SK015

Answer Exercise 7 :

K C  1.0 x10 3 At 25 oC, Kc = 7.0 x 1025 for the reaction :

K'C  1 2SO2 (g) + O2 (g) 2SO3 (g)
KC

1 What is the value of Kc for the reaction :
 1.0 x10 3

K'C  31.62 SO3 (g) SO2 (g) + ½O2 (g)

38 39

Answer Heterogeneous Equilibrium
• A heterogeneous equilibrium is achieved from
K C  7.0 x1025
a reversible reaction involving reactants and
K'C  1 product that are in different phases.
KC • The concentration of pure substances, liquid
or solid can be neglected or constant, so its
1 concentration is not included in the
 7.0 x1025 equilibrium expression, Kc and Kp.

K'C  1.2 x10 13 41

40

Heterogeneous equilibrium • Concentration of a substance is define as amount
(in mol) per unit volume
 In a heterogeneous system, all substances are in
different phases. • For pure solid or liquid, concentration correspond
to its density.
Example :
• Density of solid or liquid is almost constant at any
temperature

In the following reaction, • Since the concentration of pure substances i.e
liquid or solid is constant, its concentration is
CaCO (s) CaO(s) + CO (g) not included in the equilibrium expression
3 2

The system involves a gas, CO2 in equilibrium • So, the equilibrium constant expression, K are :
with two solids , CaCO3 and CaO
Kc = [CO2] and Kp = PCO2

42 43

198

Exercise 8: Answer
Write the equilibrium constant expression in
terms of concentration and pressure for the i. Kc  CO2   Kp  PCO2
following reactions.
 Kp
ii. Kc  CO2   PCO 2
CO
PCO 

i. CaCO3 (s) CaO (s) + CO2 (g) iii. Kc CO PCO 
CO2  H2 
PCO2 PH2
    Kp 

ii. FeO (s) + CO (g) Fe (s) + CO2 (g)

iii. CO2 (g) + H2 (g) CO (g) + H2O (l)

44 45

Exercise 9 : Answer

Consider the following equilibrium at 295 K :   KP  PNH3 PH2S

NH4HS (s) NH3 (g) + H2S (g)  0.265 0.265 

The partial pressure of each gas is 0.265 atm.  0.07
Calculate Kp for the reaction.

46 47

Exercise 10 : Answer

At equilibrium, the pressure of reacting mixture  KP  PCO 2

CaCO3 (s) CaO (s) + CO2 (g)  0.105

is 0.105 atm at 350oC. Calculate Kp for this
reaction.

48 49

199

LEARNING OUTCOMES 6.2 The Relationship between Kc and Kp
• For general reaction :
• At the end of this topic, students should be able
to : aA (g) + bB (g) cC (g) + dD (g)

(c) use the equations, Kp = Kc(RT)∆n to solve K c  C c D d and Kp  PC c PD d
equilibrium problems. A a B b PA a PB b

(d) calculate Kc, Kp or quantities of species • The relationship is derived from the PV = nRT
present at equilibrium.
equation. PB  nVB RT  [B] RT

PA  nVA RT  [A] RT

PC  nVC RT  [C] RT PD  nD RT  [D] RT
V

50 51

• Substituting this relationship into the expression for Exercise 11 :
Kp,
PC c PD d RT c RT d
Kp  PA a PB b  [C] RT a [D] RT b The equilibrium constant Kp for the reaction
[A] [B]

[C] c [D] d x RT c RT d 4NH3 (g) + 7O2 (g) 4NO2 (g) + 6H2O (g)
 [A] a [B] b RT a RT b

 Kc (RT)(cd) - (ab) is 1.8 x 10-5 at 350°C. What mis olK-1c for this
reaction ? ( R= 0.08206 L atm K-1 )
 Kp  Kc (RT)Δn

n = (number of moles gaseous products) -
(number of moles of gaseous reactants)

= (c+d) – (a+b)

52 53

Answer Exercise 12 :

KP  K C RT n Calculate 1K0c-4f.or the following reaction at STP if
1.8 x 10 5  K C 0.08206 x 623 10 11 Kp= 2.1 x

K c  9.2 x 10 4 CaCO3 (s) CaO (s) + CO2 (g)

( R = 0.08206 L atm mol-1 K-1 )

54 55

200

Answer Determination of the Kp, Kc & quantities of
species at equilibrium.
KP  K C RT n
2.1x 10 4  K C 0.08206 x 273 1 For a reversible reaction :

K c  9.37 x 10 6 A (g) + 2B (g) 2C (g)

56 If a moles of reactant A and b moles of reactant B
was allowed to achieve equilibrium in V liter
container.

… the ‘ICE΄ Table may help..

Let x = amount (mol) of reactant that dissociates (change)
n eq = n initial –n change

57

A (g) + 2B (g) 2C (g) OR

Initial mole (mol) Initial A (g) + 2B (g) 2C (g)
concentration (M)
Changes (mol) Changes (M)

Mole at Concentration at
equilibrium (mol) equilibrium (M)

Concentration at
equilibrium (M)

58 59

Example 1 SOLUTION : CH4 (g) + H2O(g) CO (g) + 3H2(g)
CO 3H2
Natural gas methane, CH4 reacts with steam to produce ninitial CH4 H2O
hydrogen gas: n 00

CH4 (g) + H2O(g) CO (g) + 3H2(g) 3.00 0.60 + x + 3x

In an experiment, 3.00 mol of methane and 0.60 of steam

were introduced into a 1 dm3 vessel. At equilibrium, 0.50 mol -x -x

of carbon monoxide is present in the mixture.

a) Calculate the number of moles of CH4 , H2O and H2 in the nat equilibrium 3.00 - x 0.60 - x x = 0.50 3x
equilibrium mixture.
[ ]at equilibrium 3.00 - x 0.60 – x x = 0.50 3x
b) Write an expression for the equilibrium constant, Kc for 1 dm3 1 dm3 1 dm3 1 dm3
the reaction.

c) Calculate the value of Kc at the temperature of the
experiment.

60 61

201

a) Calculate the number of moles of CH4 , H2O and Example 2
H2 in the equilibrium mixture.
Ans : CH4 = 2.5 mol ; H2O = 0.10 mol and H2 = 1.5 0.020 mol of SO3 is placed into a closed 1.52L
mol vacuum container and heated to 900K until
equilibrium is achieved. 0.0142 mol of SO3
b) Write an expression for the equilibrium constant, exist at equilibrium. Calculate Kc and Kp at
Kc for the reaction. 900K.

c) Calculate the value of Kc at the temperature of the 2SO (g) 2SO (g) + O (g)
experiment. 3 22
Ans : Kc = 6.75
63
62

SOLUTION : Calculate Kc
Given at equilibrium
2SO3 2SO2 O2
neq SO3 = 0.0142
n initial 0.020 00 0.020-2x = 0.0142
n
nat equilibrium -2x +2x +x x = 2.9 x 10-3
[ ]at equilibrium
0.0142 +2x +x KC  SO2 2 O2 
SO3 2

 0.0142  2x  x    3.8158 x 10-3 M 2 1.9079 x 10-3 M
1.52 1.52 1.52   9.3421 x103 M 2
x 2.9 x 10-3
9.3421x103 M  2 1.52  2.9 x 10-3
1.52
 3.8158 x 10-3M 1.9079 x 10-3M

 3.183x10 4

64 65

Calculate Kp Exercise

K P  KC RT n 1. The equal amount of hydrogen and iodine are

 3.1830 x104 (0.08206 x 900 )1 injected into a 1.50 L reaction flask at a fixed
 2.35 x 102
temperature.
66
H2(g) + I2(g) 2HI(g)

At equilibrium, analysis shows that the flask
contains 1.80 mole of H2, 1.80 mole of I2, and
0.520 mole of HI. Calculate Kc

Ans : 0.083

67

202

SOLUTION :

H2(g) + I2(g) 2HI(g) 2. At a fixed temperature, 0.200 mole of
hydrogen halide is injected into a 2.00L of
nat equilibrium 1.80 1.80 0.520 reaction flask and form hydrogen and iodine
[ ]at equilibrium gas.
1.8 1.8 0.520
 1.5  1.5  1.5 2HI(g) H2(g) + I2(g)

 1.2 M  1.2 M  0.3467 M

KC  HI2 If [ HI ]=0.078 M at equilibrium calculate the
H2 I2  value of Kc.

 0.3467 2 Ans : 0.019
1.21.2

 0.083

68 69

SOLUTION : SOLUTION :

2HI(g) H2(g) + I2(g) 0.2  2x  0.078 KC  H2 I2 
2.0 HI2
x  0.022
n initial 0.2 0 0
n -2x
nat equilibrium 0.2 – 2x +x +x  0.0112
[ ]at equilibrium 0.078 2
0.078 x x

H2   x I2   x  0.019
2 2
0.022 0.022
2 2
 0.011  0.011

70 71

3. At 440C, the equilibrium constant Kc for SOLUTION :
reaction,
H2(g) + I2(g) 2HI(g)
H2(g) + I2(g) 2HI(g) ninitial
0.200 0.200 0
-x +2x
n -x 2x
0.2-x
has a value of 49.5. If 0.200 mole of H2 and nequilibrium 0.2 – 2x
0.200 mole of I2 are placed into a 10.0 L
vessel and permitted to react at this [ ]equilibrium H2   0.2  x I2   0.2  x HI  2x
temperature, what will be the concentration 10 10 10
of each substance at equilibrium ?

Ans : [H ] = [I ] = 4.42 X 10-3 M
[HI] = 0.031 M

72 73

203

SOLUTION :

K C  49.5 H 2  0.2  0.1558
10
HI2 4. inAtopresasucrleosoefd0v.4e6s4sealtmthaot fcConOta2 iins introduced
H2 I2  graphite at
49 .5   4.42 x 10 3 M 1000 K. The reaction occurs as follow :

 2x  2
 10 
I2  0.2  0.1558 CO2 (g) + C (s) 2CO (g)
10
49 .5  0.2  x 2
10 
  4.42 x 10 3 M When the system reached equilibrium, the
 pressure in the vessel was found to be 0.746
2x atm. Calculate:
49 .5  0.2  x HI  2(0.1558 ) a) the partial pressure of each species at
10
7.0356  2x equilibrium
0.2  x  0.031 M b) the equilibrium constant, Kp

1.4071  7.0325 x  2x

2x  7.0325 x  1.4071

x  0.1558

74 75

SOLUTION : + C (s) 2CO (g) Answer
- 0
CO2 (g) - +x a) the partial pressure of each species
- +x at equilibrium

a) PT  0.746atm

Pinitial 0.464 PT  PCO2  PCO
P -x
 0.464x 2x0.746 b) the equilibrium constant, Kp

Pequilibrium 0.464-x x  0.282 b)  KP  PCO 2
PCO2
PCO2  0.464x
 0.4640.282  0.5642
 0.182atm
0.182
PCO  2x
 2x 0.282  1.75
 0.564atm

76 77

5. Nitric oxide, NO decomposes according to the SOLUTION : N2(g) + O2 (g)
following reaction, for which Kc = 83.3 at 4000oC.
2NO (g) 0 0
+x +x
2NO (g) N2 (g) + O2 (g) n initial 0.1333 x x

n  -2x

If 4.0 g of NO (g) is placed in a 2.0 L vessel and nat equilibrium 0.1333 – 2x
allowed to come to equilibrium, calculate :

i. the equilibrium concentrations of NO, N2 and  [ ]at equilibrium 0.1333  2x N2  x O2  x
O2. NO  2  2  2

ii. the equilibrium partial pressures for each gas.

iii. the mass of O2 in gram which is formed at
equilibrium.

78 79

204

SOLUTION : SOLUTION :

a) nNO 4g N2  0.0632 b) nO2 (at equilibrium)  nN2 (at equilibrium)  0.0632 mol
(14  16)gmol 1 2

 0.1333 mol PN2 V  nN2RT

K C  83.3  3.16 x 10 2 M PN2 (2)  0.0632 (0.08206)(4273.15)

83 .3  N2 O2  O2  0.0632 PN2  0.0632 (0.08206)(4273.15)
NO 2 2 2
PO2  PN2 11.08 atm
 x  2
 2 
83 .3  3.16 x 10 2 M

 0.1333  2x 2 nNO(at equilibrium)  6.9x103mol
2 
 HI  0.1333  2(0.0632 ) PNOV  nNORT
 2
x
83 .3  0.1333  2x  3.45 x 10 3 M PNO (2)  6.9x103 (0.08206)(4273.15)

9.1269  x PN2  6.9x 103 (0.08206 ) ( 4273 .15)
0.1333 2
 2x
PNO 1.21atm
1.2166  18.2538 x  x

x  18.2538 x  1.2166

x  0.0632 mol

80 81

SOLUTION : 6. A certain amount of C1a0C0O0o3Cundinergaoes0.t6h5e4rmaLl
decomposition at
c ) Mass of O2  0.0632 x 16 container until the system reached
 2.02 g equilibrium. The reaction established as
follow:

CaCO3 (s) CaO (s) + CO2 (g)

I3f.9thx1e0-e2,qucilaiblcruiulamtecothnestamnat,ssKP(inat 1000oC is
produced at equilibrium. g) of CaO

82 83

SOLUTION : LEARNING OUTCOMES 6.2
• At the end of this topic, students should be able
nNO 4g
(14  16)gmol 1 to :
(e) Determine the degree of dissociation, α .
 0.1333 mol (f) Predict the direction of net reaction by

KP  PCO2  3.9 x 10 2 comparing the values of Q and K.

nCO 2  PV 85
RT
3.9 x 102 0.654  205
 0.08206 1273 

 2.44 x 10 4 mol

mass CaO  2.44 x 104 56 

 0.0137g

84

Degree of dissociation, α • If α = 1 or 100% : complete dissociation
occurs
• Dissociation reaction : a molecule is broken
down into smaller molecules, atoms or ions. • If α less than 1 or 100% : incomplete or
partial dissociation occur.
• Degree of dissociation, α : fraction of molecules
that dissociates.

α= Amount dissociates = x
Initial amount c

• Percentage of dissociation : percentage of
molecules that dissociates.
% α = Amount dissociates x 100%

Initial amount

86 87

Exercise 1 : SOLUTION :

If 3.00 moles of A are placed in 1.00 L container at [ ]initial A (g) 2B(g)
[ ] 3 0
298 K, calculate the degree of dissociation, α, of A +2x
for the following reaction : -x
2x
A (g) 2B (g) [ ]equilibrium 3- x

The equilibrium constant, Kc at 298 K is 0.10. KC  B2
A 

0.10   2x2
3  x 

x  0.2616 M

  0.2616
3
 0.0872

88 89

Exercise 2 : SOLUTION :

At a pressure of 4.4 x 105 Pa and a PCl5 (g) PCl3 (g) + Cl2 (g)
temperature of 150 oC, phosphorus
pentachloride, PCl5 is 25% dissociated. n initial 1 00
Calculate the partial pressure equilibrium n -0.25
constant for this reaction. +0.25 +0.25

PCl5 (g) PCl3 (g) + Cl2 (g) n at equilibrium 1- 0.25 0.25 0.25

90 91

206

Answer   b)
a) PT  0.746atm KP  PPCl3 PCl2 Exercise 3
PCl5
nT  nPCl5  nPCl5  nCl2 The degree of dissociation of dinitrogen tetroxide at
 1 0.25 0.250.25  8.8 x104 2 250°C and 1 atm is 0.15. Calculate the degree of
1.25mol dissociation at 250°C and 5 atm.
 2.64x104
PPCl5  XPCl5PT
N2O4 (g) 2NO2 (g)
0.75 x 4.4 x105  2.9x104
 1.25

 2.64x105 Pa

PPCl5  XPCl3PT

 0.25 x 4.4 x105
1.25
 8.8 x104 Pa

PCl2  PPCl3 8.8 x104 Pa 93

92

SOLUTION : Answer

ninitial N2O4 (g) 2NO2 (g) Let the degree of dissociation at 1 atm = x
n 1 0
nequilibrium +2x   0.15  x  KP  PNO2 2
-x 2x PN2O4
PNO2  XNO2PT
1- x 0.3
1.15 x 1  0.2612

 0.261atm 0.739
 9.21x10-2 atm

PPCl5  XPCl5PT
0.85
 1.15 x 1

 0.739 atm

94 95

Answer   0.15  b

SOLUTION : 97 PNO2  X NO2 PT
Let the degree of dissociation at 5 atm = b 2b
207 b
 1 2b x 5

N2O4 (g) 2NO2 (g) PN2O4  X PCl5 PT

ninitial 1 0  1b x 5
n +2b 1 b  2b
-b
2
nequilibrium 1- b 2b  1  2b 2b x 5 
b
KP 
1b x 5
1 b  2b

 9.21 x 102

b  0.068

96

Equilibrium Calculations When Kc is Very Small If ; % α = x x 100 is < 5%
• The concentration change, x can often be [A]initial
neglected if Kc is very small.
• By assuming x is very small than [A]initial , assumption is allowed
∴ [A]equilibrium = [A]initial - x
≈ [A]initial If ; % α = x x 100% is > 5%
• Note : Check the assumption is valid or not : [A]initial

98  assumption is NOT allowed
 use quadratic formula

99

Reaction Quotient (Q) • For the reaction that have not reached
equilibrium :
• The equilibrium constant is used to predict the
direction in which a reaction mixture will aP (g) + bQ (g) cR (g) + dS (g)
proceed to achieve equilibrium.

• For the reaction at equilibrium : Q  Rc Sd  at non-equilibrium
Pa Qb conditions

aP (g) + bQ (g) cR (g) + dS (g) • To predict the direction of reaction, compare
the values of Keq and Q.
K eq  R c S d
P a Q b  at equilibrium

100 101

• 3 possible cases for a system are as follow : Exercise 1 :

The ratio of initial concentration of products to At 199C, the initial moles of mHo2l , Nre2spaencdtiNveHly3
reactants is too large. To reach equilibrium, are 2.0 mol, 1.0 mol and 2.0
i. Q > Keq products must be converted to reactants. was placed in a 1is.00L.10st5ainaltesthsi-ssteteeml. pIfertahteurKec,
 [Product]  and [Reactant]  for the reaction
 The system proceeds from right to left. decide whether the system is at equilibrium. If it
The initial concentrations are equilibrium is not, predict which way will the reaction
ii. Q = Keq concentration. The system is already at proceed.
equilibrium.
The ratio of initial concentrations of products 103
to reactants is too small. To reach equilibrium,
iii. Q < Keq reactants must be converted to products.
 [Product]  and [Reactant] 
 The system proceeds from left to right.

102

208

Answer Exercise 2 :

3H 2 ( g )  N 2 ( g )  2 NH 3 ( g )

NH 3   2 M The equilibrium cforonmstannitt,ricKcofxoidr ethaenfdormmoalteiocnulaorf
nitrosyl chloride
H 2   2 M chloride
N2   1M

QC  NH 3 2 2NO (g) + Cl2 (g) 2NOCl (g)
H 2 3 N 2 

is 6.5 x 106 at 35°C. In a certain experiment, 2.0 x
10-2 mol NO, 12.3 x 10-1 mol ICnl2w, ahnicdh6d.8iremcotiloNn OwCilll
22 are mixed in a 2.0 L flask.
the system proceed to reach equilibrium?
 23 1

 0.5

Qc K c
 equilibriu m shift to the left .

104 105

Answer Exercise 3:
nNO  2.0 x 102 mol , NO  0.01M An equilibrium mixture in a 1.00 L vessel at 490 K contains
0.228 mol H2(g), 0.228 mol I2(g) and 1.544 mol HI(g). An
nCl2  12.3 x101 mol , Cl2   0.615 M additional 0.200 mol H2(g) is then pumped into the container
and allowed to achieve to the new equilibrium.
nNOCl  6.8 mol , NOCl  3.4 M

V  2.0 L i. Calculate the Kc of the following reaction :

QC  NOCl 2 H2 (g) + I2 (g) 2HI (g)
NO 2 Cl2 

ii. Calculate the value of Q when H2(g) is added before the
new equilibrium is established.
3.42
iii. What is the effect of increasing the amount of H2(g) to the
 0.012 0.615 equilibrium position ?

 1.88 x 105

Qc Kc iv. Calculate the new equilibrium concentrations for each gas.
 equilibrium shift to the right.

106 107

SOLUTION : SOLUTION :

H2(g) + I2(g) 2HI(g) ii. When 0.2 mol of H2 is added

nequilibrium 0.228 0.228 1.544 H2(g) + I2(g) 2HI(g)
[ ]equilibrium
0.228 0.228 1.544 n 0.228 + 0.2 = 0.228 1.544
1 1 1 0.428
0.228 1.544
 0.228  0.228 1.544 [ ] 0.428 1 1
1
HI2  0.288 1.544
H2 I2   0.428

i. KC  HI2
H2 I2 
1.544 2 QC  iii. Since the value of Qc < Kc ,
0.228 2 the equilibrium position shift to the right
49.5  1.544 2
0.428 2
 45.9 49 .5 

 24.4

108 109

209

SOLUTION : iV HI2
H2  I2 
H2(g) + I2(g) 2HI(g) KC 

ninitial 0.428 0.228 1.544 49.5  1.544  2x2
n +2x 0.428  x  0.228
-x -x 1.544 + 2x  x

nequilibrium 0.428 –x 0.228-x x  0.0622 M
[ ]equilibrium
 0.428  x  0.228 x 1.544 2x HI 1.668 M
1 1 1
 HI22   0.166 M
  0.366 M

110 111

Exercise 4 : Answer
Consider the following equilibrium process at 686°C:
H2O CO
i. KC  H2 CO2 

CO2 (g) + H2 (g) CO (g) + H2O (g) 0.04 0.05
0.0450.086
The equilibrium concentrations of the reacting species 
are [CO] = 0.050 M, [H2] = 0.045 M, [CO2] = 0.086 M,
and [H2O] = 0.040 M.  0.5168

i. Calculate Kc for the reaction at 686°C.

ii. If we add CO2 to increase its concentration to 0.50
mol L-1, what will the concentrations of all the gases
be when equilibrium is reestablished ?

112 113

SOLUTION : SOLUTION :

ii. ii. K C  0.52
CO H 2 O 
0.52  CO 2  H2 

CO2 (g) + H2 (g) CO (g) + H2O (g) 0.52  0.05  X0.04  X
0.5  X0.045  X
0.50 0.045
[ ]initial -x -x 0.050 0.04 x  0.02516 M
[ ]
+x +x CO 2  0.5  X

[ ]equilibrium 0.50 – x 0.045-x 0.051 + x 0.04 + x  0.47 M

H2  0.045  X

 0.020 M

CO  0.05  X

 0.075 M

H2O  0.040  X

 0.065 M

114 115

210

6.3 LEARNING OUTCOMES
Le Chatelier’s Principle At the end of this topic, students should be able to :

116 (a) State Le Chatelier’s Principle.

Le Chatelier’s Principle (b) Apply the Le Chatelier’s Principle to explain
The principle states that: the effect of the following factors on a system at
equilibrium:
If a system at equilibrium is disturbed by an external
change (concentration, pressure or temperature), the i. concentration of reacting species
system will shift its equilibrium position to reduce the ii. pressure by changing volume
effect of the change. iii. addition of inert gas at constant volume
•The position of equilibrium in a system is affected by: iv. addition of inert gas at constant pressure
v. temperature
a) concentration vi. catalyst
b) pressure / volume
c) temperature 117
The effect of each can be predicted qualitatively by Le
Chatelier’s Principle. Factors Affecting Equilibrium System

118 • The main factors that affect the equilibrium
system are :
(a) Changes in Concentrations (a) Concentration
(b) Pressure / volume
• Adding a reactant or product shifts the (c) Temperature
equilibrium away from the increase.
119
• Removing a reactant or product shifts the
equilibrium towards the decrease. N2(g) + 3H2(g) 2NH3(g)

120 121

211

Example 1: H2 (g) + I2 (g) 2HI (g) Example 1: H2 (g) + I2 (g) 2HI (g)

Changes in Effect to Equilibrium Position Changes in
Concentration Concentration
Effect to Equilibrium Position
 The system must consume the H2 and
produce product.
If H2 (reactant) is  The position of equilibrium will
added to a  The position of equilibrium will shift to shift to the left in order to
system. the right in order to decrease [H2]. decrease [HI].

 Therefore, If HI (product) is
added to a system.  HI decompose to produce H2
[HI] increases and [I2] & [H2] decrease.
and I2
If H2 (reactant) is  The position of equilibrium will shift to  Therefore, [H2] & [I2] increase
removed to a
system. the left in order to increase [H2]. and [HI] decreases.
 Therefore,

[HI] decreases and [I2] & [H2] increase.

122 123

Example: SOLUTION
At equilibrium the colour of the solution is light
Consider the reaction below:
blood red
Fe 3+ + 2SCN- Fe (SCN)+2(aq)
(aq) (Blood red) i) When Fe3+ ion is added
(Yellowish) - The system will reduce the effect of the additional

Discuss the effect of adding the following concentration of Fe3+ion. More Fe3+ ions react
substances to the equilibrium mixture. with SCN- ions

i. iron(III) or Fe3+ - Thus more products, Fe (SCN)+2 will be formed
and the colour intensity of blood red increase
ii. sodium hydroxide, NaOH
- The equilibrium position shifts forward
124
.

125

ii) When NaOH is added b) Effect of Pressure

- NaOH reacts with Fe3+ to form Fe(OH)3 The pressure of a system may be changed by:
i) changing the moles of reactant or product
3NaOH + Fe(NO3)3 Fe(OH)3 + 3NaNO3 ii) changing the volume
iii) adding inert gas to the system
- Thus decreasing the concentration of Fe3+.
Affects only the reversible reaction involving
- The system will overcome the change by increasing gaseous reactants/products
the concentration of Fe3+.
127
- So the reaction shifted to the left (reaction moves
backward or reverse) to achieve a new equilibrium

.
- Therefore more Fe3+ ions will be formed and the

solution becomes more yellowish

126

212

i) Changing moles of reactant or product ii) Changes in Volume:

Adding or removing the gaseous reactant or Note that : Pα 1
product at constant volume has the same effect as V
changing the concentration
(V increase) (V decrease)

Shift the position of equilibrium Shift the position of equilibrium

Example: in the direction to increase the in the direction to decrease the
Consider the following system at equlibrium
number of moles of gas. number of moles of gas.

2SO2(g) + O2(g) 2SO3(g) • Example : N2 (g) + 3H2 (g) 2NH3 (g)
 when V decrease
• When SO2 gas is added to the system, the partial
pressure of SO2 increases (which means [SO2]  the position of equilibrium will shift in the
increases) direction to the lower number of moles of gas
particles.
• The position of equilibrium shifts to the right to
reduce the effect of high concentration of SO2.  the position of equilibrium will shift to the right

128 129

iii) Effect of Adding Inert Gases ii. At Constant Pressure
i. At Constant Volume • Adding inert gas at to the system a constant
• The total number of gaseous molecules increase.
• Total pressure increase of the equilibrium system external pressure increase the volume of the
container.
increases. • The partial pressure for the gases in the system
• The volume and the quantity of the reacting species are lowered.
• have the same effect as P  or V 
remain constant so the partial pressure of each •  the position of equilibrium will shift in the
gases in the equilibrium system remains unchanged. direction to increase the number of moles or
•  does not change the position of equilibrium molecules

and equilibrium constant 131

130

Exercise : (c) Changes in Temperature
Consider the gas-phase reaction :
• Increasing temperature favors the endothermic
2CO (g) + O2 (g) 2CO2 (g) reaction

• Decreasing temperature favors the exothermic
reaction

• Kc and Kp changes in value as the T changes.

Predict the shift in equilibrium position when i. Exothermic Reaction
helium gas is added to the equilibrium mixture

• Example :

i. at constant pressure N2 (g) + 3H2 (g) 2NH3 (g) ΔH = - 92.83 kJ
ii. at constant volume

132 133

213

i. Exothermic Reaction 2NH3 (g) ΔH =92.83 kJ ii. Endothermic Reaction 2NO (g) ΔH = +ve
Example : • Example : N2 (g) + O2 (g)
N2 (g) + 3H2 (g)

Temperature increase Temperature decrease Temperature increase Temperature decrease

• Shift the position of equilibrium in • Shift the position of equilibrium in

• Shift the position of equilibrium in the the direction as to decrease the the direction as to increase the

direction as to decrease the heat • Shift the position of equilibrium in the heat supplied. heat.

supplied. direction as to increase the heat. • So, the position of equilibrium will • So, the position of equilibrium will

• So, the position of equilibrium will shift • So, the position of equilibrium will shift shift to the right in order to absorb shift to the left in order to release

to the left in order to absorb the heat to the right in order to release the heat. the heat supplied. the heat.

supplied.

• More ammonia will dissociate. • More ammonia will produce. • More NO will produce. • More NO will dissociate

• Therefore, [NO] increases, [N2] & • Therefore, [NO] decreases, [N2] &
[O2] decrease. [O2] increase.
• Therefore, [NH3] decreases, [H2] & [N2] • Therefore, [NH3] increases, [H2] & [N2]
increase. decrease.
• The value of Kc or Kp increases. • The value of Kc or Kp decreases.

• The value of Kc or Kp decreases. • The value of Kc or Kp increases.

134 135

• Note: Exercise :
Increasing temperature favors the endothermic
reaction The graph below shows the percentage of C at 100K
Decreasing temperature favors the exothermic and 200K for reaction:
reaction
A2 (g) + B2 (g) 2C (g)
• Adding heat (i.e. heating the vessel) favors away
from the increase: %C
– if H < 0, adding heat favors the reverse reaction. 100K
200K
• Removing heat (i.e. cooling the vessel), favors
towards the decrease: time
– if H < 0, cooling favors the forward reaction.
i. Write the equilibrium constant expression Kp for
136 the reaction.

ii. Based on the graph, explain whether the forward
reaction is endothermic or exothermic.

iii. Explain whether the value of Kp is high at 100K or
200K.

137

Solution: (d). Effect of Catalyst

ii. When temperature increases, less product C is formed, therefore the • A catalyst increase the rate of a chemical reaction by
forward reaction is exothermic providing an alternative pathway with lower activation
energy, Ea.
ii. At 100K more product C is formed, therefore the value of Kp is high.
• Activation energy, Ea is the minimum amount of
138 energy required by reactant molecules to initiate a
chemical reaction.

• The presence of catalyst does not :
i. change the composition of equilibrium mixture
ii. affect the position of equilibrium
iii. affect the value of equilibrium constant, K

• Time taken to achieve equilibrium is shortened.

139

214

A (g) + B (g) C (g) + D (g)

uncatalyzed catalyzed

140

215

CHAPTER 6: CHEMICAL
EQUILIBRIUM

QUESTIONS

TUTORIAL.MEKA.KUMBE

lecture notes & questions

216

CHEMISTRY
TUTORIAL TOPIC 6: CHEMICAL EQUILIBRIUM

6.1: Dynamic Equilibrium & 6.2: Equilibrium Constant

1. a) Define reversible reaction.
b) State characteristics of a system when it reaches dynamic equilibrium.
c) The curve of concentration versus time when dinitrogen tetraoxide, N2O4
decomposes to form nitrogen dioxide, NO2 is as follows;
N2O4 (g) ⇋ 2NO2(g)

Concentration, M

[NO 2]

[N2O4]
t1 Time, t

Based on the curve, make two conclusions about concentration and rate of
reaction after t1.

2. Write the expressions of Kc and Kp for the following reactions. In each case, state
whether the reaction is homogenous or heterogeneous.
a) H2CO3(l) + H2O(l) ⇋ HCO3- (aq) + H3O+(aq)
b) 2NaCl(aq) + CaCO3(s) ⇋ Na2 CO3(s) + CaCl2(s)
c) Hb(CO)4(aq) + 4O2(g) ⇋ Hb(O2)4(aq) + 4CO(g)
d) C2H2(g) + 2Br2(g) ⇋ C2H2Br4(g)
e) Br2(l) ⇋ Br2(g)
f) 3Fe(s) + 4H2O (l) ⇋ Fe3O4(s) + 4H2(g)

3. The synthesis of NH3(g) at 298 K gives Kc = 3.6×108 at equilibrium,
3H2(g) + N2(g) ⇋ 2NH3(g)

What is the value of Kc at 298 K for the following reaction?
NH3(g)⇋ ½ N2(g) + 3/2 H2

4. The equilibrium constant, Kp for the reaction
2SO2(g) + O2(g) ⇋ 2SO3(g)

at 25oC is 3.0×1025 . Calculate Kp and Kc at the same temperature for the reaction:
SO2(g) + ½ O2(g) ⇋ SO3(g)

217

CHEMISTRY
TUTORIAL TOPIC 6: CHEMICAL EQUILIBRIUM

5. At 500 °C, the formation of NH3 is according to the following reaction:

N2(g) + 3H2(g) ⇌ 2NH3(g)

Calculate the value of the equilibrium constant, if an equilibrium mixture of NH3(g),
H2(g), and N2(g) at 500°C was found to contain 1.35M of H2, 1.15 M of N2, and
0.412M NH3.

6. The equilibrium 2NO(g) + Cl2(g) ⇋ 2NOCl(g) is established at 500 K. An
equilibrium mixture of the three gases has partial pressures of 0.05 atm, 0.30 atm and
1.20 atm for NO, Cl2 and NOCl respectively. Calculate Kp for this reaction at 500 K.

7. Iodine molecules react reversibly with iodide ions to produce triiodide ions.
I2(aq) + I−(aq) ⇌ I3−(aq)

If 1.0×10−3M of iodine and 1.0×10−3M of iodide ion is introduced in a 1.0-L vessel to
give 6.61×10−4M of iodide ions at equilibrium. Determine the equilibrium constant for
the reaction.

8. Cobalt metal can be prepared by reducing cobalt(II) oxide with carbon monoxide.

CoO(s) + CO(g) ⇌ Co(s) + CO2(g) Kc = 4.90 ×102

At equilibrium, if [CO2] = 0.100 M, determine the concentration of carbon monoxide.

9. In one experiment, 2.00 M of PCl5 was decomposed to produce chlorine gas and
phosphorus trichloride. Calculate the concentrations of PCl5, PCl3, and Cl2 at
equilibrium.
PCl5(g) ⇌ PCl3(g) + Cl2(g) Kc = 0.0211

10. A sample of NOCl with a pressure of 0.50 atm comes to equilibrium according to the
reaction below:
2NOCl(g) ⇌ 2NO(g) + Cl2(g) KP = 4.0×10−4

Calculate the partial pressures of NOCl, NO, and Cl2 at equilibrium.

11. The formation of sulphonyl chloride, SO2Cl2 is an exorthemic reaction:
SO2(g) + Cl2(g) ⇋ SO2Cl2(g)

Initially, 0.150 mol of SO2 and 0.350 mol of Cl2 are placed in a 2.0 L container at 25C.
At equilibrium, there is 0.047 mol of SO2Cl2 formed. Calculate Kp of the reaction at
25C.

12. Initially, 1.0 mol of NO2 gas is allowed to reach equilibrium in a 1.0 L flask. Calculate
the concentration of NO2 and N2 O4 gas at equilibrium if the Kc for the following
reaction is 8.3.

218

CHEMISTRY
TUTORIAL TOPIC 6: CHEMICAL EQUILIBRIUM

13. Solid NH4SH is introduced into an evacuated flask at 24C. The following reaction
takes place:
NH4SH(s) ⇋ NH3(g) + H2S(g)
At equilibrium the total pressure is 0.307 atm. What is the Kp for this equilibrium at
24C?

14. Nitrosyl bromide, NOBr, decomposes to form nitrogen monoxide and bromine.
2NOBr(g) ⇋ 2NO(g) + Br2(g)

If 1.0 mol of nitrosyl bromide in a 1.0 L flask decomposes to give 0.045 mol of
bromine at equilibrium and the temperature is 77C, determine the
a) equilibrium constant Kc and Kp.
b) degree of dissociation of NOBr.

15. When heated at high temperature, iodine vapour dissociates as follows:
I2(g) ⇋ 2I(g)

In one experiment, a chemist found that when 0.027 mol of I2 was placed in a flask of
volume 0.24 L at 587 K, the degree of dissociation was 0.0126. Calculate Kc and Kp
for the reaction at this temperature.

16. The equilibrium constant, Kc for the formation of nitrosyl chloride, NOCl an orange-
yellow compound is 6.5×104 at 35C.
2NO(g) + Cl2(g) ⇋ 2NOCl(g)
In an experiment, 4.0×10‒2 mol of NO, 1.66×10‒2 mol of Cl2 and 6.8 mol of NOCl are
mixed in a 1.0 L container. In which direction will the reaction proceed to reach
equilibrium? Explain.

17. For the synthesis of ammonia
3H2(g) + N2(g) ⇋ 2NH3(g)

the equilibrium constant Kc at 375C is 1.2. Starting with [H2]0 = 0.64 M,
[N2]0 = 0.35 M, and [NH3]0 = 0.25 M, which of the gases will increase/decrease in its
concentration as the mixture reaches equilibrium? Explain.

18. Consider the following equilibrium process at 686C:
CO2(g) + H2(g) ⇋ CO(g) + H2O(g)

The equilibrium concentrations of the reacting species are [CO] = 0.025 M,
[H2] = 0.035 M, [CO2] = 0.066 M, and [H2O] = 0.030 M.
a) Calculate Kc for the reaction at 686C.
b) If we add H2 to increase its concentration to 0.5 M, what will the

concentrations of all gases be when equilibrium is re-established?

219

CHEMISTRY
TUTORIAL TOPIC 6: CHEMICAL EQUILIBRIUM

6.3: Le Chatelier's Principle

1. Ethane, C2H6 can be cracked at high temperature to yield ethene, C2H4 and hydrogen
gas according to the equation:
C2H6(g) ⇋ C2H4(g) + H2(g)
The standard enthalpy of formation of ethene is positive whereas that of ethane is
negative. State the effect on the equilibrium constant, Kp due to the changes of:
a) the temperature
b) the pressure

2. Given the reaction:

Fe2O3(s) + 3H2(g) ⇋ 2Fe(s) 3H2O(g)

A mixture containing 2.00 mol of Fe2O3 and 0.20 mol of hydrogen gas was allowed to
react in a 3.0 L flask at 613 K. At equilibrium, the value of Kc obtained was 0.064.
a) Calculate the concentration of H2O and H2 at equilibrium.
b) What is the value of Kp at 613 K.
c) Predict the direction of the reaction if

i. some hydrogen gas is added to the equilibrium mixture.
ii. the volume of the flask is decreased.
iii. 0.70 mol of Fe2O3 is added to the mixture at equilibrium.

3. The values of Kp at certain temperatures for the following reaction
2A(g) + B2(g) ⇋ 2AB(g)

are given in the table below:

Temperature Kp (atm‒1)

700 K 3×104

1000 K 13×10‒1

Deduce whether the forward reaction is endothermic or exothermic.

4. Nitrogen(IV) oxide dimerised as follows;

2NO2(g) ⇋ N2O4(g) ΔH = ‒ve

Based on Le Chatelier's principle, suggest ways that could increase the amount of

product.

220

CHEMISTRY
KUMBE TOPIC 6: CHEMICAL EQUILIBRIUM

1. Sulphuryl chloride, SO2Cl2 is a colourless liquid that boils at 69C. At a higher

temperature, the vapour dissociates into sulphur dioxide and chlorine.

2 2( ) ⇋ 2( ) + 2( )
An experiment was conducted by placing 3.174 g of SO2Cl2 into a 1.00 L flask which

was then sealed and heated to 100C. At equilibrium, the total pressure in the flask

was 1.30 atm.

a) Calculate the partial pressure of each of the three gasses present.

[0.140 atm, 0.580 atm]

b) Calculate the equilibrium constant, Kc at the temperature. [0.0780]

2. Nitrogen and hydrogen in the mole ratio of 1:3 is allowed to achieve equilibrium at

1000 kPa and 550C. The equilibrium mixture was found to contain 15% of ammonia.

2( ) + 3 2( ) ⇋ 2 3( ) ΔH = -ve

Calculate Kp for the equilibrium at the same temperature.
[4.09×10−7]

3. The equilibrium constant, Kc for phosphorus pentachloride, PCl5 is 2.63 at 250oC.
3( ) + 2( ) ⇋ 5( )

If in an experiment 0.205 g PCl5, 0.120 g PCl3 and 2.020 g Cl2 are mixed in a 2.5 L

container,

a) in which direction will the system proceed to reach equilibrium?

b) determine the concentration of each reacting species at equilibrium.
[[PCl3] = 7.2×10−4 M, [Cl2] = 0.012 M, [PCl5] = 2.2×10−5 M]

4. Consider the equilibrium

CaCO3(s) ⇋ CaO(s) + CO2(g) ∆H = +ve

a) Calculate the values of Kp and Kc for the system at 525oC, if the total pressure

at equilibrium is 0.220 atm. [0.220, 3.36×10−3]

b) Predict the effect on the equilibrium position when the following changes are

made to the system;

i. some of the CaO is removed from the system.

ii. the pressure of the system is increased by decreasing the volume.

iii. the temperature of the system is raised.

iv. helium is added at constant volume.

v. helium is added at constant pressure.

221

CHEMISTRY
KUMBE TOPIC 6: CHEMICAL EQUILIBRIUM

5. At 500oC, a mixture of N2, H2 and NH3 are allowed to reach equilibrium.

2( ) + 3 2( ) ⇋ 2 3( ) ΔH = -ve

a) Given Kp=1.50×10−5, calculate the value of Kc. [0.0604]

b) What is the value of Kc at 500oC for the following reaction? [4.07]

3( ) ⇋ 1 2( ) + 3 2 ( )
2 2

c) The equilibrium constant, Kc for the formation of NH3 at 298 K is 3.6×108.

i. In an experiment, 0.10 mol N2, 0.30 mol H2, and 0.80 mol NH3 are

mixed in a 0.50 L container. In which direction will the reaction

proceed to reach equilibrium.

ii. State the effect of increasing the volume of the container for the system

that has reached equilibrium, E

222

CHEMISTRY
MEKA TOPIC 6: CHEMICAL EQUILIBRIUM

1. Write the equilibrium constant expressions, Kc and KP if applicable, for each of the
following reactions at equilibrium and state whether the equilibrium is homogenous
or heterogenous.
a) N2(g) + 3H2(g) ⇋ 2NH3(g)
b) HCl(aq) + H2O(l) ⇋ H3O+(aq) + Cl−(aq)
c) CaCO3(s) ⇋ CaO(s) + CO2(g)
d) 2NH3(g) + CO2(g) ⇋ NH4CO2NH2(s)

2. What is the relationship between K1 and K2 in each of the following pairs of reactions?

K1 and K2 are equilibrium constant.

a) 3H2(g) + N2(g) ⇋ 2NH3(g), K1

1 + 3 H 2 (g) ⇋ NH3(g), K2
2 N2 (g) 2

b) H2(g) + I2(g) ⇋ 2HI(g), K1

2HI(g) ⇋ H2(g) + I2(g), K2

3. Calculate the equilibrium constant, Kc for the reaction
H2(g) + I2(g) ⇋ 2HI(g)

if the concentrations at equilibrium for each species are;
[H2] = 0.302 M, [I2] = 0.508 M, [HI] = 0.770 M

4. Based on the following reaction,
2CO(g) ⇋ C(s) + CO2(g)

a) derive an equation that relates Kp to Kc.
b) Calculate the Kp for the above reaction if Kc is 7.14×10−2 at 700oC.

5. Consider the following reaction at 500°C.
2HI(g) ⇋ H2(g) + I2(g)

When 4.0 mole of HI gas is placed in a 2.0 L flask, the percentage of dissociation of HI
is 20%,
a) determine the concentration of HI, H2 and I2 after equilibrium is achieved.
b) calculate the equilibrium constant, Kc.

6. If 0.024 mole of N2O4 is allowed to reach equilibrium with NO2 in a 0.372 L flask at

25oC,

N2O4(g) 2NO2(g) Kc= 4.61 x 10-3

a) Calculate the concentration of N2O4 and NO2 presence at equilibrium.

b) Calculate the degree of dissociation of N2O4.

223

CHEMISTRY
MEKA TOPIC 6: CHEMICAL EQUILIBRIUM

7. At 425oC, Kp = 4.18×10‒9 for the reaction
2HBr(g) ⇋ H2(g) + Br2(g)

In one experiment, 0.20 atm of HBr, 0.010 atm of H2 and 0.010 atm of Br2 are
introduced into a container.
a) Calculate Qp.
b) Is the reaction at equilibrium? If not, in which direction will it proceed?

8. Decomposition of ethane produces ethene and hydrogen gas.

C2H6(g) C2H4(g) + H2(g) ΔH = +ve

Use the Le Chatelier’s principle to explain the shift (if any) that would occur in the

following equilibrium system if;

a) the concentration of hydrogen gas is decreased.

b) the temperature is lowered.

c) a catalyst is added.

d) C2H6 is removed from the system.

e) the volume of the container is increased.

f) adding Xenon gas at constant pressure.

224

CHAPTER 7: IONIC EQUILIBRIA

LECTURE NOTE

lecture notes & questions

225

7.0 7.1
IONIC EQUILIBRIA ACIDS AND BASES

7.1 Acid And Base 2
7.2 Acid-base Titration
7.3 Solubility Equilibria

11

At the end of this topic , students should be able to: g) Perform calculations involving pH, dissociation constant, initial
concentration, equilibrium concentration and the degree of
dissociation, .

a) Define acid and base according to Arrhenius and h) Explain qualitatively salt hydrolysis using hydrolysis equations and
Bronsted-Lowry classify the salts formed from the reaction between:
b) Identify conjugate acid and conjugate base according to i. strong acid and strong base
Bronsted-Lowry theory. ii. strong acid and weak base
c) Relate pH and pOH to the ionic product of water, Kw at iii. Weak acid and strong base
25 C.
d) Define strong acid and base, weak acid and base, pH and i) Define buffer solution.

pOH and pKw j) Descibe how a buffer solution controls its pH.
e) Calculate the pH values of a strong acid and base.
f) Relate the strength of a weak acid and a weak base to k) Write the Henderson-Hasselbalch equation.

the respective dissociation constant, Ka and Kb. 3 l) Calculate the pH of buffer solutions using the Henderson-Hasselbalch
before and after the addition of strong acid or base. 4

Theory of Acids and Bases Theory of Acids and Bases

Three important theories 1. Arrhenius Theory

1. Arrhenius theory Acid:
2. Bronsted-Lowry’s theory any substance that dissociates in water to produce
hydrogen ions (H+) or hydronium ions (H3O+).
Example:

HCl (aq)  H+ (aq) + Cl (aq)
HCl (aq) + H2O (l)  H3O+(aq) + Cl (aq)

56

226

Base: 2. Bronsted-Lowry Theory
any substance that dissociates in water to
produce hydroxide ions (OH-). Acid:
Example: any substance that can donate a proton (H+) to
other substances
NaOH (aq)  Na+ (aq) + OH (aq)
Example:
7
HNO3 (aq) + H2O (l)  NO3 (aq) + H3O+ (aq)

acid

NH4+ (aq) + H2O (l) NH3 (aq) + H3O+ (aq)

acid

8

Base: Conjugate Acid-base Pairs
any substance that can accept a proton from
other substances Conjugate base
A species that remains when one proton (H+) has
Example: NH4+ (aq) + OH (aq) been removed from the Bronsted acid.

NH3 (aq) + H2O (l) Conjugate acid
A species that remains when one proton (H+) has
base been added to the Bronsted base.

CO32- (aq) + H2O (l) HCO3- (aq) + OH- (aq) 10

base

H2O is able to act as an acid or a base : AMPHOTERIC
9

Example Exercise

HCO3 (aq) + H2O (l) CO32 (aq) + H3O+(aq) 1.In the following reactions, identify the acid-base conjugate
pairs.

acid base conjugate conjugate a. ClO (aq) + H2O (l) HOCl (aq) + OH (aq)

base acid conjugate conjugate
acid base
base acid

Note: b. CH3NH2 (aq) + H2PO4- (aq) CH3NH3+(aq) + HPO42 (aq)

 A weaker acid has a stronger conjugate base. base acid conjugate conjugate
acid base
 It is due to the ability of the acid to transfer its proton.

 The acid donates its proton less readily because its
conjugate base holds it more strongly.

11 12

227

c. HBr(aq) + NH3(aq) NH4+(aq) + Br (aq) 2. Write the conjugate base formula for
the following acids.
acid base conjugate conjugate

acid base

d. PO43(aq) + H2O(l) HPO42(aq) + OH(aq) a) HS d) H3PO4
b) HCN e) H3O+
base acid conjugate conjugate c) N2H5+ f) CH3NH3+
acid base

e. CH3COOH(aq) + H2O(l) CH3COO(aq) + H3O+(aq)

acid base conjugate conjugate
base acid

13 14

Answer : conjugate base 3. Write the conjugate acid formula for
acid the following bases.

a) HS S2- + H+ a) HS- d) NH3
b) HCN CN- + H+
c) N2H5+ N2H4 + H+ b) C6H5COO e) H2O
d) H3PO4 H2PO4- + H+
e) H3O+ H2O + H+ c) OH f) H2PO4

f) CH3NH3+ CH3NH2 + H+

15 16

Answer : conjugate acid Keep in mind !!!!!
bases ACID BASE DEFINITION
H2S
a) HS- + H+ C6H5COOH THEORY ACID BASE
b) C6H5COO- + H+ H2O
c) OH- + H+ NH4+ ARRHENIUS substance that produces substance that produces OH-
H3O+ H+/ H3O+ when it when it dissociates in water.
d) NH3 + H+ BRONSTED- dissociates in water.
e) H2O + H+ H3PO4 LOWRY NaOH → Na+ + OH-
HCl → H+ + Cl- base
f) H2PO4- + H+ acid

species that donates a species that accept a proton
proton to another species to another species

NH4+ + H2O NH3 + H3O+
acid base conjugate conjugate
base acid

17 18

228

Strength of Acids and Bases

Strength of acids and bases depend on the extent of

dissociation

[ ]dissociated  100%
[ ]initial
% dissociation =

Degree of dissociation,  = [ ]dissociated
[ ]initial

• The stronger the acid, the weaker its conjugate base. 19 20
• The stronger the base, the weaker its conjugate acid.

Strong Acid Example : Dissociation of 0.10 M HCl

Strong acid is an acid that ionizes completely in HCl (aq) + H2O (l)  Cl (aq) + H3O+ (aq)
water.

3 ways to write the equation for acid dissociation: [ ]i 0.10 0 0
Δ - 0.10 + 0.10 + 0.10
a) HCl (aq)  H+ (aq) + Cl (aq) [ ]f 0.00
0.10 0.10

b) HCl (aq) H2O(l) H+ (aq) + Cl (aq)

c) HCl (aq) + H2O (l) H3O+ (aq) + Cl (aq)

21 22

% dissociation = [HCl]dissociated  100% The strong acids are ...
[HCl]initial ◦Hydrohalic acids HCl, HBr and HI
◦Oxoacids in which the number of O atoms exceeds the
= 0.10 100% number of ionizable protons by two or more
0.10 ◦e.g: HNO3 and HClO4

= 100 % 24

Degree of dissociation,  = [HCl]dissociated
[HCl]initial

= 0.10
0.10

= 1.00

23

229

Strong Base % dissociation = [ NaOH ]dissociated  100%
[NaOH ]initial
Strong base is a base that ionizes completely in
water. = 0.50 100%
0.50

Example : Dissociation of 0.50 M NaOH = 100 %

NaOH (aq) H2O Na+ (aq) + OH (aq) Degree of dissociation,  = [NaOH ]dissociated
[ NaOH ]initial
[ ]i 0.50 00
Δ - 0.50 + 0.50 + 0.50 = 0.50
[ ]f 0.00 0.50
0.50 0.50
= 1.00

25 26

The strong bases are... Weak Acid
◦M2O or MOH, where M is group 1 element :
Weak acid is an acid that only ionizes partially in
Li, Na, K, Rb, Cs water
◦RO or R(OH)2, where R is Ca, Sr, Ba
Example: CH3COOH, HCOOH
27
Generally for any weak acid, HA the dissociation
reaction is:

HA (aq) + H2O (l) ⇋ A (aq) + H3O+(aq)

% dissociation < 100% 28
 < 1

Remark: ⇋ indicates partial dissociation

The weak acids are ....

Acid dissociation constant expression, Ka i) Hydrofluoric acid, HF

Ka = [A  ][H3O ] ii) Acids in which H is not bonded to O or to
[HA] halogen

e.g: HCN and H2S

Ka value depends on temperature iii) Oxoacids in which the number of O atoms
Normally Ka is measured at 25 C equals or exceeds by one the number of ionizable
Value of Ka ↓ ;  ↓ strength of acid ↓ protons

e.g: HNO2 and HClO

29 iv) Organic acids (general formula RCOOH) 30
e.g: CH3COOH

230

Weak Base Base dissociation constant expression, Kb

Weak base is a base that only ionizes partially in Kb = [BH ][OH ]
water [B]

Example: NH3 and N2H4

Generally for any weak base, B the dissociation
reaction is:

B (aq) + H2O (l) ⇋ BH+ (aq) + OH- (aq) Kb value depends on temperature
% dissociation < 100% Normally Kb is measured at 25 C
Value of Kb ↓,  ↓, strength of base ↓
 < 1
32
31

Examples of weak bases are ... Example:
At 25 C, 4.2% of 0.10 M formic acid,
i) Ammonia (NH3) HCOOH dissociated in aqueous solution.
Calculate the acid dissociation constant, Ka
ii) Amines for formic acid.

 RNH2 34

eg: CH3CH2NH2 ethanamine

 R2NH

eg: (CH3)2NH N-methylmethanamine

 R3N

eg: (CH3)3N N, N-dimethylmethanamine

33

Solution: Ka  [HCOO - ][H3O ]
[HCOOH]
[HCOOH] dissociated  4.2 0.10M
100  (0.0042 M ) 2
 (0.0958M)
 0.0042 M

HCOOH(aq) + H2O(l) H3O+ (aq) + HCOO- (aq)  1.810-4 M
00
[ ]i 0.10 Ka for HCOOH is 1.8 x 10-4 M
Δ -0.0042 +0.0042 +0.0042
[ ]eq 0.0958 0.0042 0.0042 K↓a↓o,r Kb for many weak acids or weak bases < 105,

To simplify the calculation for ionic equilibrium the
following assumption can be made

[ ]initial  [ ]dissociated  [ ]initial

35 36

231

The assumption will give correct answer to at least Try this…
2 significant figures if:
1. Acid dissociation constant, Ka for hydrofluoric
 value of Ka or Kb < 105 or acid HF at 25 C is 6.8104 M. For a solution
% dissociation < 5% of 0.20 M HF, calculate:
a. the concentration of hydronium ion at
Ka = [H3O ] [HCOO  ] equilibrium
[HCOOH] b. degree of dissociation

[ ]initial  [ ]dissociated  [ ]initial 2. Bsoalsuetiodnis, sNoHci3aatiton25coCnsitsan1t.,8K1b0fo5r ammonia
M. For a
= (0.0042 M) 2 solution of 0.50 M NH3, calculate:
0.10M a. the concentration of hydroxide ion at
equilibrium
= 1.8  104 M
b. % dissociation 38
37

Exercise 1 Exercise 2

[ ]i HF(aq) + H2O(l) H3O+(aq) + F-(aq) [ ]i NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
 
[ ]eq 0.2 -- [ ]eq 0.5 --
-x +x +x
0.2 - x +x +x 0.5-x xx

0.2 - x xx

a) [H 3O  ] [F  ] b) degree of dissociati on,  :
[HF]
Ka  Kb [NH 4 ][OH  ] degree of dissociati on,  :
[NH 3 ]
(x)( x)   [dissociat ed]  [dissociat ed]
0.2 - x [initial] [initial]
6.8x10 4  (x)( x)   x100
0.5  x
1.36x10 -4 - 6.8x10 -4 x - x2  0  0.0117 1.8x10 5  3.0 x10 3
x  0.2, (0.2 – x)  0.2 0.2 0.5
since x  0.5, (0.5 - x)  0.5  x100

x2  0.0583 x2  0.6%
0.2 0.5
6.8x10 4  1.8x10 -5 

1.36x10 4  x 2 x  3.0x10 3

x  [H3 O  ]  0.0117 39 x  [OH  ]  3.0x10 3 M 40

Water and the pH Scale 2H2O (l) H3O+ (aq) + OH (aq)

a.The Water Ionization Constant, Kw The equilibrium constant expression can be
written as follows:
When water molecules ionized, transferring a
proton from one water molecule to another Kw = H3O+ OH-
producing a hydroxonium ion, H3O+ and Kw is the ionization constant for water at 25C
hydroxide ion, OH-.

2H2O (l) H3O+ (aq) + OH (aq)

41 42

232

For pure water, at 25C, b. The pH Scale
H3O+ = OH = 1 x 10-7 M
Kw = H3O+ OH ◦ The pH scale is used to express acidity.
= (1 x 10-7) (1 x 10-7)
 Kw = 1 x 10-14 ◦ The pH of a solution is defined as the negative
logarithm (log) of the hydroxonium ion
Note: (hydrogen ion) concentration.
Temperature ↑, Kw ↑ because the dissociation of
water is an endothermic process pH = - log H3O+
or
43
pH = - log [H+]

44

 In similar way, pOH is the negative In pure water,
logarithm of the hydroxide ion
concentration. Kw = H3O+ OH = 1 x 10-14
pOH =  log OH
H3O+ = OH = 1 x 10-7 M
45
pH = pOH =  log (1 x 10-7 )
pH = pOH = 7

pH + pOH = 14

46

At 25 oC, Example of weak acids & weak bases
◦Acidic solution : pH < 7.0 ; pOH > 7.0
◦Basic solution : pH > 7.0 ; pOH < 7.0
◦Neutral solution : pH = 7.0 ; pOH = 7.0

47 48

233

Example 1 Answer

In a NaOH solution, [OH-] is 2.9 x 10-4 M. In a NaOH solution, [OH-] is 2.9 x 10-4 M.
Calculate the pH of the solution at 25 oC. Calculate the pH of the solution at 25 oC.

(10.46) pOH   log[OH  ]
pH  - log (2.9 x 10 -4 )
49  3.54
 14 - pOH
 14 - 3.54
 10.46

50

Example 2 Answer

The pH of rainwater in a certain region was 5.68. The pH of rainwater in a certain region was 5.68.
Calculate the H+ ion concentration of the rainwater. Calculate the H+ ion concentration of the rainwater.

(2.089 x 10-6) pH  log[H  ]
5.68  - log [H  ]
51 [H  ]  2.089 x 10 -6 M

52

Example 3 Answer

Calculate the pH of a 0.15 M acetic acid Calculate the pH of a 0.15 M acetic acid
(CH3COOH) solution, Ka = 1.8 x 10-5. (CH3COOH) solution, Ka = 1.8 x 10-5.

(2.78) CH3COOH (aq) + H2O(l) H3COO- (aq)+H+(aq)

[ ]i 0.15 - - -

Δ -x - +x +x

[ ]eq 0.15-x - x x

53 54

234

Answer Example 4

Calculate the pH of a 0.15 M acetic acid What is the pH of a 0.25 M ammonia solution?
(CH3COOH) solution, Ka = 1.8 x 10-5.

Kb = 1.8 x 10-5. (11.33)

Ka  [CH 3 COO  ][H ] Answer
[CH 3COOH]

x  0 .15 , (0 .15  x )  0 .15 NH3(aq) + H2O (l) NH4+ (aq) + OH-(aq)

1.8x10 5  x2 pH  -log [H ] [ ]i 0.25 - - -
0 .15   log1.64x103
Δ -X - +x +x
x  1 .64 x10  3 , x  [H  ]

pH  2.78 [ ]eq 0.25-X - x x

55 56

Answer Example 5

What is the pH of a 0.25 M ammonia solution? The pH of 0.06 M solution of formic acid
Kb = 1.8 x 10-5. (HCOOH) is 3.44. Calculate the dissociation
constant, Ka of the acid.
Kb  [NH4 ][OH ] pOH  -log [OH ]
[NH3 ]   log 2.12x103 (2.211 x 10-6)

since x  0.25, (0.25 - x)  0.25 pH  2.673 58

1.8x10-5  x2 pH  pOH  14
0.25 pH  14 - pOH

x  2.12x103, x  [OH- ]  14 - 2.673

 11.33 57

Answer Answer

The pH of 0.06 M solution of formic acid The pH of 0.06 M solution of formic acid (HCOOH)
(HCOOH) is 3.44. Calculate the dissociation is 3.44. Calculate the dissociation constant, Ka of
constant, Ka of the acid. the acid.

HCOOH(aq) HCOO- (aq) +H+(aq) pH  -log [H ] Ka  [HCOO ][H ]
3.44   log[H ] [HCOOH]
[ ]i 0.06 - - [H ]  3.63x104
Δ -x +x +x  (3.63x104 )(3.63x104 )
[]eq 0.06-x x x 0.0596

 2.21x10-6

59 60

235

Exercises Solution for exercise 1

1. The Ka for benzoic acid, (C6H5COOH) is 6.5x10-5. C6H6COOH(aq) C6H6COO- (aq) + H+(aq)
Calculate the pH of a 0.25 M benzoic acid solution.
(2.39) [ ]i 0.25 - -

2.The pH of an acid solution is 6.20. Calculate the [ ]eq 0.25-x x x
dissociation constant, Ka of the acid. The acid
concentration is 0.01M. 62
(3.98 x 10-11)

3. Calculate the pH for 0.5M C5H5N. Kb = 1.7x10-9
(9.4671)

Ka  [C6H6COO ][H ] Solution for exercise 2
[C6H6COOH] pH  -log [H ]
6.2   log [H ]
since x  0.25, (0.25 - x)  0.25 [H ]  6.31x10-7
HA
6.5x10-5  x2 H+ + A-
0.25 [ ]i 0.1 - -
x x
x  4.03x103 pH  -log [H ] [ ]eq 0.1-x
  log 4.03 x 10-3
x  [H ]

 2.39

63 64

Solution for exercise 3

[H ][A ] [ ]i C5H5N (aq) + H2O(l) C5H5NH+(aq) + OH-(aq)
Ka = [HA] Δ
[ ]eq 0.5 - --
(6.31x10−7)(6.31x10−7)
= (0.01−6.31x10−7) -x - +x +x
= 3.98x10−11
0.5-x - x x
65
66

236

Kb  1.7x109 pOH   log[OH ] Exercise (Extra)
pOH   log 2.92x105 4.Calculate pH for the following solution.
Kb [C5H5NH ][OH ] pOH  4.53
 [C5 H 5 N] pH  9.47

x  0.5, 0.5 - x  0.5 a.0.55 M CH3COOH (Ka = 1.8x10-5) Answer
b.0.23 M NH3 (Kb = 1.8x10-5) 2.5
1.7x10-9  x2 c.0.15 M HCl 11.31
0.5 0.82
d.0.20 M KOH 13.3
x  2.92x105 4.8
e.0.45 M HCN (Ka = 4.9 x 10-10)

67 68

Solution for exercise 4 (Extra)

b) NH3 (aq) + H2O(l) NH4+ (aq) + OH- (aq)

a) CH3COOH(aq) CH3COO- (aq) + H+ [ ]eq 0.55-x xx
[ ]eq 0.55-x
xx [NH ][OH ] pOH  log[OH ]
K = [NH ]  2.69
[CH COO ][H ] pH  log[H ]
K = [CH COOH] pH  14  pOH
0.55  -log3.15x10-3 1.8x10 = 0.23 14 - 2.69
assume <<< 0.55, 0.55− ≈  2.5 assume <<< 0.23, 0.23−  11.31
= 0.23
1.8x10−5 x = ⦏OH ] = 2.03x10
0.55
=

= ⦏H ] = 3.15x10

69 70

c) HCl (aq) + H2O (l) H3O+ (aq) + Cl- (aq) e) HCN (aq) H+ (aq)+ CN-(aq)
0.15 0.15 [ ]eq 0.45-x xx
[ ]f

pH   log 0.15 [H ][CN ]
 0.82 K = [HCN]

d) KOH (aq)  K+ (aq) + OH-(aq) assume x <<< 0.45, (0.45 − )
0.2 0.2
[ ]f ≈ 0.45

pOH   log[OH ] 4.9x10−10 = x
  log 0.2 0.45
 0.69897 = 2.205x10

pH  14  0.69897 = H = 1.485x10
 13.3
pH = −log(1.485x10−5)
= 4.8

71 72

237

SALTS 1) Neutral salts

Produced when acid reacts with Produced when strong acid reacts with
base. strong base. e.g

HCl + NaOH NaCl + H2O
neutral salt
 3 types of salts are :
1) neutral salts - Na+ comes from strong base
2) basic salts
3) acidic salts - Na+ does not react with water (does not
hydrolyzed)

- Cl- comes from strong acid

73 - Cl- does not react with water (does not 74
hydrolyzed)

So, pH of a solution depends on the SALT HYDROLYSIS
ionization of water
Salt hydrolysis is a chemical reaction between
H2O (l) ⇋ H+ (aq) + OH- (aq) anion or cation of a salt with water molecules
Kw = H3O+ OH = 1 x 10-14 M2 that produces OH- ion or H3O+ ion.

[H+ ] = [ OH- ] = 1.0 x 10-7 M The pH value of a solution depends on whether
OH- or H3O+ ion is produced during hydrolysis.
pH = 7
Acidic salts and basic salts have ions that
75 undergo hydrolysis in aqueous solution.

76

2) BASIC SALTS Hydrolysis of CH3COO- :
CH3COO-(aq) + H2O (l) ⇋ CH3COOH(aq) + OH-(aq)
A basic salt is produced from the reaction of a The solution is basic because the hydrolysis of
weak acid and a strong base and yields OH- ion CH3COO- produces OH-
when hydrolyzed. pH of the basic salt solution is > 7.0

Dissociation of salt 78

CH3COONa(aq) Na+ (aq) + CH3COO- (aq)

Na+ : comes from strong base
(does not hydrolyzed)

CH3COO- : comes from weak acid and 77
undergoes hydrolysis

238

Example 3) ACIDIC SALTS

Sodium cyanide, NaCN is a salt formed when a An acidic salt is produced from the reaction of a
strong base, NaOH reacts with a weak acid, HCN. strong acid and a weak base and yields H3O+
when hydrolysed.
a) Write a balanced equation for the formation of NaCN. Dissociation of salt

a) Determine the type of salt and its pH. NH4Cl(s)  NH4+(aq) + Cl(aq)
Cl- : comes from a strong acid
formation of salt: NaOH (aq) + HCN(aq)  NaCN(aq) + H2O(l)
salt dissociation : NaCN(aq)  Na+ (aq) + CN-(aq) (does not hydrolyzed)
Hydrolysis of salt: CN- + H2O ⇋ HCN + OH- NH4+ : comes from weak base and

OH- produced shows basic salt 79 undergoes hydrolysis
pH > 7
80

Hydrolysis of NH4+ : Keep in mind !!!
NH4+(aq) + H2O(l) ⇋ NH3(aq) + H3O+(aq) SALT HYDROLYSIS
The solution is acidic because hydrolysis
of NH4+ produces H3O+ • Salts formed when acids react with bases (neutralisation)
pH of the acidic salt solution is < 7.0 • Acidity of aqueous solution of salts results from hydrolysis
• Hydrolysis is the reaction of cations/anions (or both) from
81
salts with water molecules to produce H+ or OH-.
• Types of salts; neutral, acidic & basic salts

SALTS Neutral salt Basic salt Acidic salt
(pH=7) (pH >7) (pH<7)

example NaCl CH3COONa NH4Cl

formed NaOH + HCl CH3COOH + NaOH NH3 + HCl
from strong strong weak strong Weak strong
rxn of base acid acid base Base acid

82

Example BUFFER SOLUTION

CH3NH3Cl is a salt formed when a weak base,CH3NH2 A solution that maintains its pH when a small
reacts with a strong acid, HCl. amount of a strong acid or a strong base is
added to it.
a) Write a balanced equation to show the formation of
CH3NH3Cl. It contains a weak acid or a weak base with salt
that has its conjugate pair.
b) Determine the type of salt and its pH.
answer 2 types of buffer solution:
a. acidic buffer solution (pH < 7)
a) Salt formation :CH3NH2 (aq) + HCl (aq)  CH3NH3Cl(aq) b. basic buffer solution (pH >7)

Salt dissociation: CH3NH3Cl(aq) ⇋ CH3NH3+ (aq) + Cl-(aq) 84
Hydrolysis of salt:CH3NH3+(aq) + H2O (l) ⇋ CH3NH2(aq) + H3O+ (aq)

b) The H3O+ produced shows acidic salt. pH < 7

83

239

a) ACIDIC BUFFER SOLUTION CH3COO- will disturb the equilibrium of the
dissociation of ethanoic acid, CH3COOH.
• An acidic buffer solution is prepared by mixing a
weak acid and its salt (its conjugate base). Equilibrium of acid shifts backward, less
CH3COOH dissociates.
Example : CH3COOH and CH3COONa
Solution now has high concentrations of
• Reaction that occurs in an acidic buffer solution : CH3COOH (from weak base) and its
conjugate base ion CH3COO- (from salt).
CH3COOH(aq) CH3COO-(aq) + H+(aq)
86
(acid) (conjugate base)

CH3COONa(aq)  CH3COO-(aq) + Na+(aq)

The high concentration of CH3COO- in the solution is
mainly from the complete dissociation of CH3COON8a5.

ACTION of an ACIDIC BUFFER SOLUTION ACTION of an ACIDIC BUFFER SOLUTION

• When a small amount of strong acid is added • When a small amount of strong base is added

CH3COOH(aq) CH3COO-(aq) + H3O+(aq) CH3COOH(aq) CH3COO-(aq) + H+(aq)

CH3COONa(aq)  CH3COO-(aq) + Na+(aq) CH3COONa(aq)  CH3COO-(aq) + Na+(aq)
• [H3O+] increases. System try to reduce [H3O+] by
• CH3COOH will neutralise the added OH- and water
consuming it. is formed.

• Equilibrium shift to the left. CH3COOH (aq) + OH- (aq) (added)  CH3COO- (aq) + H2O
• Most of the added OH is removed this way.
• Most of the H+ added will be consumed by the • As a result, there will be only a small change in pH

CH3COO- to form CH3COOH.

CH3COO- (aq) + H3O+ (aq) (added)  CH3COOH (aq) +H2O

• As a result, there will be only a small change in pH

and [CH3COO-] , [CH3COOH] 87

88

BUFFER’S ACTION (more explanation) (ii) Adding a small amount of base to the solution :
◦the acid, CH3COOH will neutralize it.
Since buffer solution contains CH3COOH that acts
as acid and the conjugate ion CH3COO- that acts CH3COOH(aq) + OH-(aq)  CH3COO-(aq) + H2O(l)
as base, buffer solution will maintain its pH by
performing reactions as follows: ◦The amount of weak acid, CH3COOH decreased
a little but addition amount CH3COO- will cause the
(i) Adding a small amount of acid to the solution: equilibrium position to move to the left and
◦CH3COO- (conjugate base) will neutralize it. replace the amount of acid used

◦CH3COO-(aq) + H3O+ (aq)  CH3COOH(aq) + H2O(l) ◦So,the pH of the solution is not much affected.

◦The amount of weak acid, CH3COOH increased a 90
little but since the dissociation of acid is small, the

pH of the solution is not much affected. 89

240

CALCULATION OF pH OF BUFFER SOLUTION We can write the acidic concentration constant,

The pH is obtained by referring to the Ka  [H ][A ] or [H  ]  Ka[HA]
equilibrium dissociation of a weak acid, HA. [HA] [A  ]

Consider buffer solution containing HA and By applying –log on both sides, we have
its conjugate, A- [HA]

-log [H+] = -log Ka + ( - log [A ] )

HA ⇋ H+ + A-

Comes from Comes from the pH = pKa + log [A  ]
weak acids ionisation of a salt [HA]

Henderson-Hasselbalch equation

91 92

pH of ACIDIC BUFFER SOLUTION Example

CH3COOH(aq) CH3COO-(aq) + H+(aq) a. Calculate the pH of a 1.00 L solution containing
0.30 M CH3OOH and 0.10 M CH3COONa.
CH3COONa(aq)  CH3COO-(aq) + Na+(aq)
(Ka CH3COOH = 1.8 x 10-5)
Ka = [ CH3COO- ][H+] [H+] = Ka [ CH3COOH
[ CH3COOH ][ CH3COO- ]
] pH  pKa  log [conjugate base]
- log [H+] [ CH3COOH [weak acid]
= - log Ka - log ][ CH3COO- ]
[CH3COO - ]
+ log [ CH3COO- ] pH  - log Ka  log [CH3COOH]
[ CH3COOH]
pH = pKa (0.1)
(0.3)
log [ conjugate base]  - log 1.8 x 10-5  log
[ weak acid ]
pH = pKa +

Henderson-Hasselbalch Equation 93  4.27 94

b. What is the pH when 10 mL of 0.01 M HCl is c. What is the pH when 1 mL of 0.1 M NaOH is added
added to the buffer solution in (a)? to the buffer solution in (a)?

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

ni 0.1 1x10-4 0.3 - ni 0.3 1x10-4 0.1 -

Δ -1x10-4 -1x10-4 +1x10-4 - Δ -1x10-4 -1x10-4 +1x10-4 -

nf 0.1-1x10-4 0 0.3+1x10-4 - nf 0.3-1x10-4 0 0.1+1x10-4 -
=0.0999 0 =0.3001 =0.2999 0 =0.1001

[ ] 0.0999/1.01 0.3001/1.01 - [ ] 0.2999/1.001 0 0.1001/1.001 -

=9.89x10-2 0 =0.297 =0.2996 =0.1000

V=1000 mL+10 mL pH  -log Ka  log [conjugate base] V=1000 mL+1 pH  -log Ka  log [conjugate base]
=1.01 L [weak acid] mL [weak acid]

 - log 1.8x10-5  log (9.89x10-2 ) =1.001 L  - log 1.8x10-5  log 0.100
0.297 0.2996

 4.27 95  4.27 96

241

b) BASIC BUFFER SOLUTION  NH4Cl dissociates completely and produces high
concentration of NH4+ ions
• A basic buffer solution is prepared by mixing a
weak base and its salt ( its conjugate acid ).  The high concentrations of NH4+ disturb the
equilibrium of the dissociation of NH3.
• Example : NH3 and NH4Cl.
• Reactions that occur in a basic buffer solution :  Equilibrium of base shift backwards, less NH3
dissociate.
NH3 (aq) + H2O (l) NH4+(aq) + OH-(aq) Partially
dissociated  Solution now has high concentrations of NH3 and
base Conjugate acid its conjugate acid ion NH4+, originated from salt.

NH4Cl (aq)  NH4+(aq) + Cl- (aq) completely 98
dissociated

• The amount of NH4+ found in the mixture is mainly

from the complete dissociation of the salt, NH4Cl. 97

ACTION of a BASIC BUFFER SOLUTION ACTION of a BASIC BUFFER SOLUTION

• When a small amount of strong acid is added • When a small amount of strong base is added

NH3 (aq) + H2O (l) NH4+(aq) + OH-(aq) NH3 (aq) + H2O (l) NH4+(aq) + OH-(aq)

NH4Cl (aq)  NH4+(aq) + Cl- (aq) NH4Cl (aq)  NH4+(aq) + Cl- (aq)

• The H+ will be consumed by ammonia, NH3 to form • The OH- will be removed by the ammonium ion,
NH4+. NH4+

NH3 (aq) + H3O+(aq) (added)  NH4+(aq) + H2O • The NH4+ react with OH- to form NH3 and H2O.

• Most of the acid added (H+) will be removed this NH4+(aq) + OH- (aq) (added)  NH3 (aq) + H2O (aq)
way.
• Most of the OH- added is removed, hence, the pH
• As a result, there will only be a small change in pH. change only slightly.

• [NH3] and [NH4+] 99 • [NH4+] and [NH3] 100

BUFFER’S ACTION (more explanation) (ii) Adding a small amount of base to the solution
(i) Adding a small amount of acid to the solution ◦NH4+ (conjugate acid) will neutralize it.

◦NH3 (base) will neutralize it. NH4+(aq) + OH-(aq) → NH3 (aq) + H2O(l)
◦The amount of NH3 will increase a little but since
NH3(aq) + H3O+(aq) → NH4+(aq) + H2O the dissociation of NH3 is small, the pH of solution
is not much affected
◦The amount of NH4+ will increase a little but this
will cause the equilibrium to shift to the left and 102

replace the NH3 used, the pH of solution is not
much affected

101

242

CALCULATION OF pH OF BUFFER The base dissociation constant, Kb
SOLUTION
Kb  [BH ][OH ] or [OH  ]  K [B]
The pOH and pH can be calculated by using [B] [BH 
the Henderson-Hasselbalch equation. b ]

Consider the following base dissociation By applying – log on both sides:
reaction:
-log [OH] = -log Kb + ( -log [B] )
B + H2O ⇋ BH+ + OH- [BH  ]

103 pOH = pKb + log [BH ]
[B ]

Henderson-Hasselbalch equation

104

pH of a BASIC BUFFER SOLUTION Example

NH3 (aq) + H2O (l) NH4+(aq) + OH-(aq) A buffer solution is prepared by mixing 400mL 1.5 M NH4Cl
NH4Cl (aq)  NH4+(aq) + Cl- (aq) solution with 600 mL of 0.10 M NH3 solution.

Kb = [NH4+] [OH-] [OH-] = Kb [ NH3 ] [ Kb = 1.8 x 10-5 ]
[ NH3 ] [ NH4+ ] a)Calculate the pH of buffer

V = 400+600 = 1000mL of buffer

- log [OH-] = - log Kb - log [ NH3 ] nNH4Cl = 0.6 mol nNH3 = 0.06 mol
[ NH4+] [NH4Cl] = 0.6 M [NH3] = 0.06 M

pOH = pKb + log [ NH4+ ] pOH  -log Kb  log [salt] pH  14  pOH
[ NH3 ] [base]  14  5.74
 8.26
pOH = pKb+ log [ conjugate acid] pOH  log1.8x10 5  log 0.6
[ weak base ] 0.06

Henderson-Hasselbalch Equation 105  5.74 106

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

pOH  - log K b  log [salt]
[base]
H3O+ (aq) + NH3 (aq)  NH4+ (aq) + H2O(l)
- 0.6002/1.002
ni 2x10-4 0.06 0.6  -log 1.8x10 - 5  log 0.0598/1.002
-
Δ -2x10-4 -2x10-4 +2x10-4  5.74
-
nf 0 0.0598 0.6002

V = 1000 mL + 2 mL pH 14 - 5.74
= 1.002 L  8.26

107 108

243

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

OH- (aq) + NH4+ (aq)  NH3 (aq) + H2O(l) pOH  - log Kb  log [salt]
[base]
ni 3x10-4 0.6 0.06 -

Δ -3x10-4 -3x10-4 +3x10-4 -  -log 1.8x10 - 5  log 0.5997/1.002
0.0603/1.002
nf 0 0.5997 0.0603 -

 5.74

V = 1000 mL + 2 mL pH 14 - 5.74
= 1.002 L  8.26

109 110

Exercise 1 Solution for exercise 1:

a) Calculate the pH of a 1 L buffer solution

containing 1.5 M CH3COOH and 0.20 M of a) Acidic buffer [salt]
[acid]
CH3COONa. pH = 3.87 pH  -log Ka  log

b) Calculate the new pH of the buffer after the

addition of 2.0 mL of 0.1 M HCl. pH = 3.87  -log 1.8 x10-5  log 0.2
1.5
c) Calculate the new pH of the buffer after the
addition of 2.0 mL of 0.2 M NaOH. pH = 3.87  3.87

[ Ka = 1.8 x 10-5 ]

111 112

b) Addition of 2.0 mL of 0.1 M HCl b) Addition of 2.0 mL of 0.2 M NaOH

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

ni 2x10-4 0.2 1.5 - OH- (aq)+CH3COOH (aq)  CH3COO-(aq) + H2O(l)

Δ -2x10-4 -2x10-4 +2x10-4 - ni 4x10-4 1.5 0.2 -

nf 0 0.1998 1.5002 - Δ -4x10-4 -x10-4 +4x10-4 -
[ ]f 0 0.1994 1.4972 -

V = 1000 mL + 2 mL pH  -log Ka  log [conjugate base] nf 0 1.4996 0.2004 -
= 1.002 L [weak acid] [ ]f 0 1.4966 0.2000 -

 - log 1.8x10-5  log 0.1994 V = 1000 mL + 2 mL
1.4972 = 1.002 L

 3.87 113 114

244

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

a) Calculate the pH of the solution prepared by mixing

[conjugate base] 500 mL 0.1 M hydrazinium chloride, N2H5Cl with
pH = −log Ka + log [weak acid]
500 mL 0.2 M hydrazine, N2H4. pH = 8.53
0.2000
= −log 1.8x10−5 + log 1.4966

= 3.87 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 ]

115 116

Solution for exercise 2: b) Addition of 2.0 mL of 0.2 M NaOH

a) Preparation of basic buffer solution OH- (aq) + N2H5+ (aq)  N2H4(aq) + H2O(l)

V = 500+500 = 1000mL of buffer solution

nNH4Cl = 0.05 mol nNH3 = 0.10 mol ni 4x10-4 0.05 0.1 -
[NH4Cl] = 0.05 M [NH3] = 0.10 M
Δ -4x10-4 -4x10-4 +4x10-4 -

[salt] nf 0 0.0496 0.1004 -
pOH = −log K + log [base]
[ ]f 0.0495 0.1002 -

= −log 1.7 x10−6 + log 0.05
0.1
= 5.47 V = 1000 mL + 2 mL
= 1.002 L

pH = 14 − 5.47 117 118
= 8.53

c) Addition of 2.0 mL of 0.2 M NaOH

pOH  - log K b  log [N2H5 ]
[N2H4 ]
0.0495
 -log 1.7x10-6  log 0.1002

 5.47

pH  14 - 5.47
 8.53

119

245

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

a) Describe the titration process

7.2 b) Distinguish between the end point and equivalence point.
ACID-BASE TITRATIONS
c) Sketch the titrations curves between
120
i. strong acid-strong base

ii. strong acid-weak base

iii. weak acid-strong base

d) and interpret the variation of pH against titrant volume in (c)

e) Identify suitable indicators for acid-base titrations.

(Experiment 6: pH measurements and its applications

121

7.2 ACID-BASE TITRATIONS Titration apparatus

Titration Titrant
 A method for determining the concentration of
Analyte
a solution using another solution (known 123
concentration), called standard solution.

Standard Solution
 A solution of accurately known concentration.

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

122

The equivalence point Indicators
 the pH at which the number of moles of  is a weak organic acids or bases that change
colour over a range of pH values.
OH- ions added to a solution is equal
stoichiometrically to the number of Table 1: Some Common Acid-Base Indicators
moles of H+ ions originally present.
INDICATOR COLOUR pH range
The end point
 is the pH when the indicator just Thymol blue In Acid In Base
Bromophenol blue
changes colour. Methyl orange Red Yellow 1.2 – 2.8
Methyl red
124 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

246

How to determine the suitable indicator for How to determine the suitable indicator for
titrations? titrations

Choose an indicator which the endpoint pH Type pH Suitable Indicator
range lies on the steep portion of the titration jump
curve.
Strong acid - Strong 3 – 11 Any indicator except thymol
This choice ensures that the pH at the base blue
equivalent point will fall within the range over
which indicator changes color. Strong acid – Weak 3 – 7 Methyl orange, methyl red,
base chlorophenol blue,
126 bromophenol blue

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

127

TYPES OF TITRATIONS Titration Curves

1.Strong acid - strong base titrations Strong Acid - Strong Base Titration
2. Strong acid – weak base titrations
3. Weak acid – strong base titrations NaOH

128 HCl

129 129

Titration Curves Titration Curves

Strong Base - Strong Acid Titration Strong Acid and Strong Base Titration

HCl

NaOH

130 130 131 131

247