Name : _________________________( ) Class : _________

ACID BASE

Note (2) CALCULATION

- Ch 15 : Concentration and Molarity
- Ch 17 : pH value
- Ch 18 : Application of Neutralization
and Preparation of salts
- Ch 19 : Titration and Back titration
 Chapter 15 Concentration of solutions

Concentration is expressed in molarity (or molar concentration)

1dm3 (~L) = 1000cm3 (~mL)
It is the number of moles of solute per dm3 of the solution

EXERCISE 1 + TB p.65 LQ7

Complete the following table

(Relative atomic masses : H = 1.0, N = 14.0, O = 16.0,x S = 32.1, K = 39.1, Pb = 207.2)

Solution Mass Mole Volume Molarity

H2SO4 9.8 g 0.1 2.0 dm3 0.05M

Pb(NO3)2 6.62 g 0.02 250.0 cm3 5 x 10-3 M

CHEM / S.4 Mole(2) + Titration /P.2

A. Conversion between concentration and molarity

Q1. 250.0 cm3 of potassium chloride solution contains 0.50 mol of potassium chloride. Calculate
(a) the molarity of the solution;
(b) the concentration of the solution in g dm–3.
(Relative atomic masses: Cl = 35.5, K = 39.1)

0.50 mol
(a) Molarity of the solution = = 2.0 mol dm−3
250.0 3
dm
1000
(b) Concentration of solution in g dm−3
= 2.0 mol dm−3 × (39.1 + 35.5) g mol−1 = 149.2 g dm−3

Q2. 100.0 cm3 of ammonium nitrate solution contains 8.0 g of ammonium nitrate. Calculate the
molarity of the solution.
(Relative atomic masses: H = 1.0, N = 14.0, O = 16.0)

Molar mass of NH4NO3 = (14.0 × 2 + 1.0 × 4 + 16.0 × 3) g mol−1 = 80.0 g mol−1

80 g dm −3
Molarity of the solution = = 1.0 mol dm−3
80.0 g mol −1

TB p.54 Self test 15.3 + 15.4 , p.69 LQ12 + WB LQ 47

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B. Conversion of Density and Mass

Q3. A sample of 1.0 dm3 of concentrated sulphuric acid has a density of 1.83 g cm–3. If the sample
contains 98.0% of sulphuric acid, calculate the molarity of the acid.
(Relative atomic masses: H = 1.0, O = 16.0, S = 32.0)

Mass of the 1.0 dm3 sample of concentrated sulphuric acid

= 1.83 g cm−3 × (1.0 × 1000) cm3 = 1830 g

Mass of H2SO4 in the 1.0 dm3 sample = 1830 g × 98.0% = 1793.4 g

Number of moles of H2SO4 in the acid sample

1793.4 g
= = 18.3 mol
(1.0  2 + 32.1 + 16.0  4) g mol −1

18 .3 mol
Molarity of the concentrated sulphuric acid = = 18.3 mol dm−3
1.0 dm 3

TB p.69 LQ13 + WB LQ42

C. Mole = MV (based on TB Self test 15.6-15.9 + TB p.69 LQ14)

(Mole)
Q4. Calculate the number of moles of copper(II) sulphate present in 500.0 cm3 of 0.5 M copper(II)
sulphate solution.
500 .0
Number of moles of CuSO4 present = 0.5 mol dm−3 × dm 3 = 0.25 mol
1000
(Volume)
Q5. A 0.50 M sodium sulphate solution contains 0.40 mol of sodium sulphate. What is the volume
of the solution?
0.40 mol
Volume of Na2SO4 solution = −3
= 0.8 dm3 (or 800 cm3)
0.50 mol dm

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 (Mass)
Q6. Calculate the mass of potassium dichromate in 100.0 cm3 of 0.25 M potassium dichromate.
(Relative atomic masses: O = 16.0, K = 39.1, Cr = 52.0)
100 .0
Number of moles of K2Cr2O7 = 0.25 mol dm−3 × dm 3 = 0.025 mol
1000
Mass of K2Cr2O7 = 0.025 mol × (39.1 × 2 + 52.0 × 2 + 16.0 × 7) g mol−1 = 7.36 g

(ions)
Q7. What is the molarity of the constituent ions in a 0.5 M sodium phosphate (Na3PO4) solution?
1 mol of Na3PO4 contains 3 mol of Na+ ions and 1 mol of PO43− ions.
∴molarity of Na+ ions = 0.5 M × 3 = 1.5 M ; molarity of PO43− ions = 0.5 M × 1 = 0.5 M

Q8. 100.0 cm3 of 0.50 M KCl(aq) is mixed with 200.0 cm3 of 0.25 M CaCl2(aq). Calculate the
molar concentration of Cl– (aq) ions in the resultant mixture.
Number of moles of KCl = 0.05mol ; Number of moles of Cl– = 0.05mol
Number of moles of CaCl2 = 0.05mol ; ; Number of moles of Cl– = 0.1mol
Molarity of Cl– = (0.05+0.1) / (100+200)÷1000 = 0.5M

Q9. In an experiment, 0.635 g of Cu dissolves completely in 100 cm3 of 0.5 M AgNO3(aq).

(a) Write the ionic equation for the reaction involved.
(b) State TWO other observable changes in this experiment.
(c) Calculate the concentration of copper(II) ions and that of silver ions in the resultant solution.
(Relative atomic mass: Cu = 63.5)
(a) Cu(s) + 2Ag+ (aq) → Cu2+(aq) + 2Ag(s)
(b) Some shiny silvery solid deposits on the copper surface.
The solution changes from colourless to blue.
(c) Number of moles of Cu used = 0.635/63.5 mol = 0.0100 mol
Number of moles of Ag+ ions = 0.5 × 100/1000 mol = 0.05 mol

(0.0100 mol of Cu V.S. 0.025 mol of Ag+ after 1:1 fair comparison)
Hence, Cu is the limiting reactant.

Molarity of Cu2+ (aq) ions = 0.01 ÷ 100/1000 = 0.1 mol dm–3

Molarity of Ag+ (aq) ions = (0.05 – 0.02) ÷ 100/1000 = 0.3 mol dm–3

MC107,108,109,111,115,122,123,126

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D. Dilution (based on TB Self test 15.10, 15.11)

Moles before and after dilution are the SAME

M1V1 = M2V2

Q8. 50.0 cm3 of 5.0 M hydrochloric acid is diluted to 500.0 cm3 . Calculate the molarity of the
diluted acid.
Number of moles of HCl (before dilution) = Number of moles of HCl (after dilution)
M1V1 = M2V2
50 .0 500 .0
5.0 × = M2 ×
1000 1000
M2 = 0.5
∴the molarity of the diluted acid is 0.5 M.

(Add water)
Q9. Calculate the volume of water that requires to dilute 100.0 cm3 of 3.0 M copper(II) sulphate
solution to 0.5 M.
Number of moles of CuSO4 (before dilution) = Number of moles of CuSO4 (after dilution)
M1V1 = M2V2
100 .0
3.0 × = 0.5 × V2
1000
V2 = 0.6
Volume of diluted CuSO4(aq) obtained = 0.6 dm3 (or 600.0 cm3)
∴volume of water to be added = (600.0 − 100.0) cm3 = 500.0 cm3

TB p.70 LQ16 + WB MC110,112

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 Chapter 16 Indicators and pH
A. pH scale

pH = -log[H+] only based on molarity of H+

*[ ] = molarity *no unit for pH

 Acidic: [H+] > [OH−]

 Alkaline: [H+] < [OH−]
 Neutral: [H+] = [OH−]
 In pure water, pH = 7 ; [H+] = 10-7 M

 [H+] [OH−] = 10-14 M2

Q1. Calculate the pH of each of the following aqueous solutions:
(a) 0.01 M hydrochloric acid =2
(b) 0.10 M sulphuric acid = 0.7
(c) 1.0 M nitric acid =0
(d) 10.0M hydrochloric acid = -1

Hence, based on concentration of HCl in (a) and (d), student estimated that their difference in pH
is 1000 times. Do you agree? Why or Why not?
No. 3 times by a factor of 10 (TB p.77)

Q2. Calculate the concentration of H+ (aq) ions in a solution with

(a) pH = 3.1 = 7.9 x 10-4 M
(b) pH = 8.0 = 1.0 x 10-8 M

Hence, Calculate the concentration of OH- (aq) ions in (b).

(a) = 1.27 x 10-11 M (b) = 1.0 x 10-6 M

Q3. A sample of sulphuric acid has a pH of 1.2. If this sample is diluted by 100 times, calculate the
pH of the diluted sample.
[H+] = 0.063M in the original sample ; diluted 100 times [H+] = 0.00063M
pH of diluted sample = 3.2

WB MC 130,133,138,139,141,150,151

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 Chapter 18.6 General methods of preparing salts

Titration

(I) Insoluble salt (e.g. PbSO4(s))

 Mixing two solutions (BOTH soluble)
 Procedure:
1)Mix two solutions in a beaker with stirring.
2)Collect the precipitate by _____________________.
3)Wash the precipitate with ______________________ to remove any _____________ impurities.
4)Dry the solid with _________________ or an _______________.

PbSO4(s)
PbSO4(s)
Filter paper

(Filter)
Wash bottle containing
funnel
deionized water

Pb(NO3)2 (aq) __________ PbSO4(s)

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(II) Soluble salt
 Adding acid on metal / insoluble base / insoluble carbonate (ONE insoluble in EXCESS)
 _________ + H2SO4(aq) → ZnSO4(aq) + H2(g)
 _________ + H2SO4(aq) → ZnSO4(aq) + H2O(l)
 _________ + H2SO4(aq) → ZnSO4(aq) + H2O(l) + CO2(g)
 Procedure:
(1) Add dilute sulphuric acid in a beaker.
(2) Add excess metal / insoluble base / insoluble carbonate to the acid.
(3) Remove the solid left by filtration
(4) Concentrate the solution by evaporation
(5) Collect the crystals by crystallization. (Leave the hot concentrated solution to cool slowly at room
temperature.)
(6) Filter the crystals, wash the crystal with a little, cold distilled water and dry with filter paper or an
oven.

Excess Zn(s)
Zn(s) left

Filter paper __________

Funnel
H2SO4(aq) Steam
bath

Excess Zn(s) Filtrate heat

heat
_____________

_________

ZnSO4(s)

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 Adding acid on alkali / soluble carbonate by titration

 _________ + HCl(aq) → NaCl(aq) + H2O(l)

 _________ + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)

 Procedure:
(1) (if ay) Dilute NaOH(aq) by 250 cm3 volumetric flask.
(2) Transfer 25.0 cm3 of NaOH(aq) into a conical flask by a pipette and add a few drops of methyl orange
as an indicator.
(3) Titrate it against standard HCl(aq) until the colour changes from yellow to persistent orange. Record
the volume of HCl(aq) used.
(4) Remove the indicator by adding a little activated charcoal (活性炭) and filter the mixture.

Or
(5) Repeat the experiment with exactly the same volumes of acid and alkali, without adding the indicator
this time.
(6) Separating and purifying the salt

TB p.145 LQ28 + WB LQ60, 61, 68

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 Chapter 19 Volumetric analysis involving acids and alkalis
A. Volumetric analysis (VA)
 It is used to determine the amount of a substance by measuring volumes of substances.
 Titration is used to determine the concentration of a solution.
 A reaction of two aqueous reactants
 Solution A with known volume and known concentration.
 Solution B with known volume and unknown concentration.

B. Standard Solutions
 Standard solution is a solution of accurately known molarity.

- Primary standard : known molarity of weighing

(anhydrous Na2CO3 and ethanedioic acid )

✓ Highly pure

✓ Stable in air over long period of time (not decompose/not volatile/no reaction with CO 2)

✓ Able to dissolve in water

✓ Do not release / absorb water easily

✓ Better to have higher molar mass (weighing error can be minimized)

× weak acid and alkali

- Secondary standard : known molarity by titrated with primary standard

NaOH / KOH : absorb water and react with CO2 in air

Ca(OH)2 : only slightly soluble in water

HCl : gas

H2SO4 : absorb water

HNO3 : light decomposition

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C. Preparation of Standard Solutions
(1) By dissolving pure solid
Procedure: 0.1 M of 250 cm3 sodium carbonate solution
 Calculate the mass of the solid anhydrous sodium carbonate required.
 Weigh the solid anhydrous sodium carbonate with an electronic balance.
 Dissolve all solid in a beaker in about 100cm3 deionized / distilled water.
 Wash a 250.0 cm3 volumetric flask with deionized / distilled water.
 Pour the solution into the 250.0 cm3 volumetric flask and wash the beaker 2-3 times so that all the
remaining solution are transferred to the flask.
 Add deionized / distilled water into 250.0 cm3 volumetric flask until it reaches graduation mark.

Glass
rod
100 cm3 Glass
Distilled water rod
Funnel

Na2CO3(aq)
Volumetric
Na2CO3
flask

dropper

Graduation
mark

Rinse the beaker 2-3 times & Make up to 250.0 cm3

transfer the remaining solution Stopper the flask and mix the
into the V. flask content thoroughly

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 (2) By dilution
Procedure: 0.01M of 250 cm3 sodium carbonate solution from 0.1M sodium carbonate solution
 Wash the pipette with distilled water and then the solution to be held.
 Pipette 25.0 cm3 the original solution and transfer it into a 250.0 cm3 volumetric flask.
 Make up the solution to 250.0 cm3 in a volumetric flask.

Pipette filler

Distilled water;
0.1M Na2CO3(aq)
0.1M
Na2CO3(aq)

dropper
Distilled water

Graduation
Mark

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D. Acid-alkali titrations 

Q : In a titration experiment, when solution A is titrated with/against solution B.

- Solution A is contained in the conical flask (with indicator) and its concentration is unknown

- Solution B is delivered from a burette and its concentration is known, i.e. it is a standard solution

Step 1 : Rinse all the apparatus

(Rinse with water only : volumetric flask and conical flask)

- Otherwise, solution A remained in the flask requires more solution B to complete the reaction)

(Rinse with water and then solution A : pipette and burette)

- Otherwise, water remained in the apparatus dilutes the solution and affects the concentration)

(Precaution : rinse more times and even wash the apparatus with cleanser)

// solid Sample A’//

Step 2 : Weigh the given amount of sample A (s) into a beaker and Add known volume of distilled/deionized water to

dissolve the solid sample A

Step 3 : Pour all sample A solution into 250cm3 volumetric flask

(rinse the beaker with distilled/deionized water several time to ensure all sample is rinsed into volumetric flask)

// aqueous Sample A’//

Step 2 : Pour some sample A’ (aq) into a beaker

Step 3 : Transfer 25cm3 sample A’ (aq) into 250cm3 volumetric flask

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Step 4 : Dilute with distilled/deionized water to the graduation mark

(stopper the volumetric flask and invert several times to ensure the solution is completely mixed)

Step 5 : Pipette 25cm3 of the diluted sample A solution into the conical flask

(conical flask is used instead of beaker to prevent splashing out the solution when stirring)

CHEM / S.4 Mole(2) + Titration /P.15

Step 6 : Add 2-3 drops of methyl orange/phenolphthalein indicator into conical flask

Step 7 : Pour the known molarity of standard solution B into the burette with a funnel

(make sure no air bubbles inside the jet, and the whole burette is filled completely with solution, especially the opening)

Step 8 : Record the initial reading (2 d.p.) of the burette

(make sure the funnel is taken away, to avoid the solution remained in funnel falls back into the burette)

CHEM / S.4 Mole(2) + Titration /P.16

Step 9 : Titrate and record the final reading (2 d.p.) of the burette when the colour of indicator changes

(Put a white tile below for clear vision of colour change)

(Beware of the position of hand controlling the opening of burette)

Step 10 : Repeat the titration three times to calculate the average volume of standard solution B added

(Possible errors : incomplete stirring / some solution left on the inner wall of the conical flask)

Indicator Colour in acid solution Colour in alkaline End point

solution

Methyl orange Red Yellow Orange (3.1-4.4)

Phenolphthalein Colourless Pink Very pale pink (8.3-10)

 Equivalence point: an acid and an alkali have just completely reacted with each other
 End point : the acid-base indicator changes its colour

End point (pH range of indicator’s colour change)

is within or equal to
Equivalence point (vertical point of the pH curve)

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EXERCISE 2 + WB (exp steps)

The graph below shows the variation of pH when 40 cm3 of an aqueous solution of a monobasic acid HA is
titrated against 0.100 M NaOH(aq)

(a) Where is the equivalent point ?

pH range between 8-11

(b) Calculate the concentration of the HA(aq) used.

No. of mol of NaOH = 0.1 x (25/1000) = 0.0025 mol (1)
Mole ratio NaOH : HA = 1:1
No. of mol of HA = 0.0025 mol
Concentration of HA = 0.0025/ (40/1000) = 0.0625M (1)

(c) The table lists the pH range for colour change of three acid-base indicators:
Indicator pH range
Methyl orange 3.2 to 4.4
Bromothymol blue 6.0 to 7.6
Phenolphthalelin 8.2 to 10.0
(i) Name the pH range for colour change of indicators
End point

(ii) Choose from the above table, a suitable indicator for the titration. Explain your answer.
phenolphthalein (1), pH range of phenolphthalein is within the vertical part of pH curve

(d) Is HA strong or weak acid ? Hence, if NaOH is replaced by ammonia, can the titration be used?
Weak acid (1), The equivalent point is not sharp enough to choose a suitable indicator.
CHEM / S.4 Mole(2) + Titration /P.18
 Strong Acid + Strong Alkali
 Choice of indicator

HCl(aq) NaOH(aq)

burette burette

NaOH(aq) +Indicator HCl(aq) +Indicator

White tile White tile

Indicator: Methyl orange Indicator: Methyl orange

From__________ to ____________ From__________ to ____________

Indicator: Phenolphthalein Indicator: Phenolphthalein

From__________ to ____________ From__________ to ____________

Strong Acid + Weak Alkali

HCl(aq)
NH3(aq)

burette burette

NH3(aq) +Indicator HCl(aq) +Indicator

White tile
White tile

Indicator: Methyl orange Indicator: Methyl orange

From__________ to ____________ From__________ to ____________

Weak Acid + Strong Alkali

CH3COOH(aq) KOH(aq)

burette burette

CH3COOH(aq) +Indicator
KOH(aq) +Indicator

White tile White tile

Indicator: Phenolphthalein Indicator: Phenolphthalein

CHEM / S.4 Mole(2) + Titration /P.19
From__________ to ____________ From__________ to ____________
EXERCISE 3 + TB p.175 (Class Practice 19.7) , 179 (Class Practice 19.8) + WB (normal titration)

Normal titration
Q1. 2.65 g of sodium carbonate were dissolved in water and made up to a 250.0 cm3 solution. 25.0
cm3 of the solution required 20.00 cm3 of a hydrochloric acid for complete reaction. Find the molarity
of the hydrochloric acid.
(Relative atomic masses: C = 12.0, O = 16.0, Na = 23.0)

CHEM / S.4 Mole(2) + Titration /P.20

Step 1 : highlight all data with unit

Step 2 : highlight all chemicals

Step 3 : Draw the set-up

B
Original Sample A Diluted A A
15cm3 250cm3 25cm3

Step 4 : Find mole 齊資料 (usually reversed order)

Mole of B

Step 5 : Find mole of another chemicals by mole ratio (balanced chemical equation)

Mole of A in 25cm3

Step 6 : (if any) any extraction ???

Mole of A in 250cm3

Step 7 : Find ______ of the original sample (A) (See the last sentence of the question)

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Precipitation titration by Electrical Conductivity

- In the beginning (A), electrical conductivity decreased due to the decrease in

concentration of mobile ions as they react to form ppt.
- At E, minimum electrical conductivity is reached at the end point when almost all the mobile ions in the
solution are precipitated, i.e. reaction was completed
The electrical conductivity at the end point is not zero as the solution still contains mobile ions (water)
- At B, the addition of excess solution will increase the electrical conductivity due to the increase in
concentration of mobile ions (from the solution added)

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EXERCISE 4 + WB LQ 74

25.0 cm3 of a 0.10 M solution of barium hydroxide were placed in a beaker. The electrical conductivity of the
solution was measured. Sulphuric acid was then added to the beaker, 5.0 cm3 at a time. The conductivity was
measured after each addition. A precipitate formed during the titration and the reaction was represented by the
equation: Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + 2H2O(l)

The experimental results are shown in the graph :

(a) Account for the shape of the graph.

(b) What is the volume of acid required to reach the equivalence point of the titration?

(c) Calculate the molarity of the sulphuric acid.

CHEM / S.4 Mole(2) + Titration /P.23

Thermometric titration

At A, temperature rises as acid is adding into alkali because neutralization is exothermic

At B, it is the maximum temperature rise where the reaction is completed
At C, temperature drops as excess cold acid is added into reaction mixture

WB LQ 83

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E. Back Titration
For cases that normal titration does not work :
(1) Sample is insoluble
(2) Sample is impure
(3) Reaction is too slow
(4) Equivalent point is not sharp enough for colour change

Back titration is used in which one more chemical (C) is added.

✓strong acid/alkali ✓known molarity ✓able to react with BOTH A and B

Excess C B

A C left
EXERCISE 5 + WB (back titration)

Back titration
A brand of antacid tablet contains magnesium hydroxide as the only ingredient. An experiment was
performed to determine the amount of magnesium hydroxide contained in the antacid tablet. An
antacid tablet was dissolved in 150.0 cm+ of 1.00 mol dm–3 HCl(aq) to form a solution. The solution
was then diluted to 250.0 cm3 with distilled water. 25.0 cm3 of the diluted solution was titrated with
0.16 mol dm–3 KOH(aq), using a suitable indicator. 26.30 cm3 of potassium hydroxide solution was
needed to reach the end point. Calculate the mass of magnesium hydroxide in the antacid tablet.
(Relative atomic masses: H = 1.0, O = 16.0, Mg = 24.3)

no. of moles of KOH = 0.16(26.3/1000) = 4.208×10-3 mol

no. of moles of HCl reacted = (1)(150/1000) – 4.208×10-3(250/25) = 0.10792 mol

mass of Mg(OH)2 = [(1/2)(0.10792)]×(24.3 + 2×17) = 3.15g

CHEM / S.4 Mole(2) + Titration /P.25

Step 1 + 2 : highlight all data with unit AND all chemicals

Step 3 : Draw the set-up

Excess C B

Original Sample A A C left

15cm3 250cm3 25cm3

Step 4 : Find mole 齊資料 (reversed order)

Mole of B

Mole of excess C

Step 5 : Find mole of C left by mole ratio (B + C → BC)

Mole of C left in 25cm3

Step 6 : (if any) any extraction ???

Mole of C left in 250cm3

**Step 7 : Find the mole of C reacted

Mole of C reacted = Excess – Left in 250cm3

Step 8 : Find mole of A by mole ratio (A + C → AC)

Mole of A

Step 9 : Find ______ of the original sample (A) (See the last sentence of the question)

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Note-taking for WB LQ
Normal titration
Q79,88,117(find n H2O) Q110, 125 (unknown M) / 84*, 85*, 86, 87, 89, 90, 92, 93, 94, 97, 112, 113,
115, 116, 120, 121, 126

Back titration
Q91 (metal ion + OH-) Q107(metal + H+) / 82, 95, 110*, 114*, 119, 122

Exp steps
Q98, 99, 102, 106, 118, 123

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