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Calculate the pH of 0.08M solution of hydrochlorous acid HOCl. The ionization constant of the acid is 0.000025. Determine the percentage dissociation of HOCl




Ka = {[H3O+][ClO-] / [HOCl]}
= x2 / (0.08 – x)
As x<<0.08, therefore 0.08 – x = 0.08
x2 / 0.08 = 2.5 x 10-5
x2 = 2.0 x 10-6, thus, x = 1.41 x 10-34
[H+] = 1.41 x 10-3 M.
Therefore,
Percent dissociation = {[HOCl]dissociated / [HOCl]initial } x 100 = 1.41 x 10-3 / 0.08 = 1.76 %.
pH = -log(1.41 x 10-3) = 2.85.


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