pH of 0.08 mol dm^–3 HOCl solution is 2.85. Calculate its ionization constant.

Given, pH =2.85 and C =0.08 mol dm^-3

Now since HOCl is a weak acid its dissociation will be given as-

We know that;

pH = -log [H⁺]

-2.85 log [H⁺]

[H⁺] = antilog (-2.85)

[H⁺] = 1.41 × 10^-3

We also know that, for a weak monobasic acid-

H⁺ =

[H⁺]² =K_aC (On squaring both sides)

K_a = 2.5 × 10^-5

Hence the ionization constant of HOCl will be 2.5 × 10^-5