Answer
Molar solubility $Mg(OH)_2$ = $1.3 \times 10^{-4}M$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ Mg(OH)_2(s) \lt -- \gt 1Mg^{2+}(aq) + 2OH^{-}(aq)$
$8.9 \times 10^{-12} = [Mg^{2+}]^ 1[OH^{-}]^ 2$
2. Considering a pure solution: $[Mg^{2+}] = 1x$ and $[OH^{-}] = 2x$
$8.9 \times 10^{-12}= ( 1x)^ 1 \times ( 2x)^ 2$
$8.9 \times 10^{-12} = 4x^ 3$
$2.225 \times 10^{-12} = x^ 3$
$ \sqrt [ 3] {2.225 \times 10^{-12}} = x$
$1.305 \times 10^{-4} = x$
- This is the molar solubility value for this compound.