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Page 1

IIT/TYP/U2/Chemistry/07 Sri Chaitanya: The Final Word in IIT-JEE

Chapter 3

Ionic Equilibrium

Gilbert Newton Lewis

Born 23 October 1875

Died 23 March 1 946

Nationality American

Field Chemistry

Known for Acid, Base Theory

There is no scientist in American history who has contributed more extensively to all fields in chemistry than Gilbert

Newton Lewis. His thinking was far ahead of his time and his theories have had profound influence on modern
chemistry

Lewis researched standard electrode potentials, conductivity, free energy and other thermodynamic constants for the
elements. These tables are still being used. His ability to organize and apply the scattered laws of thermodynamics

brought about the evolution of physical chemistry into the science as it is known today. Lewis once defined physical
chemistry as encompassing "everything that is interesting. Lewis work on acid base theory is most generalized theory.

Lewis did not believe only that an electron completely transfers from one atom to another, as in the positive-negative
theory. He describes the partial transfer of two electrons, one from each of the two bonding atoms, so that there is a

shared pair of electrons between them. This eliminates the need for the formation of oppositely charged atoms when
there was no indication of individually charged atoms (ions) in a compound. This was the first description of covalent

bonding.

Lewis' research on isotopes is an example of his wide-ranging and prolific interests. He published twenty-six papers

on heavy hydrogen and heavy water, isotopes of lithium, and neutron physics. He predicted the existence of naturally
occurring heavy water before he isolated it.

Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10.

Web: srichaitanya.edu.in, E-mail: [email protected]
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http://en.wikipedia.org/wiki/Physicist
http://en.wikipedia.org/wiki/Germany
http://en.wikipedia.org/wiki/1887
http://en.wikipedia.org/wiki/October_17
http://en.wikipedia.org/wiki/March_12

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Sri Chaitanya: The Final Word in IIT-JEE Chapter-3 / Ionic Equilibrium

3.1 Concepts of Acid and Base
Arrhenius concept

Acid is a substance which is capable of furnishing H+ ions in aqueous solution. e.g. - HCl, H
2
SO

4
etc. While

base is a substance which furnishes OH– ions e.g. - NaOH, NH
4
OH etc.

Limitations. Arrhenius theory failed to explain.

(i) behaviour of acids/bases in non aqueous solutions.

(ii) neutralization reaction giving salt in absence of a solvent.

(iii) acid character of certain salts like AlCl
3
,BF

3
etc.

(iv) existence of H+ in water.

Bronsted Lowery Concept

Acid is a substance which is capable of donating a proton while base is a substance which is capable of
accepting a proton. This is also called protonic theory of acids-bases.

HCl H
2
O ˆ ˆ †‡ ˆ ˆ H3O

+ + Cl–

acid base conjugate acid conjugate base

H
2
O + NH

3 ˆ ˆ †‡ ˆ ˆ 4NH
+ + OH–

acid base conjugate acid conjugate base

conjugate base of an acid is a species formed by the loss of a proton from acid.

Acid → H+ + conjugated base
Similarly conjugate acid is formed from a base by gain of H+.

Base + H+ → conjugated acid
Weak acid has a strong conjugate base and vice-versa.

A Bronsted - Lowery acid base reaction always proceeds in the direction from the stronger to the weaker acid
base combination. e.g.

HI + OH– → H
2
O + I–

Strong Strong Weak Weak

acid base acid base

Lewis concept

Acid is a substance which can accept a pair of electrons while base is a substance which can donate a pair of
electrons.

Hence Lewis acids are

(i) Molecules in which central atom has incomplete octet e.g. BF
3
, AlCl

3
and FeCl

3
etc.

(ii) Simple cations like Ag+, H+ etc.

(iii) Molecules in which central atom has vacant d-orbitals e.g.- SiF
4
, SnCl

4
etc.

(iv) Molecules in which atoms of different electronegativities are joined by multiple bond. e.g. CO
2
, SO

3
etc.

(v) In carbonyl complexes, metal atoms act as Lew’s acids e.g. Ni in Ni(CO)
4
etc.

And Lewis bases are

(i) Neutral molecules like NH
3
, RNH

2
etc.

(ii) Negatively charged anions. e.g. CN–, Cl– etc.

-2-
Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh, New Delhi-110 026. Phones: 25226309/10.

Web: srichaitanya.edu.in, E-mail: [email protected]

Page 33

IIT/TYP/U2/Chemistry/07 Sri Chaitanya: The Final Word in IIT-JEE

Calculate the pH of a solution obtained by mixing equal volumes of 0.10N ammonium nitrate and 0.02N ammonium

hydroxide, Kb for NH4OH is 1.8 × 10
–5.

Solution

Before mixing

[NH4NO3] = 0.10N = 0.10M [NH4OH] = 0.02N = 0.02M

After mixing. Since the volume after mixing becomes double to that of either of the individual component, its
concentration in the resulting solution becomes half, i.e.,

4 3

0.10
[NH NO ] 0.05M

2
= = 4

0.02
[NH OH] 0.01M

2
= =

Now substituting the values in the following equation

pOH = pKb + log
[Salt]

[Base]
= – log 1.8 × 10–5 + log

= 4.7447 + 0.6990 = 5.4437

Now since pH = 14 – pOH = 14 – 5.4437 = 8.5563 ≈ 8.56

Example 7

What is [H+] in a solution obtained when 0.1M ammonia solution is just neutralised by a strong HCl? The dissociation

constant of ammonia is 1.8 × 10–5.

Solution

Conc. of NH3 solution = 0.1 M,

4 2 4NH H O NH OH H

c(1 h) ch ch

+ ++ → +


14
4w

15
b

K 10
h 0.745 10

K c 1.8 10 0.1


−= = = ×

× × ×

[H+] = ch = 0.1 × 0.745 × 10–4 = 7.45 × 10–6

Example 8

In 0.3M solution of NH4Cl, H
+ ion concentration is 1.3 × 10–5M. What is the dissociation constant of NH3?

Solution

4 4NH Cl NH Cl
+ −→ +

4 2 3 3NH (aq) H O NH (aq) H O

(0.3 x)M xM xM

+ ++ → +


5x [H ] 1.3 10+ −= = ×

w 3 3
h

b 4

K [NH ] [H O ]
K

K [NH ]

+

+= =

Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10.

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Page 34

Sri Chaitanya: The Final Word in IIT-JEE Chapter-3 / Ionic Equilibrium

14 5 2

b

10 x x (1.3 10 )

K 0.3 0.3

− −× ×
= =

14

b 5 2

0.3 10
K

(1.3 10 )




×

= =
×

1.8 × 10–5

Example 9

The dissociation constant of CH3COOH is 1.6 × 10
–5. The degree of dissociation d, of 0.01 molar CH3COOH in presence

of 0.1 M HCl is equal to

(1) 0.016 (2) 0.16 (3) 0.04 (4) 0.4

Solution

3 3CH COOH CH COO H

at start1 0 0

at equilibriumc(1 ) c c

0.01 (1 ) 0.01 0.01

[In presence of 0.1M HCl]0.01 (1 ) 0.01 (0.01 0.1)

− +→ +

− α α α
− α α α
− α α α +

3

3

[CH COO ] [H ]
K

[CH COOH]

− +

=

5 0.01 (0.01 0.1)1.6 10
0.1 (1 )

− α× α +× =
− α

On solving, α = 0.016%

Example 10

Find the percentage ionisation of 0.2M acetic acid solution whose dissociation constant is

1.8 × 10–5. Also determine concentration of H3O+, CH3COO– and CH3COOH at equilibrium.

Solution

From the expression derived earlier we know that

K = (2c

∴ α =
5

5 4K 1.8 10 9 10 0.9 10
c 0.2


− −×= = × = ×

= 2 20.910 0.95 10− −= ×

(% Ionisation = 0.95 × 10–2 × 100 = 0.95%

We know that concentration of [H+] or [H3O+] c( = 0.2 × 0.95 × 10–2 = 0.190 × 10–2

= 1.9 × 10–3

Also concentration of [CH3COO–] = c( = 1.9 × 10–3

Concentration of unionised CH3COOH = c (1 – .0095) = 0.2 (1 – 0.0095) = 0.2 × 0.9905

= 0.198 M

-34-
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Web: srichaitanya.edu.in, E-mail: [email protected]

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