Question

$K_{sp}$ of $Mg(OH{)}_2$ is $4.0\times {10}^{-6}$. At what minimum $pH,$ $M{g}^{2+}$ ions starts precipitating $0.01$ $MgCI.$

• A. $2+log2$
• B. $2-log2$
• C. $12+log2$
• D. $12-log2$

Answer 1

The correct option is C $12+log2$

According to given data, $\left[O{H}^{-}\right]$ needed for precipitation of

$Mg(OH{)}_2=(\frac{4\times {10}^{-6}}{{10}^{-2}}{)}^{1/2}=(4\times {10}^{-4}{)}^{1/2}M=2\times {10}^{-2}M$

$pOH=-log\left[O{H}^{-}\right]=2-log2$

So $pH=14-pOH=12+log2$