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 =KaC (On squaring both sides)



Ka = 2.5 × 10-5


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


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