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Question
\[\ce{pH}\] of 0.08 mol dm–3 \[\ce{HOCl}\] solution is 2.85. Calculate its ionisation constant
Solution
\[\ce{pH}\] of \[\ce{HOCl}\] = 2.85
But, \[\ce{– pH = log [H+]}\]
∴ \[\ce{– 2.85 = log [H+]}\]
⇒ `bar(3) .15` = \[\ce{log [H+ ]}\]
⇒ \[\ce{[H+]}\] = anti log `bar(3) .15`
⇒ \[\ce{[H+]}\] = 1.413 × 10–3
For weak mono basic acid \[\ce{[H+]}\] = `sqrt(K_a xx C)`
`K_a = ([H^+]^2)/C = (1.413 xx 10^-3)^2/0.08`
= 24.957 × 10–6 = 2.4957 × 10–5.
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