Calcualte the solubility of `M_(2)X_(3)` in pure water, assuming that neither kind of ion reacts with `H_(2)O`. The solubility product of `M_(2)X_(3), K_(sp) = 1.1 xx 10^(-23)`.

Explanation

It `S` is the solubility of `M_(2)X_(3)`,
`M_(2)X_(3)(s)+aq hArr underset(2S)(2M^(3+)(aq))+ underset(3S)(3X^(2-)(aq))`
`:. [M^(3+)] = 2S, [X^(2-)] =3S`
Therefore, `K_(sp) = (2S)^(2) (3S)^(2) = 108S^(5) = 1.1 xx 10^(-23)`
Thus, `S^(5) = 1 xx 1^(-25)`
`S = 1.0 xx 10^(-5) mol L`