Question

Solubility product of $Mg{\left(OH\right)}_2$ is $1\times {10}^{-11}$. At what $pH$, precipitation of $Mg{\left(OH\right)}_2$ will begin from $0.1MM{g}^{2+}$ solution :-

• A. $9$
• B. $5$
• C. $3$
• D. $7$

Answer 1

The correct option is A $9$

Given

$K_{sp}=1\ast {10}^{-11}$

Conc. of $M{g}^{2+}=0.1M$

Solution

$Mg(OH{)}_2=M{g}^{2+}+2(OH{)}^{-}$

For every mole of $M{g}^{2+}thereare2molesof$[OH]^{-}$

Therefore

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

${10}^{-11}=0.1\ast [OH-{]}^2$

$[OH{]}^{-}={10}^{-5}$

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

$\left[H^{+}\right]=\frac{{10}^{-14}}{{10}^{-5}}$

$\left[H^{+}\right]={10}^{-9}$

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

$pH=-log\left[{10}^{-9}\right]$

pH=9

The correct option is A