Question

The $K_{sp}$ of $Mg{\left(OH\right)}_2$ is $1\times {10}^{-12}$ $0.10M$ $Mg{\left(OH\right)}_2$ will precipitate at the limiting pH:

• A. $3$
• B. $9$
• C. $5$
• D. $8$

Answer 1

Answer: (B)

$9$

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

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

$0.1[O{H}^{-}{]}^2={10}^{-12}$

$\left[O{H}^{-}\right]=0.316\times {10}^{-5}$

$pOH=5.5$ or $pH=14-5.5=8.5$

So it will precipitate when $pH>8.5$.