Question

At ${25}^{\circ }C$, the solubility product of $Mg(OH{)}_2$ is $1.0\times {10}^{-11}$. At which $pH$, will $M{g}^{2+}$ ions start precipitating in the form of $Mg(OH{)}_2$ from a solution of $0.001MM{g}^{2+}ions$?

• A. $8$
• B. $9$
• C. $10$
• D. $11$

Answer 1

The correct option is C $10$

Given

Solubility of $Mg(OH{)}_2$ = $1.0\ast {10}^{-11}$

Solution molarity is 0.001

Solution

$Mg\left(OH{)}_2\right(s)\to M{g}^{2+}+2O{H}^{-1}$

Moles of $Mg\left(OH{)}_2\right(s)$ =moles of$\left[M{g}^{2+}\right]$

Ksp=$\left[M{g}^{2+}\right]\ast [O{H}^{-1}{]}^2$

$1\ast {10}^{-11}=0.001\ast [O{H}^{-1}{]}^2$

$\left[O{H}^{-1}\right]={10}^{-4}$M

$[H{]}^{+}$=$\frac{{10}^{-14}}{\left[O{H}^{-1}\right]}$

=$\frac{{10}^{-14}}{{10}^{-4}}$

$={10}^{-10}$M

$pH=-log[H{]}^{+}$

putting values

pH=10

The correct option is C