Question

The solubility product constant Ksp of $Mg(OH{)}_2$ is $9.0\times {10}^{-12}$. If a solution is $0.010$ M with respect to $M{g}^{2+}$ ion, what is the maximum hydroxide ion concentration which could be present without causing the precipitation of $Mg(OH{)}_2$.

• A. $1.5\times {10}^{-7}$M
• B. $3.0\times {10}^{-7}$ M
• C. $1.5\times {10}^{-5}$M
• D. $3.0\times {10}^{-5}$ M

Answer 1

The correct option is D $3.0\times {10}^{-5}$ M

$mg(OH{)}_2\to m{g}^{+2}+2O{H}^{-}$

$K_{sp}=9.0\times {10}^{-12}=\left[m{g}^{2+}\right][O{H}^{-}{]}^2$

$9\times {10}^{-12}=\left[{10}^{-2}\right][O{H}^{-}{]}^2$ (given $\left[m{g}^{+2}\right]={10}^{-2}m$)

$\left[O{H}^{-}\right]=3\times {10}^{-5}M$