InterviewSolution
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InterviewSolution
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`H_(2)S` behaves as a weak diprotic acid in aqueous solution. Which of the following is the correct explanantion for `pH` of a solution of `H_(2)S` in terms of its `pK_(1), pK_(2), [H_(2)S]` and `[S^(2-)]`A. `pH = (1)/(2) (pK_(1) + pK_(2))`B. `pH = (1)/(2) (pK_(1) + pK_(2) - "log"([S^(2-)])/([H_(2)S]))`C. `pH = (1)/(2)(pK_(1) + pK_(2) + "log"([S^(2-)])/([H_(2)S]))`D. `pH = (1)/(2) (pK_(1) - pK_(2) + "log"([H_(2-)S])/([S^(2-)]))` |
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Answer» Correct Answer - C Refer to illustration `H_(2)S hArr H^(o+) + hS^(Theta) (K_(1))` `HS^(Theta) hArr H^(o+) + S^(2-) (K_(2))` `K_(1) = ([H^(o+)][HS^(Theta)])/([H_(2)S])` `K_(2) = ([H^(o+)][S^(Theta)])/([HS^(Theta)])` `K_(1) xx K_(2) = ([H^(o+)]^(2)[S^(2-)])/([H_(2)S])` `[H^(o+)]^(2) = (K_(1)xxK_(2)xx[H_(2)S])/([S^(2-)])` Taking negative logarithms to both sides. `-2log [H^(o+)] =- log K_(1) - log K_(2)` `=- log [H_(2)S] + log [S^(2-)]` `:. pH = (1)/(2) (pK_(1) + pK_(2) + "log"([S^(2-)])/([H_(2)S]))` |
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