Question

A diprotic acid, H2A, has Ka1 = 3.4 x 10-4 and Ka2 = 6.7 x 10-9. What is the pH of a 0.36 M. solution of H2A?

Answer 1

Solution :

The dissociation of diprotic acid (H2A) can be written as,

H2A + H2O ===> H3O+ + HA-

Ka1 = 3.4 x 10^-4

HA- + H2O ===> H3O+ + A^2-

Ka2 = 6.7 x 10^-9

Since, ka2 is very less than ka1,.hence pH of the acid is only depends upon first dissociation.

H2A + H2O ===> H3O+ + HA-

0.36 M ---------------- 0 --------------0 (initial)

0.36 -X ---------------- X ------------- X (final)

Therefore,

Ka1 = [H3O+] [HA-] / [H2A]

3.4 x 10^-4 = X . X / (0.36 - X)

3.4 x 10^-4 = X^2 / (0.36 -X)

Again, since ka1 is very less, hence it is a weak acid, therefore, X can be neglected from denominator.

Thus,

3.4 x 10^-4 = X^2 / 0.36

X^2 = 1.224 x 10^-4

X = 1.106 x 10^-2

Thus,

[H3O+] = X = 1.106 x 10^-2 M

pH = - log [H+] = - log [H3O+]

pH = - log 1.106 x 10^-2

pH = 2 - log 1.106 = 2 - 0.04 = 1.96

Hence,

pH = 1.96