InterviewSolution
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InterviewSolution
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pH of 0.08 mol `dm^(-3)` HOCl solution is 2.85. Calculate its ionisation constant. |
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Answer» 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)`. |
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