Question

determine the pH of a solution that is 0.15 M HClO2 (Ka=1.1 x 10^-2) and 0.15 M HClO (Ka=2.9 x 10^-8).

Answer 1

It is given that,

Concentration of HClO₂ = 0.15 M,

Ka (HClO₂) = 1.1 x 10^-2

Concentration of HClO = 0.15 M,

Ka (HClO) = 2.9 x 10^-8

Since, Ka (HClO₂) = 1.1 x 10^-2 which is greater than the Ka (HClO) = 2.9 x 10^-8

So, HClO₂ is a strong acid than the HClO. Then neglecting the HClO and considering HClO₂.

Reaction is given by:-

HClO₂ H⁺ + ClO₂^-

So, Ka = {[H+] x [ClO₂^-]} / HClO₂ ........ (1)

Consider [H+] = [ClO₂^-] = X. Then HClO₂ = 0.15 - X

Since, X is very small then (0.15 - X) will be almost equal to 0.15.

Then from equation (1), we have

Ka = X²/(0.15) = 1.1 x 10^-2

or X² = 0.15 x 1.1 x 10^-2 = 0.165 x 10^-2

or X = 0.041 M

So, [H+] = [ClO₂^-] = 0.041 M.

Then pH = -log(0.041) = 1.39

So, the pH of a solution will be, pH = 1.39