueous sodium chloride
added to the resultant solution drop wise until in excess.
(i) State what was observed.
(ii) Write equation for the reaction.
(d) Chlorine gas was passed through a solution of the product in (b).
(i) State what was observed.
(ii) Write an ionic equation for the reaction.
(e) Name one reagent that can be used to test for
(i) The cation formed in (d). (ii) The anion formed in (d).
In each cases state what is observed when the reagent you have named is used.

160
 CHAPTER EIGHT
ION CHEMISTRY
INTRODUCTION
Chemistry Practical paper tests the knowledge of Candidates in two major areas, namely:
- Qualitative Analysis and
- Volumetric Analysis.

QUALITATIVE ANALYSIS

Qualitative analysis deals with the identification of cations (positively charged ions) and
anions (negatively charged ions) from unknown compound(s) usually a mixture of salts.
However, reaching the ions involves several steps before arriving at the right ions. The steps
act as a pointer thus enabling the candidate to predict the unknown ions present in the
unknown compound.

The main steps involved are the following: -

 Preliminary Examination.
 Action of heat.
 Action of water (Solubility).
 Action of acids.
 Action of Soluble salts.
 Action of Sodium hydroxide.
 Action of Ammonia Solution.
All the deductions/ predictions made as a result of the above steps are always
confirmed by carrying out specific test known as confirmatory test.

1. Preliminary Examination
The unknown compound is examined by applying some of the sense organs
(excluding the tongue).
Note the: - - Appearance
- Colour
- Smell
- Feel

161
 Summary of the Preliminary Test

OBSERVATION DEDUCTION
Coloured substance Salt of transition metal.
E. g. Copper salt, iron (II) or iron (III) salt
White powder or crystal Salt of group I, II, III, or Zinc
Green crystal Iron (II) or copper (II) salt.
Brown solid Iron (III) salt.
Smell of ammonia Ammonium salt

2. Action of Heat
A little (usually a spatula end- full) of the unknown compound is heated
strongly in an ignition tube or in a small dry test tube until no further
change occurs.
Action of heat enables the candidates to search for: -
- Evolution of gases.
- Evolution of water vapour.
- Change in colour of the oxide formed.

Summary of the Possible Observations and Deductions

OBSERVATION DEDUCTION
Water vapour condenses on the cooler part of Water of crystallization from
the test tube a hydrated salt of :-
HCO3-, HSO4-, SO42-
A gas that turns lime water milky Carbonate (CO32-),
given off. Hydrogen Carbonate (HCO3-)
A colourless gas that turns red litmus paper Ammonium salt
blue given off.
Brown fumes of a gas given off. Nitrate (NO3-)
Cracking sound and brown fumes of a gas Lead (II) nitrate
given off.
A gas that turns moist blue litmus paper red Sulphite( SO32-)
given off. or certain sulphates (SO42-)
Colour change of the oxide
formed: - Yellow when hot and
- White when cold. Zinc oxide
- Reddish brown when hot. Lead (II) oxide
- Yellow when cold.

162
3. Action of Acids
Acids have two major roles to play when added to the unknown compound in a test
tube.
These are: - to dissolve an insoluble salt. E. g A carbonate or a sulphite of
group II or a transition metal.
- to acidify the solution for the next text.

(a) Addition of dilute Hydrochloric acid or dilute nitric acid

OBNSERVATION DEDUCTION
A colourless gas that turns lime water CO32-, HCO3-
milky given off.
A colourless gas that turns moist blue Sulphite (SO32-)
litmus paper red given off.
A colourless gas with a smell of rotten egg. Sulphide (S2- )

(b) Addition of Concentrated Sulphuric acid

OBSERVATION DEDUCTION
A gas that turns lime water milky given off. CO32-, HCO3- ions.
A gas that turns moist blue litmus paper Sulphite (SO32-) ions
red given off.
A gas which smells like a rotten egg. Sulphide (S2-) ions

NB: Both acids evolve the same gases i. e. CO2 , SO2 , & H2S, except NO2 , which
is only evolved by concentrated Sulphuric acid.

4. IDENTIFICATION OF GASES
Once a gas is evolved under the action of heat or an acid, candidates are advised to
carryout the specific tests to identify the gas so as to draw a concrete deductions of
the anion present.

Gases are identified according to their: -

- colour.
- smell.
- action on moist litmus paper.
- action with glowing splint.
- action on lime water.
163
The table below shows the common gases that may be encountered during examinations

ACTION ON ACTION ON
GAS COLOUR SMELL LITMUS GLOWING
PAPER SPLINT
Oxygen (O2) None None None Relights a
glowing splint
Carbon dioxide (CO2) None None Blue – faint Does not burn
red.
Hydrogen sulphide (H2S) None Rotten egg Blue – faint Burns with blue
red. flame.

Sulphur dioxide (SO2) None Burning Blue – Red Does not burn.
sulphur
Ammonia (NH3) None Choking Red – blue Does not burn.
characteristic
smell

NB: Ammonia is the ONLY gas that turns red moist litmus blue.

5. Solubility (Action of water)

Candidates are often required to add a small amount of the unknown compound in a
given amount of water. The main aim of this step is to free the ions of the soluble salt.
Action of water therefore tests for the solubility of the salt.
Usually a mixture of soluble and insoluble salts is provided. Therefore candidates are
always instructed to add a spatula end- full of the unknown compound to some
amount of water, shake well, filter and keep both filtrate and the residue.
Candidates should therefore know that the filtrate contains ions of a soluble salt and
the residue contains the insoluble salt.

The table below shows the summary of the solubility of the common salts.

SOLUBLE SALTS INSOLUBLE SALTS

All group I/ Ammonium salts.
All nitrates.
All Chlorides except: Lead (II) & Silver chlorides.
NB: Lead( II) chloride is soluble in hot water.
All Sulphates except: Lead( II) & Barium sulphates
NB: Calcium sulphate is only slightly soluble.
All group I/ Ammonium All other carbonates.
carbonates.

164
6. IDENTIFICATION OF ANIONS
Anions are negatively charged ions. The common anions include the following:-
 HCO3-, CO32-
 HSO4-, SO42-and SO32-
 NO2-, NO3-
 Cl-

All anions are tested by using the knowledge of solubility of salts. Suitable cations
with which they show characteristic precipitates are added in to their aqueous states.
The common cations used are:
 Ag+,
 Ba2+ and
 Pb2+
They are obtained from their soluble salts.
Ag+ and Pb2+ ions are obtained from their nitrates, AgNO3, and Pb(NO3)2,
While Ba2+ ions are obtained from Ba(NO3)2 or BaCl2

Stages involved in Testing for Anions

(a) Dry Testing
The substance in the solid form is strongly heated in a dry test tube or ignition
tube until no further change.
Summary of Effects, observations and Deductions

EFFECT OBSERVATION DEDUCTION

Decompose A colourless gas that relights Nitrate of heavy metal
a glowing splint (oxygen) or Nitrate of group I
given off.
Decompose A gas that turns lime water Carbonate of heavy metal.
milky (CO2 ) given off.
Decompose A colourless gas that turns Hydrogen carbonate
limewater milky (CO2 ) and of heavy metal
water vapour given off.
Does not No observable change. Carbonate of group I metal.
decompose
Decompose White choking fumes of Hydrogen Sulphate (HSO4-)
sulphur trioxide (SO3 ) and
water vapour.

165
(b) Wet Testing
This test involves addition of a solution to solution or solution to solid.
It makes the use of precipitation or redox reaction involving colour changes for the
test purpose.
This test is used:
- To test for a carbonate.
- To test for a metal oxide or hydroxide and
- To dissolve insoluble substance.

(i) Solution to solid reaction

A solution of a known reagent is added to the solid. The possible
observations and deductions are then made.

SUMMARY OF RESULTS

REAGENT ADDED OBSERVATION DEDUCTION

Effervescence released. Carbonate (CO32- )

Dilute nitric acid (HNO3) A gas that turns lime water
milky given off.
Solid dissolves with Metal oxide or hydroxide.
effervescence and releases
out heat.

NB: Dilute nitric acid is usually used because the nitrates formed are soluble in water,
therefore dissolution continues to the end.
When the salt formed is insoluble in water, it tends to precipitate and coat the solid,
rendering it unapproachable and the reaction ceases (stops) at an earlier stage.
This is why dilute hydrochloric acid and sulphuric acid may not be used to dissolve
lead carbonate. Lead chloride and lead sulphate are insoluble and would coat the
carbonate or chloride to be dissolved thus cutting it off from the acid.

(ii) Solution to solution reaction

Solutions containing suitable metal ions are used to precipitate out the anions.
The observations and deductions are made.
The above deductions are confirmed by carrying out specific tests called
Confirmatory Test

166
 Summary of results Confirmatory Test
REAGENT ADDED OBSERVATION DEDUCTION
Cations of heavy metals Precipitates dissolve with
e. g. Pb2+ , Zn2+ , Cu2+ etc. in their effervescence releasing a CO32- , or
soluble salts and then followed by dilute colourless gas that turns HCO3- ions.
nitric acid. limewater milky given off.

(i) Barium nitrate solution - White precipitate insoluble

followed by nitric acid. in the acid. SO42- ions.

(ii) Lead (II) nitrate solution -White precipitate which

followed by barium nitrate persists on heating.
and heat.
-Add equal volume of freshly prepared Brown ring formed in
Iron (II) Sulphate solution to the between the two layers. NO3-
solution of the unknown compound.
Tilt the test tube and then pour conc.
Sulphuric acid slowly down the side of
the test tube.
(i) Silver nitrate solution followed by - White precipitate,
dilute nitric acid. insoluble in the acid. Cl–

(ii) Silver nitrate solution followed by - White precipitate that turns

ammonia solution. to grey on standing in light
forms.
(i) Lead (II) nitrate solution and heat. - White precipitate, soluble
on heating.

167
 Confirmation Tests for Anions
A confirmatory test is a specific test to identify specific ions
ANION TEST OBSERVATION
(i) Add barium nitrate solution
2- followed by dilute nitric acid or
SO4 White precipitate insoluble in the acid.
add barium chloride followed by Ba2+ (aq) + SO42-(aq)  BaSO4 (s)
dilute hydrochloric acid.
(ii) add lead (II) nitrate solution White precipitate formed.
and heat the mixture. Precipitate insoluble on heating
Pb2+ (aq) + SO42-(aq)  PbSO4 (s)
(i) Add silver nitrate solution White precipitate insoluble in the acid.
Cl - followed by dilute nitric acid. Ag+ (aq) + Cl-(aq)  AgCl (s)

(ii) add silver nitrate solution White precipitate formed.

followed by ammonia solution. White precipitate soluble in ammonia solution
forming a colourless solution.
Ag+ (aq) + Cl-(aq)  AgCl (s)
AgCl (s) + 2NH3(aq)  Ag(NH3)2+(aq)+ Cl-(aq)

(iii) Add lead (II) nitrate White precipitate formed.

solution and heat the mixture. Precipitate dissolves on heating, and reappears
on cooling.
Pb2+ (aq) + 2Cl-(aq)  PbCl2 (s)
Add equal volume of freshly Two layers form.
- prepared iron (II) sulphate A brown ring forms between the two layers.
NO3
solution to the solution of
unknown compound.
Tilt the test tube and then pour
conc. sulphuric acid slowly on
the side of the test tube.

(i) Add dilute nitric acid Effervescence of a colourless gas that turns
2- lime water milky given off.
CO3
CO32-(aq) + 2H+(aq)  H2O (l) + CO2 (g)
(ii) Add barium nitrate solution White precipitate formed.
followed by dilute nitric acid. Precipitate dissolves with effervescence in the
acid giving off a colourless gas that turns lime
water milky.
Ba2+ (aq) + CO32-(aq)  BaCO3 (s)
CO32- (s) + 2H+(aq)  H2O (l) + CO2(g)
168
 (iii) Add barium nitrate or White precipitate formed.
barium chloride solution Precipitate dissolves with effervescence in the
followed by dilute hydrochloric acid giving off a colourless gas that turns lime
acid. water milky.
Ba2+ (aq) + CO32-(aq)  BaCO3 (s)
CO32- (aq) + 2H+(aq)  H2O (l) + CO2(g)

7. IDENTIFICATION OF CATIONS
The common cations encountered by candidates during examinations include the
following:-
GROUP CATION
I Na+, K+,
II Ca2+and Mg2+
III Al3+
IV Pb2+
Transition cations:
- Colourless Zn2+
- Coloured Fe2+, Fe3+ and Cu2+
Others Ag+ and NH4+

The reagents used to test for the cations are:-

 Sodium hydroxide (NaOH) solution.
 Ammonium hydroxide (NH4OH) solution.
(a) Solution to Solution Reaction
(i) Action of sodium hydroxide solution.
To the solution of the unknown compound, add sodium hydroxide solution
drop wise until in excess. If no observable change, warm gently for ammonia.

169
 Summary of Observations and Deductions
OBSERVATION DEDUCTION
No observable change and no ammonia gas given off. Na+, K+, Ag+ions probably present.
The solution remains colourless. NH4+ ions probably present.
A gas that turns red moist litmus paper blue is given
off on warming.
White precipitate soluble in excess sodium hydroxide Zn2+, Pb2+and Al3+ ions probably present
forming a colourless solution.
White precipitate insoluble in excess sodium Ca2+, Mg2+ions. probably present.
hydroxide solution.
Blue precipitate insoluble in excess sodium hydroxide Cu2+ions probably present.
solution.
Green precipitate insoluble in excess sodium Fe2+ ions probably present.
hydroxide solution.
Reddish brown precipitate insoluble in excess Fe3+ ions probably present
sodium hydroxide solution.
NB: The hydroxides of Pb2+ , Zn2+ and Al3+ are amphoteric and dissolve in excess
sodium hydroxide solution to give their respective complexes.
Zn(OH)2(s) + 2OH- (aq)  Zn(OH)42-(aq) Zincate ion.
Pb(OH)2(s) + 2OH-(aq)  Pb(OH)42-(aq) Plumbate ion.
Al(OH)3(s) + 3OH- (aq)  Al(OH)63-(aq) Aluminate ion.
(ii) Action of Ammonia solution
Add ammonia solution to the solution of the unknown compound dropwise until in excess.
Summary of Observations and Deductions

OBSERVATION DEDUCTION
No observable change. Na+, K+, Ag+ ions probably present.
White precipitate soluble in excess ammonia solution. Zn2+ions probably present.
White precipitate insoluble in excess ammonia solution. Pb2+, Ca2+, Mg2+, & Al3+ ions probably present.
Pale blue precipitate soluble in excess ammonia solution Cu2+ ions confirmed.
forming a deep blue solution.
Green precipitate insoluble in excess ammonia solution Fe2+ ions probably present.
Reddish brown precipitate insoluble in excess ammonia Fe3+ ions probably present.
solution

NB: Copper (II) hydroxide and zinc hydroxide dissolve in excess ammonia solution to
form their respective complexes, tetra amino copper and tetra amino zinc.
Cu(OH)2(s) + 4NH3(aq)  Cu(NH3)42+(aq) + 2OH- (aq)
Zn(OH)2(aq) + 4NH3(aq)  Zn(NH3)42+(aq) + 2OH- (aq)

170
 CONFIRMATORY TESTS FOR COMMON CATIONS
CATION TEST OBSERVATION
(i) Add sodium hydroxide No observable change
Na+ & K+ solution and warm gently.
(ii) Add ammonia solution. No observable change

Add sodium hydroxide A gas that turns red moist litmus

NH4+ solution and warm. paper blue is given off or smell of
ammonia.
(i) Add sodium hydroxide White precipitate forms.
Zn2+
solution dropwise until in Precipitate soluble in excess sodium
excess. hydroxide.

(ii) Add ammonia solution White precipitate forms.

dropwise until in excess. Precipitate soluble in excess
ammonia solution.
(i) Add dilute sulphuric acid. White precipitate forms.
Pb2+ Pb2+(aq) + SO42- (aq)  PbSO4(s)

(ii) Add conc. Hydrochloric White precipitate forms.

acid. Heat the mixture and Precipitate dissolves on warming
allow to cool. and reappears on cooling.
(iii) Add potassium iodide A yellow precipitate forms.
solution. Pb2+(aq) + 2I- (aq)  PI2(s)
(iv) Add Potassium Chromate A yellow precipitate forms..
(VI) solution.
Cu 2+ (i) Add ammonium hydroxide Deep blue solution formed.
drop wise until in excess.
(i) Add Potassium hexacy- Light blue or nearly white
2+
Fe noferrate (II) solution. precipitate forms.
(ii) Add Potassiumhexacy- Dark blue precipitate forms.
noferrate (III) solution.
(iii) Add Potassium thiocyanate Traces of iron (II) salt give colour.

(i) Add Potassium hexacy- Dark blue precipitate forms.

Fe3+ noferrate (II) solution.

(ii) Add Potassium hexacy- Dark solution forms.

noferrate (III).

(iii) Add Potassium thiocynate. Deep red solution forms.

171
 (a) Flame Test
Place a little of the substance on watch glass and moisten it with pure conc.
Hydrochloric acid. Heat a little of the mixture on clean platinum or nichrome wire
and note the colour of the flame.

OBSERVATION DEDUCTION

Brilliant yellow Sodium

Lilac Potassium

Red Calcium

Blue Lead

Blue- green Copper

METHOD OF WRITING RESULTS

A sample of a substance W containing two cations and two anions was analyzed and the
observations were recorded in the table below.

Complete the column for the deductions.

TESTS OBSERVATIONS DEDUCTIONS

1. Add 10 cm3 of distilled - Salt partial soluble.

water to two spatula endfuls - Forming a Colourless Filtrate
of W . Shake well and filter. - Leaving white residue.
Keep both the filtrate and residue.
2. Divide the filtrate into 3 - White ppt formed.
parts and treat each part as - Precipitate soluble in excess sodium
follows:- hydroxide forming
(i) Add sodium hydroxide - a colourless solution.
solution dropwise
until in excess.

(ii) Add ammonia solution - White precipitate formed.

dropwise until in excess. -Precipitate soluble in excess ammonia
solution forming
- a colourless solution.
(iii) Add barium chloride - White precipitate formed.
solution followed by - Precipitate insoluble in the acid.
dilute hydrochloric acid.

172
3. Wash the residue with - Residue dissolved with
distilled water then add effervescence giving off a
4 cm3 of dilute nitric acid. - a colourless gas that turns limewater
Divide the solution into milky .
2 parts. - A colourless solution was formed.

(i) To the first part add - White precipitate formed.

dilute hydrochloric acid - Precipitate soluble on heating.
and heat gently to the - The precipitate reformed on
boiling then allow to cool. cooling.

(ii) To the second part add - Bright yellow precipitate was

potassium iodide solution. formed

The cations in W are: (i) ……………………………

(ii) ……………………………
The anions in W are: (i) ……………………………
(ii) ……………………………

SELF-CHECK 8.1
1. (2005 Q16)
Which one of the following reagents can be used to distinguish between Zn2+(aq) and
Al3+(aq)?
A. Lead (II) nitrate B. Sodium hydroxide.
C. Potassium iodine. D. Ammonia solution.

2. (2005 Q.29)
The hydroxide that dissolves in excess aqueous ammonia but does not in sodium
hydroxide solution is
A. Lead (II) hydroxide. B. Zinc hydroxide.
C. Aluminium hydroxide. D. Copper (II) hydroxide.

3. (2005 Q.35)
The cation that forms a green precipitate with sodium hydroxide is
A. Cu2+(aq). B. Fe2+(aq).
C. Al3+(aq). D. Fe3+(aq).

173
4. (2005 Q.39)
The hydroxide which turns brown when exposed to air from the list below is
A. copper (II) hydroxide. B. iron (II) hydroxide.
C. lead (II) hydroxide. D. iron (III) hydroxide.

5. (2004 Q.15)
Which one of the following ions reacts with NH4 (aq) to form a precipitate that
dissolves in excess ammonia solution?
A. Pb2+ (aq) B. Fe2+ (aq)
2+
C. Ca (aq) D. Cu2+ (aq)

6. (2004 Q.19)
A white salt, X, reacts with dilute hydrochloric acid producing brown fumes. When
dilute Sulphuric acid is added to the resulting solution, a white precipitate is
observed. Salt X is
A. Barium nitrate. B. Sodium nitrate.
C. Potassium nitrate. D. Sodium bromide.
7. (2004 Q.33)
The solution that could be containing zinc ions is one that forms a
A. reddish – brown precipitate with magnesium.
B. green precipitate with aqueous ammonia.
C. white precipitate that is soluble in excess sodium hydroxide solution.
D. white precipitate with dilute Sulphuric acid.

8. (2004 Q.37)
When sodium hydroxide solution was added to an aqueous solution of salt, X, a white
precipitate insoluble in excess alkali was formed. X contained
A. lead (II) ions. B. magnesium ions.
C. zinc ions. D. aluminium ions.
9. (2003Q.7)
Which one of the following solutions is used to test for sulphate ions in acid
condition?
A. Barium nitrate. B. Lead (II) chloride.
C. Silver nitrate. D. Silver chloride.

10. (2003Q Q.24)

Which one of the following ions reacts with ammonia to form a precipitate which
dissolves in excess ammonia to form a colourless solution?
A. Zn2+. B. Mg2+. C. Cu2+. D. Fe2+.

11. (2002 Q.7)

Which one of the following anions does not form a precipitate with Pb2+ (aq)?
2
A. CO (aq). B. OH-(aq). C. NO  (aq). D. SO 2  (aq).
3 3 4

174
12. (2002 Q.8)
Which one of the following ions reacts with Cl- (aq) to form a precipitate which
dissolves on heating?
A. Cu2+(aq). B. Fe2+ (aq). C. Pb2+ (aq). D. Ca2+ (aq).

13. (2002 Q.9)

A colourless solution reacts with dilute hydrochloric acid to give a white precipitate. The white precipitate dissolves in warm water.
The colourless solution is

A. zinc chloride. B. lead nitrate.

C. magnesium sulphate. D. sodium carbonate.

14. (2002 Q.20)

Which one of the following will occur when aluminium is added to an aqueous
solution of silver nitrate?
A. Heat will be absorbed. B. Aluminium hydroxide will be formed.
C. Silver will precipitate out. D. Aluminium will form an alloy with silver.

15. (2001 Q.27)

A white precipitate was formed when an aqueous solution of a salt was reacted with
aqueous barium nitrate. The white precipitate dissolved in nitric acid. The anion in
the salt is
A. SO 2  . B. NO  .
3 3
2- -
C. SO4 . D. Cl .

16. (2001 Q.40)

Lead and silver nitrates are frequently used as reagents because
A. they are the most suitable.
B. most nitrates of metal never decompose on heating.
C. other salts of lead and silver are insoluble.
D. other metal nitrates are insoluble.

17. (1999 Q.12)

Which one of the following salts when in solution will form a white precipitate with
acidified barium nitrate solution?
A. ZnSO4. B. Na2SO3.
C. Na2CO3. D. ZnCl2.

18. (1999 Q.22)

Which one of the following ions can be confirmed by the brown ring test?
A. Cl-. B. NO3-.
2-
C. CO3 . D. SO42-.
19. (1999 Q.28)
Which one of the following oxides below is soluble in both excess sodium hydroxide
solution and aqueous ammonia?
A. Al2O3. B. ZnO.
C. PbO. D. Fe2O3.
175
20. (1999 Q.36)
Which one of the following ions forms a green precipitate with excess sodium
hydroxide?
A. Fe3+. B. Fe2+.
C. Cu2+. D. Zn2+.
21. (1998 Q.12)
When dilute nitric acid followed by silver nitrate solution were added to a certain solution, a white precipitate which darkens when
exposed to sunlight was formed. The solution contained a

A. sulphate. B. chloride.
C. nitrate. D. carbonate.
22. (1998 Q.32)
Which one of the following ions when reacted with ammonia will form a blue
precipitate that dissolves to give a deep blue solution?
A. Fe2+ B. Cu2+
C. Fe3+ D. Zn2+
23. 1997 (Q.23)
Which one of the following mixtures would not form a precipitate?
A. Barium nitrate and sodium sulphate. B. Lead nitrate and potassium iodine.
B. Copper nitrate and sodium sulphate. D. Silver nitrate and potassium bromine.

24. (1997 Q.26)

Which one of the following hydroxides will dissolve in excess ammonia?

A. Pb(OH)2. B. Cu(OH) 2.
C. Fe(OH) 2. D. Al(OH) 3.
25. (1995 Q.19)
When silver nitrate was added to a solution followed by dilute nitric acid, a white
precipitate was formed. The solution contained
A. sulphate ions. B. carbonate ions.
C. chloride ions. D. sulphite ions.
SELF-CHECK 8.2
QUALITATIVE ANALYSIS
1. (2004 Q.3)
A mixture containing copper (II) sulphate and copper (II) carbonate was shaken with excess water and filtered.

(a) Identify the residue.

(b) The dry residue was heated strongly.
(i) State what was observed.
(ii) Write an equation for the reaction.
(c) (i) Name a reagent that can be used to identify the anion in the filtrate.
(ii) Write an ionic equation for the anion and the reagent you have named
in (c)(i).

176
2. (2004 Q.8)
The table below shows some tests that were carried out on a certain substance, Z, and
the observations made.
Test Observation
A colourless gas evolved which turned
1. Solid Z was heated.
limewater milky.
2. Aqueous sodium hydroxide was added
No apparent change.
to aqueous solution of Z.
3. Dilute hydrochloric acid was added to Effervescence and a gas that turned
a solution of Z. limewater milky evolved.
4. (i) Aqueous magnesium chloride was
No apparent change.
added to a solution of Z.
(ii) Resultant solution from 4(i) was White precipitate formed.
heated.

(a) What deduction can you make concerning the solubility of the hydroxide of
the metal ion in Z?
(b) State the: (i) likely anions present in Z.
(ii) anion actually present in Z.
(c) (i) Identify the white precipitate in test 4(ii).
(ii) Write an equation to show the reaction resulting in the formation of the
substance you have identified in (c)(i).
3. (2002 Q.10)
State one reagent that can be used to distinguish between each of the following pairs
of ions and in each case, state what would be observed if each ion is treated with the
reagent.
(a) Pb2+(aq) and Al3+(aq).
(b) SO 2  (aq) and CO32- (aq).
4
4. (2000 Q.14)
Iron forms compounds in which it shows a valency of two and three.

(a) State the general colour of iron compounds in which iron is:
(i) divalent.
(ii) trivalent.
(b) Write a formula and the name of the sulphates of iron in which iron is:
(i) divalent.
(ii) trivalent.
(c) (i) Name a reagent that can be used to distinguish between the
sulphates in (b)(i) and (ii).
(ii) State what would be observed if each of the iron sulphate is reacted
with the reagent you have named in (c)(i).
(iii) Write equations for the reaction in (c)(ii).

177
 (d) Starting from iron wool, state how the anhydrous chloride of iron (II) can be
prepared and write equation to illustrate your answer. (Diagrams not
required).

5. (1999 Q. 9)
The table below shows the results of tests carried out on an aqueous solution of a
salt. Study the table and answer the questions that follow.
Tests Observation
(i) With dilute sodium White precipitate in excess alkali.
hydroxide solution.
(ii) With aqueous ammonia. A white precipitate insoluble in excess
ammonia.
(iii) With aqueous potassium iodine. A bright yellow precipitate.

(i) Identify the cation.

(ii) Write an ionic equation for the reaction in test (ii) in the table.
(iii) Explain the observation in test (ii).

6. (1992 Q.12)
(a) Name one reagent that can be used to distinguish between each of the
following pairs of species.
In each case state what would be observed if each member of the pair is
treated with the reagent and write equation for the reaction that takes place.
(i) Lead (II) ions and zinc ions.
(ii) Carbonate ions and chloride ions.

178
 CHAPTER NINE
OXIDATION AND REDUCTION
(a) Oxidation as the addition of oxygen or removal of hydrogen

(i) Oxidation is the addition of oxygen to a substance or the removal of

hydrogen from a substance.

An oxidizing agent is a substance which transfers oxygen to another

substance, or removes hydrogen from that substance.

Examples
Consider the following reactions:

1. CuO (s) + H2 (g)  Cu (s) + H2O (l)

(black) (Brown)

In this reaction, the hydrogen has been oxidized to water (since oxygen has been
added to it).
The copper (II) oxide is described as an oxidizing agent.

2. H2S(g) + Cl2(g)  2HCl(s) + S(s)

In this reaction although oxygen does not appear in any of the substances involved in
the reaction. The reaction is still called REDOX reaction. This is because the chlorine
molecule removes hydrogen from the hydrogen sulphide (H2S). The oxidizing agent
in this case is chlorine.

(ii) Reduction
Reduction is the addition of hydrogen to a substance or the removal of oxygen
from a substance.

A reducing agent is a substance which transfers hydrogen to another

substance, or removes oxygen from another substance.

Examples
Consider examples 1 and 2 above.
In example (1), the copper (II) oxide has been reduced to copper (since oxygen has
been removed from it). The hydrogen that caused the reduction is called a reducing
agent.

179
In example (2), the hydrogen sulphide (H2S) has added hydrogen to chlorine,
therefore, reduction has occurred. Thus, the hydrogen sulphide is called a reducing
agent.

Note that:
Reduction and oxidation occur simultaneously in a reaction; therefore, in short, such
a reaction is called REDOX reaction.

Exercise
Identify the oxidizing agent and the reducing agent in the following reaction.
2Fe(s) + 4H2O(g)  Fe2O4(s) + 4H2(s)

Answer
Oxidising agent: Water (H2O)
Reducing agent: Iron (Fe)

(b) Oxidation as the addition of electronegative element

(i) Oxidation
Oxidation is the addition of an electronegative element to a
substance
Or the removal of an electropositive element from a
substance.
(ii) Reduction
Reduction is the addition of electropositive element to a substance.
or the removal of an electronegative element from a
substance.

NB: Electronegative elements are elements that tend to attract electrons to

themselves (to become ultimately, negative ions).
E.g. Non-metallic elements (F, O, C, N, Br, S).
Electropositive elements are elements that tend to donate electrons to other
substances (to become ultimately, positive ions).
E.g. Metallic elements (K, Na, Ca, Mg, Al, Zn)

The electro-negativity of some elements

(i) Non-metals

180
 (ii) Metals

Examples
1. C2H4(g) + Cl2(g)  C2H4Cl2(g)

The ethane (C2H4) has been oxidized because chlorine (electronegative) has been
added to it.
2. Mg(g) + 2HCl(aq)  MgCl2(aq) + H2(g)

Magnesium is oxidized by addition of chlorine and hydrochloric acid (HCl) is

reduced by removing chlorine from it. Therefore, magnesium is a reducing agent,
while the hydrogen chloride is an oxidizing agent.

3. S(s) + Cl2(g)  SCl2(l)

Both sulphur and chlorine are electronegative elements, but because chlorine is much
more electronegative than sulphur, then it is regarded as the oxidizing agent, while
the sulphur is reducing agent.

(c) Oxidation and Reduction in terms of a change in oxidation number

(i) Oxidation
Oxidation is an increase in oxidation number.

(ii) Reduction
Reduction is the decrease in oxidation number.

Simple rules concerning the idea of oxidation number

(i) All elements in the free state (i.e. uncombined with any other elements)
have oxidation number of zero.
(ii) In the case of a simple ion the element has oxidation number with the same
size and sign of the charge on the ion.

Ion Oxidation number/sign

Cu2+ +2
Fe2+ +2
Fe3+ +3
S2- -2
O2- -2

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2. The sum of all the oxidation numbers of the elements in a compound is zero.

The rules enable us to workout the oxidation number of an element of which

we are uncertain.

Example
Workout the oxidation number of sulphur in the compound FeSO4.

Let the oxidation number of S =x

Oxidation number of oxygen = -2
Oxidation number of iron (II) = +2

+2 + x + (4 x –2) =0
+2 + x + -8 =0
x- 6 =0
x = +6
 The oxidation number of S = +6

With these simple rules in mind the concept of oxidation and reduction may be
redefined.
Oxidation: when oxidation occurs the oxidation number of the element increases.
Reduction: when reduction occurs the oxidation number of the element decreases.

Exercise
Find the oxidation number of Fe and Ca in the following compounds.
(i) FeCl3 (ii) CaCl2 (Given that oxidation number of Cl = -1)

Answers
(i) +3 (ii) +2

Example
Now consider the reaction below.
FeCl2(s) + ½Cl2(g)  FeCl3 (s)
The oxidation number of Fe changes from +2 in FeCl2 to +3 in FeCl3 and therefore,
FeCl2 has been oxidized to FeCl3 by chlorine, which is therefore the oxidizing agent.

In the similar way the oxidation number of chlorine changes from 0 in the Free State
to –1 in FeCl3 the chlorine has therefore been reduced by the FeCl2.

182
(d) Oxidation and Reduction in terms of electron transfer

From the electronic point of view the following are the most recent definitions of
oxidation and reduction.
(i) Oxidation is the loss of electrons
(ii) Reduction is the gain of electrons

An oxidizing agent is an electron accepter.

A reducing agent is an electron donor.

Memory aid:
O Oxidation
I Is
L Loss of electrons

R Reduction
I Is
G Gain of electrons

NB: (i) Oxidation and reduction always occur simultaneously.

They are complementary processes of electron loss and electron gain
respectively.
(ii) The electrons lost by the reducing agent (electron donor) must be
accepted by the oxidizing agent (electron accepter) present.

Examples
1. When magnesium is oxidized by combination with oxygen, the metal is
oxidized by losing two electrons per atom.
These electrons are accepted by oxygen atoms, which are reduced as a result:
Magnesium (giving out electrons) is the reducing agent; while oxygen
(accepting electrons) is the oxidizing agent.
Mg - 2e  Mg 2+
½O2 + 2e  O2-
I.e. Mg (s) + ½O2 (g)  MgO(s)
2. If a metallic ion, e.g, the iron (II) ion, Fe2+, is so treated that it loses a further
electron, it is oxidized and is a reducing agent. The process is:
Fe2+ - e-  Fe3+
An iron (III) ion is formed. An agent must be present, e.g., chlorine, to accept
the electrons made available by the ferrous ions. It acts as the oxidizing agent
(electron acceptor) and is reduced.
½Cl2 + e-  Cl-

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 The complete reaction can be represented:
Fe2+(aq) + ½Cl2 (g)  Fe3+(aq) + Cl-(aq)

It will be observed that ‘valency” of the metal increases from 2 to 3 during the
oxidation.

3. The ‘removal of hydrogen’ aspect of oxidation is interpreted in the following

way.
Consider the oxidation of hydrogen sulphide by chlorine. Hydroen sulphide is
slightly ionized as:
H2S 2H+ + S2-

The sulphide ion parts with its two electrons and is, therefore, oxidized,,
acting as a reducing agent.
S2- - 2e-  S

The electrons are accepted by chlorine atoms, so that chlorine acts as an

oxidizing agent and is reduced.
Cl2 + 2e-  2Cl-
Adding the two equations, we have:
S2-(g) + Cl2 (g)  S (s) + 2Cl-(g)

The hydrogen ion of the hydrogen sulphide is unchanged.

4. The reduction of hot copper (II) oxide by hydrogen is given as:

Cu2+ + O2- (s) + H2 (g)  Cu (s) + H2O (g)

It is clear that the copper (II) ion is reduced by the reaction between the oxide
ion O2- and hydrogen:
O2- (s) + H2 (g)  H2O + 2e-

By combining with oxygen in this way and supplying electrons to the metallic
ion, hydrogen exercises reducing properties. The oxide ion is oxidized by
electron loss and the oxygen atom remains in combination with hydrogen as
water.

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 SELF-CHECK 9.1
1. Which one of the following processes is not an example of oxidation?
A. The burning of methane in air. B. The rusting of iron.
C. The melting of a candle wax. D. The smouldering of phosphorus.

2. Which one of the following equations does not represent oxidation of the first
substance?
A. Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g).
B. 2H2S(aq) + O2 (g)  2S(s) + 2H2O(l).
C. S(s) + O2(g)  SO2(g).
D. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s).

3. Which one of the following equations does not represent reduction reaction?
A. N2(g) + 3H2 (g)  2NH3(g). B. Fe(s)  Fe3+(aq) + 3e.
C. Cl2(g) + 2e  2Cl-(aq). D. Cu2+(aq) + 2e  Cu(s).

4. Which one of the following equations represents an oxidation – reduction reaction?

A. H+(aq) + OH-(aq)  H2O(l).
B. Pb2+(aq) + SO42- (aq)  PbSO4(s).
C. Cu2+(aq) + Fe(s)  Cu(s) + Fe2+(aq).
D. Ca2+(aq) + CO 2  (aq)
3
 CaCO3(s).

5. Which one of the following represents a reduction-oxidation reaction?

A. 2NaOH(aq) + CuCl2(aq)  Cu(OH)2(s) + 2NaCl(aq).
B. 2FeCl2(aq) + Cl2(g)  2FeCl3(aq).
C. 2NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2H2O(l).
D. ZnCO3(s) + 2HNO3(aq)  Zn(NO3)2(aq) + H2O(l) + CO2(g).

6. Which one of the underlined substances in the equation below is being reduced?
A. CuO(s) + H2 (g)  Cu(s) + H2O(l)
B. 2Fe2+ (aq) + Cl2 (g)  2Fe3+ (aq) + 2Cl- (aq)
C. MnO2 (s) + 4HCl (aq)  MnCl2(aq) + 2H2O(l) + Cl2 (g)
D. H2SO4 (l) + C(s)  CO2 (g) + 2SO2 (g) + 2H2O (l)

185
 In each questions 7 to 8 one or more of the answers given may be correct. Read each
question carefully and then indicate the correct answer A, B, C or D according to the
following:
A. if 1, 2, 3 only are correct. B. if 1, 3 only are correct
C. if 2, 4 only are correct. D. if 4 only is correct.
INSTRUCTIONS SUMMARISED:

Instruction Summarised
A B C D
1, 2, 3 1, 3 2, 4 4
only correct only correct only correct only correct

7. Which of the following equation(s) show(s) a redox reaction?

1. NH3(g) + HCl(g)  NH4Cl(s).
2. Mg(s) + Cl2(g)  MgCl2(s).
3. 2HNO3(aq) + Mg(s)  Mg(NO3)2(aq) + H2(g).
4. 3CuO(s) + 2NH3(g)  3Cu(s) + N2(g) + 3H2O(l).

8. Oxidation is a reaction in which

1. oxygen is removed from a substance.
2. hydrogen is removed from a substance.
3. hydrogen is added to a substance.
4. electron is lost from a substance.

SELF-CHECK 9.2
1. (a) Define oxidation in terms of electrons.
(b) Write the equations of the following half reactions and in each case state
whether the reaction is a reduction or an oxidation reaction.
(i) The conversion of hydrogen ions (H+) to hydrogen molecules (H2).
(ii) The conversion of iron (II) ions (Fe2+) to iron (III) ions (Fe3+).
(iii) The conversion of chlorine molecules (Cl2) to chloride ions (2Cl-).

(c) Which elements in the second and third periods of the periodic table are strong
(i) Oxidizing agents?
(ii) Reducing agents?
(iii) What properties do strong oxidizing agents have?
(iv) What properties do strong reducing agents have?

186
 CHAPTER TEN

ENERGY CHANGES IN CHEMICAL REACTIONS

Introduction:
Energy is either given out to or absorbed from the surrounding whenever a chemical
reaction takes place.
The change in energy is the driving force for a chemical reaction.
There are several forms of energy. The most common form of energy in chemical
reaction is the heat energy also called enthalpy.

Enthalpy and enthalpy changes

(i) Enthalpy
Enthalpy is the energy (heat) content of a substance which is stored in its bonds.
Its symbol is H.

(ii) Enthalpy change

When a chemical reaction takes place, the enthalpy of the reactants change.
The change in enthalpy is denoted by ∆H.

Bond breaking and Bond formation

When a chemical reaction occurs, bonds in the reactants break and new bonds are
formed in the products.
E.g. A-B + C-D → A-C + B-D
Bonds break Bonds form

Bond between AB and CD break and new bonds form between AC and BD.

During bond breaking, energy is absorbed from the surrounding while during bond
formation, energy is released to the surrounding.

It is the energy difference in bond breaking and bond formation that we measure. This
can be shown as a temperature rise or fall in the thermometer.

187
The Joule (J)
The unit of energy is the joule, J.

Definition:
The Joule is the energy that raises the temperature of 1g of water through 0.24 K.
The larger unit is Kilo Joule, KJ. 1KJ = 1000 J

Enthalpy changes in chemical reactions refer to one mole of a particular

reactant or product in a balanced equation.
The enthalpy changes involved are normally large. Therefore, KJ is preferred
to joule which is too small.

Types of Chemical Reactions

There are two types of reactions, namely:
(i) Exothermic reaction
(ii) Endothermic reaction

(a) Exothermic Reaction

Exothermic reaction is the reaction in which heat energy is given out
to the surroundings.

In exothermic reaction there is temperature rise in the thermometer and the

temperature rise is a measure of this energy.
The energy of the products is less than the energy of the reactants.

 The enthalpy change, ∆H has a negative sign.

Examples of exothermic reaction include:

H2 (g) + ½ O2 (g)  H2O (l) ∆H = -286 KJ
C (s) + O2 (g)  CO2 (g) ∆H = -393 KJ
N2 (g) + H2 (g)  NH3 (g) ∆H = -92 KJ

(b) Endothermic Reaction

Endothermic Reaction is a reaction in which heat energy is absorbed
from the surroundings.

There is temperature drop in the thermometer and the temperature

drop is a measure of this energy.

The energy of the products is more than the energy of the reactants.

188
 Therefore, ∆H is positive.
Examples of endothermic reaction
C(s) + 2S (s)  CS2(s) ∆H = +117 KJ
N2 (g) + ½ O2 (g)  NO(g) ∆H = + 90.3 KJ

Enthalpy change 
= Enthalpy of  
Enthalpy of
products  reactants 
∆H = H (Products) - H (reactants)
Types of Enthalpy change
The heat change/enthalpy change which occurs in a reaction is named after the
type of reaction in which it occurs. They include the following:-
(i) Enthalpy of combustion (Heat of combustion).
(ii) Enthalpy of formation (Heat of formation).
(iii) Enthalpy of neutralization (Heat of Neutralization).
(iv) Enthalpy of solution (Heat of solution).

(a) Enthalpy of Combustion

Enthalpy of Combustion is the enthalpy change that occurs when one mole
of a substance is completely burnt in (pure) oxygen.

Or is the heat liberated when one mole of a substance is completely

burnt in oxygen.

Determination of Enthalpy of Combustion of Liquid eg ethanol

Apparatus:

Retort stand/clamp, a thin walled tin can, thermometer, water, ethanol, and
a spirit lamb.

Procedure:
 Weigh a spirit burner containing pure ethanol.
 Fill a thin-walled tin can with a known volume of water.
 Clamp the tin can with its content above the lamp as shown in the diagram below.
 Insert the thermometer into the water.
 Read and record the initial temperature of the water.

189
 Heat the water while carefully stirring with the thermometer.

 When the thermometer shows a convenient temperature rise by 25 C, put out the
flame and read and record the final temperature.

 Reweigh the lamb and its content.

Results:
Volume of cold water = V1 cm3
Initial temperature of cold water = t1 C
Final temperature of water = t2 C
Mass of spirit burner before burning = m1 g
Mass of spirit burner after burning = m2 g
Specific heat capacity of water = 4.2 J/g C

Calculation
Heat gained by water = mct

= Mass of  
Sp. ht. cap  
Change in
water x of water x Temperature 
= (Vol. x Density) x 4.2 x (t2 - t1)

= (V1 x 1 x 4.2 x (t2 - t1)

= (V1 x 4.2 x (t2 - t1) Joules

190
Heat produced/lost by ethanol is calculated as follows:
The relative molecular mass of ethanol, C2H5OH,
= (2 x 12) + (1 x 6) + 16
= 24 + 6 + 16
= 46 g

Mass of ethanol used up (burnt) 

= Mass 
of spirit lamp Mass of spirit lamp
before burning  after burning 
= m1 - m2

Assuming heat gained by water = Heat given out by ethanol

(m1 - m2) g of ethanol produce, V1  4.2  (t2 - t1) joules of energy

46
46 g of ethanol produce = x (t2t1)  (Vol.  4.2) KJ mol-1
m1  m2

Or in terms of moles

Mass of ethanol used

Moles of ethanol used up =
Relative molecular mass
m1  m2
=
46

m1  m2
If moles of ethanol produce V1  4.2  (t2 - t1) joules of KJ energy
46
 46 
Then 1 mole of ethanol produces    (V1  4.2) x (t2t1) KJ mol-1
 m1  m2 

This is the heat of combustion of ethanol.

Example:
An S.4 student in TLA carried out an experiment to determine the enthalpy of
combustion of ethanol in the laboratory and obtained the following results.

Volume of water = 120 cm3

Initial temperature of water = 20 C
Final temperature of water = 46 C
Mass of the spirit burner before = 33.0 g
Mass of the spirit burner after = 32.5 g

191
 (a) Using the above results calculate the heat of combustion of ethanol.
(b) State any precautions that could be taken to obtain accurate result.
(specific heat capacity of water = 4.2 g-IK-I, and = 1g/cm3)

Solution
(a) Heat energy gained by water = mct


= Mass of  
Sp. ht cap  
Change in
water  of water  Temperature 

= Vol. x Density x Sp . ht cap 
Change in
of water  Temperature 
= 120  1  4.2  (t2 - t1)
= 120  4.2  (46 – 20)
= 120  4.2  26
= 13104 J

Mass of ethanol burned 

= Mass 
of spirit lamp Mass of spirit lamp
before burning  after burning 
= 33 - 32.5
= 0.5 g

Formula mass of C2H5OH, = 2  12 + 1  6 + 16

= 46 g

0.5 g of ethanol liberates 13104 J of heat energy.

13104
1 g of ethanol liberates J of heat energy.
0 .5
13104
46 g of ethanol will liberate x 46 = 1205568 J
0 .5
= 1205.568 KJ mol-1

The enthalpy of combustion of ethanol = -1205.568 KJ mol-1

(b) This experiment gives less value because some heat was lost to the surroundings
and in heating the can itself.

The flame should be shielded from wind by fixing a hard cardboard around the
setup.

192
 Enthalpy of neutralization
Enthalpy of neutralization is the enthalpy change that occurs when an
acid reacts with an alkali to produce a salt and a mole of water.
Or is the enthalpy change that occurs when one mole of aqueous hydrogen ions
reacts with one mole of aqueous hydroxide ions to form 1 mole of water.

E.g. NaOH (aq) + HCl (aq)  NaCl (aq) + H2O (g)

Or OH-(aq) + H+ (aq)  H2O (l) H = -57.3 KJ mol-1

NaOH(aq) + ½H2SO4 (aq)  ½Na2SO4 (aq) + H2O (l)

Or OH- (aq) + H+ (aq)  H2O (l) H = -57.3 KJ mol-1

NB:
 The basicity of H2SO4 is two i.e. one mole of H2SO4 supplies two moles of H+ ions.
 Therefore to get one mole of H+ ions half a mole of H2SO4 is required.
 The enthalpy of neutralization of any strong acid and alkali is -57.3 KJmol-1.
 Strong acids and alkalis dissociate completely into ions when dissolved in water.
 For weak acid and alkali, the enthalpy of neutralization is less than -57.3 KJ mol-1.
 This is because the acid or alkali is slightly dissolved into ions when in aqueous
solution. Some heat is absorbed in order for complete dissociation to take place.

Determination of the enthalpy of neutralization

Experiment
To determine the enthalpy of neutralization of sodium hydroxide by
hydrochloric acid

Apparatus:
Thermometer, a plastic cup or beaker, stirrer, lagging material.

Procedure:
- Place hydrochloric acid of known volume and molarity into the plastic cup.
- Read and record the initial temperature t1, of the acid.
- Fill sodium hydroxide of the same volume and molarity as for the acid in the
burette.
- Read and record the initial temperature t2, of the alkali.
t t
- Determine the initial temperature = average temperature, 1 2 , of the reactants.
2

193
- As rapidly as you can transfer the base solution to the plastic cup using burette
and stir the mixture using the thermometer

- Read and record the maximum temperature attained.

Specimen results
Volume of 2M of hydrochloric acid solution = 20.0 cm3
Volume of 2M of sodium hydroxide solution = 20.0 cm3
Initial temperature hydrochloric acid acid = 15.0 0C
Initial temperature of sodium hydroxide = 15.4 0C
Final maximum temperature of mixture = 28.2 0C
Density of water = 1g/cm3

Calculations:
Hint: (i) In the calculations, it is assumed that the total volume of the
reactants is equal to the volume of water formed when the acid reacts
with the base.
(ii) All the heat energy liberated is absorbed by the water formed.
(iii) The mass of water formed obtained by multiplying the density of
water and the volume.

Final volume of solution = Vol. of acid + Vol. of base

= (20 + 20) cm3
= 40 cm3

Mass of solution = Volume  density

= 40  1
= 40 g
Initial temperature = average of initial temp. of acid base

194
 15.0  15.4
=
2
0
= 15.2 C
Final temperature of mixture = 28.2 0C

Change in temperature, t = (Final - Initial) temperature

= 28.2-15.2
= 13 0C

Given the spirit heat capacity of water is 4.2 Jg-1,

Heat evolved = Gained by the water formed
= Mass  heat cap  temperature rise
= 40  4.2  13.0 J
 Heat evolved = 2184 J

Moles of hydrochloric acid = Molarity  volume in litres

2 x 20
=
1000
= 0.04 moles of acid

From NaOH (aq) + HCl (aq)  NCl (aq) + H2O (l)

And since one mole of hydrochloric acid produces one mole of water, 0.04 moles
will produce 0.04 moles of water.

Hence 0.04 moles of water is produced with the evolution of 2184 J heat energy.
2184 x 1
1 mole of water will be produced with the evolution of
0.04
= 54600J
= 54.6 KJ mol-1.
Therefore the heat of neutralization is -54.6 KJ mol-1 water formed.

NB: The value obtained is less than the actual value -57.3 KJ mol-1
due to experimental errors.

195
Enthalpy change of solution

Enthalpy change of solution is the enthalpy change that occurs when one mole of a
solute/compound completely dissolves in a stated amount of solvent/water.

NB: Heat of solution depends on the quantity of water used.

Determination of heat of solution

Experiment: To determine heat of solution of:

1. Sulphuric acid

Add 53.5 cm3 of concentrated sulphuric acid (i.e 98 g or 1 mole) carefully to

900 cm3 of water in a large, thin glass or polythene beaker.
Stir well and record the rise in temperature (170C)
(The final solution is about M concentration, its mass is about 1000 g and its
specific heat capacity is about 4.2 Jg0C).

Heat of solution = Mass  specific heat capacity  temperature rise

= 1000  4.2  17
= 71.400
= 71.4 KJ mol-1

H2SO4 (aq) + (aq)  M H2SO4 (aq) H = -70 KJ mol-1

196
2. Heat of solution of Sodium hydroxide
Add 40 g sodium hydroxide (1 mol) pellets to 960 cm3 of water in a beaker as
in 1 above.
Stir well and record the temperature rise, .
Calculate the heat of solution of sodium hydroxide.

Heat of solution = Mass  sp. heat capacity  temperature rise

= 1000  4.2  
= 4200 x  KJ mol-1

3. Heat of solution Ammonium nitrate

Add 80 g (1 mol) of the finely powdered Ammonium nitrate to 920 cm3 of
water.
Record the temperature fall (θ 0C).
The concentration of the final solution I s M, its mass is 1000 g and its
specific heat capacity is 4.2 J/g0C.
Use the above equation to calculate the heat of solution of the salt.
NH4NO3 (s) + (aq)  M NH4NO3 (aq) H = +25 KJ/Mol-
1

Heat of solution = Mass  sp. heat capacity  temperature rise

= 1000  4.2  
= 4200 x  KJ mol-1

4. Enthalpy of displacement of a metal ion in solution by another metal

When ion fillings are added to copper (II) sulphate solution, the blue solution
turns pale green and a brown solid is deposited.
The pale green colour is due to ion (II) sulphate solution and the brown solid
is copper.

CuSO4 (aq) + Fe (s)  FeSO4 (aq) + Cu (s)

Or ionically
Cu2+(aq) + Fe (s)  Fe2+(aq) + Cu (s)

In this reaction, ion displaces copper (II) ions.

The reaction is exothermic and a thermometer would show a temperature rise.

197
 Example
Candidates of TLA carried out an experiment to determine the enthalpy change of
displacement. They added 1.0 g of zinc powder to 50 cm3 of 0.2M copper (II)
sulphate solution into a plastic beaker and stirred the mixture well. The temperature at
the beginning of the experiment was 20 C and it rose to a maximum of 27 C.
(a) Why was a plastic beaker used in this experiment?
(b) (i) State what was observed.
(ii) Write an ionic equation for the reaction which took place.
(c) Calculate:
(i) The heat produced during the reaction.
(ii) The enthalpy change of the reaction.
(iii) Would the value of H have been bigger or smaller if iron
fillings were used instead of zinc?
Solution:
(a) Plactic beaker is a good heat insulator and absorbs hardly any of the
heat produced in the reaction.
(b) (i) The blue colour of copper (II) sulphate solution turned to
Colourless and
A brown precipitate was formed.
(ii) Cu2+ (aq) + Zn (s)  Zn2+ (aq) + Cu (s)

(c) (i) Volume of solution = 50 cm3, sp. ht. cap. = 4.2 J/g/K,
Density of water = 1 g/cm3.

Heat produced = mc

= Vol. x  x c x (t2 – t1)
= 50 x 1 x 4.2 x (27 – 20)
= 50 x 4.2 x 7
= 1470 J
 Heat produced = 1.47 KJ

(ii) 1000 cm3 of copper (II) sulphate solution contains 0.2 moles.
0.2 x 50
50 cm3 of copper (II) sulphate solution contains
1000
= 0.01 moles.
0.01 moles of copper liberates 1470 KJ of heat energy.
1 x 1470
1 mole of copper will liberate = 147 KJ mol-1
0.01
(iii) Smaller: this is because iron is less reactive than zinc

198
 (i.e. iron is below zinc in the reactivity series) and
would require more energy to break the bonds in copper
(II) sulphate and in the formation of the new bonds.

Enthalpy of formation
Enthalpy of formation is the heat liberated or absorbed when 1 mole of substance is
formed from its constituent elements under standard conditions.

Example
1. (a) Define the heat of formation
(b) Iodine and chlorine react as follows
I2 (S) + Cl2(g)  2ICl (s) H = -68 KJ

Calculate the heat of formation of iodine monochloride. From the equation;

Formation of 2 moles of ICl liberates 68 KJ.

Solution
(a) Enthalpy of formation is the heat liberated or absorbed when 1 mole of
substance is formed from its constituent elements under standard conditions.

1 x 68
(b) The heat of formation of 1 mole of ICl =
2
= 34 KJ mol-1

SELF-CHECK 9.1
1. (2005 Q.36)
10 g of methanol, CH3OH, burns in air to liberate 226 kJ of heat. The amount of heat
liberated when 1 mole of methanol is burnt in air is [H = 1; C = 12]
32 x 226 10 x 32
A. B.
10 226

10 10 x 226
C. D.
32 x 226 32

2. (2005 Q.12)
Butane burns in excess air according to the following equation:
2C4H10(g) + 13O2(g)  8CO2(g) + 10H2O ∆H = -5760 kJ.
3
The quantity of heat evolved when 1.6 dm of butane is burnt at room temperature
Is [1 mole of gas occupies 24 dm3 at room temperature]

199
 5760 x 116 5760 x 1.6
A. B.
2 x 24 2 x 24
5760 x 1.6 5760 x 116
C. D.
24 24

3. (2004 Q.5)
Carbon burns in oxygen according to the following equation:
C(s) + O2 (g)  CO2 (g)

The amount of the heat evolved when 480g of carbon is burnt completely in oxygen
is [Molar heat of combustion of carbon is 2.2 x 10-7 kJ mol-1, C = 12]

480 x 12
A. 480 x 12 x 2.2 x 10-7 B.
7
2.2 x 10

2.2 x 107 x 12 2.2 x 107 x 480

C. D.
480 12
4. (2003 Q.12)
When 1 g of methanol was burnt in excess air, 22.6 kJ of heat were liberated.
What was the quantity of heat in kJ liberated when 1 mole of methanol was
burnt under similar conditions?
A. 22.6 B. 32.0
C. 723.2 D. 777.8

5. (2003 Q.27)
5.3 kJ of heat energy are required to vaporise 13 g of a liquid X (X = 78).
The molar heat of vaporisation of X in kJ/mole is
5.3 x 78 80
A. B.
13 1.88 x 188
188
C. 13 x 5.3 x 78 D.
1.88 x 80
6. (2003 Q.37)
When 0.4 g ethanol was burnt, it raised the temperature of 0.1 kg of water by
20 oC. The heat of combustion of ethanol is
[Specific heat capacity of water = 4.2 kJ/kg/oC, C2H5OH = 46]

4.2 x 20 x 46 0.4 x 4.2 x 20

A. kJ mol-1 B. kJ mol-1
0.4 x 0.1 46 x 0.1
0.1 x 4.2 x 20 x 46 0.1 x 4.2 x 20
C. kJ mol-1 D. kJ mol-1
0 .4 46 x 0.4

200
7. (2002 Q.13)
The amount of heat evolved when 6.0 g of metal, M was displaced from a solution
was 28.8 kJ. The amount of heat produced when 0.5 mole of M was displaced is
28.8 x 54.9 x 0.5 54.9 x 6.0
A. B.
6.0 28.8 x 0.5
28.8 x 54.9
C. D. 28.8 x 54.9 x 6.0 x 0.5
6.0 x 0.5

8. (2002 Q.30)
Glucose burns in oxygen at 25 oC according to the equation below, giving out
2802 kJ mol-1 of heat energy.
C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(g).
The amount of heat produced when 18.0 g of glucose is burnt in oxygen at the
same temperature is (H = 1, C = 12, O = 16)
2802 x 18.0 180
A. B.
180 x 25 2802 x 18.0
180 x 25 x 18.0 2802 x 18.0
C. D.
2802 180

9. (2001 Q.3)
Methanol burns in excess air according to the equation
2CH3OH(l) + 3O2(g)  2CO2(g) + 4H2O(l). ΔH = -730kJ mol-1.
The amount of heat liberated when 3.2 g of methanol, (Mr = 32.0) is completely
burnt is
A. 73 kJ. B. 730 kJ. C. 1416 kJ. D. 2929 kJ.

10. (2001 Q.38)

Carbon burns in excess oxygen according to the equation:
C(s) + O2(g)  CO2(g). ΔH = -393 kJ mol-1.
What mass of carbon in grams would produce 750 kJ of energy?
393 x 12 750 x 12
A. B.
750 393000
750 x 12 750 x 393
C. D.
393 12

201
11. (2000 Q.32)
13.70 kJ of heat was evolved when 4.0 g of copper was displaced from copper(II)
sulphate solution by zinc. The amount of heat evolved when one mole of copper was
displaced is
63.5 x 4 13.7 x 63.5
A. B.
13.7 4
13.7 x 4 63.5 x 4
C. D.
63.5 13.7

12. (1998 Q.14)

When 8 grams of a salt was dissolved in 100 g of water the temperature decreased by
10oC. The drop in temperature when 2 grams of the salt is dissolved in 100 g of water
would be
A. 10 oC. B. 8.5 oC. C. 5.0 oC. D. 2.5 oC.

13. (1998 Q.36)

Ethanol burns in oxygen according to the equation
7
C2H5OH + O2(g)  2CO2(g) + 3H2(g) ∆H = -1185 kJ mol-1..
2
Calculate the amount of heat given out when 0.2 moles of ethanol is burned
completely.
A. –237 kJ. B. –592.5 kJ. C. –1185 kJ. D. –2370 kJ.
14. (1997 Q.17)
Carbon reacts with sulphur according to the following equation:
C(s) + 2S(s)  CS2(l) H = 116 kJ mol-1.
The amount of heat absorbed when 16 g of sulphur reacts with excess carbon is
(C = 12, S = 32)
A. 7 kJ. B. 29 kJ. C. 58 kJ. D. 116 kJ.
15. (1996 Q.8)
Methane burns in air according to the equation:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) + Heat (H = 890 kJ mol-1.)
The energy liberated when 4 g of methane is burnt in air: (C = 12, H = 1)
A. 222,5 kJ. B. 445.0 kJ. C. 1780.0 kJ. D. 3560.0 kJ.

202
 SELF-CHECK 10.2
1. (a) Define enthalpy of combustion.
(b) Describe an experiment that can be carried out to determine the heat of
combustion of propanol. Draw a diagram to illustrate your answer.
(c) When 0.54 g of propanol was burnt, the heat evolved raised the temperature
of 150 cm3 of water from by 21.5 oC. Calculate the heat of combustion of
propanol. (C = 12, H = 1, O = 16, Specific heat capacity of water is 4.2 kJ g-1,
density of water = 1 gcm3,).
2. (2001 Q.8)
(a) Define the term enthalpy (heat) of neutralisation.
(b) When 50 cm3 of a 0.5M hydrochloric acid was added to 50 cm3 of a 0.5 M
potassium hydroxide in a calorimeter, there was a temperature rise from 27.5
o
C to 30.8 oC. Calculate the enthalpy of the reaction.
(Density of water is 1 g dm-3 and its heat capacity is 4.2 Jg-1)

EXTRA SELF-CHECK QUESTIONS

1. (1994 Q.14)
(a) Describe an experiment that can be carried out to determine the heat of
combustion of ethanol. Draw a diagram to illustrate your answer.
(b) Would you expect the heat of combustion as determined in the experiment in
(a) to be greater than, lower than or equal to the theoretical value? Give a
reason.
(c) When 0.382 g of ethanol was burnt, the heat evolved raised the temperature of
100 g of water from 16.5oC to 43.5oC.
Calculate the heat of combustion of ethanol.
(Heat capacity of water is 4.2 kJ g-1)
(d) Name two products, other than water, of incomplete combustion of ethanol.

2. (2003 Q. 8)
(a) What is meant by the term enthalpy of neutralisation?
(b) When 50.0 cm3 of a 1M sulphuric acid was added to 50.0 cm3 of a 2M sodium
hydroxide, the temperature of the resultant mixture rose by 13.6 oC.
(i) Write an ionic equation for the reaction that took place.
(ii) Calculate the enthalpy of neutralisation of sodium hydroxide. [Specific
heat capacity of water = 4.2 J g-10C-1, density of water = 1 gcm-3]

203
 CHAPTER ELEVEN
EXTRACTION OF METALS
(Mineral Resources)
Introduction
The reactivity or electrochemical series is a valuable guide in the study of metals and their
compounds. Only the metals low in the activity series such as gold, silver, mercury and
copper which are unreactive metals, occur naturally. All the other metals occur as
compounds in mineral ore.
The reactive metals are extracted by electrolysis, while the metals in the middle of the series
(zinc and iron) are obtained by reducing their oxides by coke or carbon monoxide.
NB: - The method of extraction of metals depends on the stability of the metal.
- the extraction of metals is a reduction process.
- The metals are in their ores as positive ions and extracted by supplying
the ions with electrons.
- The cathode supplies electrons to the metallic ions during electrolysis and
- The coke combines with oxide ions of the oxides to release electrons which
reduce the metallic ions.

Metal Chief ores Method of Extraction

Rock salt, NaCl
Na Chile saltpetre, NaNO3 Electrolysis
Bauxite (Al2O3.2H2O)
Al Cryolite, Na3AlF6 Electrolysis of oxide
Corundum (Al2O3)
Mica and China clays
Zinc blende, Zinc sulphide (ZnS)
Zn Calamine, Zinc carbonate (ZnCO3)
Hematite, Iron (III) oxide (Fe2O3) Reduction of oxide by
Magnetite, Fe3O4 carbon or carbon monoxide
Siderite, Iron (II) carbonate, FeCO3
Fe
Pyrite, iron disulphide, FeS2

Copper pyrite, iron disulphide (CuFeS2) Roast in air

Cu Cuprite, Cu2O
Chalcocite, Cu2S
Malachite, CuCO3.Cu(OH)2

204
 EXTRACTION OF IRON

Occurrence:
Iron is the second most abundant metal after aluminium. It forms 4% of the earth’s
crust.
It exists in form of ores. The chief ores being:
- Hematite - (Fe2O3) - found in United States, Australia & Russia.
- Magnetite - (Fe3O4) -found in Sweden & North America.
- Siderite - (FeCO3). - found in Britain.
These iron ores are mined from their deposits and transported to the plants for
extraction.

Concentration of the ore

The richer ores are picked from the earthly impurities, crushed and then roasted in air
to remove water vapour and non metallic impurities such as sulphur and phosphorous
into their respective oxides according to the equations:
S (s) + O2 (g)  SO2 (g)
4P (s) + 3O2 (g)  2P2O3 (g)

The removal of these non metals leaves Silicon (IV) oxide (SiO2) as the main impurity.
Extraction of Iron from Hematite
The mixture of the iron ore, coke and limestone is fed into a blast furnace from the top.

Preheated air is injected into the furnace through the pipes called tuyeres at the lower
part of the furnace. The air oxidizes the hot coke to carbon dioxide.
C (s) + O2 (g)  CO2 (g)
C (s) + CO2 (g)  2 CO (g)

NB: The reaction is exothermic hence large quantity of heat is liberated.

Reduction of the ore Fe2O3 by Carbon monoxide (CO) and Carbon (C)
The carbon monoxide and the unchanged carbon reduce the iron (II) oxide to iron at
temperature ranges of (500 – 800) ˚C according to the equations:

Fe2O3 (s) + 3CO (g)  2Fe (l) + 3CO2 (g)

2Fe2O3 (s) + 3C (s)  4Fe (l) + 3CO2 (g)

The hot iron formed absorbs carbon from the coke and lowers its melting point. It
then sinks and collects at the bottom of the furnace in molten form, where it is taped
off and solidified into blocks called “pig iron” or cast iron.

205
 The diagram showing the blast furnace

The Role of the Limestone

- The purpose of the limestone is to remove Silcon (IV) oxide which is the main
impurity.
- It first decomposes to calcium oxide (quick lime CaO) according to the equation:

CaCO3 (s)  CaO (s) + CO2 (g)

The calcium oxide then combines with Silcon (IV) oxide to form Calcium
silicate(CaSiO3), the slag, according to the eqation:

CaO (s) + SiO2 (s)  CaSiO3 (s)

The slag floats on the molten iron and is tapped off separately.

206
Purification of Cast Iron (Puddling)
- The cast iron obtained from the blast furnace is 90 - 95% pure. The main
impurities are carbon, silicon, manganese, sulphur and phosphorus depending on
the original ore.
- Pure iron (wrought iron) is obtained by heating a mixture of pig iron with iron (II)
oxide (FeO) as an oxidizing agent and lime stone in the blast furnace.
- The oxygen of the iron (II) oxide oxidizes the impurities as follows:
- Carbon and sulphur oxidized to gaseous oxide (CO2 and SO2) which escape
as waste gases at the top of the furnace,
- Phosphorus to phosphate and
- Silicon to silicate.

Use of Iron
(a) Cast Iron
Contains about 3-5% carbon, 1% silcon and 2% phosphorus.
It is brittle but extremely hard.
(i) It is used in making:
- Furnace, grates, Bunsen burner bases, drainage pipes, iron boxes etc.
(ii) It is used in the manufacture of wrought iron and steel.

(b) Wrought Iron

It contains about 0.1% carbon. It is malleable and can be welded. As a result it
is used to make:
- Iron nails,
- iron sheets,
- horse shoes and
- agricultural in puts.
(c) Steel
Steel is manufactured by blowing oxygen into molten pig iron in presence of
lime. The impurities carbon, sulphur and phosphorus present are oxidized by
the oxygen as gaseous oxides. The lime serves to form slag and is removed.
Calculated amount of carbon is added to the iron formed. One or more metals
such as chromium, nickel, manganese and tungsten are also added in small
calculated amounts. This gives steel. It is hard, tough and strong.

Uses of steel
Steel is used for making the following:
 Cutlery i.e. stainless forks, spoons and knives.
 Drilling and high speed cutting tools.
 Electromagnets.
 Rock drills and railway lines
 Spanners.
 Tin is used in food preservation for canning fruits, meat and fish.
207
3. Extraction of reactive metals e.g. Potassium, Sodium, Calcium, Magnesium
and Aluminium.
Extraction of reactive metals Sodium from Sodium Chloride.
Sodium is extracted by the electrolysis of molten sodium chloride in the Downs Cell.
The cell consists of a cylindrical steel cathode and graphite anode.
Sodium chloride is heated and melts at 801ºC. Calcium chloride is added as an
impurity to lower the melting point to about 600 ºC.

The Diagram of Downs Cell showing the Extraction of Sodium from Molten
Sodium chloride

Explanation:
Ions present: From NaCl(s)  Na+(l) + Cl-(l)
From CaCl2 (s)  Ca(l) + 2Cl-(l)
Reaction equation
At anode: 2 Cl-(l)  Cl2 (g) + 2e -
At cathode: Na+(l) + e-  Na (l)

The products Sodium and Chlorine produced are kept apart to prevent them from
reacting and reforming sodium chloride.
The liquid Sodium is collected under dry nitrogen gas. This prevents the metal from
reacting with the atmosphere.

208
 SELF-CHECK 11.1
(1996 Q.37) Which one of the following methods can be used to extract magnesium
from its ore?
A. Decomposition by heat. B. Electrolysis.
C. Reduction with carbon monoxide. D. Crystallisation.

2. (2004 Q.9) Which one of the following metals is extracted by electrolysis?

A. Zinc. B. Lead. C. Sodium. D. Copper.

3. (2002 Q.1) Which one of the following ores is a valuable ore of copper?
A. Malachite. B. Siderite. C. Haematite. D. Dolomite.

4. (2002 Q.33) Which one of the following gases is used to extract iron from its ore?
A. Chloride. B. Nitrogen monoxide.
C. Carbon monoxide. D. Sulphur trioxide.

5. (2001 Q.4) The role of coke in the extraction of iron in the blast furnace is to
A. produce carbon monoxide which reduces the oxides.
B. produce quick lime which combines with silica.
C. combine with iron to form steel.
D. reduce excessive heat produced in the furnace.

6. (1998Q.18) Which one of the following substances is not used in the extraction of
iron?
A. Coke. B. Air. C. Silica. D. Limestone.

7. (1996 Q.32) Which of the following metals can be extracted by reduction of the oxide with
carbon?
A. Potassium. B. Aluminium.
C. Zinc. D. Magnesium.

8. (1996 Q.35) Which one of the following reactions does not take place in the
extraction of iron in the Blast furnace?
A. Carbon monoxide, reduces iron (II) oxide to iron.
C. Coke burns in air forming carbon dioxide.
B. Limestone decomposes to form calcium oxide.
D. Limestone reduces iron (II) oxide to iron

209
 SELF-CHECK 11.2
1. 1988 Q.12
(a) (i) Name one ore of each of the following metals; sodium and iron.
(ii) Briefly describe how sodium and iron are extracted from their ores.
Explain why the method you have described can be used to extract the
metal from the ore.
(b) State the conditions under which sodium and iron react with water. Write
equations for the reaction in each case.
2. 1995 Q.14
(a) Name one ore from which sodium can be extracted.
(b) Describe how sodium is extracted from the ore you have named in (a).
Your answer should include the following:
(i) Names of the materials used as the electrodes.
(ii) Equations for the reactions that took place at the electrodes.
(iii) Method of collecting the sodium produced.
[A diagram is not required]
(c) State two factors that you would consider if you were to choose a place in
Uganda where a plant for extracting sodium could be built.
(d) A piece of sodium was heated and plunged into a jar of chlorine.
(i) State what was observed.
(ii) Write the equation for the reaction which took place.
(e) When aqueous silver nitrate was added to a solution of the product in (d), a
white precipitate was formed.
Write equation for the reaction that took place.
3. (2005 Q.12)
In the exctraction of cast iron using a blast furnance, spathic iron ore, which contains some
impurities, is first roasted in air. It is then mixed with some other substances and finally
introduced into the blast furnace. Cast iron can be obtained from iron (II) carbonate ore.
(a) Name the major impurity in the iron ore.
(b) (i) Give the chemical name of the spathic iron ore.
(ii) Write an equation for the reaction which takes place when iron (II)
carbonate is roasted in air.
(c) Name the substances that are fed into the blast furnace:
(i) from the top.
(ii) from the bottom.
(d) Outline the reactions leading to:
(i) the formation of cast iron.
(ii) the removal of the major impurity you have named in (a).
(e) State the major components of steel.

210
4. (1993 Q.11)
In the extraction of iron ore, coke and limestone are fed into a blast furnace and hot
air is blown into the mixture.
(a) Name and give the formula of one ore of iron.
(b) Why is limestone added to the mixture?
(c) Write equations for the reactions that lead to the formation of iron.
(d) Describe what happens when
(i) dilute hydrochloric acid is added to iron fillings.
(ii) steam is passed over heated iron fillings.
Write equation for the reaction that takes place in each case.

211
 ANSWERS TO SELF CHECK QUESTIONS
CHAPTER ONE
THE MOLE CONCEPT
SELF-CHECK 1.1
1. B. 2. A. 3. B. 4. D. 5. A.
6. B. 7. C. 8. D. 9. B. 10. A.
11. B. 12. B. 13. C. 14. B. 15. B.
16. C. 17. A. 18. D. 19. B. 20. C.

SELF-CHECK 1.2
1. (a) (i) Percentage of hydrogen = 20%, Percentage of carbon = (100 – 20)%
= 80%
Elements C H
Percentage mass 80 20
80 20
Moles
12 1
6.667 20
6.667 20
Mole ratios
6.667 6.667
1 3
The empirical formula of X is CH3

(ii) Mass of X = 7.5 g, Volume occupied by 7.5 g of X at stp = 5.6 dm3

Molar mass = ?, Molar gas volume = 22.4 dm3

7.5 g of X occupies 5.6 dm3 at stp.

5 .6 3
1 g of X occupies dm .
7 .5
The mass of X (molar mass) that will occupy 22.4 dm3 at stp
7.5 x 22.4
=
5.6
= 30.4
 The molar mass of X = 30 g

(iii) (Empirical formula) n = Molecular mass

(CH3) n = 30
[12 + (1 x 3)] n = 30
15 n = 30

212
 30
n =
12
=2
(CH3)2 = C2H6
 The molecular formula is C2H6

(b) (i) X is ethane.

(ii) CH3CH3

CHAPTER TWO
ELECTROLYSIS
SELF-CHECK 2.1
1. (a) Cu2+ (aq) + 2e-  Cu (s)
(b) (i) I = 0.20 A, t = 20 min x 60 = 1200 s, 1F = 96500 C, Cu = 64

Q = It
= 0.20 x 1200
= 240 C

(ii) 1 mole of coulombs = 96500 C.

Then the number of moles of coulombs in 240 C
1 x 240
=
96500
= 0.0025 moles

(iii) Equation at cathode:

Cu2+ (aq) + 2e-  Cu (s)
2F 1 mole (64)
If (2 x 96500) C liberates 64 g of copper,

64
1 coulomb liberates g of copper.
2 x 96500
64
240 C will liberates x 240 = 0.08 g
2 x 96500

Or Alternatively:
2 moles of coulombs of electricity liberates 1 mole of copper (64) g,
64
1 mole of coulomb liberates g of copper.
2

213
 64
0.0025 moles of coulombs liberates x 0.0025 = 0.08 g
2

2. (a) Anode.
(b) Equation at anode:
4OH-(aq) - 4e-  2H2O (l) + O2 (g)
4mol 1mol (32g)
(4F = 4 x 96500 C)

(c) t = 35 min = 35 x 60 = 2100 s, I = 0.65 A, F = 96500 C, Q = ?

Q = It
= 0.65 x 2100
= 1365 C

(4 x 96500) C of electricity liberates 32 g of oxygen,

32
1 Coulomb of electricity liberates g of oxygen gas.
4 x 96500
32
1365 C of electricity will liberate x 1365 = 0. 113 g
4 x 96500

3. (a) (i) No observable change (i.e the bulb did not glow).
(ii) The lamp glowed brightly.
Reddish brown colouration was seen around the anode.

(b) Before the lead (II) bromide was melted, the Pb2+ ions and Br- were not free to move.
But after melting the ions were free to move in the electrolyte, hence conducting
current, leading to the glowing of the lamp.
(c) (i) P = Anode: 2Br-(l) - 2e  Br2 (g)

or 2Br-(l)  Br2 (g) + 2e

(ii) Q = Cathode: Pb2+(l) + 2e  Pb (l)

4. (a) (i) P cathode Q. anode

R cathode S . anode
(ii) Bubbles of a colourless gas that burns with a pop sound are evolved.
(b) P. Cu2+ (aq) + 2e-  Cu (s)
(c) S. 4OH -(aq) - 4e-  2H2O (l) + O2 (g)

Cu2+ (aq) + 2e-  Cu (s)

1 mol 2 mol 1mol
If 2F produce 1 mol of copper, then the number of Faraday that produce 0.02 moles
of copper = 2 x 0.02

214
 = 0.04 F
Equation for reaction at S.
4OH -(aq) - 4e-  2H2O (l) + O2 (g)
4 mol 1 mol
4F
If 4F liberate 1 mol of oxygen, then the number of moles of oxygen liberated by 0.04
1 x 0.04
F =
4
= 0.01 moles

5. (a) (i) At Anode: - Bubles of a colourless gas that rekindles a burning splint
were formedt.
(ii) At Cathode: - A brown solid deposit.

(b) Cu2+ (aq) + 2e-  Cu (s)

The positively charged ions Cu and H+ ions move to the cathode. The Cu2+ ions
2+

which are lower in the electrochemical series are preferentially discharged by gain
of electrons from the cathode according to the equation above:

6. (a) No observable change.

a. The lamp glowed brightly.
Bubles of reddish brown gas were seen around the anode.
(b) Before the lead (II) bromide was melted, the Pb2+ ions and Br- were not free to move.
But after melting the ions were free to move in the electrolyte, hence conducting
current, leading to the glowing of the lamp.

7. (a) (i) At Anode: - Reddish brown colouration of bromine gas.

At Cathode : A grey melt solid formed melts and sinks to the bottom of the
crucible.
(ii) At anode: 2Br-(l) - 2e  Br2 (g)

or 2Br-(l)  Br2 (g) + 2e

At cathode: Pb2+(l) + 2e  Pb (l)

(b) t = 1½ hours = 90 x 60 = 5400 s, I = 2 A, Q = ?

Q = It
= 2 x 5400
= 10,800 C

Pb2+(l) + 2e  Pb (s)
2F 1 mole

(2 x 96500) C of electricity liberates 207g of lead,

215
 207
1 coulomb of electricity liberates g of lead.
2 x 96500
207
10800 C of electricity will liberate x 10800 = 11.58 g
2 x 96500

8. (a) (i) Impure copper.

(ii) Pure copper.

(b) Copper (II) sulphate solution. or any soluble salt of copper.

(c) (i) Cu (s)  Cu2+(aq) + 2e-

(ii) Cu (aq) + 2e  Cu (s).
2+ -

9. (a) In solid state, lead (II) bromide the ions Pb2+ ions and Br- are not free to
move. But in molten state the ions are free to move in the electrolyte, hence are able
to conduct electricity between the electrodes.
(b) (i) Reddish brown colouration of bromine gas.
(ii) A grey solid formed melts and sinks to the bottom of the crucible.
(c) 2Br-(l) - 2e  Br2 (g)

or 2Br-(l)  Br2 (g) + 2e

10. (a) (i) X - Oxygen

Y - Hydrogen
(ii) 4OH-(aq) - 4e  2H2O (l) + O2 (g)
(iii) Electron flow is from anode to cathode.
(iv) At anode: 4OH -(aq) - 4e-  2H2O (l) + O2 (g)
4mol 4mol 2mol 1mol (32g) or
(4F = 4 x 96000C) (24000cm3 at rt)

(b) (i) t = 10 min = 10 x 60 = 600 s, I = 1 A, F = 96000 C, Q = ?

Q = It
= 1 x 600
= 600 C
If (4 x 96000) C liberates 24000 cm3 of oxygen gas at room temperature,
24000
1 Coulomb of electricity liberates of oxygen gas.
4 x 96000
24000
600 C liberates x 600 = 375 cm3
4 x 96000
(ii) Electroplating, Extraction of reactive metals and Purification of copper
(iii) Chlorine is a poisonous gas. Its likage to the environment would poison the workers.

216
 ELECTROCHEMICAL CELL
SELF-CHECK 2.2
1. (a)

(b) To complete the circuit between the zinc sulphate and coppr (II) sulphate solutions.
(c) (i) - At anode: Zn (s)  Zn 2+(aq) + 2e- (Oxidation)
(ii) - At cathode: Cu (aq) + 2e  Cu(s) (Reduction)
2+ -

(d) E = 0.9 V, n = 2 (Valence of Zn and Cu = 2)

Energy = EnF
= 0.90 x 2 x 96500
= 173.7 KJ
(e) The voltmeter reading would be greater than 0.90 V
(f) Because magnesium is higher than zinc in the reactivity series.

2. (a) (i) Zn (s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)

(ii) There will be a complete discharge of Cu2+ ions and dissolution of zinc rod
leading to electrolysis in which the zinc rod is dissolved and Cu2+ ions are
completely discharged and so the reaction ceases.

(b) E = 1.10 V, n = 2 (Valence of Zn and Cu = 2)

Energy = EnF
= 1.10 x 2 x 96500
= 212.3 KJ
3. (a) and (b) The diagram showing Daniel cell

(c) (i) - At anode: Zn (s)  Zn 2+(aq) + 2e- (Oxidation)

- At cathode: Cu (aq) + 2e  Cu(s) (Reduction)
2+ -

(ii) Zn (s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)

217
4. (a) (i) - At cathode: Cu 2+(aq) + 2e-  Cu(s) (Reduction)
- At anode: Zn (s)  Zn 2+(aq) + 2e- (Oxidation)
(ii) Zn (s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)
(b)

5. (a) (i) Zinc sulphate solution (ii) Lead (II) nitrate.

(b) Zinc rod or plate.
(c) (i) At the anode i.e the metal higher in the E.C.S, Zinc
Zn (s)  Zn 2+(aq) + 2e- (Oxidation)
(ii) At the cathode: i.e the metal lower in the E.C.S, Lead
Pb 2+(aq) + 2e-  Pb (s) (Reduction)
(d) Zn (s) + Pb 2+ (aq)  Zn 2+ (aq) + Pb(s)

CHAPTER THREE
REACTION RATES
SELF-CHECK 3.1
1. C. 2. C. 3. B. 4. A. 5. A.

SELF-CHECK 3.2
1. (i) Graph B.
Reason - Because the particles are larger, the total surface area is less for B
compared to A.
NB: - Reaction rate decreases with decrease in surface area.
(ii) The hydrogen peroxide is becoming dilute as more water molecules are produced in
the reaction.
NB: - Reaction rate decreases with decrease in concentration.
(iii) See the broken curve on the graph (Page 91)

218
 Reason: Reaction rate decreases with decrease with drop in temperature.
(iv) In each case the same quantity of hydrogen peroxide is used, the same volumes of
oxygen gas is produced at the end of the reaction. So all the graphs begin to level at
the same value. And the difference in the slopes is due to the different factors of rate
of reaction affecting the reaction.

CHAPTER FOUR

ORGANIC CHEMISTRY
SELF-CHECK 4.1
1. D. 2. C. 3. C. 4. B. 5. D.

6. C. 7. A. 8. D. 9. D. 10. C.

11. B. 12. B. 13. D. 14. C. 15. B.

16. B. 17. B. 18. B. 19. D. 20. B.

21. B 22. A 23. C 24. D 25. D

26. A

SELF-CHECK 4.2

1. (a) CH2=CH2 or H2C=CH2

or
(b) (i) The reddish brown colour of bromine becomes colourless
The brown colour of bromine water becomes colourless.
C2H4 (g) + 2 Br2 (aq)  C2H4Br2 (aq)

Or C2H4 (g) + 2 Br2 (aq)  CH2Br CH2Br (aq)

(ii) Acidified potassium manganate (VII).

(c) Ethanol.

2. (a) X - Vegetable oil

(c) Saponification
(d) Brine (Concentrated sodium chloride).
(e) White precipitate.

219
3. (a) Mıllet flour, cassava flour, maıze flour etc.
(b) Maize, cassava and millet flour are mixed with water. The product is soaked in a tin
or pot or buried in the ground for about a week to allow starch to be converted into
sugar.
The mixture is then removed, roasted and dried. The product is then allowed to
ferment for two to three days in suitable container. During this process, the sugar
formed is converted to crude ethanol.
The formation of the crude ethanol from the carbohydrates (starch) involves the
participation of enzymes.
The relevant equations for the main reactions involved are given below.

2 C6H10O5 (aq) + H2O (l) Diastase C12H22O11 (aq)

C12H22O11 (aq) + H2O (l) Maltase 2 C6H12O6 (aq)

C6H12O6 (aq) Zymase 2 C2H5OH (l) + CO2 (g)

(c) The ethanol is fractionally distilled at 78 – 82C to a liquid containing about

95% of ethanol and 5% water. The water can be absorbed by absorption
using quicklime (CaO).
(d) (ı) C2H5OH (l)  C2H4 (g) + H2O (l)
- The sulphuric acid must be concentrated.
- The mixture of ethanol and concentrated sulphuric acid must
be heated to a temperature of 170C.
(ii) - beverage.
- fuel.
- thermometric liquid in thermometers.
- A solvent for organic compounds

4. (a) (i) Oils occur naturally in plants while fats occur naturally in animals only.
Fats are solid at room temperature while oils are liquid at room temperature.
(ii) Examples of oil include:
- Sunflower oils, Groundnut oils, simsim oils, corn oil etc.
Examples of fats include:
- Lard in pigs, butter from milk.
(b) Soap is manufactured by a process called saponofocation.
A mixture of vegetable oil and concentrated sodium hydroxide solution is boiled.
Oil or Fat + Sodium hydroxide  Soap + Glycerol
When the process is complete, brine (concentrated sodium chloride) is added to the
mixture to precipitate (to solidify) the soap. The soap, which floats above the liquid,
is removed and pressed into bars.
(c) White precipitate.

220
 (d) The Mg2+ ions in the magnesium hydrogen carbonate reacts with soap to form white
solid called scum.
(e) There would be no observable change.
(f) They cause water pollution in rivers and lakes.

5. (a) C6H12O6 (aq) Zymase 2 C2H5OH (l) + CO2 (g)

(b) (i) Ethene
(ii) CH2=CH2 or H2C=CH2

or
(iii) Bromine water.
Or Acidified potassium manganate (VII).

6. (a) (i) Decomposition (ii) Zymase.

C6H12O6 (aq) Zymase 2 C2H5OH (l) + CO2 (g)

(b) The ethanol is fractionally distilled at 78 – 82C to a liquid containing about 95% of
ethanol and 5% water. The water can be absorbed by absorption using quicklime
(CaO).

7. (a) (i) Polymerization is the joining together of simple molecules called

monomers to form a complex molecule called polymer.

(ii) Natural polymers:

Starch, proteins, Fats, Cellulose, Glycogen, Rubber, Wool, Silk.

Synthetic polymer:
Polythene, Polypropene, Perspex, Nylon, Bakelite,
Synthetic rubber, Poly Vinyl Chloride (PVC), Nylon
(b) CH3CH=CH2 (Propene).

(c) - Thermoplastics are plastics which soften on heating and harden

on cooling.
- Softening and hardening are reversible.
- As a result they can be molded or remolded under hot conditions.
- Thermo sets (Thermosetting plastics) are plastics, which soften or
melt on heating during manufacture and take the shape of the mold in
which they are processed on cooling.
- They cannot be remolded by heating after manufacture.
- At high temperatures, they decompose.

(d) Cracking is the process by which long chain alkanes are broken down to
produce shorter chain hydrocarbons. E.g. cracking of paraffin in the lab.

221
 (e) See diagram on page 108

8. (a) A - ethanol
(b) (i) Ethene

(ii) CH2=CH2 or H2C=CH2

or

(iii) Bromine water.

Or Acidified potassium manganate (VII).

(iv) With bromine water:

The reddish brown colour of bromine becomes colourless.

With acidified potassium manganate (VII):

The pink colour of the solution becomes colourless.

CHAPTER FIVE

NITROGEN AND ITS COMPOUNDS

SELF-CHECH 5.1
1. C. 2. D. 3. D. 4. A. 5. B.
6. D. 7. A. 8. A. 9. B. 10. C.
11. B. 12. C. 13. B. 14. D. 15. C.
16. A. 17. B. 18. A. 19. B. 20. D.
21. C. 22. A 23. C. 24. B. 25. B.

SELF-CHECH 5.2
1. (a) An alkali is a base that is very soluble in water to give hydroxide ions.
Or An alkali is a base which gives hydroxide ions (OH-) in solution.

(b) (i) Ammonia is a basic gas that is very soluble in water thus it would
dissolve in water instead of being collected.

222
 (ii) Ammonia is less dense than air or lighter than air and would not therefore
displace air from the gas jar.

(c) Hydroxide ion (OH -) ions.

(d) Volume of ammonia at room temperature = 240 cm3

Molar Gas Volume any gas at rt = 24000 cm3

Volume of at root temperature

Number of moles of a gas =
Molar Gas Volume
240
=
24000
= 0.01 moles

(e) (i) The ammonium sulphate solution should be heated to evaporate the
excess water or to concentrate the solution slowly to crystallization point,
allowed to cool until crystals form, then filtered, dried using filter paper.
(ii) From the equation, 2 moles of ammonia produce 1 mole of ammonium
sulphate. But from (d) moles of ammonia = 0.01.
Therefore, moles of ammonium sulphate produced by 0.01 moles
0.01
=
2
= 0.005
Formula mass (Molar mass) of ammonium sulphate, (NH4)2SO4
= 3(14 + 4 x 1) + 32 + 4 x 16
= 3 x 18 + 32 + 64
= 36 + 32 + 64
= 132 g
1 mole of ammonium sulphate = 132 g
0.005
0.005 moles of ammonium = x 132 g
1
= 0.66 g

2. (a) (i) Gas A is ammonia.

Liquid L is water.
(ii) Black solid, Copper (II) oxide changes to reddish brown solid because it is
reduced to copper by ammonia.

(iii) NH3 (g) + 3CuO (s)  3Cu (s) + N2 (g) + 3H2O (l)

(iv) I. From the equation, 2 moles of ammonia produce 1 mole of

nitrogen gas.

223
 Therefore, the volume of nitrogen produced by 320 cm3 of
ammonia
1
= x 320
2
= 160 cm3
Volume of at room temperature
II. Number of moles of ammonia =
Molar Gas Volume
320
=
24000
= 0.0133
From the equation, 2 moles of ammonia reacts with 3 moles of
Copper (II) oxide, then the number of moles of Copper (II) oxide that
reacts with 0.0133 moles of ammonia
3
= x 0.0133
2
= 0.01995
= 0.02 moles

Relative Formula Mass of Copper (II) oxide, CuO,

= 63.5 + 16.0
= 79.5 g

Mass of subs tan ce

From Number of moles =
Re lative Formula Mass
Mass of Copper (II) oxide = Number of moles x RFM
= 0.02 x 79.5
= 1.59 g

III The excess ammonia from the reaction dissolves in the water in the
beaker to form ammonium hydroxide which is a weak alkali or base
of pH about 10.

(b) The burning splint would be extinguished.

(c) Because it is cheaper and ammonia is made from nitrogen.

3. (a) A dry sample of ammonia is prepared in the laboratory by heating a

mixture of ammonium chloride and calcium hydroxide. The gas is dried
by passing it through calcium oxide (quick lime) and then is collected

224
 by upward delivery method since it is lighter (less dense) than air.

(b) Ammonia is tested by placing a moist red litmus paper in a stream of ammonia. The
red litmus paper turns blue.
(c) (i) See diagram on page 124
(ii) 4 NH3 (g) + 3 O2(g)  2 N2 (g) + 6 H2O (g)
(d) (i) The black solid copper (II) oxide turned brown and
A colourless liquid that turned anhydrous copper (II) sulphate blue was
formed.

(ii) 2 NH3 (g) + 3 CuO (s)  3 Cu (s) + 3 H2O (g) + N2 (g)

Black Brown

4. (a) (i) See diagram on page 123

(ii) Ca(OH)2 (s) + 2NH4Cl (s)  CaCl2 (s) + 2 H2O (l) + 2 NH3 (g)
(b) (i) The black solid turned to grey and
A colourless liquid that turned anhydrous copper (II) sulphate blue was
formed.
(ii) 2 NH3 (g) + 3 PbO (s)  3 Pb (s) + 3 H2O (g) + N2 (g)

(c) Oxidation of Ammonia

Ammonia from Haber process is mixed with excess air and the mixture is passed over
platinum catalyst heated at about 800C and 8 atmospheres pressure in the catalytic
chamber. It is oxidized to nitrogen monoxide according to the equation:
4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g) + Heat

Oxidation of nitrogen monoxide to nitrogen dioxide

The nitrogen monoxide formed is rapidly cooled and combines with the oxygen from
excess air to form nitrogen dioxide.
2 NO (g) + O2 (g)  2 NO2 (g)

Oxidation of nitrogen dioxide and absorption

The nitrogen dioxide is oxidized by more air and then absorbed in hot water to form
nitric acid. 4 NO2 (g) + O2(g) + 2 H2O (g)  4 HNO3 (g)

5. (a) (i) Hydrogen and nitrogen

(ii) N2 (g) + 3 H2 (g) 2NH3 (g) + Heat

(b) (i) High pressure increases the percentage yield of ammonia.
(ii) The percentage yield increases with increase in temperature.
(c) Presence of a catalyst

225
 (d) The black solid in the combustion tube turned brown and
A colourless liquid that turns anhydrous copper (II) sulphate blue was formed.
The ammonia reduces the copper (II) oxide to solid copper which is brown in colour.
6. (a) See diagram on page 123.
(b) The fountain experiment

- Fill a large, dry, round – bottomed flask with dry ammonia.

- Place the mouth of the flask under water, coloured with litmus in a trough.
- Clamp it firmly, open the clip attached at the end of the long glass tube.
- So much ammonia is dissolved in the jet such that a partial vacuum is created in the
flask.
- Air pressure forces water rapidly up the tube and enters the flask as a fountain. The
litmus turns blue.
(c) (i) The Copper coil glows red hot, reddish brown fumes of NO2 are
seen.
(ii) This is because of the catalytic oxidation of ammonia by oxygen. The catalyst
is the copper. The initial products are steam and nitrogen oxide, but the
latter is immediately oxidized to NO2 – the brown fumes.
4NH3 (g) + 5O2 (g)  4NO(g) + 6H2O (g)
2NO (g) + O2 (g)  2NO2 – reddish brown

(d) R.f.m of PbO = 207 + 16 = 223.

2NH3 (g) + 3PbO (s)  3Cu (s) + N2 (g) + 3H2O (l)

From the equation,

3 mols of PbO require 2 mols of NH3

669g (2 x 223) of PbO require 2 mols of NH3
48
1g of PbO requires dm3 of NH3
669
48
2.5g of PbO will require x 2.5 = 0.1794 dm3 of NH3
669

226
7. (a) A Ammonia
(b) (i) The black copper oxide turns brown (copper), a colourless vapour
is seen.
(ii) 2NH3 (g) + 3CuO (s)  3Cu (s) + N2 (g) + 3H2O (g)
(c) (i) Water
(ii) White anhydrous Copper (II) sulphate

8. (a) (i) A = Water

(ii) B = Nitrogen gas, N2
(iii) Water

(b) 2NH3 (g) + 3CuO (s)  3Cu (s) + N2 (g) + 3H2O (g)
(c) Ammonia is a highly soluble gas and dissolve, in the water.
(d) Lead (II) Oxide, PbO.

9. (a) The industrial manufacture of nitric acid is by the Ostwald process.

- Ammonia (NH3) is mixed with air; the two are cleaned and heated in a heat
exchanger.
- The pure gas mixture is conveyed to a hot Platinum alloy catalyst in a catalytic
chamber, where ammonia and Oxygen combine to form Nitrogen oxide and water.
4NH3 (g) + 5O2 (g)  4NO (g) + 6H2O (g)
- In the heat exchanger, more air oxidizes Nitrogen oxide to Nitrogen dioxide
2NO (g) + O2 (g)  2NO2 (g)
- The Nitrogen dioxide NO2 is mixed with excess air and conveyed to towers,
down which hot water flows. The combination of nitrogen dioxide and air
and hot water gives nitric acid.
4NO2 (g) + O2 (g) + 2H2O (l)  4HNO3 (aq)

(b) When conc HNO3 is added to copper, a reddish brown nitrogen dioxide and a green
solution of copper (II) nitrate are formed.
The reaction between copper and conc HNO3 is vigorous. Copper is oxidized by
losing two electrons to conc HNO3. the latter is reduced to NO2, NO3- and H2O.
Cu (s)  Cu2+ (aq) + 2e
2HNO3 (aq) + e  NO2 (g) + NO3- (aq) + H2O (l)

(c) Confirmatory test for NO3- ion

To a solution of a nitrate, add an equal volume of freshly prepared Iron (II) sulphate
solution.

227
 Slope the test tube and carefully pour conc H2SO4 down the side. The dense acid sinks to the
bottom. A brown ring forms where the two layers meet.
This confirms the presence of a NO3- ion.
OR
To a solution of a nitrate, add excess NaOH solution followed by Aluminium or Zinc powder
or Devarda’s alloy (Alloy of zinc and aluminium). A vigorous reaction takes place,
effervescence of a colourless alkaline gas (NH3) confirms the presence of NO3- ion.

CHAPTER SIX
SULPHUR AND ITS COMPOUNDS
SELF-CHECK 6.1

1. B 2. C. 3. D. 4. A. 5. B.

6. B. 7. B. 8. A. 9. C. 10. D.

SELF-CHECK 6.2

1. (a) See diagram on page 142

- In the Frasch process for the extraction of sulphur, 3 concentric pipes are sunk
to the sulphur deposit.
- Super heated water at 170oC is pumped down through the outer pipe while hot
compressed air is pumped down through the central pipe.
- As a result of froth of air, water and molten suplhur is forced up through the
middle pipe and is led off to tanks where sulphur separates out.
(b) S (s) + O2 (g)  SO2 (g)
2SO2 (g) + O2 (g)  2SO3 (g)
SO3 (g) + H2SO4(l)  H2S2O7 (l)
H2S2O7 (l) + H2O (l)  2H2SO4 (l)
(c) When concentrated sulphuric acid is added to sugar, heat is evolved; the acid turns
sugar into black solid mass of carbon.

(d) Any two fo the following:

Manufacture drugs and skin ointments.
Production of:
- Matches,
- Vulcanized rubber,
- Dyes, and
- Explosives e.g Gunpowder

228
2. (a) SO3 2-(s) + 2 H+ (aq)  H2O (l) + SO2 (g)
(b) (i) Potassium dichromate or acidified potassium permanganate
solution.

(ii) - With potassium dichromate

The solution turns from orange to green.

(ii) - With acidified potassium permanganate solution.

The purple colour of the solution becomes colourless.
(c) (i) The red colour of the flower becomes colourless.
(ii) The bleaching property of the sulphur dioxide bleaches the colour.

3. (a) (i) Sodium sulphite.

(ii) The hydrochloric acid must be dilute.
The mixture of the dilute hydrochloric acid and sodium sulphite must be
heated.
(iii) Concentrated sulphuric acid.
(iv) Na2SO3 (s) + 2 HCl (aq)  2 NaCl (aq) + H2O (l) + SO2 (g)

(b) (i) The solution turns from orange to green.

(ii) The colour of the dye turns to colourless.
(c) Oxidation of sulphur to sulphur dioxide by air
The sulphur dioxide is mixed with air. The mixture is dried and passed over
vanadium (V) oxide catalyst heated at 450C – 500C and pressure of 1 atmosphere.
The sulphur dioxide is oxidized to sulphur trioxide.
2 SO2 (g) + O2 (g)  2SO3 (g)
Absorption of Sulphur trioxide
The sulphur trioxide formed is absorbed into concentrated sulphuric acid forming a
fuming liquid called ‘oleum’.
SO3 (g) + H2SO4 (l)  H2S2O7 (l)
(oleum)
Absorption, cooling and Dilution
The oleum formed is absorbed, cooled and carefully diluted with the correct amount
of water to form ordinary, sulphuric acid which is 98%.
H2S2O7 (l) + H2O (l)  2 H2SO4 (aq)
4. (a) Heat is evolved, the sugar is changed into a black porous mass of carbon.
This is an exothermic reaction in which concentrated sulphuric acid dehydrates the
sugar i.e. removes the elements of water from the sugar.
C12H22O11 (s)  12C (s) + 11H2O (g)
(b) The two react with effervescence of a colourless and weakly acid gas that turns lime
water milky; the resultant solution is colourless
ZnCO3 (s) + H2SO4 (aq)  ZnSO4 (aq) + H2O (l) + CO2 (g)
229
 (c) – The sulphuric acid must be concentrated
- The mixture must be heated.
Cu (s) + 2H2SO4 (l)  CuSO4 (s) + 2H2O (l) + SO2 (g)
(d) By use of acidified Barium nitrate or Barium chloride solution. When a little Barium
nitrate or Barium chloride is added to sulphuric acid, a white precipitate is formed
which is insoluble in the acid.
(e) - In the manufacture of fertilizers e.g. (NH4)2SO4
- In the production of synthetic fibres and plastics
- In car batteries
- To make detergents

5. (a) S (s) + O2 (g)  SO2 (g)

(b) Step II is necessary to remove any dust which would “poison” the catalyst i.e. make
it inactive and also to purify the two gases SO2 and O2

(c) (i) Drying agent in step III – Concentrated sulphuric acid.

(ii) Catalyst in step IV: – Vanadium pentaoxide (V2O5)
(d) The sulphur trioxide formed is absorbed or passed into Concentrated sulphuric acid
to form fuming sulphuric acid (H2S2O7) – oleum.
Oleum is then diluted with water to form the ordinary conc. sulphuric acid.
SO3 (g) + H2SO4 (l)  H2S2O7 (l)
H2S2O7 (l) + H2O (l)  2H2SO4 (l)

(e) (i) Three conditions for the maximum yield of sulphur trioxide.
- Low temperature (450oC)
- High pressure
- Catalyst of Vanadium pentaoxide (V2O5)
(ii) 2SO2 (g) + O2 (g)  2SO3 (g)
From the equation, 22 moles of sulphur dıoxıde produces 2 moles of sulphur
trıoxıde, the volume of sulphur trıoxıdeproduced by 20 cm3of sulphur dıoxıde
2 x 20
=
2
= 20 cm3
(f) Sulphuric acid is used in the manufacture of:
- Fertilizers
- Explosives
- drugs
- dyes
- detergents
- plastics

230
 2 x 2000
(g) Mols of dilute H2SO4 = = 4 moles
1000
18 moles are in 1000cm3
1000 3
1 mole is in cm
18
1000 x 4
4 moles are in = 222.2cm3
18
.: Required volume of sulphuric acid = 222.2 cm3

6. (a) Sulphur dioxide is prepared in the laboratory by heating a mixture of

dilute hydrochloric acid and sodium sulphite in a round bottom flak and the gas is
collected by downward delivery method. If the gas is required dry, it is passed
through concentrated sulphuric acid.

See diagram on page 144

(b) Na2SO3 (s) + 2HCl (aq)  2 NaCl (aq) + H2O (l) + SO2 (g)
(c) The presence of sulphur dioxide can be tested by as follows:
(i) A filter paper soaked in acidified solution of potassium dichromate is placed
in a stream of sulphur dioxide.
Observation:
The solution turns from orange to green.
(ii) Sulphur dioxide is bubbled through acidified potassium permanganate solution.
Observation:
The purple colour of the solution becomes colourless.
(d) SO2(g) + NaOH(aq)  Na2SO3 (aq) + H2O (l)
Na2SO3 (aq) + H2O (l) + SO2(g)  NaHSO3 (aq)

(e) The reaction equation

2NaHCO3(aq) + H2SO4(aq)  Na2SO4(aq) + 2H2O(l) + 2CO2(l)

1000 cm3 of sodium hydrogen carbonate solution contains 0.1 moles.

1 x 0 .1
1.00 cm3 of sodium hydrogen carbonate solution contains moles.
1000
25.0 x 0.1
25.0 cm3 of sodium hydrogen carbonate solution contains
1000
= 0.025 moles
From the mole ration 1 mole of acid : 2 moles of base,
The number of moles of sulphuric acid that reacted with sodium hydrogen carbonate
= ½ the number of moles of the base
= ½ x 0.025
= 0.0125 moles

231
 27.8 cm3 of sulphuric acid contains 0.0125 moles of the acid.
0.0125
1.0 cm3 of sulphuric acid contains moles.
27.80
0.0125 x 1000
1000 cm3 of sulphuric acid contains = 0.45 M
27.80

CHAPTER SEVEN

CHLORINE AND ITS COMPOUNDS

SELF-CHECK 7.1

1. B. 2. C. 3. D. 4. A. 5. B.

6. A. 7. C. 8. D. 9. D.

SELF-CHECK 7.2
1. (a) (i) A. Potassium Permanganate
B. Water
C. Concentrated Sulphuric acid
(ii) To remove unchanged hydrogen chloride.
(b) The hydrochloric acid must be concentrated.
(c) 2KMnO4(s) + 16HCl(aq)  2KCl (aq) + 2MnCl2 (aq) + 8 H2O(l) + 5Cl2(g)
2. (a) (i) Potassium permanganate
(ii) 2KMnO4(s) + 16HCl(aq)  2KCl (aq) + 2MnCl(aq) + 8H2O(l) + 5Cl2(g)
(b) (i) A pale yellow solution is observed.
(ii) Cl2(g) + 2NaOH(aq)  NaOCl (aq) + NaCl(aq) + H2O (l)

3. (a) (i) See diagram on page 154 or 155

- Chlorine is prepared in the laboratory by dropping conc. hydrochloric acid on to
Potassium manganate crystals in a flat bottom flask.
- The chlorine produced is passed through water to remove nay unchanged
hydrogen chloride, and then through concentrated sulphuric acid to dry it. The
gas is then collected by downward delivery.
(ii) 2KMnO4(s) + 16HCl(aq)  2KCl(aq) + 2MnCl2(aq) + 8H2O(l) + 5Cl2(g)

(iii) – As a bleaching agent in the pulp and textile industries

- In the treatment of sewerage and purification of water
- In the manufacture of plastics e.g. P.V.C. (Polyvinyl chloride)

232
 (b) (i) The green solution of Iron (II) chloride turns to yellow/ brown solution due to
oxidation of Iron (II) chloride to Iron (III) chloride solution by chlorine.
2FeCl2 (aq) + Cl2 (g)  2FeCl3 (aq)
(ii) The colourless potassium iodide solution turns to purple/ violet due to the
displacement of I - ions by chlorine to give Iodine.

2I - (aq) + Cl2 (aq)  2Cl - (aq) + I2 (aq)

(c) (i) Sodium continues to burn in chlorine, producing white fumes of sodium chloride.

2Na (s) + Cl2 (g)  2 NaCl (s) - white solid.

The reaction is vigorous because both sodium and chlorine are very reactive
elements with sodium being a strong reducing agent while chlorine is a strong
oxidizing agent.
(ii) 2Na (s) + Cl2(g)  2NaCl(s)

4. (a) - To remove fumes of hydrogen chloride.

- To dry the gas.
(b) (i) Cl2 (g) + H2O (aq) HOCl(aq) + HCl(aq)
(ii) Chlorine dissolved in water is a bleaching agent. The bleaching action of
chlorine is due to the ion OCl - in the hypochlorous acid. The hypochlorous
acid is a very reactive compound and readily gives up its oxygen.

2HOCl (aq)  2HCl (aq) + O2 (g)

The oxygen given off reacts with the dye, to form a colourless compound.
2Dye + O2 (g)  2(Dye + O)
Colourless compound
(c) (i) The green solution of iron (II) sulphate would turns to yellow in colour due
to the formation of iron (III) chloride solution.
(ii) Cl2 (g) + 2Fe2+(aq)  2Cl - (aq) + 2Fe3+(aq)
5. (a) (i) Chlorine.
(ii) 2PbO(s) + 8HCl(aq)  2PbCl2 (aq) + 4H2O (l) + 2Cl2 (g)
(b) (i) 2KBr(aq) + Cl2(g)  2KCl(aq) + Br2(aq)
(ii) Red colour develops much faster than in (b) (i).
(iii) 2Br-(aq) + F2(g)  Br2(aq) + 2F-(aq).
6. (a) MnO2 (s) + 4HCl(aq)  MnCl2 (aq) + 2H2O (l) + Cl2 (g)
(b) (i) 2H2O(l) + Cl2(g)  MnCl2(aq) + 2H2O(l) + Cl2(g)
(ii) The Chloric (I) acid, HClO formed is decomposed by sunlight, giving
off oxygen.

233
 2HClO(aq)  2HCl(aq) + O2(g)
(c) (i) 2KOH(aq) + Cl2(g)  KClO(aq) + KCl(aq) + H2O(l)
Or 2OH-(aq) + Cl2(g)  ClO -(aq) + C l-(aq) + H2O(l)
(ii) Potassium chlorate (I) (Hypochlorite); potassium chloride.
(iii) It is used as a mild antiseptic.

SELF-CHECK 7.3
7. (a) H2SO4(aq) + NaCl (aq)  NaHSO4 (aq) + HCl (g)
(b) See diagram on page 159
(c) (i) Effervescence of a colourless gas that turns lime water milky
would be given off.
CaCO3(s) + 2HCl (aq)  CaCl2(aq) + H2O (l) + CO2(g)
(ii) White precipitate would be formed
AgNO3(s) + HCl (aq)  AgCl(s)) + HNO3 (aq)
(iii) Effervescence of a colourless gas that burns with a pop sound would be
evolved.
Mg (s) + 2HCl (s)  MgCl2 (aq) + H2(g)
(d) Copper is below hydrogen in the reactivity series. So can not displace hydrogen from
hydrogen chloride.

8. (a) A dry sample of hydrogen chloride can be prepared by action of conc.

Sulphuric acid on rock salt (impure sodium chloride).
The gas hydrogen chloride gas is dried by passing it through concentrated Sulphuric
acid and then collected by downward delivery method.
Equation: H2SO4 (l) + NaCl (s)  NaHSO4 (s) + HCl(g)
(b) Hydrochloric acid
(c) (i) Acidified Silver nitrate solution
(ii) A white precipitate is formed insoluble in the acid.
(d) H (aq) + HCO3- (aq)  H2O (l) + CO2 (g)
+

(e) Pb2+ (aq) + 2Cl- (aq)  PbCl2 (s)

25 x 0.2
Moles of Pb(NO3)2 that reacted =  0.005
1000
Moles of PbCl2 formed = 0.005 since the mole ratio Pb(NO3)2 : PbCl2
= 1 : 1
R.f.m of PbCl2 = 207 + (35.5 x 2) = 278
Mass of PbCl2 formed = 278 x 0.005 = 1.39g
9. (a) See diagram on page 161
(b) Fe (s) + 2HCl (g)  FeCl2 (s) + H2 (g)

234
 (c) (i) When aqueous sodium hydroxide was added to a solution of Iron (II)
chloride, a green precipitate is formed which is insoluble in excess sodium
hydroxide solution.
(ii) FeCl2 (aq) + 2 NaOH (aq)  Fe(OH)2 (s) + 2NaCl (aq)

Or Fe2+(aq) + 2OH - (aq)  Fe(OH)2 (s)

(d) (i) The green solution turned to a yellow solution when chlorine was
passed through Iron (II) chloride solution because of the oxidation of

Fe2+ (green) to Fe3+ (yellow/brown)

(ii) Equation: 2Fe2+(aq) + Cl2 (g)  2Fe3+ (aq) + 2Cl -(aq)
(e) (i) To test for Fe3+ ions, use potassium hexacyanofferate (II) – a dark blue
precipitate is formed or use sodium hydroxide solution- a brown precipitate
is formed.
(ii) To test for Cl - ions, use acidified silver nitrate solution – a white precipitate
is formed, soluble in aqueous ammonia.

CHAPTER EIGHT
ION CHEMISTRY
SELF-CHECK 8.1
1. B. 2. D. 3. B. 4. B. 5. D.

6. A. 7. C. 8. B. 9. A. 10. A.

11. C. 12. C. 13. B. 14. C. 15. C.

16. C. 17. A. 18. B. 19. B. 20. B.

21. B. 22. B. 23. C 24. B. 25. C.

SELF-CHECK 8.2
1. (a) Copeer (II) carbonate
(b) (i) - Black solid formed
- A colourless gas that turns lime water milky was evolved.
(ii) CuCO3(s)  CuO (s) + CO2 (g)
(c) (i) Barium nitrate solution followed by dilute nitric acid.
(ii) Ba2+ (aq) + SO42-(aq)  BaSO4(s)
2. (a) The hydroxide is highly soluble.
(b) (i) HCO3- , CO32- (ii) HCO3-
(c) (i) MgCO3
235
 (d) (ii) Mg(HCO3)2 (aq)  MgCO3 (s) + CO2 (g) + H2O (g)

3. (a) Dilute sulphuric acid.

With: Pb2+ - White precipitate,
Al3+ - No observable change.
(b) Dilute nitric acid:
With: SO42- - No observable change.
CO32+ - A colourless gas that turns lime water milky evolved.

4. (a) (i) Green. (ii) Brown.

(b) (i) FeSO4. (ii) Fe2(SO4)3.
(c) (i) Dilute sodium hydroxide solution.
(ii) With: - Fe2+ - Green precipitate formed.
- Fe3+ - Brown precipitate formed.
(iii) - Fe2+ (aq) + 2OH - (aq)  Fe(OH)2(s)
3+
- Fe (aq) + 3OH (aq) 
-
Fe(OH)3(s)
(d) Pass dry hydrogen chloride over heated iron wool in a combustion tube.

2HCl (g) + Fe (s)  FeCl2(s) + H2(g)

5. (i) Pb2+ ions (ii) Pb2+ (aq) + 2OH -(aq)  Pb(OH)2 (s)
(ii) Lead (II) hydroxide forms white precipitate which is insoluble in excess
ammonium hydroxide solution.

6. (a) (i) Ammonium hydroxide:

With: Pb2+ ion - White precipitate insoluble in excess
ammonium hydroxide.
Zn2+ ion - White precipitate soluble in excess
ammonium hydroxide forming a colourless
solution.
(ii) Silver nitrate:
With: CO32- ion: - No observable change.
Cl- ion: - White precipitate which turns to grey on
standing.

CHAPTER NINE

OXIDATION AND REDUCTION

SELF-CHECK 9.1
1. C. 2. A. 3. B. 4. C.
5. B. 6. A. 7. C 8. C

236
 SELF-CHECK 9.2
7. (a) Oxidation is the loss of electrons
(b) (i) H+ + 2e-  H2 Reduction
(ii) Fe2+ - e-  Fe3+ Oxidation
(iii) Cl2 - 2e-  2Cl - Oxidation
(c) (i) the elements on the right hand of the periodic table are good
oxidizing agents.

Second period - Oxygen and fluorine

Third period - Chlorine
(ii) The elements on the left hand of the periodic table are good
reducing agents.
Second period - lithium
Third period - sodium and magnesium.
(iii) Good oxidizing agents have almost filled valence orbitals;
Small atomic radius (less electropositive element).
(ivi) Good reducing agents have a few electrons more than an inert
gas electronic configuration; greater atomic radius (more
electro positive element)

CHAPTER TEN
ENERGY CHANGES IN CHEMICAL REACTIONS
SELF-CHECK 10.1
1. A. 2. B. 3. D. 4. C. 5. A.
6. C. 7. A. 8. D. 9. A. 10. C.
11. B. 12. D. 13. D. 14. B. 15. A.
SELF-CHECK 10.2
1. (a) Enthalpy of combustion is the enthalpy change that occurs when one mole
of a substance is completely burnt in (pure) oxygen.
(b) Apparatus:
Retort stand/clamp, a thin walled tin can, thermometer, water, ethanol, and
a spirit lamb.

Procedure:
 Weigh a spirit burner containing pure propanol.
 Fill a thin-walled tin can with a known volume of water.
 Clamp the tin can with its content above the lamp as shown in the
diagram below.
 Insert the thermometer into the water.
 Read and record the initial temperature of the water.

237
 See diagram on page 193

 Heat the water while carefully stirring with the thermometer.

 Reweigh the lamb and its content.

Results:
Volume of cold water = V1 cm3
Initial temperature of water = t1 C
Final temperature of water = t2 C
Mass of spirit burner before burning= m1 g
Mass of spirit burner after burning = m2 g
Specific heat capacity of water = 4.2 J/g C
Calculation
Heat gained by water = mct

= Mass of  
Sp. ht. cap  
Change in
water x of water x Temperature 
= (Vol. x Density) x 4.2 x (t2 - t1)

= (Vol. x 4.2 x (t2 - t1) Joules

Heat lost/ produced by propanol is calculated as follows:
The relative molecular mass of ethanol, C3H7OH,
= (3 x 12) + (1x 8) + 16
= 36 + 8 + 16
= 60 g

Mass of ethanol used up (burnt) = Mass 
of spirit lamp Mass of spirit lamp
before burning  after burning 
= (m1 - m2) g

Assuming heat gained by water = Heat given out by ethanol

(m1 - m2) g of ethanol produces, V  4.2  (t2-t1)

60
60 g of ethanol produce will produce = x (V  4.2) x (t2t1) KJ mol-1
m1  m2

(c) Mass of propanol burnt = 0.54 g, t = 21.5 0C, Volume of water = 150 cm3
Dnsity of water = 1g/cm3, sp. ht. cap of water = 4.2 Jg-1K-1

Heat energy gained by water = mct

= V  4.2  (t2-t1)
= 150 x 1  4.2  21.5
= 13545 J
= 13.545 KJ
0.54 g of propanol produces 13.545 KJ of heat energy.

238
 13.545
1 g of propanol produces KJ of heat energy.
0.54
60
60 g (I mole ) of propanol will produce x 13.545 = 1505 KJ mol-1
0.54

2. (a) Enthalpy of neutralization is the enthalpy change that occurs when an acid
reacts with an alkali to produce a salt and a mole of water.

(b) Final volume of solution = Vol. of acid + Vol. of base

= (50 + 50) cm3
= 100 cm3
Mass of solution = Volume  density
= 100  1
= 100 g
Change in temperature, t = (Final - Initial) temperature
= 30.8 - 27.5
= 3.3 0C
Given the spirit heat capacity of water is 4.2 Jg-1,
Heat evolved = Heat gained by the water formed
= Mass  sp. heat cap  temperature rise
= 100  4.2  3.3 J
 Heat evolved = 1386 J
Moles of hydrochloric acid = Molarity  volume in litres
0.5 x 50
=
1000
= 0.025 moles of acid
From KOH(aq) + HCl/aq) → KCl(aq) + H2O(l)
1 mole of acıd produces 1 mole of wqter, therefore, the number of moles of water produced
= 0.025

0.025 oles of water is produced with the evolution of 1386 J of heat energy.
1386 x 1
1 mole of water will be produced with the evolution of
0.025
= 55440 J
= 55.44 KJ mol-1.

CHAPTER ELEVEN
EXTRACTION OF METALS

239
 SELF-CHECK 11.1
1. B. 2. C. 3. A. 4. C.
5. A 6. C. 7. C. 8. D.

SELF-CHECK 11.2
1. (a) (i) Sodium - Sodium Chloride (NaCl) and
Iron - Hematite (Fe2O3)

(ii) Extraction of Sodium by electrolysis

Sodium is extracted by the electrolysis of molten sodium chloride in the Downs Cell.
The cell consists of a cylindrical steel cathode and graphite anode.
Sodium chloride is heated and melts at 801 ºC. Calcium chloride is added as an
impurity to lower the melting point to about 600 ºC. A very high current is passed
through the electrolyte to keep it in molten state.

Ions present: From NaCl (s)  Na+(l) + Cl-(l)

From CaCl2 (s)  Ca2+(l) + Cl -(l)
Equations at the electrodes.
At the anode: 2Cl -(l)  Cl2 (g) + 2e-
Or 2Cl -(l) - 2e-  Cl2 (g)
At the cathode: Na+(l) + e  Na (l)
The liquid Sodium is collected under dry nitrogen gas. This prevents the metal from
reacting with the atmosphere.

Extraction of Iron
By reduction of its ore hematite (Fe2O3) in a blast furnace by carbon monoxide
A mixture of iron ore (concentrated, crushed and roasted), coke and limestone is fed
into a blast furnace from the top of the furnace.

A blast of preheated air is injected into the furnace through the pipes called tuyeres
at the lower part of the furnace. The air oxidizes the hot coke to carbon monoxide.
2C (s) + O2 (g)  2CO (g)
At the upper part of the furnace, the carbon monoxide formed reduces the iron (II)
oxide to iron according to:
. Fe2O3 (s) + 3CO (g) 2Fe (l) + 3CO2 (g)
The iron absorbs carbon from the coke, its melting point is lowered. It melts and then
sinks and collects at the bottom of the furnace, where it is taped off and solidified into
blocks called “pig iron” or cast iron.

(b) Sodium reacts with water at ordinary room temperature.

2Na (s) + 2H2O (l)  2NaOH (aq) + H2 (g)
Iron does not react with cold water, but reacts with steam.
240
 3Fe (s) + 4H2O (g)  Fe3O4 (s) + H2 (g)

2. (a) Sodium chloride (NaCl)

(b) Extraction of sodium
Sodium is extracted by electrolysis of molten or fused sodium chloride in the Down’s
cell using of steel cathode and carbon anode.
A very high current is passed to keep the electrolyte in liquid state.
 Calcium chloride is added to lower the melting point of sodium chloride.
 Chlorine gas is liberated at the anode as a byproduct.
2Cl – (l)  Cl2 (g) + 2e.
Molten sodium is produced at the cathode.
Na+ (l) + e  Na (s) .
 The molten sodium is collected under nitrogen to prevent it from reacting with
air.

(c) Source of the raw material,

Availability of: - Power,
- Market and
- labour
(d) (i) The greenish colour of chlorine disappears and
A white solid formed.
(ii) 2Na (s) + Cl2(g)  2 NaCl (s)

(e) NaCl (aq) + AgNO3(aq)  NaNO3 (aq) + AgCl (s)

3. (a) Silicon (IV) oxide.

(b) (i) Siderite (Iron (II)carbonate.
(ii) FeCO3 (s)  FeO (s) + CO2(g)
(c) (i) Iron ore, lime and coke.
(ii) A blast of hot air
(d) (i) 2C (s) + O2(g)  2CO2 (g)
FeO (s) + CO (g)  2Fe (l) + CO2 (g)
2FeO (s) + C (s)  4Fe (l) + CO2 (g)

241
 (ii) Remember that the role of lime stone is to remove the main impurity.
It first decomposes to calcium oxide (quick lime CaO) according to:
CaCO3 (s)  CaO (s) + CO2 (g)
The calcium oxide then combines with Silcon (IV) oxide to form
Calcium silicate (CaSiO3), the slag according to the equation:
CaO (s) + SiO2 (s)  CaSiO3 (s)
(e) Iron and Carbon.

4. (a) Magnetite - Fe3O4

(b) To remove the main impurity, silcon (IV) oxide.
(c) Carbon monoxide and carbon reduce the iron (II) oxide to iron at temperature
ranges of (500 – 800) ˚C according to the equations:
Fe2O3 (s) + 3CO (g)  2Fe (l) + 3CO2 (g)
2Fe2O3 (s) + 3C (s)  4Fe (l) + 3CO2 (g)
(d) (i) A colourless gas that burns with a pop sound will be given off and
a green solution is formed.
(ii) A black solid, iron (II) oxide, will be formed and a colourless gas
that burns with a pop sound will be given off.
Fe (s) + 2 HCl (aq)  FeCl2 (aq) + H2 (g)
Fe (s) + H2O (g)  FeO (s) + H2 (g)

242