Question

How do you calculate the solubility (moles/L) of $Mg(OH{)}_2$ in water. The $K_{sp}=5.0\times {10}^{-11}$?

Answer 1

$Mg\left(OH{)}_2\right(s)\rightleftharpoons M{g}^{2+}+2O{H}^{-}$
$s$ $2s$

And $K_{sp}=\left[M{g}^{2+}\right][{}^{-}OH{]}^2$

Solubility of $Mg(OH{)}_2$, then by definition, $S=\left[M{g}^{2+}\right]$, and $2S=\left[O{H}^{-}\right]$

And thus $K_{sp}=S\times (2S{)}^2=4S^3$

$S=\sqrt[3]{3\frac{K_{sp}}{4}}$
$S=\sqrt[3]{3\frac{5.0\times {10}^{-11}}{4}}=2.32\times {10}^{-4}\dot{}mol\dot{}{L}^{-1}$