CHAPTER 7.0_IONIC EQUILIBRIA

Titration Curve: Titration Between 51
Weak Acid – Strong Base

• Titrating a weak acid with a strong base results in differences
in the titration curve at the equivalence point and excess acid
region.

• The initial pH is determined using the Ka of the weak acid.

• The pH after base is added, but before equivalence point, is
determined using the Henderson-Hasselbalch equation of an
acidic buffer.

• The pH at the equivalence point is determined using the Kb of
the conjugate base of the weak acid.

• The pH after equivalence point is dominated by the excess
strong base (the basicity from the conjugate base anion is
negligible).

Titration Curve: Titration Between Strong Acid – Strong Base
(Titrant: Strong Base; Analyte: Weak Acid)

After equivalence point (Excess of strong base)

pH at the equivalence point pH after
equivalence point
is determined using the Kb of is determined by
the conjugate base of the End point pH range the excess strong
Equivalence point; base.
weak acid
pH > 7 due to the
hydrolysis of anion
which produced OH-

Initial pH Before equivalence point pH after base is added, but
(Buffered solution) before equivalence point, is
Initial pH is determined using the
determined using Henderson-Hasselbalch
equation of an acidic buffer.
the Ka of the
weak acid. 52

Titration Curve: Titration Between Strong Acid – Weak Base
(Titrant: Strong Acid; Analyte: Weak Base)

Initial pH is Initial pH pH after acid is added, but before equivalence
point, is determined using the Henderson-
determined using Hasselbalch equation of an basic buffer.

the Kb of the Before equivalence point
weak base. (Buffered solution)

Equivalence point: pH < 7

due to the hydrolysis of

cation which produced
H3O+

End point pH range

After equivalence point

Beyond the equivalence point, the
excess strong acid determines the pH.

53

pH Calculation:
Strong Acid – Strong Base Reaction

Example:
0.1M NaOH was gradually added to 25 mL of 0.1M HCl
solution. Calculate the pH of the solution,
a) before NaOH was added.
b) after the addition of 24 mL NaOH.
c) at equivalence point.
d) after the addition of 30 mL NaOH.

54

Guideline refer slide #47.

a) Before NaOH was added.

Before titrant (a strong base) is added, pH is
determined by the strong acid alone.

HCl(aq)  H+ (aq) + Cl– (aq)

[ ]i/M 0.1

Therefore, [H+] = 0.1 M

pH = - log [H+]

= 1.0

55

b) After the addition of 24 mL NaOH.

After base has been added, but before equivalence
point, excess strong acid determines the pH.

HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)

ni/mol 0.1 x 25 0.1 x 24 0-

1000 1000
= 2.5 x 10-3 = 2.4 x 10-3

 n - 2.4 x 10-3 - 2.4 x 10-3 + 2.4 x 10-3
2.4 x 10-3
nf/mol 1 x 10-4 0

[ ]f/M 1 x 10-4
(25 + 24) x 10-3
= 2.04 x 10-3 56

HCl(aq)  H+ (aq) + Cl– (aq)

[ ]f/M 2.04 x 10-3

Therefore, [H+] = 2.04 x 10-3

pH = - log [H+]

= 2.69

57

c) At equivalence point. (nacid = nbase)
At equivalence point, neutral salt is formed, neither cation
nor anion undergoes hydrolysis, therefore pH = 7.

HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)

ni/mol 0.1 x 25 0.1 x 25 0 -

1000 1000
= 2.5 x 10-3 = 2.5 x 10-3

n - 2.5 x 10-3 - 2.5 x 10-3 + 2.5 x 10-3
nf/mol 0 0 2.5 x 10-3

At 25oC, Kw = [H+] [OH-] x = [H+] 58
1 x 10-14 = (x)(x)  pH = - log 1 x 10-7
x = 1 x 10-7
=7

d) After the addition of 30 mL NaOH.

Beyond equivalence point, excess strong base
determines the pH.

HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)

ni/mol 0.1 x 25 0.1 x 30 0-

1000 1000
= 2.5 x 10-3 = 3.0 x 10-3

n - 2.5 x 10-3 - 2.5 x 10-3 + 2.5 x 10-3
nf/mol
0 5 x 10-4 2.5 x 10-3
[ ]f/M
5 x 10-4
(25 + 30) x 10-3 59
= 9.09 x 10-3

NaOH(aq)  Na+ (aq) + OH– (aq)

[ ]f/M 9.09 x 10-3

Therefore, [OH-] = 9.09 x 10-3

pOH = - log [OH-]

= 2.04

 pH = 11.96

60

Practice exercise
What is the pH of the resultant solution upon addition of 25
mL of 0.1M NaOH to 25 mL of 0.1M H2SO4 in a titration.
Ans: 1.30

61

ACID – BASE INDICATORS

• Use of indicator: to detect the equivalence point of an
acid-base titration.

• A good choice of indicator is one at which it changes
colour coincides with the equivalence point.

• How and why an indicator changes colour can be
explained by the equilibrium principle.

HIn(aq) H+(aq) + In–(aq)
Colour A Colour B

• The weak acid HIn and its conjugate base In– have
different colours.

62

SOME COMMON ACID – BASE INDICATORS

Indicator Colour in Colour pH range of
Acid in Base colour change
Thymol blue
Bromophenol blue Red Yellow 1.2 – 2.8
Methyl orange Yellow Blue 3.0 – 4.6
Methyl red Orange Yellow 3.2 – 4.4
Bromocresol purple Red Yellow 4.8 – 6.0
Bromothymol blue Yellow Purple 5.2 – 6.8
Cresol red Yellow Blue 6.0 – 7.6
Phenolphthalein Yellow Red 7.0 – 8.8
Alizarin yellow Colorless Pink 8.2 – 10.0
Yellow Red 10.1 – 12.0

63

TYPES OF ACID-BASE TITRATIONS
AND ITS SUITABLE INDICATORS

Type of Titration End point pH Suitable Indicator
Range
Strong Acid – 3 – 10 All indicators
Strong Base
7 – 11 Phenolphthalein,
Weak Acid – Cresol Red
Strong Base 3–7
Methyl Orange,
Strong Acid – None Methyl Red
Weak Base
None
Weak Acid –
Weak Base

64

Example:
In a weak acid-strong base titration, 25.00 mL KOH
solution, is titrated with 21.45 mL of 0.10 M HCN to reach
the equivalence point.

iv) Explain suitable indicator for the titration.

Solution:

• Strong base is titrated with weak acid.
• At equivalence point, pH > 7.
• End point pH range: 7 – 11.
• Suitable indicators: cresol red, bromothymol blue,

phenolphthalein.

65

7.3
Solubility Equilibria

67

Many ionic salts are sparingly or very slightly soluble in water.
When this salt is dissolved in water, a saturated solution will

eventually form.
A saturated solution is a solution in which no more solute will

dissolve at a given temperature.

68

In a saturated solution of a slightly soluble salt:
a heterogeneous equilibrium is established between the

undissolved solid salt and its dissolved ions (also called the
solubility equilibrium).
the rate of dissolution equals the rate of precipitation.

Saturated solution dissolution
of silver sulphide

Ag+ S2– S2– Ag2S(s) 2Ag+(aq) + S2–(aq)
S2– Ag+
dissolved
Ag+ S2– Ag+ ions

solid Ag2S precipitation

69

Solubility Product Constant, Ksp

Is the product of molar concentrations of ions in a saturated solution
of a salt, each raised to the power of the stoichiometric coefficient of
the respective ions.

MxAy(s) xMy+ (aq) + yAx– (aq)

Ksp = [My+]x [Ax–]y

The higher the Ksp, the greater the solubility.

(Note: The equilibrium in a saturated solution is a heterogeneous
equilibrium. The concentration of pure solid is excluded
as it is a constant)

Solubility Product Constants of Sparingly Soluble Salts 70

Compound Ksp Compound Ksp
5.0 x 10-13 4.0 x 10-38
AgBr 1.8 x 10-10 Fe(OH)3 6.3 x 10-18
8.3 x 10-17 FeS 1.6 x 10-52
AgCl 3.1 x 10-8 1.8 x 10-11
1.3 x 10-20 HgS 8.6 x 10-5
AgI 2.0 x 10-32 1.9 x 10-13
5.0 x 10-3 Mg(OH)2 2.5 x 10-13
AgIO3 1.1 x 10-10 MgC2O4 1.0 x 10-24
Ag3PO4 1.0 x 10-97 Mn(OH)2 1.6 x 10-5
Al(OH)3 4.8 x 10-9 MnS 1.6 x 10-8
Ba(OH)2 4.0 x 10-9 8.0 x 10-28
BaSO4 1.2 x 10-6 NiS 3.2 x 10-7
Bi2S3 8.0 x 10-27 3.3 x 10-17
CaCO3 2.0 x 10-25 PbCl2 1.6 x 10-23
CaC2O4 6.3 x 10-36 PbSO4
CaSO4 PbS
CdS
SrSO4
CoS Zn(OH)2
ZnS
CuS

71

Writing Ksp Expressions

Write the solubility equilibrium equation and the Ksp
expression for each of the following sparingly soluble

salts :

a) BaCrO4 c) Ag2SO4

b) Mn(OH)2 d) Pb(IO3)2

a) BaCrO4 (s) Ba2+ (aq) + CrO42– (aq)

Ksp = [Ba2+][CrO42–]

72

b) Mn(OH)2(s) Mn2+(aq) + 2OH–(aq)

Ksp = [Mn2+] [OH–]2

c) Ag2SO4(s) 2Ag+(aq) + SO42–(aq)

Ksp = [Ag+]2 [SO42–]

d) Pb (IO3)2(s) Pb2+(aq) + 2IO3–(aq)

Ksp = [Pb2+] [IO3–]2

73

Solubility and Ksp

Solubility is the maximum amount of solute that dissolves in
a given quantity of solvent to form a saturated solution.

Molar solubility (in mol L–1 ) is the number of moles of solute
dissolved in 1 L of a saturated solution.

Solubility of Molar Molar Ksp of
compound Solubility of concentration salt
compound
(g/L) of ions
(mol/L)

74

Calculations involving Ksp can be divided into two categories:

Calculate Ksp Solubility of Molar Concentrations Ksp of
from solubility compound solubility of of cations and compound
compound
data anions

Calculate Ksp of Concentrations of Molar Solubility
solubility from compound cations and solubility of of
anions compound
Ksp compound

CALCULATIONS

INVOLVING

KSP AND
SOLUBILITY

76

3. The molar solubility of calcium phosphate, Ca3(PO4)2 in water
at 25 oC is 2.47 x 10–6 M. Calculate the solubility product
constant for this salt.

Let x = molar solubility of Ca3(PO4)2 = 2.47 x 10–6 M

Ca3(PO4)2 (s) 3Ca2+ (aq) + 2PO43– (aq)


3x 2x

Ksp = [Ca2+]3 [PO43–]2 = (3x)3 (2x)2 = 108 x5
= 108 (2.47 x 10–6)5
= 9.93 x 10–27

77

2. At 25 oC, the solubility of CaF2 in water is 1.69 x 10–2 g L–1.
What is the value of Ksp for CaF2 at this temperature ?

Molar solubility = 1.69 x 10–2 gL–1 / 78 gmol–1
= 2.16 x 10–4 mol L–1 = x

CaF2 (s) Ca2+(aq) + 2F– (aq)

x 2x

Ksp = [Ca2+] [F–]2 = (x) (2x)2
= 4 (2.16 x 10–4)3
= 4.03 x 10–11

78

5. Write the solubility equilibrium equation and calculate the
molar solubility of Hg(OH)2 which has a Ksp value of 3.2x10–26.

Let x = molar solubility of Hg(OH)2

Hg(OH)2 (s) Hg2+ (aq) + 2OH– (aq)

 
x 2x

Ksp = [Hg2+] [OH–]2
3.2 x 10–26 = (x)(2x)2

4x3 = 3.2 x 10–26
 x = 2 x 10–9 M

79

6. The Ksp value of Ba3(PO4)2 is 3.4x10–23. What is the molar
solubility of Ba3(PO4)2 in water?

Let x = molar solubility of Ba3(PO4)2

Ba3(PO4)2 (s) 3Ba2+ (aq) + 2PO43– (aq)

3x 2x

Ksp = [Ba2+]3 [PO43–]2
3.4 x 10–23 = (3x)3(2x)2

108 x5 = 3.4 x 10–23

 x = 1.26 x 10–5 M

80

When two solutions containing ions of a slightly soluble salt are
mixed, prediction can be made whether a precipitate will form,
by comparing the ion product, Q with Ksp.

Condition Description

Q < Ksp Solution is unsaturated. No precipitate forms.

Q = Ksp Solution is saturated. An equilibrium exists
between undissolved solute and its ions.

Q > Ksp Solution is supersaturated. Precipitate forms.

81

Concentration Calculate ion product, Q Compare Q
of cation from with Ksp
one solute
Q > Ksp  precipitate forms.
Concentration Q = Ksp  equilibrium between
of anion from
another solute undissolved salt and
dissolved ions.

Q < Ksp  no precipitate forms.

82

Additional Information:

Unsaturated solution contains less solute than the
solvent has the capacity to dissolve at a specific
temperature.

Saturated solution contains the maximum amount of
a solute that will dissolve in a given solvent at a
specific temperature.

Supersaturated solution contains more solute than
is present in a saturated solution at a specific
temperature.

QUESTIONS ON
PRECIPITATION

84

1) Will a precipitate of Mg(OH)2 form when 50 mL of
0.001 M NaOH is added to 150 mL of 0.01 M MgCl2?
(Ksp of Mg(OH)2 = 2 x 10–11)

2NaOH(aq) + MgCl2(aq) Mg(OH)2(?) + 2NaCl(aq)

In 200 mL mixture: (using M1V1 = M2V2) Mg(OH)2(s) Mg2+(aq) + 2OH–(aq)

[Mg2+] 150 x 0.01 10−3M Ion product, Q = [Mg2+] [OH–]2
200
= = 7.5 x = (7.5x10–3) (2.5x10–4)2

[OH−] = 50 x 0.001 = 2.5 x 10−4M = 4.7 x 10–10
200
Q > Ksp Solution is supersaturated.
Mg(OH)2 precipitate will form.

85

2) 0.02 M of CaCl2 solution is added to 0.02 M of Na2SO4
solution in equal volumes. Will CaSO4 precipitate form
in the mixture? (Ksp of CaSO4 = 2x10–4 M2)

CaCl2(aq) + Na2SO4 (aq) CaSO4(?) + 2NaCl(aq)

In 2V L solution: Ca2+(aq) + SO42–(aq)

CaSO4(s)

[Ca2+] = 0.02 x V = 0.01M
2V
Ion product, Q = [Ca2+] [SO42–]
[SO24−] = 0.02 x V = 0.01M
2V = (0.01) (0.01)
= 1 x 10–4

Q < Ksp the solution is unsaturated.
CaSO4 precipitate will not form.

3) Does a precipitate form when 1.0 x 10–2 mol of Ba(NO3)2 and 86
2.0 x 10–2 mol of NaF are dissolved in 1 Liter of solution?
(Ksp of BaF2 = 1.7 x 10–6 )

Ba(NO3)2(aq) + 2NaF(aq) BaF2 (?) + NaNO3(aq)

In 1L solution: BaF2(s) Ba2+(aq) + 2F–(aq)

0.01 mol Ion product, Q = [Ba2+] [F–]2
1L
[Ba2+] = = 0.01M = (0.01) (0.02)2

= 4 x 10–6

[F−] = 0.02 mol = 0.02M Q > Ksp solution is supersaturated
1L BaF2 precipitate will form.

4) A 0.1 M solution of AgNO3 is added dropwise to a solution 87
containing 0.1 M of NaCl and 0.01 M of K2CrO4. Which salt
will precipitate first, AgCl or Ag2CrO4 ?
(Ksp AgCl = 2.0 x 10–10 ; Ksp Ag2CrO4 = 2.4 x 10–12 )

*Precipitation begins when Q = Ksp

AgCl(s) Ag+(aq) + Cl–(aq) Ag2CrO4 (s) 2Ag+(aq) + CrO42–(aq)

Q = [Ag+] [Cl–] = Ksp Q = [Ag+]2 [CrO42–] = Ksp
[Ag+]2 (0.01) = 2.4 x 10–12
[Ag+] (0.1) = 2.0 x 10–10

[Ag+] = 2.0 x 10–9 M [Ag+] = 1.55 x 10–5 M

AgCl will precipitate first because the concentration of Ag+ ion required
for AgCl to precipitate is lower than that required by Ag2CrO4.

88

Add CrO42–

Common ion – An ion in a  [CrO42–]
solution that has been supplied
from more than one solute. PbCrO4(s) Pb2+(aq) + CrO42-(aq)

Common ion effect – The
lowering of the solubility of a
sparingly soluble salt by the
addition of a common ion.

*The presence of a common ion decreases
the solubility of the salt*

89

Solubility equilibrium in the saturated Second solute Common ion
solution of slightly soluble salt added

BaSO4 (s) Ba2+(aq) + SO42– (aq) Na2SO4 (aq) SO42–

Fe(OH)2 (s) Fe2+ (aq) + 2OH– (aq) NaOH (aq) OH–

AgBr (s) Ag+ (aq) + Br– (aq) AgNO3 (aq) Ag+

The net effect upon addition of common ions:
the solubility equilibrium shifts to the LEFT.
more precipitate forms.
solubility decreases.

90

The solubility equilibrium that exists in a saturated solution of
silver chloride is:
AgCl (s) Ag+ (aq) + Cl–(aq)

When NaCl (aq) is added to the saturated solution of AgCl, it
provides the common ion, Cl– , thus [Cl–] in the solution increases.

According to Le Chatelier’s principle, the solubility equilibrium
will shift to the LEFT to offset the increase in [Cl–].

This means that the solubility of AgCl decreases, or more AgCl
precipitate forms.

The [Ag+] decreases in the same degree as the increase in [Cl–],
hence the Ksp value is unaffected.

QUESTIONS
RELATED TO
COMMON ION

EFFECT

92

1) What is the solubility of PbSO4 , Ksp = 1.66x10–8 in a 1.00 M
solution of H2SO4?

Let s = molar solubility of PbSO4 in 1 M H2SO4

H2SO4 (aq) 2H+(aq) + SO42–(aq)
(1 M)

In 1M H2SO4: PbSO4 (s) Pb2+ (aq) + SO42– (aq)
(s) (s + 1)

Ksp = [Pb2+] [SO42–]
1.66 x 10–8 = (s)(s + 1.00)

Assumption: Since Ksp is small, s << 1.00, s + 1.00  1.00
1.66 x 10–8 = (s)(1.00)
 s = 1.66 x 10–8 M

2) If the solubility of Ag2CrO4 in water is 8.4x10–5 mol L–1, 93
what is the solubility in 0.10 M AgNO3?

Solubility of Ag2CrO4 = y = 8.4 x 10–5 M

Ag2CrO4 (s) 2Ag+(aq) + CrO42–(aq)
(2y) (y)

Ksp = [Ag+]2 [CrO42–]
= (2y)2 (y) = 4y3
= 4 x (8.4 x 10–5)3
= 2.37 x 10–12

94

Let x = solubility of Ag2CrO4 in 0.1 M AgNO3

AgNO3 (s) Ag+(aq) + NO3–(aq)
(0.1 M)

In 0.1M AgNO3 Ag2CrO4 (s) 2Ag+(aq) + CrO42–(aq)
(2x+0.1) (x)

Ksp = [Ag+]2 [CrO42–]
2.37 x 10–12 = (2x + 0.1)2 (x)

Assumption : Since Ksp is small, x << 0.1, 2x + 0.1  0.1
2.37 x 10–12 = (0.1)2 (x)
 x = 2.37 x 10–10 M

3) Calculate the molar solubility of Ag2SO4 at 25oC in 95

a) water b) 0.1 M Na2SO4 c) 0.1 M AgNO3

Explain the difference in the value of solubilities.

(Ksp of Ag2SO4 = 1.5x10–5 M3)

a) Let x = solubility of Ag2SO4 in water

Ag2SO4 (s) 2Ag+(aq) + SO42–(aq)
(2x) (x)

Ksp = [Ag+]2 [SO42–]
1.5 x 10–5 = (2x)2 (x)

4x3 = 1.5 x 10–5
 x = 1.55 x 10–2 M

b) Let y = molar solubility of Ag2SO4 in 0.1 M Na2SO4 96

Na2SO4 (aq) 2Na+(aq) + SO42–(aq)
(0.1 M)

In 0.1M Na2SO4 Ag2SO4 (s) 2Ag+ (aq) + SO42–(aq)
(2y) (y+0.1) M

Ksp = [Ag+]2 [SO42–]
1.5 x 10–5 = (2y)2 (y + 0.1)

Assumption: Since Ksp is small, y << 0.1, y + 0.1  0.1
1.5 x 10–5 = (2y)2 (0.1)
4y2 = 1.5 x 10–4
 y = 6.12 x 10–3 M

97

c) Let z = molar solubility of Ag2SO4 in 0.1 M AgNO3

AgNO3 (s) Ag+(aq) + NO3–(aq)
(0.1 M)

In 0.1M AgNO3  Ag2SO4 (s) 2Ag+(aq) + SO42–(aq)
(2z+0.1) (z)

Ksp = [Ag+]2 [SO42–]
1.5 x 10 –5 = ( 2z + 0.1 )2 (z)
Assumption: Since Ksp is small, z << 0.1, 2z + 0.1  0.1
1.5 x 10–5 = (0.1)2 (z)

 z = 1.5 x 10–3 M

98

in water Molar solubility of Ag2SO4 (M)
in Na2SO4 1.55 x 10 –2
in AgNO3 6.12 x 10 –3
1.50 x 10 –3

The solubility of Ag2SO4 in Na2SO4 and AgNO3 are lower than its
solubility in water.

The presence of common ions, SO42– in Na2SO4 and Ag+ in AgNO3
causes a shift in the solubility equilibrium to the LEFT.

Ag2SO4 (s) 2Ag+(aq) + SO42–(aq)

Hence, more Ag2SO4 precipitate form or its solubility decreases.

CHEMISTRY UNIT, KMNS

THE END