Question

The solubility product of $Mg(OH{)}_2$ at ${25}^{\circ }C$ is $1.4\times {10}^{-11}$. The solubility of $Mg(OH{)}_2$ in gm per litre is:

• A. $1.4\times {10}^{-3}$
• B. $2.8\times {10}^{-3}$
• C. $1.11\times {10}^{-6}$
• D. $8.7\times {10}^{-3}$

Answer 1

The correct option is D

$8.7\times {10}^{-3}$

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

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

Let us take the the solubility of $Mg(OH{)}_2=Smoles/litre$

$\left[M{g}^{2+}\right]=1\times S$; $\left[O{H}^{-}\right]=2S$ and

$1.4\times {10}^{-11}=4S^3$

Solving the cubic for $S$,

$S=1.5\times {10}^{-4}moles/litre$

$=8.7\times {10}^{-3}$ gram per litre
(since Molar mass $Mg(OH{)}_2=58)$