Question

The $K_{sp}$ of $Mg(OH{)}_2$ is $1\times 10^{-12};0.01MMgC{l}_2$ will precipitate at limiting $pH$ of _____.

Answer 1

Answer: 9

$\left[M{g}^{2+}\right]=0.01M,K_{sp}=\left[M{g}^{2+}\right][OH{]}^2$

$1\times 10^{-12}=0.01\times [OH{]}^2,[OH]=10^{-5}M$

$pOH=-\log 10^{-5}=5,pH=14-5=9$