Engineering Chemistry S Datta

IONIC EQUILIBRIUM 189
Example 1. Calculate pH of the following solutions:
(a) 4.9 × 10 –4 (N) acid,
(b) 0.0016 (N) base.
Sol. Both acid and base are strong.
As the acid and base are strong, the dissociation is complete. So,
(a) [H + ] = 4.9 × 10 –4 g-ion l –1
– log 10
[H + ] = – log 10
(4.9 × 10 –4 )
∴ pH = 4 – log 10
4.9 = 4 – 0.69 = 3.31
(b) [OH – ] = 1.6 × 10 –3 g-ion l –1
– log 10
[OH – ] = – log 10
(1.6 × 10 –3 )
∴ pOH = 3 – log 10
1.6 = 3 – 0.2041
= 2.7959 or 2.76
∴ pH = 14 – pOH = 14 – 2.76 = 11.24.
Example 2. Calculate [H + ] of a solution of pH 3.63.
Sol. pH = 3.63
– log 10
[H + ] = 3.63
log 10
[H + ] = – 3.63 = – 3 – 0.63 = 4 – 0.37 = 437 .
[H + ] = antilog 437 . = 2.34 × 10 –4
∴ [H + ] = 2.34 × 10 –4 g-ion litre –1 .
Example 3. Calculate [H + ], [OH – ] and pH of a weak base [0.1(N)].
Given: α = 0.013.
Sol. [OH – ] = 0.013 × 0.1 = 1.3 × 10 –3 g-ion litre –1 .
−14
10
[H + ] =
−3
= 7.7 × 10 –12 g-ion litre –1 .
13 . × 10
pH = – log 10
(7.7 × 10 –12 ) = 12 – log 7.7
= 12 – 0.8865 = 11.1135.
Relative Strengths of Acids and Bases
Tendency of donating protons in any aqueous solution is a measure of acidity of the
solution. If concentrations of two acids A and B are C A
and C B
respectively, ratio of strengths
should be C A
: C B
. There are several other processes for comparing strengths of acids. One such
method is by comparing dissociation constants of the acids.
Let, degree of dissociation of acid A 1
= α 1
Degree of dissociation of acid A 2
= α 2
The concentrations of the acids were same (C).
So, their strengths S 1
and S 2
are represented by,
S α C α KC
1 1 1 1
K
1
= = =
S α C α
2 2 2
K C K
2 2
So, the ratio of dissociation constants of two acids compare their strengths.
Buffer Solutions
Some solutions do not change their pH on addition of little acid or alkali. Such solutions
are known as buffer solutions. The arrest of pH changes of these solutions is known as buffer
action.