pH of 0.08 mol `dm^(-3)` HOCI solution is 2.85. Calculate its ionisation constant.

Explanation

pH of Hocl = 2.85, i.e, - log `[H^(+)]= 2.85 ` or log `[H^(+)]=-2.85= bar(3).15`
or `[H^(+)]`= antilog ` bar(3) .15 = 1.413 xx 10^(-3)M`
For weak monobasic acid, `HA hArr H^(+) + A^(-)`
`K_(a) = ([H^(+)]^(2))/([HA])=((1.413xx10^(-3))^(2))/(0.08) = 2.4957xx10^(-5)`.