`K_(1)` and `K_(2)` for dissociation of `H_(2)A` are `4xx10^(-3)` and `1xx10^(-5)`. Calculate concentration of `A^(2-)` ion in `0.1M H_(2)A` solution. Also report `[H^(+)]` and pH.

Explanation

`H_(2)S hArr H^(oplus) +HS^(Theta)`
`K_(1) = ([H^(o+)][HS^(Θ)])/([H_(2)S]) = 4 xx 10^(-3)`
`[H^(o+)][, Calpha [HS^(Θ)] = C alpha, [H_(2)S] = C(1-alpha)`
or `4 xx 10^(-3) = (C alpha. C alpha)/(C(1 - alpha)) = (C alpha^(2))/((1-alpha))`
or `4 xx 10^(-3) = (0.1 xx alpha^(2))/((1-alpha))` (`1-alpha` should not be neglected)
`:. alpha = 0.18 , :. [H^(o+)] = C alpha = 0.1 xx 0.18 = 0.018M`
`:. pH = 1.7447`
`[HS^(Θ)] = C alpha = 0.1 xx 0.18 = 0.018M`
`[H_(2)S] = C (1-alpha) = 0.1 (1- 0.18) = 0.082M`
Now, `HS^(Θ)` further dissociates to `H^(o+)` and `S^(2-)`,
`C_(1) = [HS^(Θ)] = 0.018M`
`{:(HS^(Theta),hArr,H^(oplus),+,S^(2-)),(1,,0,,0),((1-alpha_(1)),,alpha_(1),,alpha_(1)):}`
`:. K_(2) = 1 xx 10^(-5) = ([H^(o+)][S^(2-)])/([HS^(Θ)]`
Because `[H^(o+)]` already in solution ` = 0.018` and thus, dissociation of `HS^(Θ)` further suppersses due to common ion effect and `1 - alpha ~~1`.
`:. 1 xx 10^(-5) = (0.018 xx C_(1)alpha_(1))/(C_(1)(1-alpha_(1)) = 0.018 xx alpha_(1)`
`:. alpha_(1) = (1xx 10^(-5))/(0.018) = 5.55 xx 10^(-4)`
`:. [S^(2-)] = C_(1)alpha_(1) = 0.018 xx 5.55 xx 10^(-4) = 10^(-5)`
`:. [HS^(Θ)] = C_(1) (1-alpha_(1)) = C_(1) = 0.018M`