`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-)]`

Options

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-)]))`

Answer

C

Explanation

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]))`