CHAPTER 7 - IONIC EQUILIBRIA

CHAPTER 7
IONIC EQUILIBRIA

CHEMISTRY UNIT
KOLEJ MATRIKULASI MELAKA

SHARED BY MISS DALINA BINTI DAUD

7.0
IONIC EQUILIBRIA

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

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7.1
ACIDS AND BASES

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

a) Define acid and base according to Arrhenius and

Bronsted-Lowry

b) Identify conjugate acid and conjugate base according to

Bronsted-Lowry theory.

c) Relate pH and pOH to the ionic product of water, Kw at
25 C.

d) . Define strong acid and base, weak acid and base, pH

and pOH and pKw
e) Calculate the pH values of a strong acid and base.

f) Relate the strength of a weak acid and a weak base to

the respective dissociation constant, Ka and Kb. 4

g) Perform calculations involving pH, dissociation constant, initial
concentration, equilibrium concentration and the degree of
dissociation, .

h) Explain salt hydrolysis using hydrolysis equations and classify the salts
formed from the reaction between:
i. strong acid and strong base
ii. strong acid and weak base
iii. Weak acid and strong base

i) Define buffer solution.

j) Descibe how a buffer solution controls its pH.

k) Write the Henderson-Hasselbalch equation.

l) Calculate the pH of buffer solutions before and after the addition of 5

strong acid or base.

Theory of Acids and Bases

Three important theories
1. Arrhenius theory
2. Bronsted-Lowry’s theory

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Theory of Acids and Bases

1. Arrhenius Theory

Acid:
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)

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Base:
any substance that dissociates in water to
produce hydroxide ions (OH-).
Example:

NaOH (aq)  Na+ (aq) + OH (aq)

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2. Bronsted-Lowry Theory

Acid:
any substance that can donate a proton (H+) to
other substances

Example:

HNO3 (aq) + H2O (l)  NO3 (aq) + H3O+ (aq)

acid

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

acid

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Base:
any substance that can accept a proton from
other substances

Example: NH4+ (aq) + OH (aq)

NH3 (aq) + H2O (l)

base

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

base

H2O is able to act as an acid or a base : AMPHOTERIC
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Conjugate Acid-base Pairs

Conjugate base
A species that remains when one proton (H+) has
been removed from the Bronsted acid.

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

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Example

HCO3 (aq) + H2O (l) CO32 (aq) + H3O+(aq)

acid base conjugate conjugate

base acid

Note:
 A weaker acid has a stronger conjugate 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.

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Exercise

1.In the following reactions, identify the acid-base conjugate
pairs.

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

base acid conjugate conjugate
acid base

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

base acid conjugate conjugate
acid base

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c. HBr(aq) + NH3(aq) NH4+(aq) + Br (aq)

acid base conjugate conjugate

acid base

d. PO43(aq) + H2O(l) HPO42(aq) + OH(aq)

base acid conjugate conjugate
acid base

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

acid base conjugate conjugate
base acid

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2. Write the conjugate base formula for
the following acids.

a) HS d) H3PO4
b) HCN e) H3O+
c) N2H5+ f) CH3NH3+

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Answer : conjugate base H+
acid
S2- + H+
a) HS CN- + H+
b) HCN N2H4 + H+
c) N2H5+ H2PO4- + H+
d) H3PO4 H2O + H+
e) H3O+
CH3NH2 +
f) CH3NH3+

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3. Write the conjugate acid formula for
the following bases.

a) HS- d) NH3

b) C6H5COO e) H2O

c) OH f) H2PO4

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Answer : conjugate acid
bases
H2S
a) HS- C6H5COOH
b) C6H5COO- H2O
c) OH- NH4+
H3O+
d) NH3
e) H2O H3PO4

f) H2PO4-

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Strength of Acids and Bases

Strength of acids and bases depend on the extent of
dissociation

% dissociation = [ ][di]sisnoictiiaalted100%

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

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Strong Acid

Strong acid is an acid that ionizes completely in
water.

3 ways to write the equation for acid dissociation:

a) HCl (aq)  H+ (aq) + Cl (aq)

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

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

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Example : Dissociation of 0.10 M HCl

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

[ ]i 0.10 0 0
Δ - 0.10 + 0.10 + 0.10

[ ]f 0.00 0.10 0.10

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

0.10 100%
0.10

= 100 %

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

0.10
0.10

= 1.00

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The strong acids are ...

◦Hydrohalic acids HCl, HBr and HI
◦Oxoacids in which the number of O atoms exceeds the
number of ionizable protons by two or more
◦e.g: HNO3 and HClO4

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Strong Base

Strong base is a base that ionizes completely in
water.

Example : Dissociation of 0.50 M NaOH

NaOH (aq) H2O Na+ (aq) + OH (aq)

[ ]i 0.50 0 0
Δ - 0.50 + 0.50 + 0.50
[ ]f 0.00
0.50 0.50

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

= 0.50 100%
0.50

= 100 %

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

= 0.50
0.50

= 1.00

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The strong bases are...
◦M2O or MOH, where M is group 1 element :

Li, Na, K, Rb, Cs
◦RO or R(OH)2, where R is Ca, Sr, Ba

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Weak Acid

Weak acid is an acid that only ionizes partially in
water

Example: CH3COOH, HCOOH

Generally for any weak acid, HA the dissociation
reaction is:

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

% dissociation < 100%
 < 1

Remark: ⇋ indicates partial dissociation

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Acid dissociation constant expression, Ka

Ka = [A  ][H3O ]
[HA]

Ka value depends on temperature

Normally Ka is measured at 25 C
Value of Ka ↓ ;  ↓ strength of acid ↓

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The weak acids are ....

i) Hydrofluoric acid, HF

ii) Acids in which H is not bonded to O or to
halogen

e.g: HCN and H2S

iii) Oxoacids in which the number of O atoms

equals or exceeds by one the number of

ionizable protons

e.g: HNO2 and HClO

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

Weak Base

Weak base is a base that only ionizes partially in
water

Example: NH3 and N2H4

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

B (aq) + H2O (l) ⇋ BH+ (aq) + OH- (aq)
% dissociation < 100%

 < 1

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Base dissociation constant expression, Kb

Kb = [BH ][OH ]
[B]

Kb value depends on temperature
Normally Kb is measured at 25 C
Value of Kb ↓,  ↓, strength of base ↓

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Examples of weak bases are ...

i) Ammonia (NH3)

ii) Amines

oRNH2

eg: CH3CH2NH2 ethanamine
oR2NH

eg: (CH3)2NH N-methylmethanamine
oR3N

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

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Example:
At 25 C, 4.2% of 0.10 M formic acid,
HCOOH dissociated in aqueous solution.
Calculate the acid dissociation constant, Ka
for formic acid.

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Solution:

[HCOOH]dissociatde  4.2  0.10M
100

 0.0042 M

HCOOH + H2O H3O+ + HCOO-
00
[ ]i 0.10
Δ -0.0042 +0.0042 +0.0042
[ ] 0.0958 0.0042 0.0042

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Ka  [HCOO- ][H3O ]
[HCOOH]

 (0.0042M)2
(0.0958M)

 1.810-4 M

Ka for HCOOH is 1.8 x 10-4 M

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

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The assumption will give correct answer to at least
2 significant figures if:

 value of Ka or Kb < 105 or
% dissociation < 5%

Ka = [H3O ] [HCOO ]
[HCOOH]

[ ]initial  [ ]dissociated  [ ]initial

= (0.0042 M)2
0.10M

= 1.8  104 M

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Try this…

1. Acid dissociation constant, Ka for hydrofluoric
acid HF at 25 C is 6.8104 M. For a solution
of 0.20 M HF, calculate:

a. the concentration of hydronium ion at
equilibrium

b. degree of dissociation

2. Base dissociation constant, Kb for ammonia
solution, NH3 at 25 C is 1.8105 M. For a
solution of 0.50 M NH3, calculate:

a. the concentration of hydroxide ion at

equilibrium

b. % dissociation 37

Exercise 1

HF(aq) + H2O(l) H3O+(aq)+ F-(aq)

[ ]i 0.2 --
 0.2 - x xx
xx
[ ]eq 0.2 - x

a) Ka  [H3O ] [F ] b) degree of dissociation,  :
[HF]   [dissociated]

6.8x104  (x)(x) [initial]
0.2 - x  0.0117

1.36x10 -4 - 6.8x10 -4 x - x2  0 0.2
x  0.2, 0.2 – x  0.2  0.0583

6.8x104  x2 38
0.2

1.36x104  x2

x  [H3O ]  0.017M

Exercise 2 NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

[ ]i 0.5 --
 -x +x +x
[ ]eq 0.5-x xx

Kb  [NH4 ][OH ] degree of dissociation,  :
[ NH3 ]   [dissociated] x100

1.8x105  (x)(x) [initial]
0.5  x  3.0x103 x100

since x  0.5, 0.5 - x  0.5 0.5
 0.6%
1.8x10 -5  x2
0.5 39

x  3.0x103

x  [OH ]  3.0x103 M

Water and the pH Scale

a.The Water Ionization Constant, Kw
◦ When water molecules ionized, transferring a

proton from one water molecule to another
producing a hydroxonium ion, H3O+ and
hydroxide ion, OH-.

2H2O (l) H3O+ (aq) + OH (aq)

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2H2O (l) H3O+ (aq) + OH (aq)

The equilibrium constant expression can be
written as follows:

Kw = H3O+ OH-
Kw is the ionization constant for water at 25C

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For pure water, at 25C,
H3O+ = OH = 1 x 10-7 M
Kw = H3O+ OH
= (1 x 10-7) (1 x 10-7)
 Kw = 1 x 10-14

Note:
Temperature ↑, Kw ↑ because the dissociation of
water is an endothermic process

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b. The pH Scale

◦ The pH scale is used to express acidity.

◦ The pH of a solution is defined as the negative
logarithm (log) of the hydroxonium ion
(hydrogen ion) concentration.

pH = - log H3O+
or

pH = - log [H+]

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 In similar way, pOH is the negative
logarithm of the hydroxide ion
concentration.
pOH =  log OH

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In pure water,

Kw = H3O+ OH = 1 x 10-14

H3O+ = OH = 1 x 10-7 M

pH = pOH =  log (1 x 10-7 )
pH = pOH = 7

pH + pOH = 14

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At 25 oC,
◦Acidic solution : pH < 7.0 ; pOH > 7.0
◦Basic solution : pH > 7.0 ; pOH < 7.0
◦Neutral solution : pH = 7.0 ; pOH = 7.0

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Example 1

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

(10.46)

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Answer

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

(10.46)

pOH  log[OH  ]
 - log (2.9 x 10-4 )
 3.54

pH  14 - pOH
 14 - 3.54
 10.46

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Example 2

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

(2.089 x 10-6)

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Answer

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

(2.089 x 10-6)

pH  log[H  ]
5.68  - log [H ]
[H ]  2.09 x 10 -6 M

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