Indicator used: Bromocresol green
Observation Table 4.2: Determination of bromocresol green end point, BE
S.No. Volume of Burette reading Titre value (in cm3)
orthophosphoric acid Initial Final (Final-initial reading)
(in cm3)
1
2
3
Concordant reading
d) Determination of mixed indicator end point, ME by titrating against sodium
hydroxide
Volume of given sample of orthophosphoric acid taken, VP = 10 cm3
Solution in the burette: NaOH
Indicator used: mixed indicator
Observation Table 4.3: Determination of mixed indicator end point, ME
S.No. Volume of Burette reading Titre value (in cm3)
orthophosphoric acid
(in cm3) Initial Final (Final-initial reading)
Concordant reading
e) Observation Table 4.4: Determination of pKa values
S.No. volume of volume of NaOH pH
orthophosphoric acid solution used in the
solution (in cm3) titration (in cm3)
1 Vp
[BE /2] = pKa1
2 Vp
[BE + (ME-BE)/2] = pKa2
3 Vp
[2BE + (ME-BE)/2] = pKa3
4.7 RESULT
The pKa values of orthophosphoric acid are determined to be:
pKa1 =
pKa2=
pKa3=
EXPT. 5 DETERMINATION OF TOTAL
ALKALINITY OF A GIVEN WATER
SAMPLE
Structure
34
5.1 Introduction
Objectives
5.2 Principle
5.3 Requirements
5.4 Solutions Provided
5.5 Procedure
5.6 Observations and Calculations
5.7 Result
5.1 INTRODUCTION
The pure water is neutral; however the natural waters may be acidic or alkaline. The
pH of most natural waters is usually in the range of pH of 4 – 9. The majority of water
samples are slightly basic because of the presence of carbonates and bicarbonates of
the alkali and alkaline earth metals. Sometimes, borates, silicates and phosphates may
also be present as the sources of alkalinity. The alkalinity may be defined as a measure
of the capacity of water to neutralize acids. The total of all of the alkaline species
found in a water sample contribute to the alkalinity of the water sample. However, for
the sake of simplicity, the alkalinity of water samples is expressed in terms of mg of
CaCO3/dm3. Depending on the possible constituents responsible for alkalinity in the
water sample there are five different types of alkalinities of water. These are:
• Hydroxides only alkalinity
• Carbonates only alkalinity
• Bicarbonates only alkalinity
• Hydroxides and carbonates alkalinity
• Carbonates and bicarbonates alkalinity
It is essential to assess the nature and extent of alkalinity present in the water sample to
ascertain its suitability for different applications. In this experiment you would learn
about the principle of the determination of total alkalinity of a water sample based on
acid base titration and also perform such a determination. This method of titration will
help you in understanding the basic principle of some important industrial analyses The hydroxides and
bicarbonates can not exist
such as that of soda ash, sodium bicarbonate, mixture of sodium carbonate and sodium together because they
bicarbonate, commercial caustic soda, washing soda, etc. In the next experiment you combine with each other, to
would learn about the determination of hardness of water by complexometric titration form the respective carbonates
Objectives
After studying and performing this experiment you should be able to:
• state and explain the principle of acid-base titration with special reference to the
determination of alkalinity of a water sample,
• prepare a standard solution of sodium carbonate,
• standardise the given solution of hydrochloric acid and use it in determining the
alkalinity of water sample, and
• determine the hydroxide only, carbonate only and hydroxide and carbonate only
alkalinity in the given water sample
5.2 PRINCIPLE
Let us first understand the way the carbonate and bicarbonate ions cause alkalinity in
the water sample; the alkalinity due to hydroxide ions is obvious. The carbonate and
bicarbonate ions cause alkalinity due to hydrolysis as per the following equations:
35
CO32 + H2 O HCO3 + OH .....5.1
HCO3 + H2O H2CO3 + OH .....5.2
Here, the carbonate ions are hydrolysed to bicarbonate ions which in turn get
hydrolysed to carbonic acid. The hydroxide ions so produced are responsible for the
alkalinity of the carbonate and bicarbonate ions. You may note here that the carbonate
ions get hydrolysed in two steps. The hydroxide ions so produced may be determined
with the help of a titration with a standard solution of hydrochloric acid.
The alkalinity of waters is a As there are more than one species causing the alkalinity, we cannot determine them
function of the concentrations
of carbonate, bicarbonate and
by a single titration. The double indicator method for the determination of the
hydroxide ions. alkalinity of water exploits the titration behaviour of the alkaline ions with
hydrochloric acid. The reactions of hydroxide and carbonate ions with a monoprotic
acid like hydrochloric acid can be represented as
Hydroxide ions:
OH + H+ H2 O ...5.3
Carbonate ions:
2
CO3 + H+ HCO3 + OH ...5.4
HCO3 + H+ CO2 + H2 O ...5.5
The hydroxide ions act as a strong base and titrate with the strong acid as indicated in
Equation 5.3. The carbonate ions titrate with acid in two steps; in the first step these
react with one mole of hydrogen ions and generate an equivalent amount of
bicarbonate ions, Eq. 5.4, whereas in the second step these bicarbonate ions react with
another mole of hydrogen ions to give carbon dioxide and water, Eq. 5.5. Thus in case
of carbonate ions there are two half titrations, the titration of carbonate to bicarbonate
ions is a strong base Vs strong acid titration like the titration of hydroxide ions. The
titration of bicarbonate ions so obtained, on the other hand is a weak base Vs strong
acid titration. The bicarbonate present (if any) in the solution, in addition to the one
obtained from the carbonate also titrates in the same way. Thus, in a water sample
containing hydroxide, carbonate, and bicarbonate ions the hydroxide ions and the
carbonate ions (upto half neutralisation) titrate as strong bases and the carbonate ions
(in the second half neutralisation) and the bicarbonate ions titrate as weak bases.
When the hydroxide ions and the carbonate ions (upto half neutralisation) are titrated
with a strong acid like HCl, the pH of the solution drops to about 8. On the other hand
when all the bases present in the water sample are completely neutralised by acid the
pH drops to the range of 2-3 depending on the concentration of the acid. This fact is
exploited in determining the amounts of the different ions responsible for the alkalinity
of the water sample.
In the double indicator method, two titrations of the given sample are performed using
phenolphthalein and methyl orange respectively as the indicators. In the titration using
phenolphthalein as the indicator, the end point is obtained when the hydroxide ions are
fully neutralised and the carbonate ions are half neutralised. On the other hand in the
titration using methyl orange as the indicator, the end point is obtained when all the
ions are fully neutralised. Thus, the titre value at the phenolphthalein end point, PE
36
equals the sum of the volume of hydrochloric acid required for the complete
neutralisation of hydroxide ions and half neutralisation of carbonate ions.
PE= Vhydroxide + ½ Vcarbonate … 5.6
Where, Vhydroxide and Vcarbonate are the volumes of hydrochloric acid required for the
neutralisation of hydroxide and carbonate ions respectively. Similarly, the titre value at
the methyl orange end point, ME equals the sum of the volume of hydrochloric acid
required for the complete neutralisation of hydroxide ions, and carbonate ions.
ME= Vhydroxide + Vcarbonate ... 5.7
The volumes of hydrochloric acid required individually for neutralising the hydroxide
ions and the carbonate ions can be calculated by using the titre values for the
phenolphthalein and methyl orange end points as explained below.
Subtracting the PE from ME we get the volume required for half neutralization of
carbonate ions,
1 1
M E − PE = (Vhydroxide + Vcarbonate ) − (Vhydroxide + Vcarbonate ) = Vcarbonate …5.8
2 2
Thus, Vcarbonate = 2 (M E − PE ) ...5.9
Similarly, we can show that
Vhydroxide = 2 PE − M E …5.10
Knowing, the volumes of the hydrochloric acid required individually for the
neutralization of the hydroxide ions and the carbonate ions we can calculate their
individual amounts.
5.3 REQUIREMENTS
Apparatus Chemicals
Volumetric flasks (100 cm3) 1 Hydrochloric acid
Burette (50 cm3) 1 Sodium carbonate
Pipettes (10 cm3) 1 Ethyl alcohol
Weighing bottle 1 Phenolphthalein
Burette stand with clamp 1 Methyl orange
Conical flask 1 Water sample
Funnel 1
5.4 SOLUTIONS PROVIDED
Phenolphthalein indicator solution: It is prepared by dissolving 1 g of the reagent in
100 cm3 of ethanol and adding 100 cm3 of water. If a precipitate is formed, it is
filtered.
Methyl orange indicator solution: It is prepared by dissolving 0.5 g of free acid
form of the indicator in 100 cm3 of water.
Hydrochloric acid solution (~0.1 M): It is prepared by slowly adding 10 cm3 of
conc. HCI to about 100 cm3 of distilled water taken in a 1 dm3 volumetric flask and
diluting the acid up to the mark with distilled water.
5.5 PROCEDURE
The procedure for the determination of the alkalinity of a water sample by double
indicator method consists of the following steps:
a) Preparation of standard solution of sodium carbonate
37
b) Standardisation of hydrochloric acid solution by titrating against standard
sodium carbonate solution.
c) Determination of phenolphthalein end point by titrating against standardised
hydrochloric acid
d) Determination of methyl orange end point by titrating against standardised
hydrochloric acid
e) Calculation of the concentrations of hydroxide and carbonate ions.
Follow the procedure given below in a step wise manner to determine the alkalinity of
a given water sample.
a) Preparation of 100 cm3 of standard solution of 0.05 M sodium carbonate
i) To prepare the standard solution of sodium carbonate we have to first
calculate the amount of sodium carbonate required. As you know that the
mass (m) of the solute required to prepare V cm3 of M molar solution of a
substance having a molar mass Mm is given as
MVM m
m (in g ) =
1000
The amount of sodium carbonate (Mm = 106 g mol-1) required to prepare
100 cm3 of 0.05 M solution would be:
0.05 × 100 × 106
m (in g ) = = 0.53 g
1000
ii) Weigh about 0.55 g of dry anhydrous sodium carbonate on a rough
weighing balance and transfer it to a clean dry weighing bottle and
accurately weigh the weighing bottle with sodium carbonate.
iii) Transfer the sodium carbonate to a clean volumetric flask of 100 cm3
capacity through a glass funnel.
iv) Weigh the weighing bottle containing sodium carbonate left (if any) and
find the exact mass of sodium carbonate transferred by subtracting this
mass from the mass of the weighing bottle plus sodium carbonate.
v) Dissolve sodium carbonate in about 30-40 cm3 of distilled water taken in
the volumetric flask. Once dissolved, make the volume up to the mark
with distilled water.
b) Standardisation of hydrochloric acid solution by titrating against standard
sodium carbonate solution
i) Fill up the burette with hydrochloric acid solution with the help of a
funnel and mount it on the burette stand. Note the reading on the burette
and record it in the observation Table 5.1 under the initial reading column.
ii) Carefully pipette out 10 cm3 of the standard sodium carbonate solution
and transfer to a clean 100 cm3 conical flask. Add two to three drops of
methyl orange indicator.
iii) Titrate the yellow coloured solution with constant swirling against a white
background (you may use a porcelain tile for the purpose) till a red colour
is obtained. Record the reading in the Observation Table I under the final
reading column.
iv) Repeat the titration to get at least two concordant readings and record the
same in the Observation Table 5.1.
38
c) Determination of phenolphthalein end point, PE by titrating against
standardised hydrochloric acid
i) Transfer 50 cm3 of the water sample into a conical flask; add 1 drop of
phenolphthalein indicator, and titrate the sample with the standardised
HCl solution till a faint pink colour persists. Record the initial and final
burette readings in Observation Table 5.2.
ii) Repeat the titration to get at least two concordant readings and record the
same in the Observation Table 5.2.
d) Determination of methyl orange end point, ME by titrating against
standardised hydrochloric acid
i) Transfer 50 cm3 of the water sample into a conical flask; add 2-3 drops of
methyl orange indicator, and titrate the sample with the standardised HCl
solution till you obtain a red colour. Record the initial and final burette
readings in Observation Table 5.3.
ii) Repeat the titration to get at least two concordant readings and record the
same in the Observation Table 5.3
Note: Alternatively, step (c) and (d ) given above can be combined. In this method,
50 cm3 of the water sample taken in a conical is titrated with the standard HCl solution
using phenolphthalein indicator, till a faint pink colour persists. The titre value is
recorded as the phenolphthalein end point, PE and 2 to 3 drops of methyl orange
indicator are added to the same solution and the titration is continued until a red
coloured end point is obtained. The titre value from the beginning of the titration is
recorded as the methyl orange end point, ME.
5.6 Observations and Calculations
a) Preparation of standard solution of sodium carbonate
Mass of weighing bottle + anhydrous sodium carbonate = m1 g = ..............g
Mass of weighing bottle (after transferring the salt) = m2 g = .............. g
Amount of sodium carbonate transferred =(m1 – m2) g = m g = ............. g
Molar mass (Mm) of sodium carbonate = 106 g mol-1
Volume of sodium carbonate prepared = 100 cm3
Molarity of sodium carbonate solution =
m×1000 10 m
M Na 2CO 3 = = =...........M
106 ×100 106
b) Standardisation of hydrochloric acid solution by titrating against standard
sodium carbonate.
Volume of standard Na2CO3 solution taken in conical flask, VN = …cm3 Solution
in the burette: HCl
Indicator used: Methyl Orange
Observation Table 5.1: Standardisation of Hydrochloric acid
S.No. Volume of sodium Burette reading Titre value (in cm3)
carbonate (in cm3) Initial Final (Final-initial reading)
1
2
3
Concordant reading
39
The concentration of the given hydrochloric acid solution can be determined as
follows.
The reaction involved:
Na2CO3 + 2HCl 2NaCl + H2O + CO2
Molarity equation: 2MNVN = MHVH
2 M NV N
MH =
VH
2 M N × 10
As the volume of sodium carbonate taken was 10 cm3, we get: M H =
VH
2. × ...... × 10
Substituting the values, the molarity of HCl= M H = M
...........
The molarity of given hydrochloric acid solution is = ……….M
c) Determination of phenolphthalein end point, PE by titrating against
standardised hydrochloric acid
Volume of given sample of water taken in conical flask, Vw = 50 cm3
Solution in the burette: HCl
Indicator used: Phenolphthalein
Observation Table 5.2: Determination of phenolphthalein end point, PE
S.No. Volume of water Burette reading Titre value (in cm3)
sample (in cm3) Initial Final (Final-initial reading)
1
2
3
Concordant reading
d) Determination of methyl orange end point, ME by titrating against
standardised hydrochloric acid
Volume of given sample of water taken in conical flask, Vw = 50 cm3
Solution in the burette: HCl
Indicator used: Methyl Orange
40
Observation Table 5.3: Determination of methyl orange end point, ME
S.No. Volume of water Burette reading Titre value (in cm3)
sample (in cm3) Initial Final (Final-initial reading)
1
2
3
Concordant reading
e) Calculation of the concentrations of hydroxide ions and carbonate ions
As explained in Section 5.2 the phenolphthalein end point and methyl orange
end point titre values can be used to determine the volumes of hydrochloric acid
required individually for the neutralising the hydroxide ions and the carbonate
ions.
Vcarbonate = 2 (M E − PE ) = 2 ( …….−…….) = ……… cm3
Vhydroxide = 2 PE − M E = 2…….−……. = ……… cm3
Hydroxide alkalinity
The molarity of given hydrochloric acid solution is = ……….M (From ‘b’ above) Vhydroxide= the volume of HCl
Volume of water solution (hydroxide ions) taken, Vw = …….. cm3 used for the neutralization of
hydroxide ions
Volume of hydrochloric acid used for the neutralisation of hydroxide, Vhydroxide=…. cm3
Molarity of hydroxide ions = Mhydroxide
The reaction involved:
NaOH + HCl NaCl + H2O
Molarity equation, Mhydroxide × Vhydroxide = MHCl × VHCl
In our case it can be written as
Mhydroxide × Vwater = MHCl × Vhydroxide
Thus, substituting the values,
M HCl × Vhydroxide ........ × ........
M hydroxide = = = ..........M
Vw ...........
Hydroxide alkalinity expressed in terms of milligrams of calcium carbonate per litre
can be calculated as under:
One mole of carbonate ions consume two moles of HCl whereas one mole of
hydroxide ions consume 1 mole of HCl. The carbonate equivalent of hydroxide
alkalinity would be 50% of the hydroxide alkalinity.
M hydroxide
M carbonate =
2
M hydroxide
Hydroxide alkalinity (Equivalent of CaCO 3 ) = ×100 (g mol −1 ) × 1000 (mg g −1 ) = .......... ..mg dm -3
2
Hydroxide alkalinity = M hydroxide ×50,000=............mg dm -3 = .........ppm CaCO3
Carbonate alkalinity
The molarity of given hydrochloric acid solution is = ……….M (From ‘b’ above)
Volume of water solution (carbonate ions) taken, Vw = …….. cm3
Volume of hydrochloric acid used for the neutralisation of carbonate, Vcarbonate=…. cm3
Molarity of carbonate ions = Mcarbonate
41
The reaction involved:
Na2CO3 + 2HCl 2NaCl + H2O + CO2
Molarity equation: 2McarbonateVcarbonate = MHClVHCl
In our case it can be written as: 2McarbonateVw = MHCl Vcarbonate
M HClVcarbonate
M carbonate =
2Vw
Rearranging and substituting the values, we get
M HClVcarbonate .......... × ........
M carbonate = = = .........M
2Vw 2 × ........
Carbonate alkalinity expressed in terms of milligrams of calcium carbonate per litre
can be calculated as under:
Carbonate alkalinity = M carbonate ×100 (g mol −1 ) × 1000 (mg g −1 ) = ............mg dm -3
Carbonate alkalinity = M carbonate ×100 ,000 = ............mg dm -3 = ........ppm
Total alkalinity
Total alkalinity = Hydroxide alkalinity + Carbonate alkalinity
Total alkalinity = [ ( M carbonate × 100 ,000) + ( M hydroxide × 50,000)]
= [ ( ......... × 100 ,000) + (........ × 50,000) ] = .............mg.
5.7 RESULT
i) Hydroxide alkalinity = …………mg of CaCO3 = …..ppm
ii) Carbonate alkalinity = …………mg of CaCO3 = …..ppm
iii) Total alkalinity = …………mg of CaCO3 = …..ppm
42
