Question

Calculate the concentration (in $mol{L}^{-1}$) of hydroxide ion necessary to precipitate $Mg(OH{)}_2$ from $0.1M$ solution of $MgS{O}_4$.

$[K_{sp}$ of $Mg(OH{)}_2=5.5\times {10}^{-12}]$

• A. $7.4\times {10}^{-6}$
• B. $6.8\times {10}^{-5}$
• C. $7.4\times {10}^{-7}$
• D. $6.8\times {10}^{-8}$

Answer 1

Answer: (A)

$7.4\times {10}^{-6}$

The concentration of magnesium ion in $0.1M$ solution of magnesium sulphate is $0.1M$.

The solubility product of magnesium hydroxide is $5.5\times {10}^{-12}$.

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

Substituting values in the above expression, we get

$5.5\times {10}^{-12}=\left(0.1\right)[O{H}^{-}{]}^2$

Hence, $[O{H}^{-}{]}^2=\frac{5.5\times {10}^{-12}}{0.1}=5.5\times {10}^{-11}$

$\left[O{H}^{-}\right]=\sqrt{5.5\times {10}^{-11}}=7.4\times {10}^{-6}$

Hence, the minimum concentration of hydroxide ion necessary for the precipitation of $Mg(OH{)}_2$ is $7.4\times {10}^{-6}M$.