The solubility of `TI_(2)S` in pure `CO_(2)` -free water is `6.3 xx 10^(-6)M`. Assume that the dissolved `S^(2-)` ion hydrolyses almost completely to `HS^(Θ)` and that the further hydrolysis to `H_(2)S` is neglected. What is the `K_(sp).(K_(2)(H_(2)S) = 10^(-14))`.

Explanation

`TI_(2)S hArr 2TI^(oplus) +S^(2-), K_(sp) = [TI^(oplus)]^(2)][S^(2-)]`
`S^(2-) +H_(2)O hArr HS^(Θ) + overset(Θ)OH`
`K_(h) = (K_(w))/(K_(2)) = (10^(-14))/(10^(-14)) = 1.0`
`[TI^(oplus)] = 2 (6.3 xx 10^(-6)), [overset(Θ)OH] = [HS^(Θ)] = 6.3 xx 10^(-6)`
`K_(h) = ((6.3 xx 10^(-6))^(2))/([S^(2-)]) = 1.0, :. [S^(2-)] = (6.3 xx 10^(-6))^(2)`
`K_(sp) = (6.3 xx 10^(-6))^(2) [2(6.3 xx 10^(-6))]^(2) = 6.3 xx 10^(-21)`