Transcript Ionic Equilibria

Acid-base Theory
1. Arrhenius (1880s) : Applied to aqueous solutions only
Acids : Substances that produce hydrogen ions, H+(aq)
when dissolved in water. E.g. HCl
Bases : Substances that produce hydroxide ions,
OH(aq) when dissolved in water. E.g. NaOH
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2. Brnsted / Lowry (1923) : Applied to aqueous and non-aqueous solutions
Acids : Proton donors E.g.
Molecule – HCl, H2O, CH3COOH
Cation – H3O+, NH4+
Anion – HSO42
Bases : Proton acceptors
e.g.
Molecule - H2O, NH3
Anion - Cl, OH, CH3COO, SO42
Each contains at least one lone pair to form a
dative bond with a proton (H+)
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3. Lewis (1930s) : More widely applied to systems with and
without solvents
Acids : electron pair acceptors,
e.g. H+, BF3 , AlCl3 , BeCl2
(electron-deficient species)
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Bases : electron pair donors
e.g. NH3, OH
species containing lone pair(s)
H+(aq) + OH(aq)  H
BF3
Lewis acids
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+ NH3
 H3N
Lewis bases
OH
BF3
A conjugate acid-base pair is a pair of
species that can be inter-converted by
transfer of proton.
HCl(aq) + H2O(l)
acid
Loss of H+
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Cl(aq) + H3O+(aq)
Conjugate base
A conjugate acid-base pair is a pair of
species that can be inter-converted by
transfer of proton.
Gain of H+
HCl(aq) + H2O(l)
Conjugate
acid
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Cl(aq) +
base
H3O+(aq)
A conjugate acid-base pair is a pair of
species that can be inter-converted by
transfer of proton.
Gain of H+
HCl(aq) + H2O(l)
base
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Cl(aq) +
H3O+(aq)
Conjugate
acid
A conjugate acid-base pair is a pair of
species that can be inter-converted by
transfer of proton.
Loss of H+
HCl(aq) + H2O(l)
Conjugate
base
Cl(aq) +
H3O+(aq)
Acid
An acid-base equilibrium system consists of TWO
inter-convertible conjugate acid-base pairs.
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A species can behave as an acid or a base
depending on the situation.
Acid
Base
Conjugate base
Conjugate acid
HCl(aq)
+
H2O(l)
Cl(aq)
+
H3O+(aq)
H2O(l)
+
NH3(aq)
OH(aq)
+
NH4+(aq)
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1. Give the conjugate acid and conjugate
base of each of the following species
H2SO4(aq)
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HSO4(aq)
SO42(aq)
NH4+(aq)
NH3(aq)
NH2(aq)
H2O(l)
OH(aq)
O2(aq)
Concept of Conjugate Acid-base Pairs
Acid
Base
HCl(aq)
+
H2O(l)
+
H2O(l)
NH3(aq)
CH3COOH(aq) + H2O(l)
H2O(l)
+
HSO4(aq) +
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Conjugate base
Cl(aq)
Conjugate acid
H3O+(aq)
+
OH(aq) +
NH4+(aq)
CH3COO(aq) + H3O+(aq)
H2O(l)
OH(aq)
+
H3O+(aq)
H2O(l)
SO42(aq) +
H3O+(aq)
Relative Strengths of Conjugate Acid-base Pairs
Stronger acid/base produces weaker conjugate
base/acid more completely.
stronger + stronger
acid
base
weaker + weaker
conjugate
conjugate
base
acid
The forward reaction is more complete.
The equilibrium position lies to the right.
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Relative Strengths of Conjugate Acid-base Pairs
Weaker acid/base produces stronger conjugate
base/acid less completely.
weaker + weaker
acid
base
stronger + stronger
conjugate
conjugate
base
acid
The forward reaction is less complete.
The equilibrium position lies to the left.
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Relative
strength
Acid
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Relative
strength
Conjugate
base
For a strong acid
HCl(aq) + H2O(l)
stronger acid
stronger base
Cl(aq) + H3O+(aq)
weaker base
weaker acid
The equilibrium position lies far to the right.
Strong acids at the top left of Table 1 almost
ionize completely to give H3O+(aq) ,
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For a strong acid
HCl(aq) + H2O(l)
stronger acid
∴
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stronger base
Cl(aq) + H3O+(aq)
weaker base
weaker acid
the strongest acid present in aqueous
solutions is hydronium ions, H3O+(aq)
For a weak acid
CH3COOH(aq) + H2O(l)
CH3COO(aq) + H3O+(aq)
weaker acid
stronger base
weaker base
stronger acid
The equilibrium position lies far to the left.
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For a strong base
CH3(aq) +
H2O(l)
CH4(aq)
Stronger base
stronger acid
weaker acid
+
OH-(aq)
weaker base
The equilibrium position lies far to the right.
Strong bases at the bottom right of Table 1
react almost completely with water to give
OH(aq),
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For a strong base
CH3(aq) +
weaker acid
H2O(l)
weaker base
CH4(aq)
stronger base
+
H3O+(aq)
stronger acid
∴ the strongest base present in aqueous
solutions is hydroxide ions, OH(aq).
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For a weak base
HSO4(aq) + H2O(l)
H2SO4(aq) + OH(aq)
weaker base
stronger acid
weaker acid
stronger base
The equilibrium position lies far to the left.
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Formation of hydronium ion
H
Lone pair
+
O
H
Empty 1s
orbital
H
+
H
Hydronium ion
O
H
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H
Formation of ammonium ion
H
Lone pair
H+
N
H
H
+
H
N
H
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Empty 1s
orbital
H
H
Ammonium ion
Hydrated Hydronium ion, H3O+(H2O)20
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pH and Its Calculation
Definition :
pH  log10[H3O (aq)]

pH
[H3O (aq)]  1  10

pH is the negative power of the
molarity of Hydronium ion
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Self-ionization of Water
H2O(l) + H2O(l)
H3O+(aq) + OH(aq)
[H3O (aq)][OH (aq)]
Kc 
2
[H2O(l)]
∵ [H2O(l)]  55.5 M
It is in large excess  treated as a constant
 Kc[H2O(l)]2 = Kw = [H3O+(aq)][OH(aq)]
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Kw = [H3O+(aq)][OH(aq)]
ionic product of water
At 298 K
Kw  1  10
14
mol dm
2
Temperature dependent
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6
In neutral solution,
[H3O+(aq)] = [OH(aq)]
 1.00 10
14
mol dm
2
6
 1.00 10 mol dm
∴ pH = log10(1.00107) = 7.00
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7
3
In acidic solution,
[H3O+(aq)] > [OH(aq)]
 [H3O+(aq)] > 1.00107 mol dm3
 pH < 7.00
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In alkaline solution,
[H3O+(aq)] < [OH(aq)]
 [H3O+(aq)] < 1.00107 mol dm3
 pH > 7.00
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Q.2 Given the following Kw values at different
temperatures,
Temp/oC
18.0
25.0
40.0
75.0
mol2
0.610
1014
1.00
1014
2.92
1014
16.9
1014
Kw /
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dm6
Temp/oC
18.0
25.0
40.0
75.0
Kw /
mol2
0.610
1014
1.00
1014
2.92
1014
16.9
1014
(a)
An increase in T increases the value of Kw
dm6
 the equilibrium position shifts to the right
 the system absorbs heat by shifting to
the right
 the forward reaction is endothermic
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2.(b) In neutral solution,
[H3O+(aq)] = [OH(aq)]
 2.94 1014 mol2 dm6  1.71 107 mol dm3
pH = log10(1.71107) = 6.77
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Example 1
Calculate the pH of
(a) 0.200 M HCl and (b) 0.200 M KOH at 298 K
Given : Kw at 298 K = 1.001014 mol2 dm6
(a) Assumptions :
(i) HCl ionizes completely when dissolved in water.
(ii) [H3O+(aq)] from H2O(l) is negligible.
∵ [HCl(aq)] = 0.200 M
∴ [H3O+(aq)] = 0.200 M assumption (i)
pH = log10[H3O+(aq)] = log100.200
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= 0.699
assumption (ii)
(b) Assumptions :
(i) KOH dissociates completely when dissolved in water.
(ii) [OH(aq)] from H2O(l) is negligible.
∵ [KOH(aq)] = 0.200 M
∴ [OH(aq)] = 0.200 M assumption (i)
At 298 K
Kw  [H3O (aq)][OH (aq)]  1.00 1014 mol2 dm6
14
K
1.00

10
14
w
[H3O  (aq)] 


5.00

10
M

0.200
[OH (aq)]
pH = log10(5.001014) = 13.3
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assumption (ii)
Alternatively,
[H3O (aq)][OH (aq)] = 1.001014 mol2 dm6
log10[H3O+(aq)][OH(aq)] = 14.0
log10[H3O+(aq)] + (log10[OH(aq)]) = 14.0
pH + pOH = 14.0 (at 298 K)
pH = 14.0 – pOH
= 14.0 – (log100.200) = 14.0 – (0.699) = 13.3
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Relationships of [H3O+(aq)] and [OH(aq)] at 298 K
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Q.3
Calculate the pH values of the following solutions
at 298 K.
Given : Kw at 298 K = 1.001014 mol2 dm6
(a) Assumptions :
(i) HCl ionizes completely when dissolved in water.
(ii) [H3O+(aq)] from H2O(l) is negligible.
[H3O+(aq)] = [HCl(aq)] = 2.00 M
pH = log102.00 = 0.301
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(b) Assumptions :
(i) NaOH dissociates completely when dissolved in H2O.
(ii) [OH(aq)] from H2O(l) is negligible.
[OH(aq)] = [NaOH(aq)] = 2.00 M
14
K
1.00

10
15
w
[H3O  (aq)] 


5.00

10
M

2.00
[OH (aq)]
pH = log10(5.0010-15) = 14.3
pH values can be > 14 or < 0
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(c)
[H3O+(aq)] from H2O should be considered
2H2O(l)
OH(aq)
[X(aq)]equil 1
1.0010-7
[X(aq)]  HCl
1.0010-7
[X(aq)]equil 2
(1.0010-7 – x)
+
H3O+(aq)
1.0010-7
1.0010-7 + 1.0010-9
(1.0010-7 + 1.0010-9 – x)
Kw = (1.0010-7 – x)(1.0010-7 + 1.0010-9 – x) = 1.0010-14
x = 5.0010-10 M
or
2.0010-7 M (rejected)
[H3O+(aq)] = (1.0110-7  5.0010-10) = 1.005107 M
 pH = 6.997 < 7 (acidic)
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At very low acid / base concentrations (<106 M),
the [H3O+(aq)] / [OH(aq)] from H2O(l) cannot
be ignored.
At very high acid / base concentrations (>6 M),
the ionization of strong acids / bases is no
longer complete.
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Variation of [H3O+(aq)] with [HCl(aq)]
[H3O+] = 1106 M
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Variation of [OH(aq)] with [NaOH(aq)]
[OH] = 1106 M
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Measurement of pH
TWO ways : 1. Universal indicator / pH paper
2. pH meter
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Universal indicator / pH paper
Universal indicator : a mixture of dyes which shows
different colours at different pH values.
pH paper : paper coated with universal indicator
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Procedures of using universal indicator
Apply a few drops of universal indicator to the
sample.
Apply a few drops of sample to a piece of pH
paper.
This method gives only a
rough estimation of the pH of the sample.
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A pH sensor (probe)
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pH meter
Probe : A glass electrode coupled with a
reference electrode
Glass electrode : The electrode potential varies
linearly with the pH of the
solution in which it is immersed.
At 298 K, The electrode potential is given by
E1 = C – 0.0592pH, where C is a constant
Reference electrode : An electrode with a
fixed electrode potential, E2
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E.g. Ag /AgCl electrode
E.m.f. generated between the two
electrodes,
V = E2 – E1 = C’ + 0.0592pH
the measured e.m.f. can be
converted to pH
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Set-up of the probe
V
Ag/AgCl
wire
Ag/AgCl wire
saturated
KCl(aq)
Porous plug
HCl(aq)
Glass membrane
of the glass
electrode
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Ag/AgCl reference
electrode
Optional
Q.4(a) Calculate the [OH(aq)] in lemon juice of pH 3.0
at 298 K.
(Given : Kw at 298 K = 11014 mol2 dm6)
At 298 K,
pOH = 14.0 – pH = 14.0 – 3.0 = 11.0
[OH(aq)] = 1.00  10pOH
= 1.00  1011 mol dm3
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Q.4(b) Calculate the [OH(aq)] in lime water of pH 12.0
at 298 K.
(Given : Kw at 298 K = 11014 mol2 dm6)
At 298 K,
pOH = 14.0 – pH
= 14.0 – 12.0 =
2.0
[OH(aq)] = 1.00  10pOH
= 1.00  102 mol dm3
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Q.5.(a) Calculate the pH of the solution obtained by mixing
equal volumes of two solutions of HCl of pH 3.0 and
pH 4.0 respectively at 298 K.
(Given : Kw at 298 K = 1.001014 mol2 dm6)
2H2O(l)
H3O+(aq) +
1
[X] right after mixing (103 + 104)
2
= 5.5  104
OH(aq)
1
(1011 + 1010)
2
= 5.5  1011
[H3O+(aq)][OH(aq)] = (5.5  104 )(5.5  1011 ) > Kw
 Equilibrium position shifts to the left
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2H2O(l)
[X] at equilibrium
H3O+(aq) +
5.5104 - x
OH(aq)
5.51011 - x
1.001014 = (5.5104 – x)(5.51011 – x)
Solving the quadratic equation,
x = 3.71011
or 5.5104 (rejected)
pH = -log10 (5.5104 – 3.71011 ) = 3.3
Or pH = -log10 (5.5104 – x)  -log105.5104 = 3.3
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Q.5.(b) Calculate the pH of the solution obtained by mixing
equal volumes of two solutions of NaOH of pH 11.0
and pH 12.0 respectively at 298 K.
(Given : Kw at 298 K = 1.001014 mol2 dm6)
2H2O(l)
H3O+(aq) +
OH(aq)
1
1
11
12
[X] right after mixing
(10 + 10 )
(103 + 102)
2
2
= 5.50  1012
= 5.50  103
[H3O+(aq)][OH(aq)] = (5.50  1012 )(5.50  103 ) > Kw
 Equilibrium position shifts to the left
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2H2O(l)
[X] at equilibrium
H3O+(aq) +
5.501012 - x
OH(aq)
5.50103 - x
1.001014 = (5.501012 – x)(5.50103 – x)
Solving the quadratic equation,
x = 3.681012
or ~5.50103 (rejected)
pH = -log10 (5.501012 – 3.681012 ) = 11.7
Or pH = 14.0 – pOH  14.0 - log105.50103 = 11.7
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Q.5.(c) Calculate the pH of the solution obtained by mixing
equal volumes of two solutions of NaOH of pH 12.0
and HCl of pH 2.0 respectively at 298 K.
(Given : Kw at 298 K = 1.001014 mol2 dm6)
2H2O(l)
H3O+(aq) +
1 2
[X] right after mixing (10 + 1012)
2
= 5.0  103
OH(aq)
1
(1012 + 102)
2
= 5.0  103
[H3O+(aq)][OH(aq)] = (5.0  103 )(5.0  103 ) > Kw
 Equilibrium position shifts to the left
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2H2O(l)
[X] at equilibrium
H3O+(aq) +
5.0103 - x
OH(aq)
5.0103 - x
1.001014 = (5.0103 – x)(5.0103 – x)
Solving the quadratic equation,
x = 4.9999103
pH = -log10 (5.0103 – 4.9999103 ) = 7.0
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Q.5.(d) Calculate the pH of the solution obtained by mixing
equal volumes of two solutions of NaOH of pH 13.0
and HCl of pH 2.0 respectively at 298 K.
(Given : Kw at 298 K = 1.001014 mol2 dm6)
2H2O(l)
H3O+(aq) +
1 2
[X] right after mixing (10 + 1013)
2
= 5.0  103
OH(aq)
1
(1012 + 101)
2
= 5.0  102
[H3O+(aq)][OH(aq)] = (5.0  103 )(5.0  102 ) > Kw
 Equilibrium position shifts to the left
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2H2O(l)
H3O+(aq) +
[X] at equilibrium
5.0103 - x
OH(aq)
5.0102 - x
1.001014 = (5.0103 – x)(5.0102 – x)
Solving the quadratic equation,
x = 5.0103
or 5.0102 (rejected)
pH = 14 – pOH = 14.0 – [-log10 (5.0102 – 5.0103 )]
= 14.0 – 1.35 = 12.7
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Strength of Acids and Bases
Q.6(a) A strong acid is an acid which
dissociates/ionizes completely when
dissolved in water.
Q.6(b) A weak acid is an acid which
dissociates/ionizes slightly when
dissolved in water.
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Equilibrium Positions of Strong Acids Lie to the Right
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Equilibrium Positions of Weak Acids Lie to the Left
Before
dissociation
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After dissociation,
at equilibrium
Q.6(c)
Suggest at least THREE ways to
distinguish between 2M HCl and 2M
CH3COOH.
1. Measure the pH,
2M HCl < 2M CH3COOH
2. Measure the electrical conductivi,
2M HCl > 2M CH3COOH
3. Add Mg / NaHCO3 / …etc.
2M HCl reacts more vigorously
4. Add AgNO3(aq),
only 2M HCl forms white ppt of AgCl
5. Titrate with NaOH using methyl orange
Titre : 2M HCl > 2M CH3COOH
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Q.6(d) Which one has the lower pH value :
(i) 1.0103 M HCl (100% dissociation at 298 K) or
(ii) 1.0 M CH3COOH (0.42% dissociation at 298 K)
(i) [H3O+(aq)] = [HCl(aq)] = 1.0103 M
pH = -log10 1.0103 = 3.0
(ii) [H3O+(aq)] = [CH3COOH(aq)]0.42%
= 4.2103 M
pH = -log10 4.2103 = 2.4 < 3.0
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Comparing Strengths of Acids and Bases
In terms of
1. pH (only for equimolar solutions)
2. the extent of dissociation (i.e. Ka or Kb)
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Consider the equilibrium between an acid
HA and its conjugate base A(aq) : HA(aq) + H2O(l)
H3O+(aq) + A(aq)
[H3O (aq)][A (aq)]
 Ka
[HA(aq)]


dissociation
constant of acid
The numerical values of Ka is a measure
of the extent of dissociation of the acid.
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• large Ka  strong acid.
• small Ka  weak acid.
• Using pKa to compare the strengths of
acids
pKa = log10Ka
Smaller pKa  stronger acid
Larger pKa  weaker acid
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Consider the equilibrium between a base
B and its conjugate base HB+(aq) : B(aq) + H2O(l)
HB+(aq) + OH(aq)
[HB (aq)][OH (aq)]
 Kb
[B(aq)]


dissociation
constant of base
The numerical values of Kb is a measure of
the extent of dissociation of the base.
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• large Kb  strong base.
• small Kb  weak base.
• Using pKb to compare the strengths of
bases
pKb = log10Kb
Smaller pKb  stronger base
Larger pKb  weaker base
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Q.7 (a) Show that for any conjugate acidbase pair, KaKb = Kw
Consider the conjugate acid-base pair, HA / A,
HA(aq) + H2O(l)
H3O+(aq) + A(aq)
[H3O (aq)][A (aq)]
Ka of HA =
[HA(aq)]
A(aq) + H2O(l)
HA(aq) + OH(aq)

[HA(aq)][O
H
(aq)]

Kb of A =
[A (aq)]
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Q.7 (a) Show that for any conjugate acidbase pair, KaKb = Kw
[H3O (aq)][A (aq)]
Ka of HA =
[HA(aq)]

[HA(aq)][O
H
(aq)]

Kb of A =
[A (aq)]
[H3O (aq)[A (aq)] [HA(aq)][OH (aq)]
KaKb 

[HA(aq)]
[A (aq)]
 [H3O (aq)][OH (aq)]  Kw
74
CH3COOH /CH3COO
[H3O  (aq)][CH3COO (aq)]
Ka of CH3COOH =
[CH3COOH(aq)]

[CH
COOH(aq)][
OH
(aq)]
3

Kb of CH3COO =
[CH3COO (aq)]
[H3O (aq)][CH3COO (aq)] [CH3COOH(aq)][OH (aq)]
KaKb 

[CH3COOH(aq)]
[CH3COO (aq)]
 [H3O (aq)][OH (aq)]  Kw
75
NH3 /NH4+

[NH4 (aq)][OH- (aq)]
Kb of NH3 =
[NH3 (aq)]
Ka of NH4+ =
[NH3 (aq)][H3O (aq)]

[NH4 (aq)]

[NH3 (aq)][H3O  (aq)] [NH4 (aq)][OH (aq)]
KaKb 


[NH3 (aq)]
[NH4 (aq)]
 [H3O (aq)][OH (aq)]  Kw
76
Q.7 (b) Show that
Ka of H3O+ at 298 K = 55.5 mol dm3
H3O+(aq) + H2O(l)
H2O(l) + H3O+(aq)

[H
O(l)][H
O
(aq)]
2
3
+
Ka of H3O =
[H3O  (aq)]
= [H2O(l)]
1000 g

77
18 g mol-1
3
1 dm
 55.5 mol dm
-3
Ostwald’s Dilution Law for Weak Acids
and Weak Bases(1888)
Consider the equilibrium between a weak acid HA and
its conjugate base A(aq) : HA(aq) + H2O(l)
[X]initial
[X]equil
C
C(1-)
H3O+(aq)
+
A(aq)
0 (from H2O) 0
C
C
 is the degree of dissociation of HA
78
HA(aq) + H2O(l)
[X]initial
[X]equil
H3O+(aq)
C
+
0 (from H2O) 0
C(1-)
C
C
[H3O (aq)][A (aq)]

Cα 
Cα2
Ka 


[HA(aq)]
C1  α  1  α
2


For weak acids, 1  α  1
Ka  Cα
2
  as C 
79
A(aq)
Ka
α
C
HA(aq) + H2O(l)
[X]initial
[X]equil
C
C(1-)
H3O+(aq)
+
0 (from H2O) 0
C
C
[H3O (aq)][A (aq)]

Cα 
Cα2
Ka 


[HA(aq)]
C1  α  1  α
2
For strong acids,
  100% (if C  6M)
 is independent of C
80
A(aq)
Similarly, for weak bases,
[OH (aq)][HA(aq)]

Cα 
Cα
Kb 



C1  α  1  α
[A (aq)]
2

Kb
α
C
81
2
()
82
H3O+
83
Q.8
You are given two unlabelled bottles each containing
one of the following solutions.
Hydrochloric acid of pH 3; Ethanoic acid of pH 3
Suggest how you would identify the contents of the
two bottles. Tasting the samples and chemical tests
are NOT allowed.
Test for electrical conductivity.
∵ CH3COO ions are less mobile than Cl ions,
CH3COOH(aq) is less conducting than HCl(aq).
84
Q.8
You are given two unlabelled bottles each containing
one of the following solutions.
Hydrochloric acid of pH 3; Ethanoic acid of pH 3
Suggest how you would identify the contents of the
two bottles. Tasting the samples and chemical tests
are NOT allowed.
Dilute the samples with water, followed
by pH measurement.
The pH of the weak acid, CH3COOH, is
less affected by dilution.
85
Q.9 Given that Ka of ethanoic acid at 298 K is 1.74105 mol
dm3, complete Tables 3 and 4 to show that the effect
of dilution on pH is less for ethanoic acid.
*
**
Ka
(%) can be calculated using  
C
C 2
(%) can be calculated using K a 
1
the assumption : 1    1 is not valid
when C is very small
Ka
Or more accurately when
> 103
C
86
i.e.  > 3.2%
Effect of Dilution on pH of Hydrochloric Acid
Table 3
87
[HCl(aq)]
 (%)
[H3O+(aq)]
pH
1.00
100
1.00
0
0.100
100
0.100
1
0.0100
100
0.0100
2
0.00100
100
0.00100
3
Effect of Dilution on pH of Ethanoic Acid
Table 4
88
[CH3COOH]
 (%)
[H3O+]
pH
*1.00
0.417
4.17103
2.38
*0.100
1.32
1.32103
2.88
**0.0100
4.09
4.09104
3.39
**0.00100
12.3
1.23104
3.91
Q.10 Given that Kb of ammonia at 298 K is 1.74105 mol
dm3, complete Tables 5 and 6 to show that the
effect of dilution on pH is less for ammonia.
*
**
Kb
(%) can be calculated using  
C
C 2
(%) can be calculated using K b 
1
the assumption : 1    1 is not valid
when C is very small
Kb
Or more accurately when
> 10-3
C
89
i.e.  > 3.2%
Effect of Dilution on pH of NaOH(aq) (Table 5)
90
[NaOH(aq)]
(%)
[OH(aq)]
pOH
pH
1.00
100
1.00
0
14
0.100
100
0.100
1
13
0.0100
100
0.0100
2
12
0.00100
100
0.00100
3
11
Effect of Dilution on pH of NH3(aq) (Table 6)
91
[NH3(aq)]
(%)
[OH(aq)]
pOH
pH
*1.00
0.417
4.17103
2.38
11.6
*0.100
1.32
1.32103
2.88
11.1
**0.0100
4.09
4.09104
3.39
10.6
**0.00100
12.3
1.23104
3.91
10.1
Conclusion : The effects of dilution on pH are
less for weak acids and bases
92
Calculations involving pH, Ka and Kb
A. Given Ka or Kb , calculate the pH values of weak
acids or weak bases
(refer to Tables 4 & 6)
B. Given the pH values of weak acids or weak
bases, calculate Ka or Kb
(refer to examples 1 & 2)
93
Example 1 : Calculate the Ka of 0.100 M HOCl of pH 4.23
at 298 K.
ClO(aq) + H3O+(aq)
HOCl(aq) + H2O(l)
[X]equil 0.100 – x
x
x
x  [H3O+(aq)] = 110-pH
= 110-4.23 = 5.89105 M


-5 2
x
5.89 10
8
-3
Ka 


3.47

10
mol
dm
0.100  x 0.100  5.89 10-5
2
94
Q.11(a) : Calculate the pH value of 0.100 M HOCl(aq)
at 298 K. (Ka of HOCl = 3.47108 mol dm3)
HOCl(aq) + H2O(l)
[X]equil 0.100 – x
ClO(aq) + H3O+(aq)
x
x
Ka
3.47  108
4
α

 5.89 10
C
0.100
[H3O+(aq)] from HOCl
= 0.100 M  5.89104
= 5.89105 M > 1106 M
95
H3O+ from water is
negligible
Q.11(a) : Calculate the pH value of 0.100 M HOCl(aq)
at 298 K. (Ka of HOCl = 3.47108 mol dm3)
HOCl(aq) + H2O(l)
[X]equil 0.100 – x
ClO(aq) + H3O+(aq)
x
x
2
x
x2
8
-3
Ka  3.47  10 mol dm 

0.100  x
0.100
x = 5.89105
pH = 4.23
96
Ka
  1  10 3
C
Example 2 : Calculate the Kb of 0.010 M HS of pH 9.5
at 298 K.
HS(aq) + H2O(l)
H2S(aq) + OH(aq)
[X]equil 0.010 – x
x
x
x  [OH(aq)] = 1  10-pOH = 1  10-(14 -pH)
= 1  10-(14-9.5)
= 3.2105 M


5 2
x
3.2  10
7
-3
Kb 


1.0

10
mol
dm
0.010  x 0.010  3.2  105
2
97
Q.11(b) : Calculate the pH value of 0.010 M HS(aq)
at 298 K. (Kb of HS = 1.0107 mol dm3)
HS(aq) + H2O(l)
[X]equil 0.010 – x
H2S(aq) + OH(aq)
x
x
Kb
1.0  107
-3
α

 3.16 10
C
0.010
[OH(aq)] from HS
= 0.010 M  3.16103
= 3.16105 M > 1106 M
98
OH from water is
negligible
Q.11(b) : Calculate the pH value of 0.010 M HS(aq)
at 298 K. (Kb of HS = 1.0107 mol dm3)
HS(aq) + H2O(l)
[X]equil 0.010 – x
Kb  1.0  10 7
x
x
2
x
x2
-3
mol dm 

0.010  x 0.010
x = 3.2105
pOH = 4.5
pH = 9.5
99
H2S(aq) + OH(aq)
Kb
  1  10 3
C
Q.11(c) : Calculate the pH value of 1.0105 M HCN(aq)
at 298 K. (Ka of HCN = 4.01010 mol dm3)
HCN(aq) + H2O(l)
[X]equil 1.0105 – x
CN(aq) + H3O+(aq)
x
x + 1.00107
Ka
4.0 1010
-3
α

 6.3 10
-5
C
1.0 10
[H3O+(aq)] from HCN
= 1.0105 M  6.3103
= 6.3108 M < 1106 M
100
H3O+ from water
cannot be ignored
Q.11(c) : Calculate the pH value of 1.0105 M HCN(aq)
at 298 K. (Ka of HCN = 4.01010 mol dm3)
HCN(aq) + H2O(l)
[X]equil 1.0105 – x
Ka  4.0 1010
[H3O+(aq)]equil = 1.3107 M
101
x
x + 1.00107
x(x  1.00 10-7 ) x(x  1.00 10-7 )


-5
1.0  10  x
1.0  10-5
x = 3.1108
pH = 6.9
CN(aq) + H3O+(aq)
Kb
  1  10 3
C
Q.12 Calculate the pH values of the following salt
solutions at 298 K.
(a) 0.10 M NaOCl (Ka of HOCl = 3.5108 mol dm3)
(b) 0.010 M NH4Cl (Kb of NH3 = 1.8105 mol dm3)
(c) 0.010 M CH3COONH4
(Ka of CH3COOH = Kb of NH3 = 1.8105 mol dm3)
Kw = 1.001014 mol2 dm6 at 298 K
A salt solution may be considered as an
acid, a base or a mixture of an acid and
a base.
102
Q.12(a)
NaOCl(aq)  Na+(aq) + ClO(aq)
(complete dissociation)
Na+(aq) is neither an acid nor a base
103
Q.12(a)
NaOCl(aq)  Na+(aq) + ClO(aq)
(complete dissociation)
Since HClO is a weak acid, ClO is a ‘strong’ base
ClO(aq) + H2O(l)
HClO(aq) + OH(aq)
[X]equil 0.10 – x
x
Kw 1.00  10

Kb 
8
Ka
3.5  10
14
104
x
Q.12(a)
NaOCl(aq)  Na+(aq) + ClO(aq)
(complete dissociation)
Since HClO is a weak acid, ClO is a strong base
ClO(aq) + H2O(l)
[X]equil 0.10 – x
HClO(aq) + OH(aq)
x
x
x2
Kw 1.00  1014

Kb 

8
Ka
3.5  10
0.10  x
=
105
2.9107
2.9  10-7
 0.10 
 106
0.10
2.9  10
7
x

0.10  x
2
Kb 2.9  10
 
C
0.10
2.9  10
7
7
 2.9  10
 1.0  10
x
x


0.10  x 0.10
2
pOH  log101.7  10
pH = 10.2
3
2
4
106
6
 3.8
x = 1.7104
Q.12(b)
NH4Cl(aq)  NH4+(aq) + Cl(aq)
(complete dissociation)
Since HCl is a very strong acid,
Cl(aq) is a very weak base.
The following equilibrium can be ignored.
Cl(aq) + H2O(l)
107
HCl(aq) + OH(aq)
Q.12(b)
NH4Cl(aq)  NH4+(aq) + Cl(aq)
(complete dissociation)
Since NH3 is a weak base, NH4+ is a ‘strong’ acid
NH4+(aq)
NH3(aq) + H3O+(aq)
+ H2O(l)
[X]equil 0.010 – x
x
1.00  10
Kw

Ka 
5
Kb
1.8  10
14
108
x
Q.12(b)
NH4Cl(aq)  NH4+(aq) + Cl(aq)
(complete dissociation)
Since NH3 is a weak base, NH4+ is a strong acid
NH4+(aq)
+ H2O(l)
NH3(aq) + H3O+(aq)
[X]equil 0.010 – x
x
x
x
Kw 1.00  1014


Ka 
5
Kb
1.8  10
0.010  x
2
= 5.61010
109
5.6  10-10
 0.010
 106
0.010
5.6  10
10
x

0.010  x
2
Ka 5.6  10
 
C
0.010
10
5.6  10
10
 5.6  10
 1.0  10
x2
x2


0.010  x 0.010
x  2.4  10
6
pH  log10 2.4  10
6
110
8
= 5.6
3
Q.12(c)
CH3COONH4(aq)  CH3COO(aq) + NH4+(aq)
(complete dissociation)
Since CH3COOH is a weak acid, CH3COO is a
‘strong’ base
Since NH3 is a weak base, NH4+ is a ‘strong’ acid
NH4+(aq)
+
H2O(l)
NH3(aq)
+
H3O+(aq)
CH3COO is a stronger base than H2O
NH4+ (aq) + CH3COO(aq)
111
NH3(aq) + CH3COOH(aq)
Q.12(c)
CH3COONH4(aq)  CH3COO(aq) + NH4+(aq)
(complete dissociation)
Since CH3COOH is a weak acid, CH3COO is a
‘strong’ base
Since NH3 is a weak base, NH4+ is a ‘strong’ acid
CH3COO(aq) +
H2O(l)
CH3COOH(aq) + OH(aq)
NH4+ is a stronger acid than H2O
CH3COO(aq) + NH4+ (aq)
112
CH3COOH(aq) + NH3(aq)
Q.12(c)
CH3COONH4(aq)  CH3COO(aq) + NH4+(aq)
(complete dissociation)
CH3COO(aq) + NH4+ (aq)
[X]equil 0.10 – x
Kc 

113
0.10 – x
CH3COOH(aq) + NH3(aq)
x
x
[CH3COOH(aq)][NH3 (aq)]

[CH3COO (aq)][NH4 (aq)]
[CH3COOH(aq)]
[NH3 (aq)]




[H
O
(aq)][OH
(aq)]
3



[CH3COO (aq)][H3O (aq)] [NH4 (aq)][OH (aq)]
 1  1 
     Kw
 Ka  Kb 
Kw

KaKb
Q.12(c)
CH3COO(aq) + NH4+ (aq)
[X]equil 0.10 – x
Kc 
CH3COOH(aq) + NH3(aq)
0.10 – x
x
x
[CH3COOH(aq)][NH3 (aq)]

[CH3COO (aq)][NH4 (aq)]

x


Kw
1.00  10
5




-5 2 = 3.110
(1.8  10 )
KaKb
 0.10  x 
-14
x = 5.5104
114
2
x = 5.5104
Q.12(c)
NH4+(aq)
+
H2O(l)
[X]equil 0.10 – x
NH3(aq)
x
+
H3O+(aq)
y
xy
Kw
1.00  10-14
10

Ka 
= 5.610 
-5
0.10  x
1.8  10
Kb
5.61010
(5.5  10-4 )y

-4
0.10  5.5  10
y = 1.0107
115
pH = 7.0
Q.12(c)
x = 5.5104
CH3COO(aq) +
H2O(l)
[X]equil 0.10 – x
CH3COOH(aq) + OH(aq)
x
z
Kw
1.00  10-14
xz
10

Kb 
= 5.610 
-5
0.10  x
1.8  10
Ka
5.61010
(5.5  10-4 )z

-4
0.10  5.5  10
z = 1.0107
116
pOH = 7.0
pH = 7.0
Q.13 Given that the Ka of an acid HA is 2.1  104 mol
dm3, find the change in pH when the degree of
dissociation of HA is increased from 0.10 to 0.20.
HA(aq) + H2O(l)
[X]equil
C(1-)
H3O+(aq)
+
C
A(aq)
C
∵ HA is a weak acid,
 in  from 0.10 to 0.20 is caused by dilution
On dilution,
1. C of HA decreases
2.  of HA increases
117
Net effect : C  and pH 
Q.13 Given that the Ka of an acid HA is 2.1  104 mol
dm3, find the change in pH when the degree of
dissociation of HA is increased from 0.10 to 0.20.
HA(aq) + H2O(l)
[X]equil
C(1-)
Cα 2
Ka 
1α
H3O+(aq)
C
+
A(aq)
C
[H3O (aq)]  Cα  Ka (1  α)/α
When  = 0.1, [H3O+(aq)] = C = Ka(1-0.1)/0.1
pH = 2.7
When  = 0.2, [H3O+(aq)] = C = Ka(1-0.2)/0.2 pH = 3.1
Change in pH = 3.1 – 2.7 = +0.4
118
Q.14(a)(i)
For HSO4, Ka > Kb
HSO4 acts as an acid rather than a base.
HSO4(aq) + H2O(l)
[X]equil 0.10 – x
SO42(aq) + H3O+(aq)
x
x
2
Ka 
 1.2  10
0.10  x
x
2
1.2  10-2
 0.10 
 106
0.10
x = [H3O+(aq)]equil = 0.029
119
pH = 1.5
Q.14(a)(ii)
For HCO3, Kb > Ka
2.4  10-8
 0.10 
 106
0.10
HCO3 acts as a base rather than an acid.
HCO3(aq) + H2O(l)
[X]equil
H2CO3(aq) + OH(aq)
0.10 – x
Kb  2.4  10
x
8
x
x2


0.10  x
0.10
2
x = [OH(aq)]equil = 4.9105
120
pOH = 4.3
pH = 9.7
x
Kb
  1  10 3
C
Q.14(a)(iii)
6.2  10-8
 0.10 
 106
0.10
For HSO3, Ka > Kb
HSO3 acts as an acid rather than a base
HSO3(aq) + H2O(l)
[X]equil 0.10 – x
Ka  6.2  10
SO32(aq) + H3O+(aq)
x
8
x2
x2


0.10  x 0.10
x
Ka

 1  10 3
C
x = [H3O+(aq)]equil = 7.9105 pH = 4.1
121
Q.14(b)
The pH of an acid salt solution can be > 7, < 7
or = 7 at 298 K, depending on the relative
values of Ka and Kb of the same species.
If Ka > Kb, the solution is acidic with pH < 7;
If Ka < Kb, the solution is basic with pH > 7;
If Ka = Kb, the solution is neutral with pH = 7;
122
Buffer Solutions
Definition : A buffer solution is one that
resists changes in pH upon dilution or
addition of a small amount of acid or base.
123
Types of Buffers
Acidic buffers : A mixture of a weak acid and its salt of
a strong base,
e.g. CH3COOH / CH3COONa
pH < 7 at 298 K
Or, it is a mixture of a weak acid and its
conjugate base
CH3COOH / CH3COO
124
Types of Buffers
Basic buffers : A mixture of a weak base and its salt of a
strong acid,
e.g. NH3 / NH4Cl
pH > 7 at 298 K
Or, it is a mixture of a weak base and its
conjugate acid
NH3 / NH4+
125
A buffer is essentially a solution
containing a conjugate acid-base pair.
126
Q.15(a) Give THREE examples for each type
of buffers mentioned above.
Acidic Buffers : HOCl(aq) / OCl(aq)
H3PO4 (aq) / H2PO4(aq)
HSO4 (aq) / SO42(aq)
127
Q.15(a) Give THREE examples for each type
of buffers mentioned above.
Acidic Buffers : -
HOCl(aq) / NaOCl(aq)
H3PO4 (aq) / NaH2PO4(aq)
NaHSO4 (aq) / Na2SO4(aq)
128
Q.15(a) Give THREE examples for each type of
buffers mentioned above.
Basic Buffers : SO32(aq) / HSO3(aq)
CO32(aq) / HCO3(aq)
PO43 (aq) / HPO42(aq)
129
Q.15(a) Give THREE examples for each type of
buffers mentioned above.
Basic Buffers : Na2SO3(aq) / NaHSO3(aq)
Na2CO3(aq) / NaHCO3(aq)
Na3PO4 (aq) / Na2HPO4(aq)
130
Q.15(b) Why is HSO3(aq) / H2SO3(aq)
NOT a basic buffer ?
This is an acidic buffer.
It is because Ka of H2SO3 > Kb of HSO3
131
Interpretation of Buffer Action
- A Qualitative Approach
Consider an acidic buffer HA(aq) / NaA(aq)
HA(aq) + H2O(l)
Ka 
A(aq) + H3O+(aq)
[A (aq)]equil [H3O  (aq)]equil
[HA(aq)]equil
[H3O (aq)]equil  Ka

pH  pK a - log10
132
[HA(aq)] equil
[A (aq)]equil
[HA(aq)] equil
[A (aq)]equil
pH  pK a - log10
[HA(aq)] equil
[A (aq)]equil
HA(aq) + H2O(l)
A(aq) + H3O+(aq)
Assumption 1 : In the presence of a large amount of
A(aq), the ionization HA(aq) can be
ignored.
[HA(aq)]equil  [HA(aq)]initial
[HA(aq)]initial
pH  pKa - log10
[A (aq)]equil
133
pH  pKa - log10
[HA(aq)]initial
[A (aq)]equil
Assumption 2 : In the presence of a large amount of
HA, the basic property of A can be
ignored.
A(aq) + H2O(l)
HA(aq) + OH(aq)
[A(aq)]initial  [A(aq)]equil
[HA(aq)]initial
pH  pKa - log10
[A (aq)]initial
134
[HA(aq)]initial
pH  pKa - log10
[A (aq)]initial
Assumption 3 :
The salt, NaA(aq), undergoes
complete ionization in water.
[NaA(aq)]initial  [A(aq)]initial
[HA(aq)]initial
[Acid]initial
pH  pKa  log10
 pKa  log10
[NaA(aq)]initial
[Salt]initial
135
Effect of Dilution on pH of an Acidic Buffer
[Acid]initial
pH  pKa  log10
[Salt]initial
On dilution, both [Acid]initial and [Salt]initial decrease
to the same extent.
[Acid]initial
remains unchanged
[Salt]initial
pH of the system remains unchanged
136
Q.15 Give ONE assumption made in the above argument.
Comment on the validity of the assumption.
The weak acid does not ionize on
dilution.
The assumption is valid in the
presence of an excess of A.
137
Effect of Addition of a Small Amount of
H3O+(aq)/ OH(aq) on pH of an Acidic Buffer
HA(aq) + H2O(l)
A(aq) + H3O+(aq)
excess
Addition of H3O+(aq) : -
The equilibrium position shifts to the left
138
Effect of Addition of a Small Amount of
H3O+(aq)/ OH(aq) on pH of an Acidic Buffer
HA(aq) + H2O(l)
A(aq) + H3O+(aq)
excess
Addition of H3O+(aq) : -
A(aq) which is in large excess compared with the
amount of H3O+(aq) added tends to act as a base to
convert H3O+(aq) to H2O(l) which do not ionize easily.
139
Effect of Addition of a Small Amount of
H3O+(aq)/ OH(aq) on pH of an Acidic Buffer
HA(aq) + H2O(l)
A(aq) + H3O+(aq)
excess
Addition of H3O+(aq) : -
Since [HA(aq)] and [A(aq)] >> [H3O+(aq)]added
[HA(aq)] equil 1
[A (aq)]equil 1

140

[HA(aq)] equil 2
[A (aq)]equil 2

∴ pHequil 1  pHequil 2
Effect of Addition of a Small Amount of
H3O+(aq)/ OH(aq) on pH of an Acidic Buffer
HA(aq) + H2O(l)
A(aq) + H3O+(aq)
Addition of OH(aq) : -
HA(aq) which is in large excess compared with the
amount of OH(aq) added tends to act as an acid to
convert OH(aq) to H2O(l) which does not ionize easily.
HA(aq) + OH(aq)
141
A(aq) + H2O(l)
Effect of Addition of a Small Amount of
H3O+(aq)/ OH(aq) on pH of an Acidic Buffer
HA(aq) + H2O(l)
A(aq) + H3O+(aq)
Addition of OH(aq) : Alternatively, H3O+(aq), tends to convert OH(aq) to
H2O(l) which does not ionize easily.
The equilibrium position shifts to the right to
replenish H3O+(aq)
142
Effect of Addition of a Small Amount of
H3O+(aq)/ OH(aq) on pH of an Acidic Buffer
HA(aq) + H2O(l)
A(aq) + H3O+(aq)
Addition of OH(aq) : Since [HA(aq)] and [A(aq)] >> [OH(aq)]added
[HA(aq)] equil 1
[A (aq)]equil 1

143

[HA(aq)] equil 2
[A (aq)]equil 2

∴ pHequil 1  pHequil 2
A buffer is essentially a solution
containing significant amounts of a
conjugate acid-base pair.
144
Similarly, for a basic buffer prepared by mixing a weak
base, B(aq), with its conjugate acid, BH+(aq),
B(aq)+ H2O(l)
Kb 
BH+(aq) + OH(aq)
[BH (aq)]equil [OH (aq)]equil
[B(aq)]equil
[OH (aq)]equil  Kb

pOH  pK b - log10
[B(aq)] equil
[BH  (aq)]equil
[B(aq)] equil
[BH  (aq)]equil
[Bass]initial
pOH  pKb - log10
[Salt]initial
145
Effect of Dilution on pH of a Basic Buffer
On dilution, both [B(aq)]initial and [BH+(aq)]initial
decrease to the same extent.
[Base]initial
remains unchanged
[Salt]initial
pH of the system remains unchanged
146
Effect of Addition of a Small Amount of
H3O+(aq)/ OH(aq) on pH of a Basic Buffer
B(aq)+ H2O(l)
BH+(aq) + OH(aq)
Addition of H3O+(aq) : -
B(aq) which is in large excess compared with the
amount of H3O+(aq) added tends to act as a base to
convert H3O+(aq) to H2O(l) which does not ionize
easily.
B(aq) + H3O+(aq)
147
BH+(aq) + H2O(l)
Effect of Addition of a Small Amount of
H3O+(aq)/ OH(aq) on pH of a Basic Buffer
B(aq)+ H2O(l)
BH+(aq) + OH(aq)
Addition of H3O+(aq) : Alternatively, OH(aq), tends to remove H3O+ (aq) to
give H2O(l) which does not ionize easily.
The equilibrium position shifts to the right to
replenish OH(aq)
148
Effect of Addition of a Small Amount of
H3O+(aq)/ OH(aq) on pH of a Basic Buffer
B(aq)+ H2O(l)
Addition of OH(aq) : -
BH+(aq) + OH(aq)
excess
The equilibrium position shifts to the left
149
Effect of Addition of a Small Amount of
H3O+(aq)/ OH(aq) on pH of a Basic Buffer
B(aq)+ H2O(l)
Addition of OH(aq) : -
BH+(aq) + OH(aq)
excess
BH+(aq) which is in large excess compared with the
amount of OH(aq) added tends to act as an acid to
convert OH(aq) to H2O(l) which does not ionize easily.
150
Effect of Addition of a Small Amount of
H3O+(aq)/ OH(aq) on pH of a Basic Buffer
B(aq)+ H2O(l)
Addition of OH(aq) : -
BH+(aq) + OH(aq)
excess
Since [B(aq)] and [BH+(aq)] >> [OH(aq)]added
[B(aq)] equil 1
[BH (aq)]equil 1

151

[B(aq)] equil 2
[BH (aq)]equil 2

∴ pOHequil 1  pOHequil 2
Illustration of Buffer Action – A Quantitative Approach
Example
Calculate the change in pH when 0.00100 mole
of solid NaOH is added to each of the following
systems. Assume that there is no volume
change of solution upon addition of NaOH(s).
(a)
1.0 dm3 of pure water
(b)
1.0 dm3 of an aqueous solution of 0.10 M CH3COOH
(c)
1.0 dm3 of an aqueous solution of 0.10 M CH3COONa
(d)
1.0 dm3 of an aqueous solution of 0.10 M CH3COOH
and 0.10 M CH3COONa
Given : Ka of CH3COOH at 298 K = 1.8  105 mol dm3
Kw of water at 298 K = 1.00  1014 mol2 dm6
152
(a)
pH of pure water = 7.0 at 298 K
H2O(l)
+ H2O(l)
H3O+(aq)
+
[X]equil 1
1.00  107
1.00  107
[X]addition of NaOH
1.00  107
[X]equil 2
1.00  107 – x 1.00  107 + 0.00100 – x
1.00  107 + 0.00100
pH = 14.0 – pOH
= 14.0 – (-log100.00100) = 11.0
Change in pH = 11.0 – 7.0 = +4.0
153
OH(aq)
 0.00100
Alternatively, Kw = (1.00  107 – x)( 1.00  107 + 0.00100 – x)
= 1.00  1014
x = 9.999  108 or 1.000  103 (rejected)
[H3O+(aq)]equil 2 = 1.00  107 – 9.999  108
pH = 11.0
Change in pH = 11.0 – 7.0 = +4.0
154
(b)
H2O(l) + CH3COOH(aq)
[X]equil 1
CH3COO(aq) +
0.10 – x
[X]addition of NaOH 0.10 – x – 0.00100
x
x + 0.00100
H3O+(aq)
x
x
(complete removal of OH)
[X]equil 2
155
0.10 – x – 0.00100 + y x + 0.00100 – y
x–y
= 0.099 – (x – y)
= 0.00100 + (x –y) x – y
= 0.099 – z
= 0.001 + z
z
(b)
H2O(l) + CH3COOH(aq)
[X]equil 1
CH3COO(aq) +
0.10 – x
H3O+(aq)
x
x
At equilibrium 1,
2
x
5
Ka 
 1.8 10
0.10  x
x = 1.3  103 or 1.4  103 (rejected)
pH = log10x = log10(1.3  103) = 2.9
156
(b)
H2O(l) + CH3COOH(aq)
[X]equil
0.099 – z
CH3COO(aq) +
0.00100 + z
At equilibrium 2,
(0.00100  z ) z
5
Ka 
 1.8  10
0.099  z
z = 9.2  104 or 1.9 103 (rejected)
pH   log10 z  log10 (9.2 104 )  3.0
Change in pH = 3.0 – 2.9 = +0.10
157
H3O+(aq)
z
(c)
CH3COONa(aq)  CH3COO(aq) +Na+(aq)
(Complete ionization)
[CH3COO(aq)]initial = 0.10 M
158
H2O(l ) + CH3COO(aq)
CH3COOH (aq) +
[X]equil 1 0.10 – x
x
Kw 1.00  10
Kb 

Ka
1.8  105
14
5.6  10
10
Kb

 1  10 3
0.10
x = 7.5  106
159
x
= 5.61010
x2
x2


0.10  x 0.10
pOH = 5.1
OH(aq)
pH = 8.9
H2O(l ) + CH3COO(aq)
CH3COOH (aq) + OH(aq)
[X]addition of NaOH 0.10 – x
[X]equil 2
x
0.10 – x + y
0.00100 + x
x–y
0.00100 + x – y
z
0.00100 + z
= 0.10 – (x – y) = 0.10 – z
x is very small and z = x – y  0, z is very small
160

[OH(aq)] = 0.00100 + z  0.00100

pOH = 3.0 

Change in pH = 11.0 – 8.9 = +2.1
pH = 11.0
(d)
H2O(l ) + CH3COO(aq)
CH3COOH (aq) + OH(aq)
The basic property of CH3COO is ignored because
(i)
Kb of CH3COO << Ka of CH3COOH
(ii)
CH3COOH is in large amount
p.18, assumption 2
161
Complete ionization of CH3COONa
(p.18, Assumption 3)
(d)
H2O(l) + CH3COOH(aq)
[X]initial
0.10
CH3COO(aq)
0.10
+
H3O+(aq)
x=1.00107
(0.10)(1.00107 )
Q
 1.00107  K a of CH 3COOH
0.10
Equilibrium position shifts to the right
[X]equil 1
0.10 – y
z (0.1  y ) 0.1z
Ka 

z
0.1  y
0.1
162
0.10 + y
x+y=z
K a 1.8 105


 1103
C
0.1
(d)
H2O(l) + CH3COOH(aq)
[X]equil 1
0.10 – y
z (0.1  y ) 0.1z
Ka 

z
0.1  y
0.1
CH3COO(aq)
0.10 + y
H3O+(aq)
x+y=z
p.18, Assumption 1
log10Ka = log10z = pH
pH = log101.8105 = 4.745
163
+
H2O(l)
+
CH3COOH(aq)
[X]addition of NaOH 0.10 – y – 0.00100
CH3COO(aq) + H3O+(aq)
0.10 + y + 0.00100
z
(complete removal of OH)
[X]equil 2
0.099 – y + w
0.101 + y – w
z–w
 z is very small, y=z–x & z–w>0  both w & y are very small
( z  w )( 0.101 y  w ) ( z  w )0.101
5
 Ka 

 1.8  10
0.099  y  w
0.099
164

z – w = [H3O+(aq)] = 1.76  105 M

pH = 4.753

Change in pH = 4.753 – 4.745 = +0.008
The effects of addition of 0.001 mole NaOH(s) on the pH values
165
Pure
water
0.1M
CH3COOH
0.1 M
CH3COONa
0.1M
CH3COOH +
0.1 M
CH3COONa
Final pH
11.0
3.0
11.0
4.753
Initial
pH
7.0
2.9
8.9
4.745
Change
in pH
+4.0
+0.1
+2.1
+0.008
Q.17 Calculate the change in pH when 0.00100 mole of HCl is
added to 1.0 dm3 of a mixture containing 0.10 M
CH3COOH and 0.10 M CH3COONa. Assume that there is
no volume change of the solution upon addition of HCl.
Given : Ka of CH3COOH at 298 K = 1.8  105 mol dm3
Kw of water at 298 K = 1.00  1014 mol2 dm6
166
H2O(l) + CH3COOH(aq)
[X]initial
0.10
[X]equil 1
0.10 – y
[X]addition of HCl 0.10 – y + 0.00100
[X]equil 2
167
0.101 – y – w
CH3COO(aq)
+ H3O+(aq)
0.10
x
0.10 + y
0.1 + y – 0.00100
0.099 + y + w
x+y=z
z
z+w
CH3COO(aq)
H2O(l) + CH3COOH(aq)
[X]equil 1
0.10 – y
0.10 + y
 0.10
 0.10
+ H3O+(aq)
z
p.18, Assumption 1
At equilibrium 1
0.10 z
Ka 
z
0.10
pH   log10 z   log10 Ka   log10 1.8 105  4.745
168
H2O(l) + CH3COOH(aq)
[X]equil 2
0.101 – (y + w)
CH3COO(aq)
+ H3O+(aq)
0.099 + (y + w)
z+w
At equilibrium 2, (y + w) is very small (Assumption 1)
(z  w)(0.099 y  w) (z  w)0.099
Ka 

 1.8  10 5
0.101  y  w
0.101
z + w = [H3O+(aq)] = 1.84  105 M
pH = 4.735
Change in pH = 4.735 – 4.745 = -0.01
169
Q.18 (a) (i)
CH3COOH(aq) + H2O(l)
[X]initial
[X]equil 1
0.10
0.10 – x  0.10
CH3COO (aq) + H3O+(aq)
0.0050
0.0050 + x
Ka 1.8  10 5

 1  10 3
C
0.10
Ka
1.8  105
C
 (0.10)
 1.3  103  1  106
C
0.10
170
0
x
Q.18 (a) (i)
CH3COOH(aq) + H2O(l)
[X]initial
[X]equil 1
0.10
0.10 – x  0.10
CH3COO (aq) + H3O+(aq)
0.0050
0.0050 + x
0
x
Since [X]initial = 0.10,
x cannot be ignored when compared with 0.0050
0.10
0.10
 pK a  log10
pH  pK a  log10
0.0050
0.0050  x
pH cannot be calculated directly using the formula
171
Q.18 (a) (i)
CH3COOH(aq) + H2O(l)
[X]initial
[X]equil 1
0.10
0.10 – x  0.10
x(0.0050 x)
Ka 
0.10
x = 3.4  10-4
pH = 3.47
172
CH3COO (aq) + H3O+(aq)
0.0050
0.0050 + x
0
x
CH3COOH(aq) + H2O(l)
CH3COO (aq) + H3O+(aq)
[X]equil 1
0.10
5.34103
3.4 104
[X]+NaOH
0.099
6.34103
3.4 104
[X]equil 2
0.099 + y
6.34103 – y
3.4 104 – y
 0.099
6.34103 – y
3.4 104 – y
3
4
(6.3410  y )(3.4 10  y )
Ka 
0.099
y = 5.6105, [H3O+(aq)] = 3.4 104 – 5.6105  pH = 3.55
173
Change in pH
= 3.55 – 3.47
= +0.08
174
Equilibrium 1
0.10
pH  pK a  log10
 3.47
3
5.34 10
Equilibrium 2
0.099
pH  pKa  log10
 3.55
-3
6.3410
If [Salt] is small,
The ratio [Acid]/[Salt] is affected more upon
addition of acid/base.
175
Q.18(a) (ii)
Ka
1.8  105
C
 (0.0050)
 3  104  1  106
C
0.0050
CH3COOH(aq) + H2O(l)
CH3COO (aq) + H3O+(aq)
[X]initial
0.0050
0.0050
[X]equil 1
0.0050 – x
0.0050
0.0050 + x
0.0050
0
x
Ka 1.8  105

 3.6  103  1  103
C
0.0050
However, in the presence of 0.0050 M CH3COO,
the dissociation of CH3COOH can be ignored,
176
Q.18(a) (ii)
CH3COOH(aq) + H2O(l)
[X]initial
0.0050
[X]equil 1
0.0050 – x  0.0050
0.0050x
Ka =
0.0050
CH3COO (aq) + H3O+(aq)
0.0050
0.0050 + x  0.0050
0
x
x = 1.8  10-5
pH = log101.8  105 = 4.7
Or
177
0.0050
pH  pK a  log10
 pK a = 4.7
0.0050
CH3COOH(aq) + H2O(l)
[X]equil 1
0.0050
[X]+NaOH
0.0050 – 0.0010
= 0.0040
[X]equil 2
0.0040 + y
 0.0040
CH3COO(aq) + H3O+(aq)
0.0050
x 0
0.0050 + 0.0010
= 0.0060
x 0
0.0060 – y
x – y0
 0.0060
x–y=z
Since x is very small, y must be very small
178
CH3COOH(aq) + H2O(l)
[X]equil 2
0.0040
Ka  1.8 10
0.0060
5
(0.0060)z

0.0040
z = 1.2105
pH = 4.9
179
CH3COO(aq) + H3O+(aq)
z
CH3COOH(aq) + H2O(l)
[X]equil 2
0.0040
CH3COO(aq) + H3O+(aq)
0.0060
z
Alternatively,
[Acid]
pH  pKa  log10
[Salt]
0.004
 log10 1.8 10  log10
0.006
5
= 4.9
180
Change in pH
= 4.9 – 4.7
= 0.2
181
Equilibrium 1
0.0050
pH  pK a  log10
 pK a  4.7
0.0050
Equilibrium 2
0.004
pH   log10 1.8 10  log10
 4.9
0.006
5
If both [Acid] and [Salt] are small,
The ratio [Acid]/[Salt] is affected further more
upon addition of acid/base.
182
183
0.1M
CH3COOH/0.1M
CH3COONa
0.1M
CH3COOH/0.005
M CH3COONa
0.005M
CH3COOH/0.005
M CH3COONa
Final pH
4.753
3.55
4.9
Initial pH
4.745
3.47
4.7
Change in pH
+0.008
+0.08
+0.2
Criteria necessary for an effective buffer
From 18(a)(i), effectiveness of buffer  when there
is a great difference between [Acid(aq)] and
[Salt(aq)] ,
For an effective buffer,
1. [Acid(aq)] and [Salt(aq)] should be
comparable
[Acid] [Base]
10
1
or
should be within the range of
to
[Salt]
[Salt]
1
10
Refer to ‘Range of Buffer’
184
Criteria necessary for an effective buffer
• From 18(a)(ii), effectiveness of buffer drops
further when both [Acid(aq)] and [Salt(aq)]
are small.
[Acid]
10
1
although
is within the range of
to
[Salt]
1
10
For an effective buffer,
2. [Acid(aq)] and [Salt(aq)] should be high
enough to remove the acid/base added.
Refer to ‘Buffering capacity’
185
The Range of Buffer
Definition : The pH region within which the buffer is
effective as a buffer.
186
For an effective acidic buffer,
pH = pKa  1
For an effective basic buffer,
pOH = pKb  1
pH = 14 - pKb  1
187
A buffer system with a pH value
out of this range would have
either insufficient amount of
acid, HA(aq), or its conjugate base,
A(aq), or both to deal with the
base or acid added to the system.
188
Range of
buffer
pKa = 4.5
 pKa 1
10%
189
90%
[ HA(aq)]
At 10% neutralization, pH = pKa  log10 
[ A (aq)]
= pKa  log10
0 .9
= pKa  0.95
0 .1
[ HA(aq)]
At 90% neutralization, pH = pKa  log10 
[ A (aq)]
= pKa  log10
190
0 .1
= pKa + 0.95
0 .9
Q.19 Find suitable conjugate acid-base pairs
from Table 2 in p.12 to prepare effective
buffer solutions of the following pH
values.
191
(a)
4.0
(b)
8.0
(c)
12.0
(d)
15.0
(e)
1.0
Effective Buffer of pH 4
Choose acids with pKa values within pH  1
HF / F
pKa = log107.2  104 = 3.14
HNO2 / NO2
pKa = log104.5  104 = 3.35
CH3COOH / CH3COO pKa = log101.8  105 = 4.74
192
Effective Buffer of pH 4
Choose bases with (14 – pKb) values within pH  1
F / HF
14 - pKb = 14 +log101.4  1011 = 3.14
NO2 / HNO2
14 - pKb = 14 +log102.2  1011 = 3.35
CH3COO / CH3COOH
14 - pKb = 14 +log105.6  1010 = 4.74
193
Effective Buffer of pH 8
Choose acids with pKa values within pH  1
194
H2S / HS
pKa = log101.0  107 = 7.00
H2PO4 / HPO4
pKa = log106.2  108 = 7.21
HSO3 / SO32
pKa = log106.2  108 = 7.21
HOCl / ClO
pKa = log103.5  108 = 7.46
Effective Buffer of pH 12
Choose acids with pKa values within pH  1
HPO42 / PO43
pKa = log103.6  1013 = 12.44
195
Effective Buffer of pH 15
Choose acids with pKa values within pH  1
H2O / OH
pKa = log101.8  1016 = 15.74
Choose bases with (14 – pKb) values within pH  1
OH / H2O
14 – pKb = 14 + log1055.5 = 15.74
196
Effective Buffer of pH 1
Choose acids with pKa values within pH  1
H3O+ / H2O
pKa = log1055.5 = 1.74
Choose bases with (14 – pKb) values within pH  1
H2O / H3O+
14 – pKb = 14 + log101.8  1016 = 1.74
197
Concentrated solutions of STRONG
bases and acids are effective buffers at
extreme pH values.
OH / H2O
198
H3O+ / H2O
Calculate [H3O+(aq)] required to prepare a buffer
with a pH value of –1.
[H3O  (aq)]
pH  pKa  log10
[H2O(l)]
[H3O  (aq)]
 1  log10 55.5  log10
[H2O(l)]
[H3O  (aq)]
 0.74  log10
[H2O(l)]
[H3O  (aq)]
 0.18
[H2O(l)]
[H3O+(aq)] = 0.1855.5 M = 10 M
199
170
Buffering Capacity
Definition : Buffering capacity is defined as the number of
millimoles of H3O+/OH ions required per cm3 of
buffer solution to cause a unit change (1) in pH
of the buffer solution.
200
Buffering capacity depends on the
concentrations of the buffering
components, HA and A-.
A buffer with higher concentrations of
HA(aq) and A(aq) has a higher
buffering capacity.
Refer to Q.18
201
Example (a) Calculate the pH of a solution which
contains 12.0 millimoles of CH3COOH and 8.0 millimoles of
CH3COO in 100 cm3. Given : Ka for ethanoic acid is 1.8 
105 mol dm3 at 298 K.
H2O(l) +
[X]equil 1
CH3COOH(aq)
12.0 103
0.1
pH  pKa  log10
202
CH3COO(aq) + H3O+(aq)
8.0  10 3
0.1
[CH3COOH(aq)]
[CH3COO (aq)]
12.0 10 3
0.1
 4.74  log10
 4.56
3
8.0 10
0.1
?
Example (b)
H2O(l) +
CH3COOH(aq)
12.0  103
+y
0.1
203
Add HCl(aq)
CH3COO(aq) + H3O+(aq)
8.0  10 3
y
0. 1
1  10( pH 1)
Example (b)
H2O(l) +
CH3COOH(aq)
CH3COO(aq) + H3O+(aq)
12.0  103
+y
0.1
8.0  10 3
y
0. 1
1  10( pH 1)
pH is reduced by one unit
pH = 4.56  1 = 3.56
[CH3COOH(aq)]
3.56  4.74 log10
[CH3COO (aq)]
nCH3COOH
nCH3COOH
[CH3COOH(aq)]
V

15.1 


nCH COO
[CH3COO (aq)]
nCH COO
3
204
V
3
nCH3COOH
nCH COO
nCH COO  1.24 mmol
 15.1
3
3
nCH3COOH  nCH COO  20.0 mmol
3
H3O+(aq) + CH3COO(aq)  CH3COOH(aq) + H2O(l)
mmole of H3O+ ions added
= mmole of CH3COO ions removed
= 8.0  1.24 = 6.76
buffering capacity = 6.76/100
205
= 0.0676 mmol cm-3
Q.20 Calculate the buffering capacity of the buffer in
Example above if the pH of the buffer is increased by
one unit by the addition of a certain amount of OH(aq).
H2O(l) +
CH3COOH(aq)
CH3COO(aq) + H3O+(aq)
pH = 4.56 + 1 = 5.56
[CH3COOH(aq)]
5.56  4.74 log10
[CH3COO (aq)]
[CH3COOH(aq)]
 0.15

[CH3COO (aq)]
206
nCH3COOH
nCH COO
 0.15
3
nCH3COOH  nCH COO  20.0mmol
3
nCH3COOH  2.61mmol
OH(aq) + CH3COOH(aq)  CH3COO(aq) + H3O+(aq)
mmole of OH ions added
= mmole of CH3COOH ions removed
= 12.0  2.61 = 9.39
207
buffering capacity = 9.39/100
= 0.0939 mmol cm-3
More Calculations Involving Buffer Solutions
1. Preparation of buffer solutions with desired
pH values
Example :
Calculate the volumes of equimolar solutions of
HCOOH(aq) and HCOONa(aq) required to prepare a
100 cm3 of effective buffer of pH 3.80.
(Ka for HCOOH(aq) is 1.75 x 10-4 mol dm3 at 298 K)
208
H2O(l) +
HCOO(aq) + H3O+(aq)
HCOOH(aq)
[HCOOH(aq)]
pH  pKa  log10
[HCOO (aq)]
[HCOOH(aq)]
3.8  log101.75 10  log10
[HCOO (aq)]
nHCOOH
[HCOOH(aq)]
V
0.91 

[HCOO (aq)] nHCOO
V
nHCOOH MHCOOHVHCOOH VHCOOH



MHCOO VHCOO
VHCOO
n

4
HCOO
209
Equimolar
VHCOOH
 0.91
VHCOO
VHCOOH + VHCOO = 100
VHCOOH = 47.6 cm3 ;
VHCOO = 52.4 cm3
210
Q.21 What mass of sodium hydrogen phosphate,
Na2HPO4, must be added to 500 cm3 of 0.10 M
aqueous solution of potassium dihydrogen
phosphate, KH2PO4, in order to prepare a buffer
of pH 6.8?
Assume the addition of sodium
hydrogenphosphate does not cause any volume
change.
(Kb of HPO42 = 1.6 x10-7 mol dm-3 ;
Kw of water = 1x10-14 mol2 dm-6 )
211
H2O(l)
+
H2PO4(aq)
HPO42(aq)
+
H3O+(aq)
[X(aq)]equil 1
0.10 – x
x
x
HPO42 added
0.10 – x
x+w
x
0.10 – x + y
x+w–y
x–y
[X(aq)]equil 2
 0.10
x – y = 110-6.8 = 1.6107
Kw 1.00 1014 (w  x  y)(x  y)

Ka 

7
0.10
Kb
1.6 10
212
H2O(l)
+
H2PO4(aq)
HPO42(aq)
+
H3O+(aq)
x – y = 110-6.8 = 1.6107
Kw 1.00 10
(w  x  y)(x  y)
Ka 


7
Kb
1.6 10
0.10
14
(w  1.6 10 )(1.6 10 ) 1.00 10

7
0.10
1.6 10
7
w = 0.039 M
213
7
14
H2O(l)
+
[X(aq)]equil 2
H2PO4(aq)
HPO42(aq)
0.10 – x + y
x+w–y
+
H3O+(aq)
w = 0.039 M
nNa2HPO4 added  0.039 0.500 0.020mol
Mass of Na2HPO4 added
= 0.020 mol  142 g mol1 = 2.8 g
214
x–y
2. Preparation of buffers by partial
neutralization of a weak acid by a strong
base
Example :
Calculate the pH of the buffer prepared by
mixing 200 cm3 of 0.20 M CH3COOH with
300 cm3 of 0.10 M NaOH at 298 K
Given :Ka for CH3COOH is 1.8 x 105 mol dm3
at 298 K
215
Step 1 - the stoichiometric calculation
Assuming all the OH- ions from NaOH(aq) are neutralized
by ethanoic acid.
CH3COOH(aq) +
OH(aq)
0.200.200
0.100.300
= 0.040 mol
= 0.030 mol

0.010 mol
Before reaction
0.030 mol
After reaction
216
CH3COO(aq) + H2O(l)
Step 2 - the equilibrium calculation
H2O(l) +
CH3COOH(aq)
[X(aq)]initial 0.010/0.500
= 0.020
[X(aq)]equil 0.020  x
 0.020
CH3COO(aq) +
0.030/0.500
?
= 0.060
0.060 + x
 0.060
[CH3COOH(aq)]
pH  pKa  log10
[CH3COO (aq)]
0.020
 4.7  log10
= 5.2
0.060
217
H3O+(aq)
??
Q.22(a)
HCOOH(aq) +
0.500.30
OH(aq)
0.500.15

HCOO(aq) + H2O(l)
(before reaction)
Assuming all the OH- ions from NaOH(aq) are neutralized
by HCOOH(aq).
0.500.15 (after reaction) 0.500.15
[HCOOH(aq)]
nHCOOH
pH  pKa  log
 pKa  log

[HCOO (aq)]
nHCOO
0.50  0.15
 pK a  log
 pK a = 3.76
0.50  0.15
218
Q.22(b)
Solution X is an acidic buffer which can
resist change in pH upon dilution and
addition of small amounts of acid/base.
219
3. Preparation of buffers by partial
neutralization of a weak base by a
strong acid
Q.23
1 dm3 of a solution is obtained by mixing 500 cm3
of 0.200 M HCl and 500 cm3 of 0.700 M NaCN.
Given that the dissociation constant of HCN is
4.0 x 1010 mol dm3,
calculate the pH of this solution.
220
Q.23
Step 1-the stoichiometric calculation
Assuming all the H3O+ ions from HCl(aq) are neutralized
by CN(aq).
CN(aq)
+
H3O+(aq)
0.5000.700
0.5000.200
= 0.350
= 0.100
0.350 – 0.100
= 0.250
After reaction
221

HCN(aq)
+
H2O(l)
Before reaction
0.100
Step 2 - the equilibrium calculation
H2O(l)
+
HCN(aq)
CN(aq)
+
H3O+(aq)
[X(aq)]initial
0.100
0.250
?
[X(aq)]initial
0.100 – x
0.250 + x
?
 0.100
 0.250
[HCN(aq)]
0.100
pH  pKa  log10
 pKa  log10

[CN (aq)]
0.250
0.100
pH  9.4  log10
0.250
222
= 9.8
4. Determination of Ka and Kb by measurement of pH of
buffers with known [HA]/[A-] and [B]/[BH+]
respectively (Refer to TAS Experiment 13)
[HA(aq)]
pH  pKa  log10
[A (aq)]
If [HA(aq)] = [A(aq)], pH = pKa
By mixing equal volumes of equimolar solutions
of HA and A, followed by pH measurement,
Ka of HA can be determined.
223
4. Determination of Ka and Kb by measurement of pH of
buffers with known [HA]/[A-] and [B]/[BH+]
respectively (Refer to TAS Experiment 13)
[B(aq)]
pOH  pKb  log10
[BH (aq)]
If [B(aq)] = [BH+(aq)], pOH = pKb
By mixing equal volumes of equimolar solutions
of B and BH+, followed by pH measurement,
Kb of B can be determined.
224
USES OF BUFFER SOLUTIONS
• Industrial processes : Electroplating / manufacture of dyes,
photographic materials and leather /
preservation of food, juices and medicines.
• Chemical analysis and calibration of pH meter
e.g. EDTA titration
• Maintaining pH in biological systems
225
Buffer Systems in Human Blood
H2CO3(aq)
+
H2O(l)
H3O+(aq) + HCO3(aq)
acid
conjugate base
pH range : 7.35 to 7.45
(pKa of H2CO3 = 6.38)
It would be fatal if human blood’s
pH values are out of this range.
226
Weak
base
Acid strength : -COOH > -NH3+
H2N
COO
Protein
H+
Conjugate acid
+H
Conjugate base
OH
+H
227
COO
Protein
3N
3N
H+
Protein
COOH
Weak acid
Decreasing pH
Increasing pH
OH
Weak
base
Base strength : -NH2 > -COO
H2N
COO
Protein
H+
Conjugate acid
+H
Conjugate base
OH
+H
228
COO
Protein
3N
3N
H+
Protein
COOH
Weak acid
Decreasing pH
Increasing pH
OH
Acid-Base Indicators
An acid-base indicator is a substance which
shows different colours in solutions of
different pH values.
An indicator is a weak acid or a weak base , for
which either the acid/base or its conjugate
base/acid, or both, are coloured.
229
Phenolphthlein is a weak acid that ionizes
slightly in water to give its conjugate base
+ H3O+(aq)
Colourless
230
Red
In the presence of HCl(aq)
The equilibrium position shifts to the left.
+ H3O+(aq)
Colourless
231
Red
When NaOH(aq) is added,
[H3O+(aq)] 
Equilibrium position shifts to the right.
+ H3O+(aq)
Colourless
232
Red
When pH < 8.3,
The colourless form predominates
+ H3O+(aq)
Colourless
233
Red
When pH > 10,
The red form predominates
+ H3O+(aq)
Colourless
234
Red
When 8.3 < pH < 10,
Both forms have similar concentrations  pink
+ H3O+(aq)
Colourless
Red
[In ]
pH  pKa  log10
[HIn]
-
235
Overall colour of indicator depends on
[BH (aq)]
[In (aq)]
or
[B(aq)]
[HIn(aq)]


Which in turn depend on pH of the solution
[In (aq)]
pH  pKa  log10
[HIn(aq)]
-
[BH (aq)]
pOH  pKb  log10
[B(aq)]

236
HIn(aq)
+
H2O(l)
weak acid
+
H+(aq)
conjugate base
(colour A)
If
In(aq)
(colour B)
[In (aq)]
 10
[HIn(aq)]

only the colour of In (colour B) would be
observed.
237
HIn(aq)
+
H2O(l)
weak acid
(colour A)
If
In(aq)
+
H+(aq)
conjugate base
(colour B)
[In (aq)] 1

[HIn(aq)] 10
only the colour of HIn (colour A) would be
observed.
238
HIn(aq)
+
H2O(l)
weak acid
(colour A)
If
In(aq)
+
H+(aq)
conjugate base
(colour B)
[In (aq)]
1

 10
10 [HIn(aq)]
a MIX of colour A and colour B would be
observed.
239
HIn(aq)
+
weak acid
In(aq)
H2O(l)
+
H+(aq)
conjugate base
(colour A)
(colour B)
For an indicator, HIn, that shows only colour B in
[In (aq)]
 10
aqueous solution, i.e.
[HIn(aq)]
[In (aq)]
 pKa  log1010
pH  pKa  log10
[HIn(aq)]
pH  pKa  1
240
HIn(aq)
+
In(aq)
H2O(l)
weak acid
+
H+(aq)
conjugate base
(colour A)
(colour B)
For an indicator, HIn, that shows only colour A
in aqueous solution,
[In (aq)] 1
i.e.

[HIn(aq)] 10
[In (aq)]
 pKa  log1010
pH  pKa  log10
[HIn(aq)]

pH  pKa  1
241
Q.24
HPh(aq) +
H2O(l)
colourless
Ph(aq) + H3O+(aq)
red
(a) At low pH, the equilibrium position shifts to
the left
the solution turns colourless
(b) At high pH, the equilibrium position shifts to
the right
the solution turns red
242
Q.24(c)(i)
HPh(aq) +
Ph(aq) + H3O+(aq)
H2O(l)
colourless
red
pKa  1  (log10 7 10
10
)  1  8.15
pH  3  pKa  1
Only the colour of HPh is observed,
i.e. colourless
243
Q.24(c)(ii)
HPh(aq) +
Ph(aq) + H3O+(aq)
H2O(l)
colourless
red
pKa  1  (log10 7 10
10
)  1  8.15
pH  7  pKa  1
Only the colour of HPh is observed,
i.e. colourless
244
Q.24(c)(iii)
HPh(aq) +
Ph(aq) + H3O+(aq)
H2O(l)
colourless
red
pKa  1  (log10 7 10
)  1  8.15
pKa  1  (log10 7 10
)  1  10.15
10
10
pKa  1  pH  9  pKa  1
The colour of the solution is a mix of red
and colourless, i.e. Pink
245
Q.24(c)(iv)
HPh(aq) +
Ph(aq) + H3O+(aq)
H2O(l)
colourless
red
pKa  1  (log10 7 10
10
)  1  10.15
pH  12  pKa  1
Only the colour of Ph is observed,
i.e. red
246
Q.25
Me(aq) +
MeH+(aq) + OH(aq)
H2O(l)
yellow
red
(a) At low pH, the equilibrium position shifts to
the right.
The solution turns
red
(b) At high pH, the equilibrium position shifts to
the left
The solution turns
247
yellow
Q.25(c)(i)
Me(aq) +
yellow
H2O(l)
MeH+(aq) + OH(aq)
red
[MeH (aq)]
pOH  pKb  log10
[Me(aq)]
[MeH (aq)]
14  pH  pKb  log10
[Me(aq)]
248
Q.25(c)(i)
Me(aq) +
H2O(l)
yellow
MeH+(aq) + OH(aq)
red
[MeH (aq)]
14  pH  pKb  log10
[Me(aq)]
[MeH (aq)]
pH  14  pKb  log10
[Me(aq)]
[MeH (aq)]
If
 10 Only red colour is
[Me(aq)]
observed
249
pH  (14 - pKb) - 1
Q.25(c)(i)
Me(aq) +
H2O(l)
yellow
MeH+(aq) + OH(aq)
red
[MeH (aq)]
pH  14  pKb  log10
[Me(aq)]
[MeH (aq)] 1
If

[Me(aq)] 10
Only the yellow colour is observed
pH  (14 - pKb) + 1
250
Q.25(c)(i)
Me(aq) +
MeH+(aq) + OH(aq)
H2O(l)
yellow
red
(14  pKb )  1  13  pKb
= 13 + log10510-11 = 2.7
pH  2  2.7
Only the
251
red colour can be observed
Q.25(c)(ii)
Me(aq) +
H2O(l)
MeH+(aq) + OH(aq)
yellow
red
(14  pKb )  1  2.7
(14  pKb )  1  4.7
2.7 < pH=4 < 4.7
Yellow + red = orange
252
Q.25(c)(iii)
Me(aq) +
H2O(l)
MeH+(aq) + OH(aq)
yellow
red
(14  pKb )  1  2.7
(14  pKb )  1  4.7
pH  6  4.7
Only the yellow colour can be
observed
253
The pH Ranges of Acid-Base
Indicators
The pH range over which the colour change
of an indicator is observable.
The pH range of indicator =
or
254
pKa  1
(14 – pKb)  1
Me(aq) + H2O(l)
MeH+(aq)
(yellow)
[MeH ]red
 10
[Me]yellow
Colour
Red A
+
(red)
(14  pKb )  1
[Me]yellow
[MeH ]red

Orange
14 – pKb
 10
Colour B
Yellow
pH
255
OH (aq)
The pH Ranges of Acid-Base Indicators
Red
256
Yellow
Q.26(a)
Indicators with two pH ranges : Cresol Red
Thymol blue
Alizarin
They are dibasic acids with two conjugate
acid-base pairs.
257
The pH Ranges of Acid-Base Indicators
Red
258
Yellow
Blue
Thymol blue, H2In
H2In(aq) + H2O(l)
Red
259
HIn(aq) + H3O+(aq)
Yellow
HIn(aq) + H2O(l)
In2(aq) + H3O+(aq)
Yellow
Blue
Q.26(b)
Universal indicator can be prepared
by mixing indicators each having a
different pH range over the range
of pH 0 to pH 14.
260
Acid-Base Titration
End Point and Equivalence Point
End Point and Equivalence Point are two
important but confusing concepts associated
with acid-base titration.
End point, in your Cert-level Chemistry study, is
defined as the point at which the indicator
used in a titration undergoes a colour change.
261
A titration has reached its end point when
the reaction mixture has just undergone a
change in pH across either boundaries of the
pH range of the indicator.
262
[ HIn ]colourA
 10

[ In ]colourB
pKa  1
Colour A
[ In ]colourB
 10
[ HIn ]colourA
Colour B
pKa
pH
For phenolphthalein, pKa = 9.15
The end point has been reached when the
reaction mixture has just undergone a change
in pH across pH 8.15 or pH 10.15.
263
For methyl orange, pKb = 10.3
The end point has been reached when the
reaction mixture has just undergone a change
in pH across pH 2.7 or pH 4.7.
264
It should be noted that the end point is a property of
the indicator and does NOT necessarily show that
the reaction involved in the titration is complete.
On the other hand, the equivalence point refers to
the exact pH value at which the reaction involved in a
titration is
reactant.
just complete, with no excess of either
A good indicator should be the one that gives a
sharp colour change at or close to the
equivalence point.
265
Strong acid – strong base titration
equivalence point at pH 7
Strong acid – weak base titration
equivalence point at pH < 7
Weak acid – strong base titration
equivalence point at pH > 7
Weak acid – weak base titration
Not sure, depending on Ka and Kb
266
Q.27(a)
At the equivalence point, 0.050 M NaCl
solution is obtained.
pH at equivalence point = 7
Both methyl orange and phenolphthalein are
suitable indicators for the titration.
In fact, all indicators giving a sharp colour
change over the pH range 4 to 10 are suitable.
267
Q.27(b)
At equivalence point,
0.050 M NH4Cl solution is obtained.
NH4+ is an acid that tends to donate protons
to water
NH4+(aq) + H2O(l)
NH3(aq) + H3O+(aq)
Cl is an extremely weak base, so its effect
on the pH can be ignored.
268
Q.27(b)
NH4+(aq) + H2O(l)
[X]equil0.050 – x
NH3(aq) + H3O+(aq)
x
Kw
x2
 5.6  1010
 Ka 
Kb
0.050 x
x = 5.3106
pH = 5.3
269
x
Q.27(b)
Bromcresol Green, Methyl Red
Eriochrom* Black T, Bromcresol Purple
are suitable indicators
In practice, methyl orange (pH 3.3 – 4.5) is
also suitable although the equivalence pH
(5.28) is out of its pH range.
See Q.28
270
Q.27(c)
At equivalence point,
0.050 M CH3COONa solution is obtained.
CH3COO is a base that tends to gain protons
from water
CH3COO(aq) + H2O(l)
271
CH3COOH(aq) + OH(aq)
Q.27(c)
CH3COO(aq) + H2O(l)
[X]equil 0.050 – x
CH3COOH(aq) + OH(aq)
x
Kw
x2
 5.6  1010
 Kb 
Ka
0.050 x
x = 5.3106
pOH = 5.3
pH = 8.7
272
x
Q.27(c)
Cresol Red,
Thymol Blue
o-Cresolphthalein,
Phenolphthalein
273
Q.27(d)
At equivalence point,
0.050 M CH3COONH4 solution is obtained.
274
Q.27(d)
CH3COO + NH4+
[X]equil 0.050 – x
CH3COOH
0.050 – x
+
x
Kw
x2



2
KaKb
[CH3COO ][NH4 ] (0.050 x)
[CH3COOH][NH3 ]
x
1.00  10

2
5 2
(0.050 x)
(1.8  10 )
2
x = 2.8104
275
14
NH3
x
Q.27(d)
CH3COOH + H2O
[X]equil 2.8104
Ka  1.8  105
Y = 1.0107
pH = 7.0
CH3COO
+
0.050 - 2.8104
H3O+
y
y(0.050 2.8  104 )

4
2.8  10
Refer to Q.29 &
p.40, last paragraph
Cresol red is probably the only choice because it
changes colour(yellow to orange ) at exactly pH 7
276
Q.29
Select, from the figure on p.37, a suitable indicator for
the titration of a weak base against a weak acid with an
equivalence point at pH 7.
Cresol red : pH range 7.0 – 8.8
Yellow
Orange
7.0
277
8.8
Red
Q.28(a)
Ka of phenolphthalein = 71010 mol dm3,
pH range over which phenolphthalein changes
colour
= pKa  1 = 9.15  1
the colour change at pH 8.15 is nearest to the
equivalence point.
278
Q.28(a)
Let x cm3 be the volume of NaOH added to reach the
end point (pH 8.15).
CH3COO(aq) + H3O+(aq)
CH3COOH(aq) + H2O(l)
Vinitial
25.0
0
VpH 8.15 25.0 – x
x
0
nCH3COOH
[CH3COOH(aq)]
V

pK

log
pH  pKa  log10
a
10
nCH COO
[CH3COO (aq)]
3
 pK a  log10
V
nCH3COOH
nCH COO
3
 pK a  log10
279
MCH3COOHVCH3COOH
MCH COO VCH COO
3
3
 pK a  log10
VCH3COOH
VCH COO
3
?
Q.28(a)
Let x cm3 be the volume of NaOH added to reach the
end point (pH 8.15).
CH3COOH(aq) + H2O(l)
Vinitial
25.0
VpH 8.15 25.0 – x
pH  pK a  log10
VCH3COOH
VCH COO
3
CH3COO(aq) + H3O+(aq)
0
0
x
?
25.0  x
 4.74  log10
x
25.0  x
8.15  4.74  log10
x
x = 24.99
The end point is deviated from the equivalence point
by (25.00 – 24.99) cm3 = 0.01 cm3
280
Q.28(b)
Kb of methyl orange = 5.01011 mol dm3,
pH range over which methyl orange changes
colour
= 14.0 - pKb  1 = 3.7  1
the colour change at pH 4.7 is nearest to the
equivalence point.
281
Q.28(b)
Let y cm3 be the volume of NaOH added to reach the
end point (pH 4.7).
CH3COOH(aq) + H2O(l)
Vinitial
25.0
VpH 4.7 25.0 – y
CH3COO(aq) + H3O+(aq)
0
0
y
?
VCH3COOH
[CH3COOH(aq)]
pH  pKa  log10
 pK a  log10

[CH3COO (aq)]
VCH COO
3
282
25.0  y
4.7  4.74  log10
Y = 11.9
y
The end point is deviated from the equivalence point by
(25.0 - 11.9) cm3 = 13.1 cm3
pH Titration Curves
Phenolphthalein
Equivalence point,
pH 7.00
Methyl orange
283
The Titration Curve for a Strong Acid and a Strong Base
284
pH Titration Curves
Equivalence point,
pH 8.72
Phenolphthalein
285
The Titration Curve for a Weak Acid and a Strong Base
286
The Shape of Titration Curves as a Function of Ka
287
pH Titration Curves
Equivalence point,
pH 5.28
Methyl orange
288
pH Titration Curves
Slow change in pH
around equivalence point
Most indicators are
not suitable
289
Q.30(a)
290
nH   nOH 
[ H  ( aq )]
pH
0.0025
0.0025
0.100
1.00
5.0
0.0005 0.0025
0.0020
0.0667
1.18
10.0
0.0010 0.0025
0.0015
0.0429
1.37
15.0
0.0015 0.0025
0.0010
0.0250
1.60
20.0
0.0020 0.0025
0.0005
0.0111
1.95
22.0
0.0022 0.0025
0.0003
0.0064
2.19
24.0
0.0024 0.0025
0.0001
0.00204
2.69
24.5
0.00245 0.0025
0.00005
0.00101
3.00
24.9
0.00249 0.0025
0.00001
0.000200
3.70
25.0
0.0025 0.0025
1.00107
7.00
VNaOH/cm3
nOH 
0.00
-
nH 
from water
Q.30(a)
291
VNaOH/cm3
nOH 
nH 
nOH   nH 
[ OH  ( aq )]
pH
25.1
0.00251
0.0025
0.00001
0.0001996
10.30
25.5
0.00255
0.0025 0.00005
0.0009901
11.00
26.0
0.00260
0.0025
0.001960
11.29
28.0
0.00280
0.0025 0.00030 0.005660 11.75
30.0
0.00300
0.0025 0.00050 0.009091
35.0
0.00350
0.0025
45.0
0.00450
0.0025 0.00200 0.028571 12.46
50.0
0.00500
0.0025 0.00250 0.033333 12.52
0.00010
0.00100
11.96
0.016667 12.22
31(a)
nOH  = nCH COO = x
[ CH 3COOH ] 0.0025 x

[ CH 3COO  ]
x
pH
0.0
-
-
2.87
5.0
0.0005
4.00
4.14
10.0
0.0010
1.50
4.56
15.0
0.0015
0.667
4.92
20.0
0.0020
0.250
5.34
22.0
0.0022
0.136
5.61
24.0
0.0024
0.042
6.12
24.5
0.00245
0.020
6.44
24.9
0.00249
0.0040
7.14
25.0
0.00250
-
8.72
VNaOH
292
/cm3
Q.31(a)
293
VNaOH/cm3
nOH 
nH 
nOH   nH 
[ OH  ( aq )]
pH
25.1
0.00251
0.0025
0.00001
0.0001996
10.30
25.5
0.00255
0.0025 0.00005
0.0009901
11.00
26.0
0.00260
0.0025
0.001960
11.29
28.0
0.00280
0.0025 0.00030 0.005660 11.75
30.0
0.00300
0.0025 0.00050 0.009091
35.0
0.00350
0.0025
45.0
0.00450
0.0025 0.00200 0.028571 12.46
50.0
0.00500
0.0025 0.00250 0.033333 12.52
0.00010
0.00100
11.96
0.016667 12.22
Double Indicator Titration
294
Refer to TAS experiment 2
CO32(aq) + H3O+(aq)
HCO3(aq) + H2O(l)
End point indicated by phenolphthalein
HCO3(aq) + H3O+(aq)
H2CO3(aq) + H2O(l)
End point indicated by methyl orange
295
Q.32(a)
Phenolphthalein becomes colourless when 13.0 cm3 of HCl
have been added.
296
Q.32(a)
Methyl orange turns orange when 25.0 cm3 of HCl have
been added.
297
Q.32(a)
Methyl orange turns red when 25.5 cm3 of HCl have been
added.
298
Q.32(b)
CO32(aq) + H3O+(aq)
HCO3(aq) + H2O(l)
End point indicated by phenolphthalein
299
Q.32(c)(i)
Phenolphthalein.
The colour change is sharper.
Q.32(c)(i)
Methyl orange.
The curve is steeper in this region.
Q.32(d)
(i) 9.0 (mid-point of pH range)
(ii) pKa  log10Ka
300
Ka  10pKa  1.0  109
Units of Ka : mol dm3
Q.32(e)
Blue to green when 13.5 cm3 of HCl(aq) have been added
Green to yellow when 22.5 cm3 of HCl(aq) have been added
301
Q.32(e)
This indicator is useless for this titration
because
its end-point is not sharp and, in any case,
does not correspond with either of the
equivalence points.
302
Solubility
Product
303
Meaning of Solubility Product
Consider the dissolution of a sparingly
soluble salt MX(s) (e.g. AgCl) in water
to form M+(aq) and X–(aq):
MX(s)
dissolution
precipitation
M+(aq) + X–(aq)
At equilibrium,
rate of dissolution = rate of precipitation
304
Meaning of Solubility Product
MX(s)
Kc 
dissolution
precipitation
M+(aq) + X–(aq)
[M(aq)]eq[X(aq)]eq
[MX(s)]
[MX(s)] is a constant at fixed T.
305
The amount of MX(s) present has no
effect on the equilibrium position and its
solubility
Meaning of Solubility Product
MX(s)
Kc 
dissolution
precipitation
M+(aq) + X–(aq)
[M(aq)]eq[X(aq)]eq
[MX(s)]
Kc [MX(s)] = [M+(aq)]eq [X–(aq)]eq = Ksp
Ksp = solubility product
(a constant at fixed T)
306
Meaning of Solubility Product
Generally,
MaXb(s)  aMb+(aq) + bXa–(aq)
the solubility product is given by:
Ksp = [Mb+(aq)]aeq [Xa–(aq)]beq
307
Q.33
BaSO4(s)
Ba2+(aq) + SO42–(aq)
2
2
Ksp  [Ba (aq)]eq [SO4 (aq)]eq
Ag2CO3(s)
2Ag+(aq) + CO32–(aq)
2
3
Ksp  [Ag (aq)] [CO (aq)]eq

308
2
eq
Q.33
Ca(OH)2(s)
Ca2+(aq) + 2OH–(aq)
Ksp  [Ca (aq)]eq [OH (aq)]
2
Fe(OH)3(s)
2
eq

Fe3+(aq) + 3OH–(aq)
Ksp  [Fe (aq)]eq [OH (aq)]
3
309

3
eq
Q.33
Bi2S3(s)
2Bi3+(aq) + 3S2–(aq)
Ksp  [Bi (aq)] [S (aq)]
3
310
2
eq
2
3
eq
Calculating Ksp from Molar Solubility
Example : The solubility of lead(II) chromate(PbCrO4)
is 4.5 × 10–5 g dm3. Calculate the solubility
product of the compound.
The molar mass of PbCrO4 is 323.2 g mol–1.
4.5 10 g dm
Molar solubility 
1
323.2g mol
5
3
= 1.4107 mol dm3
311
Calculating Ksp from Molar Solubility
Example : The solubility of lead(II) chromate(PbCrO4)
is 4.5 × 10–5 g dm3. Calculate the solubility
product of the compound.
Molar solubility of PbCrO4 = 1.4 × 10–7 mol dm–3
PbCrO4(s)
Pb2+(aq)
At equilibrium
1.4107 M
+
CrO42(aq)
1.4107 M
Ksp = (1.4107 M)2 = 2.01014 mol2 dm6
312
Q.34
Bi2S3(s)
2Bi3+(aq)
At equilibrium
+
21.01015
3S2(aq)
31.01015
The molar solubility of Bi2S3(s) is 1.0  10–15 mol dm–3.
One unit of Bi2S3 dissolves to give 2Bi3+ and 3S2– ions,
Ksp  [Bi (aq)] [S (aq)]
2
2
eq
2
3
eq
= (2.0  10–15)2(3.0  10–15)3
= 1.1  10–73 mol5 dm–15
313
Calculating Solubility from Ksp
The solubility product, Ksp, of silver bromide (AgBr)
is 7.7 × 10–13 mol2 dm–6. Calculate the molar
solubility of AgBr.
AgBr(s)
Ag+(aq)
+
Br(aq)
At equilibrium
x
x
Let x mol dm–3 be the molar solubility of AgBr.
One unit of AgBr dissolves to give one Ag+ ion
and one Br ion,
Ksp = x2 = 7.71013
314
x = 8.8107
Solubility of AgBr is 8.8107 mol dm3
Q.35
The solubility product of silver phosphate,
Ag3PO4, is 3.4 × 10–14 mol4 dm–12 at 298 K. Calculate
its solubility in mol dm–3 at 298 K.
Ag3PO4(s)
At equilibrium
3Ag+(aq)
3x
+
PO43(aq)
x
Let x mol dm3 be the solubility of the salt.
One unit of Ag3PO4 dissolves to give 3 Ag+ and 1 PO42
315
Q.35
Ag3PO4(s)
3Ag+(aq)
At equilibrium
3x
+
PO43(aq)
x
3
Ksp  [Ag (aq)] [PO4 (aq)]  (3x) (x)

3
3
3.41014 = 27x4
x = 1.9104
Solubility of Ag3PO4 is 1.9104 mol dm3
316
Q.36
Are compounds having a higher solubility product more
soluble than those having a smaller solubility product?
It is true ONLY when the compounds have
the same stoichiometry.
Or, when the Ksp values to be compared
bear the same units.
317
Comparison between Ionic Product
and Solubility Product
Consider the following system :
MX(s)
M+(aq)
+
X(aq)
Ionic product = [M+(aq)][X(aq)]
Solubility product = [M+(aq)]eq[X(aq)]eq
318
Comparison between Ionic Product
and Solubility Product
1. [M+(aq)] [X–(aq)] < Ksp (at a given T)
 the solution is unsaturated and the
system is NOT at equilibrium
No ppt in the system
 more MX(s) can be dissolved in it until
[M+(aq)] [X–(aq)] = Ksp
319
Comparison between Ionic Product
and Solubility Product
2. [M+(aq)] [X–(aq)] = Ksp (at a given T)
 the solution is saturated and the
system is at equilibrium
Ppt co-exists in equilibrium with the
saturated solution
 No more MX(s) can be dissolved in the
solution
320
Comparison between Ionic Product
and Solubility Product
3. [M+(aq)] [X–(aq)] > Ksp (at a given T)
 the solution is supersaturated and the
system is NOT at equilibrium
 precipitation of MX(s) occurs until
[M+(aq)] [X–(aq)] = Ksp
321
Q.37(a)
Is a precipitate expected to form at equilibrium when
50.0 cm3 of 1.00103 M BaCl2(aq) is added to 50.0 cm3 of
1.00 104 M Na2SO4(aq)?
(Given: Ksp for barium sulphate = 1.10 × 10–10 mol2 dm–6)
[Ba2+(aq)] = 1.00103 mol dm–3  2 = 5.0010–4 mol dm–3
[SO42–(aq)] = 1.00104 mol dm–3  2 = 5.0010–5 mol dm–3
BaSO4(s)
322
Ba2+(aq)
5.00104
+
SO42(aq)
5.00105
[Ba2+(aq)] [SO42–(aq)] = (5.00 × 10–4) (5.00 × 10–5)
= 2.50 × 10–8 mol2 dm–6 > Ksp
Precipitation of barium sulphate occurs until
[Ba2+(aq)][SO42(aq)] = Ksp.
Q.37(b)
Hence, calculate the concentrations of Ba2+ and
SO42 after mixing.
BaSO4(s)
At equilibrium
Ba2+(aq)
5.00104 - x
+
SO42(aq)
5.00105 - x
Ksp  1.101010  (5.00104  x)(5.00105  x)
x = 4.98105
[Ba2+(aq)] = 4.50104 M
323
[SO42(aq)] = 2.00107 M
Common Ion Effect
Consider the equilibrium system :
MX(s)
At equilibrium
M+(aq)
x mol dm3
+
X(aq)
x mol dm3
where x mol dm3 is the molar solubility
of MX in water without common ion
E.g. MX dissolves in Pure Water
At a given Temperature, Ksp = x2
324
Common Ion Effect
In the presence of common ions (M+ and/or X)
MX(s)
Initial
At equilibrium,
M+(aq)
+
X(aq)
c mol dm3
(y + c) mol dm3
y mol dm3
where y mol dm3 is the molar solubility of
MX in a solution containing c mol dm3 of M+
At the same Temperature, Ksp = x2 = (y+c)(y)
y < x The solid is less soluble in the
presence of its common ions.
325
A solid is always less soluble in the
presence of its common ions.
326
Q.38(a)
Let x mol dm3 be the molar solubility of Al(OH)3
Al(OH)3(s)
Al3+(aq)
At equilibrium
+
x
3OH(aq)
3x + 1.0107
Ksp  1.0  1033  (x)(3x 1.0  10-7 )3
 (x)(1.0 10-7 )3
x = 1.01012
Solubility of Al(OH)3 is 1.01012 mol dm3
327
Q.38(b)
Let y mol dm3 be the solubility of Al(OH)3
Al(OH)3(s)
Al3+(aq)
Initial
At equilibrium
+
3OH(aq)
1.0
y
3y + 1.0
Ksp  1.01033  (y)(3y 1.0)3  (y)(1.0)3  y
y = 1.01033
Solubility of Al(OH)3 is 1.01033 mol dm3
328
Q.38(c)
Al(OH)3(s)
Initial
Al3+(aq)
+
3OH(aq)
2.0
One unit of Al2(SO4)3 gives TWO units of
Al3+ ions
329
Q.38(c)
Let z mol dm3 be the solubility of Al(OH)3
Al(OH)3(s)
Al3+(aq)
Initial
+
3OH(aq)
2.0
At equilibrium
Ksp  1.010
33
2.0 + z
3z
 (2.0 z)(3z)
3
 (2.0)(3z)3  54z3
z = 2.61012
Solubility of Al(OH)3 is 2.61012 mol dm3
330