Question

$Mg(OH{)}_2$ has a solubility product equal to $1.2\times 10^{-11}$. A solution contains $0.10mol$ of magnesium ions per liter. The maximum $pH$ of the solution from which $Mg(OH{)}_2$ will not precipitate is

• A. 4.96
• B. 9.04
• C. 12.13
• D. 6.94

Answer 1

Answer: (B)

$Mg(OH{)}_2\rightleftharpoons M{g}^{2+}+2O{H}^{-}{K}_{sp}=1.2\times 10^{-11}$

$[O{H}^{-}]$ higher than that contained in a saturated solution will cause precipitation. Therefore, the solution must be at the point of attaining equilibrium and the concentration of ions in solution must be no greater than those required to satisfy the solubility product constant. In this solution, $\left[M{g}^{2+}\right]=0.10M$ and $\left[M{g}^{2+}\right][OH{]}^2=1.2\times 10^{-11}$

$[OH{]}^2=\frac{1.2\times 10^{-11}}{0.10}=1.2\times 10^{-10}$ or $[OH]=1.1\times 10^{-5}$

or $pOH=4.96$

or $pH=9.04$