CHAPTER 7 - IONIC EQUILIBRIA

Example 3

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

(2.78)

51

Answer

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

(2.78)`

CH3COOH CH3COO- + H+

[ ]i 0.15 - -

Δ -x +x +x

[ ]eq 0.15-x x x

52

Answer

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

(2.78)`

Ka  [CH3COO ][H ] pH  -log [H ]
[CH3COOH]   log 1.64x103

x  0.15, 0.15  x  0.15 pH  2.78

1.8x10 5  x2
0.15

x  1.64x103 , x  [H ]

53

Example 4

What is the pH of a 0.25 M ammonia solution?

Kb = 1.8 x 10-5. (11.33)

54

Answer

What is the pH of a 0.25 M ammonia solution?

Kb = 1.8 x 10-5. (11.33)

NH3 + H2O NH4+ + OH-

[ ]i 0.25 --

Δ -X +x +x

[ ]eq 0.25-X xx

55

Answer

What is the pH of a 0.25 M ammonia solution?

Kb = 1.8 x 10-5. (11.33)

Kb  [NH4 ][OH ] pOH  -log [OH ]
[NH3 ]
  log 2.12x103

since x  0.25, 0.25 - x  0.25 pH  2.673

1.8x10-5  x2 pH  pOH  14
0.25
pH  14 - pOH
x  2.12x10 3, x  [OH- ]
 14 - 2.673

 11.33

56

Example 5

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

(2.211 x 10-6)

57

Answer

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

(2.211 x 10-6)

HCOOH HCOO- + H+

[ ]i 0.06 - -
Δ -x +x +x
[]eq 0.06-x x x

58

Answer

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

(2.211 x 10-6)

pH  -log [H ] Ka  [HCOO ][H ]
3.44   log[ H ] [HCOOH]
[H ]  3.63x10 4
 (3.63x104 )(3.63x104 )
0.0596

 2.21x10 -6

59

Exercises

1. The Ka for benzoic acid, (C6H5COOH) is 6.5x10-5.
Calculate the pH of a 0.25 M benzoic acid solution.
(2.39)

2.The pH of an acid solution is 6.20. Calculate the Ka
for the acid. The acid concentration is 0.01M.
(3.98 x 10-11)

3. Calculate the pH for 0.5M C5H5N. Kb = 1.7x10-9
(9.47)

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4.Calculate pH for the following solution.

a.0.55 M CH3COOH (Ka = 1.8x10-5) Answer
b.0.23 M NH3 (Kb = 1.8x10-5) 2.5
c.0.15 M HCl 11.31
0.82
d.0.20 M KOH 13.3
e.0.45 M HCN (Ka = 4.9 x 10-10) 4.8

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Excercises (Solution)

1. C6H6COOH C6H6COO- + H+
[ ]i 0.25 --

[ ]eq 0.25-x x x

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Excercises (Solution)

Ka  [C6H6COO ][H ]
[C6H6COOH]

since x  0.25, 0.25 - x  0.25

6.5x10-5  x2 pH  -log [H ]
0.25   log 4.03 x 10-3

x  4.03x103

x  [H ]

 2.39

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2. pH  -log [H ] H+ + A-
6.2   log [H ] - -
[H ]  6.31x10-7 x x

HA
[ ]i 0.1
[ ]eq 0.1-x

64

Ka  [H ][A- ]
[HA]

 (6.31x10 -7 )(6.31x10 -7 )
(0.01- 6.31x10 -7 )

 3.98x10 -11

65

3. C5H5N + H2O C5H5NH+ + OH-
0.5 --
[ ]i -x +x +x
Δ 0.5-x xx
[ ]eq

66

Kb  1.7x109 pOH   log[ OH ]
pOH   log 2.92x105
Kb  [C5H5NH ][OH ] pOH  4.53
[C5H 5 N] pH  9.47

x  0.5, 0.5 - x  0.5

1.7x10-9  x2
0.5

x  2.92x10 5

67

4.a) CH3COOH CH3COO- + H+
[ ]eq 0.55-x
xx

Ka  [CH3COO ][H2O] pH  log[H  ]
[CH3COOH]  -log3.15x10 -3
 2.5
assume x  0.5, 0.5 - x  0.55
68
1.8x10-5  x2
0.55

x  3.15x103

4.b) NH3 + H2O NH4+ + OH-
[ ]eq 0.55-x x x

Kb  [NH4 ][OH ] pOH  log[OH  ]
[NH3 ]  2.69

1.8x105  x2 pH  14  pOH
0.23 14 - 2.69
 11.31
assume x  0.23, 0.23 - x  0.23

x  2.03x103

69

4.c) HCl + H2O H3O+ + Cl-
[ ]f 0.15 0.15

pH   log 0.15

 0.82

4.d) KOH  K+ + OH-
[ ]f 0.2 0.2

pOH   log[ OH ] 70
  log 0.2
 0.69897

pH  14  0.69897
 13.3

4.e) HCN H+ + CN-
0.45-x x x
[ ]eq
71
Ka  [H ][CN ]
[HCN]

assume x  0.45, 0.45-x  0.45

4.9x10-10  x2
0.45

x2  2.205x1010

x  1.485x105

pH  -log(1.485x 10-5 )
 4.8

SALTS
Produced when acid reacts with
base.

 3 types of salts are :
1) neutral salts
2) basic salts
3) acidic salts

72

1) Neutral salts

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

HCl + NaOH NaCl + H2O
neutral salt

- Na+ comes from strong base

- Na+ does not react with water (does not
hydrolyzed)

- Cl- comes from strong acid

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

So, pH of a solution depends on the
ionization of water

H2O (l) ⇋ H+ (aq) + OH- (aq)
Kw = H3O+ OH = 1 x 10-14 M2

[H+ ] = [ OH- ] = 1.0 x 10-7 M

pH = 7

74

SALT HYDROLYSIS

Salt hydrolysis is a chemical reaction between
anion or cation of a salt with water molecules
that produces OH- ion or H3O+ ion.

The pH value of a solution depends on whether
OH- or H3O+ ion is produced during hydrolysis.

Acidic salts and basic salts have ions that
undergo hydrolysis in aqueous solution.

75

2) BASIC SALTS

A basic salt is produced from the reaction of a
weak acid and a strong base and yields OH- ion

when hydrolyzed.

Dissociation of salt

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

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

CH3COO- : comes from weak acid and
undergoes hydrolysis

76

Hydrolysis of CH3COO- :
CH3COO-(aq) + H2O (l) ⇋ CH3COOH(aq) + OH-(aq)
The solution is basic because the hydrolysis of
CH3COO- produces OH-
pH of the basic salt solution is > 7.0

77

Example

Sodium cyanide, NaCN is a salt formed when a
strong base, NaOH reacts with a weak acid, HCN.

a) Write a balanced equation for the formation of NaCN.

a) Determine the type of salt and its pH.
formation of salt: NaOH (aq) + HCN(aq)  NaCN(aq) + H2O(l)
salt dissociation : NaCN(aq)  Na+ (aq) + CN-(aq)
Hydrolysis of salt: CN- + H2O ⇋ HCN + OH-

OH- produced shows basic salt
pH > 7

78

3) ACIDIC SALTS

An acidic salt is produced from the reaction of a
strong acid and a weak base and yields H3O+
when hydrolysed.
Dissociation of salt

NH4Cl(s)  NH4+(aq) + Cl(aq)
Cl- : comes from a strong acid

(does not hydrolyzed)
NH4+ : comes from weak base and

undergoes hydrolysis

79

Hydrolysis of NH4+ :
NH4+(aq) + H2O(l) ⇋ NH3(aq) + H3O+(aq)
The solution is acidic because hydrolysis
of NH4+ produces H3O+
pH of the acidic salt solution is < 7.0

80

Example

CH3NH3Cl is a salt formed when a weak base,CH3NH2
reacts with a strong acid, HCl.

a) Write a balanced equation to show the formation of
CH3NH3Cl.

b) Determine the type of salt and its pH.

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

The H3O+ produced shows acidic salt.

pH < 7

81

BUFFER SOLUTION

A solution that maintains its pH when a small
amount of a strong acid or a strong base is
added to it.

It contains a weak acid or a weak base with salt
that has its conjugate pair.

2 types of buffer solution:
a. acidic buffer solution
b. basic buffer solution

82

ACIDIC BUFFER SOLUTION

Has pH < 7.

An acidic buffer solution is made up of a weak
acid and its salt (containing its conjugate
base)

Example:

A solution of CH3COOH and CH3COONa

(weak acid) (salt)

83

The dissociation reactions are:

CH3COOH (aq) ⇋ H+(aq) + CH3COO-(aq)

CH3COONa(aq)  CH3COO-(aq) + Na+(aq)

CH3COONa dissociates completely and
produces high concentration of CH3COO-.

The high concentration of CH3COO- will disturb
the equilibrium of the dissociation of ethanoic acid,
CH3COOH.

84

Equilibrium of acid shifts backward, less
CH3COOH dissociates.

Solution now has high concentrations of
CH3COOH (from weak base) and its
conjugate base ion CH3COO- (from salt).

85

BUFFER’S ACTION

Since buffer solution contains CH3COOH that acts
as acid and the conjugate ion CH3COO- that acts
as base, buffer solution will maintain its pH by
performing reactions as follows:

(i) Adding a small amount of acid to the solution:
◦CH3COO- (conjugate base) will neutralize it.

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

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

pH of the solution is not much affected. 86

(ii) Adding a small amount of base to the solution :
◦the acid, CH3COOH will neutralize it.

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

◦The amount of weak acid, CH3COOH decreased
a little but addition amount CH3COO- will cause the
equilibrium position to move to the left and
replace the amount of acid used

◦So,the pH of the solution is not much affected.

87

CALCULATION OF pH OF BUFFER SOLUTION

The pH is obtained by referring to the
equilibrium dissociation of a weak acid, HA.

Consider buffer solution containing HA and
its conjugate, A-

HA ⇋ H+ + A-

Comes from Comes from the
weak acids ionisation of a salt

88

We can write the acidic concentration constant,

Ka  [H ][A ] or [H ]  Ka[HA]
[HA] [A ]

By applying –log on both sides, we have
[HA]

-log [H+] = -log Ka + ( - log [A ] )

pH = pKa + log [A  ]
[HA]

Henderson-Hasselbalch equation

89

Example

a. Calculate the pH of a 1.00 L solution containing
0.30 M CH3OOH and 0.10 M CH3COONa.
(Ka CH3COOH = 1.8 x 10-5)

pH  pKa  log [conjugate base]
[weak acid]

pH  - log Ka  log [CH3COO- ]
[CH3COOH]

 - log 1.8 x 10-5  log (0.1)
(0.3)

 4.27 90

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

CH3COO- + H3O+  CH3COOH + H2O

ni 0.1 1x10-4 0.3

Δ -1x10-4 -1x10-4 +1x10-4

nf 0.1-1x10-4 0 0.3+1x10-4

=0.0999 0 =0.3001

[ ] 0.0999/1.01 0.3001/1.01

=9.89x10-2 =0.297

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

 - log 1.8x10-5  log (9.89x10 -2 )
0.297

 4.27 91

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

CH3COOH + OH-  CH3COO- + H2O

ni 0.3 1x10-4 0.1

Δ -1x10-4 -1x10-4 +1x10-4

nf 0.3-1x10-4 0 0.1+1x10-4

=0.2999 0 =0.1001

[ ] 0.2999/1.001 0.1001/1.001

=0.2996 =0.1000

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

=1.001 L  - log 1.8x10-5  log 0.100

0.2996

 4.27 92

BASIC BUFFER SOLUTION
Has pH > 7

Basic buffer solution is made up of a weak base and
its salt (containing its conjugate)

Example: A basic buffer solution of NH3 and NH4Cl
The dissociation reactions are:

NH3(aq) + H2O(l) ⇋ NH4+(aq) + OH-(aq)
NH4Cl(aq) 
NH4+(aq) + Cl-(aq)

93

NH4Cl dissociates completely and produce high
concentration of NH4+ ions
The high concentrations of NH4+ disturb the
equilibrium of the dissociation of NH3.
Equilibrium of base shift backwards, less NH3
dissociate.
Solution now has high concentrations of NH3 and
its conjugate acid ion NH4+, originated from salt.

94

BUFFER’S ACTION
Buffer solution maintains its pH by performing
reactions as follow :
NH3 acts as base and the conjugate ion, NH4+
acts as an acid.

95

(i) Adding a small amount of acid to the solution

◦NH3 (base) will neutralize it.

NH3(aq) + H3O+(aq) → NH4+(aq) + H2O

◦The amount of NH4+ will increase a little but this
will cause the equilibrium to shift to the left and

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

96

(ii) Adding a small amount of base to the solution

◦NH4+ (conjugate acid) will neutralize it.

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

◦The amount of NH3 will increase a little but since
the dissociation of NH3 is small, the pH of solution
is not much affected

97

CALCULATION OF pH OF BUFFER
SOLUTION

The pOH and pH can be calculated by using
the Henderson-Hasselbalch equation.

Consider the following base dissociation
reaction:

B + H2O ⇋ BH+ + OH-

98

The base dissociation constant, Kb

Kb  [BH  ][OH ] or [OH ]  Kb [B]
[B] [BH
]

By applying – log on both sides:

-log [OH] = -log Kb + ( -log [B] )
[BH ]

pOH = pKb + log [BH ]
[B ]

Henderson-Hasselbalch equation

99

Example

A buffer solution is prepared by mixing 400mL 1.5 M NH4Cl
solution with 600 mL of 0.10 M NH3 solution.

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

V = 400+600 = 1000mL of buffer

nNH4Cl = 0.6 mol nNH3 = 0.06 mol
[NH4Cl] = 0.6 M [NH3] = 0.06 M

pOH  -log Kb  log [salt] pH  14  pOH
[base]  14  5.74
 8.26
pOH  log1.8x10 5  log 0.6
0.06 100

 5.74