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

`A_(2)X_(3) rarr 2A^(3+) + 3X^(2-)`
`K_(sp) = [A^(3+)]^(2) [X^(2-)]^(3) = 1.1 xx 10^(-23)`
If `S =` solubility of `A_(2)X_(3)`, then
`[A^(3+)] = 2S, [X^(2-)] = 3S`
therefore, `K_(sp) = (2S)^(2)(3S)^(3) = 108S^(5)`
`= 1.1 xx 10^(-23)`
thus, `S^(5) = 1 xx 10^(-25)`
`S = 1.0 xx 10^(-5) "mol"//"L"`.