## Chemistry 9th Edition

Molar solubility $Mg(OH)_2$ = $1.3 \times 10^{-4}M$
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.