CHAPTER 7 - IONIC EQUILIBRIA

b) Calculate the new pH of the buffer after the
addition of 2.0 mL of 0.10 M HCl

H3O+ + NH3  NH4+ + H2O

ni 2x10-4 0.06 0.6

Δ -2x10-4 -2x10-4 +2x10-4

nf 0 0.0598 0.6002

V = 1000 mL + 2 mL
= 1.002 L

101

b) Calculate the new pH of the buffer after the addition of
2.0 mL of 0.10 M HCl

pOH  - log Kb  log [salt]
[base]

 -log 1.8x10 - 5  log 0.6002/1.002
0.0598/1.002

 5.74

pH 14 - 5.74
 8.26

102

c) Calculate the new pH of the buffer after the addition of
2.0 mL of 0.15 M NaOH

OH- + NH4+  NH3 + H2O

ni 3x10-4 0.6 0.06

Δ -3x10-4 -3x10-4 +3x10-4

nf 0 0.5997 0.0603

V = 1000 mL + 2 mL
= 1.002 L

103

c) Calculate the new pH of the buffer after the addition of
2.0 mL of 0.15 M NaOH

pOH  - log Kb  log [salt]
[base]

 -log 1.8x10 - 5  log 0.5997/1.002
0.0603/1.002

 5.74

pH 14 - 5.74 104
 8.26

Exercise 1

a) Calculate the pH of a 1 L buffer solution

containing 1.5 M CH3COOH and 0.20 M of

CH3COONa. pH = 3.87

b) Calculate the new pH of the buffer after the

addition of 2.0 mL of 0.1 M HCl. pH = 3.87

c) Calculate the new pH of the buffer after the
addition of 2.0 mL of 0.2 M NaOH. pH = 3.87

[ Ka = 1.8 x 10-5 ]

105

Exercise 2

a) Calculate the pH of the solution prepared by mixing

500 mL 0.1 M hydrazinium chloride, N2H5Cl with

500 mL 0.2 M hydrazine, N2H4. pH = 8.53

b) Calculate the new pH of the buffer after the addition

of 2.0 mL of 0.1 M HCl. pH = 8.53

c) Calculate the new pH of the buffer after the addition of

2.0 mL of 0.2 M NaOH pH = 8.53

[ Kb N2H4 = 1.7 x 10-6 ]

106

Excercises (Solution)

1. a) Acidic buffer

pH  -log Ka  log [salt]
[acid]

 -log 1.8 x10-5  log 0.2
1.5

 3.87

107

1. b) Addition of 2.0 mL of 0.1 M HCl

H3O+ + CH3COO- CH3COOH + H2O

ni 2x10-4 0.2 1.5

Δ -2x10-4 -2x10-4 +2x10-4

nf 0 0.1998 1.5002

[ ]f 0.1994 1.4972

V = 1000 mL + 2 mL pH  -log Ka  log [conjugate base]
= 1.002 L [weak acid]

 - log 1.8x10-5  log 0.1994
1.4972

 3.87 108

1. b) Addition of 2.0 mL of 0.1 M HCl

H3O+ + CH3COO- CH3COOH + H2O

ni 2x10-4 0.2 1.5

Δ -2x10-4 -2x10-4 +2x10-4

nf 0 0.1998 1.5002
[ ]f 0.1994 1.4972

V = 1000 mL + 2 mL
= 1.002 L

109

1. c) Addition of 2.0 mL of 0.2 M NaOH

pH  -log Ka  log [conjugate base]
[weak acid]

 - log 1.8x10-5  log 0.2000
1.4966

 3.87

110

2. a) Basic buffer

V = 500+500 = 1000mL of buffer

nNH4Cl = 0.05 mol nNH3 = 0.10 mol
[NH4Cl] = 0.05 M [NH3] = 0.10 M

pOH  -log Kb  log [salt]
[base]

 -log 1.7 x10-6  log 0.05
0.1

 5.47

pH  14  5.47 111
 8.53

2. b) Addition of 2.0 mL of 0.1 M HCl

H3O+ + N2H4  N2H5+ + H2O

ni 2x10-4 0.1 0.05

Δ -2x10-4 -2x10-4 +2x10-4

nf 0 0.0998 0.0502
[ ]f 0.0996 0.0501

V = 1000 mL + 2 mL
= 1.002 L

112

2. b) Addition of 2.0 mL of 0.1 M HCl

pOH  - log Kb  log [N2H5 ]
[N2H4 ]

 -log 1.7x10 -6  log 0.0501
0.0996

 5.47

pH  14 - 5.47
 8.53

113

2. c) Addition of 2.0 mL of 0.2 M NaOH

OH- + N2H5+  N2H4 + H2O

ni 4x10-4 0.05 0.1

Δ -4x10-4 -4x10-4 +4x10-4

nf 0 0.0496 0.1004

[ ]f 0.0495 0.1002

V = 1000 mL + 2 mL
= 1.002 L

114

2. c) Addition of 2.0 mL of 0.2 M NaOH

pOH  - log Kb  log [N2H5 ]
[N2H4 ]

 -log 1.7x10 -6  log 0.0495
0.1002

 5.47

pH  14 - 5.47
 8.53

115

7.2
ACID-BASE TITRATIONS

116

At the end of this topic , students should be able to:
a) Describe the titration process and distinguish between

the end point and equivalence point.
b) Sketch and interpret the variation of pH against titrant

volume for titrations between
i. strong acid-strong base
ii. strong acid-weak base
iii. weak acid-strong base
c) Identify suitable indicators for acid-base titrations. 117

7.2 ACID-BASE TITRATIONS

Titration
 A method for determining the concentration of

a solution using another solution (known
concentration), called standard solution.

Standard Solution
 A solution of accurately known concentration.

Titration Curve
 a graph of pH versus volume of titrant.

118

Titration apparatus

Titrant

Analyte

119

The equivalence point
 the pH at which the number of moles of

OH- ions added to a solution is equal
stoichiometrically to the number of
moles of H+ ions originally present.

The end point
 is the pH when the indicator just

changes colour.

120

Indicators
 is a weak organic acids or bases that change

colour over a range of pH values.

Table 1: Some Common Acid-Base Indicators

INDICATOR COLOUR pH range

Thymol blue In Acid In Base
Bromophenol blue
Methyl orange Red Yellow 1.2 – 2.8
Methyl red
Chlorophenol blue Yellow Bluish Purple 3.0 – 4.6
Bromothymol blue
Cresol Red Orange Yellow 3.1 – 4.4

Red Yellow 4.2 – 6.3

Yellow Red 4.8 – 6.4

Yellow Blue 6.0 – 7.6

Yellow Red 7.2 – 8.8

Phenolphthalein Colourless Reddish Pink 8.3 – 10.0

How to determine the suitable indicator for
titrations?

Choose an indicator which the endpoint pH
range lies on the steep portion of the titration
curve.

This choice ensures that the pH at the
equivalent point will fall within the range over
which indicator changes color.

122

How to determine the suitable indicator for
titrations

Type pH Suitable Indicator
jump

Strong acid - Strong 3 – 11 Any indicator except thymol
base blue

Strong acid – Weak 3 – 7 Methyl orange, methyl red,
base chlorophenol blue,

bromophenol blue

Weak acid – strong 7 - 11 Phenolphthalein, cresol red
base

123

TYPES OF TITRATIONS

1.Strong acid - strong base titrations
2. Strong acid – weak base titrations
3. Weak acid – strong base titrations

124

Strong acid-strong base titration curves

The pH starts out
low, reflecting the
high [H3O+] of the
strong acid, and
increases gradually
as acid is being
neutralised by the
added base.

125

 The pH rises
sharply when the
mole of OH- that
have been added
nearly equal the
mole of H3O+ .

126

 An additional drop or two of 127
base neutralises the
final tiny excess of acid
and introduces a tiny
excess of base, so pH
jumps 6 to 8 unit.

 Beyond this sharp portion,
the pH increases slowly
again as more base is
added.

Weak acid-strong base titration curves

 The initial pH is 128
higher than strong
acid-strong base
titration curve
because the weak
acid dissociates only
slightly, less H3O+ is
present.

 A gradually rising
portion of the curve,
called the buffer
region, appears
before the sharp rise
to the equivalence
point.
•pH jump: 7 to 11

129

 The pH at the
equivalence point >
7.0.

The solution contains
the basic salt.

130

Weak base-strong acid titration curves

 The pH starts above 7.0
(~11) because the weak
base dissociates only
slightly.

 The pH gradually
decreases in the buffer
region.

131

 After the buffer region, 132
the curve drops
vertically to the
equivalence point.
•pH jump: 3 to 7

 Beyond the
equivalence point, the pH
decreases slowly as
excess H3O+ is added.

How to sketch a titration curve

Steps :

1. Calculate the initial pH of the solution
- Identify the analyte (solution in the conical
flask and given volume). Whether strong acid,
weak acid, strong base or weak base.

2. Determine the equivalence point :
- Volume ( use formula, MaVa  a )

MbBb b

- pH (depends on pH of salt)

133

3. pH jump (steep portion / sharp portion)
- depends on the type of the titration

Type pH jump
Strong acid - Strong base 3 – 11
Strong acid – Weak base 3–7
Weak acid – strong base 7 - 11

4. Identify the final pH

- Depends on the [ ] of the titrant (solution in the
burette).

134

Example 1 :

Sketch the titration curve of 25.0 mL 0.10 M HCl with
0.10 M NaOH.

Step 1 :

Analyte is a strong acid,
HCl – dissociates completely

HCl(aq) H+(aq) + Cl- (aq)

[H+] = [HCl] = 0.10 M
pH = - log [H+]

= 1.0

135

Step 2 : At equivalence point

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

- pH equivalence:

- the solution is NaCl (aq)
- both Na+ and Cl- do not hydrolyse
- therefore pH = 7.0

- Volume equivalence:

n HCl = n NaOH

MHCl VHCl = MNaOH VnaOH

VNaOH = 0.10 x 25 = 25 mL
0.10

136

Step 3 :

Type of titration : strong acid-strong base
pH jump : 3 – 11

Step 4 :

Titrant is a strong base, NaOH.

NaOH (aq) Na+ (aq) + OH- (aq)

[OH-] = [NaOH] = 0.10 M

pOH = - log 0.10 = 1
pH = 14 – 1

= 13

Final point approaching pH =13.

137

Sketch a titration curve

13 equivalent point
pH 11

7

3

1

25
Volume of NaOH added (mL)

138

Example 2 :

Sketch the titration curve of 25.0 mL 0.10 M NH3 and
0.10 M HCl.

Step 1 :

Analyte is a weak base, NH3.

NH3 (aq) + H2O(l) NH4+ (aq) + OH- (aq)
[OH-] = [x] = ?

1.8 x 10-5 = x2

0.10-x

[x] = 1.3 x 10-3 M

pOH = -log (1.3 x 10-3 )

= 2.9
pH = 14 – 2.9

= 11.1 139 139

Step 2 : At equivalence point NH4Cl (aq)
HCl(aq) + NH3 (aq)

- pH equivalence

- the solution is NH4Cl (aq)
- only NH4+ hydrolyses to form H3O+
- therefore pH < 7.0

- Volume equivalence

n NH3 = n HCl
M VNH3 NH3
=1
MHCl VHCl
M VNH3 NH3 1
= MHClVHCl
VNH3 = 0.10 x 25.0 = 25 mL

0.10

140

Step 3 :
Type of titration : strong acid-weak base
pH jump : 3 – 7

Step 4 :
Titrant is a strong acid, HCl.

 pH   log H3O

 log 0.1
1

Final point approaching pH = 1

141 141

Sketch a titration curve

pH
11

7 equivalent point
pH<7

3

1 142

25
Volume of HCl added
(mL)

Example 3 :

Sketch the titration curve of 25.0 mL 0.10 M CH3COOH and
0.10 M NaOH.

Step 1 :

Analyte is a weak acid, CH3COOH.

CH3COOH (aq) + H2O(l) CH3COO- (aq) + H3O+ (aq)

[H3O+] = [x] = ?

1.8 x 10-5 = x2

0.10-x
x = 1.3 x 10-3 M
pH = -log (1.3 x 10-3 )

= 2.9

143

Step 2 : At equivalence point

CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l)

- pH equivalence

- the solution is CH3COONa (aq)
- only CH3COO- hydrolyses to form OH-
- therefore pH  7.0

- Volume equivalence

n NH3 = n HCl
M VCH3COOH CH3COOH =1

M VNaOH NaOH 1

M V M V=CH3COOH CH3COOH
NaOH NaOH

VNaOH = 0.10 x 25.0 = 25 mL

0.10

144

Step 3 :
Type of titration : weak acid-strong base
pH jump : 7 - 11

Step 4 :
Titrant is a strong base, NaOH.

 pOH   log OH
  log 0.1
1

pH  14  1
 13

Final point approaching pH =13

145

Sketch a titration curve

13
pH 11

7

3

25
Volume of NaOH added (mL)

146

Exercise 1

What is the colour of the solution when 3 drops
of the below indicators are added separately to
water (pH = 7) ?

Indicator pH range Colour Change

Phenolphthalein 8.2 – 10.0 Colourless → reddish pink

Methyl orange 3.2 – 4.2 Red → Yellow
Blue
Bromothymol blue 6.0 – 7.6 Yellow →

Phenol red 6.8 – 8.4 Yellow → Red

147

Exercise 2

The pH range and the colour change for 3
indicator X, Y and Z is shown in the table
below. Determine X, Y and Z.

Indicator pH range Colour Change

X 1.2 – 2.8 Red → Yellow

Y 6.0 – 7.6 Yellow → Blue

Z 8.3 – 10.5 Colourless → Reddish pink

148

7.3
SOLUBILITY EQUILIBRIA

149

At the end of this topic , students should be able to:

a)Define solubility, molar solubility and solubility

product, Ksp.
b)Calculate Ksp from concentration of ions and vice

versa.

c) Predict the possibility of precipitation of slightly

soluble ionic compounds by comparing the values

of ion-product, Q to Ksp.
d)Define and explain the common ion effect.

e)Perform calculations related to common ion

effect. 150