Question

$K_{sp}$ of salt $Ni(OH{)}_2$ is $2\times {10}^{-15}$, then molar solubility of $Ni(OH{)}_2$ in $0.01MNaOH$ is:

• A. $2\times {10}^{-15}M$
• B. $2^{\frac{1}{3}}\times {10}^{-5}M$
• C. ${10}^{-7}M$
• D. $2\times {10}^{-11}M$

Answer 1

The correct option is D $2\times {10}^{-11}M$

$Ni(OH{)}_2\rightleftharpoons N{i}^{2+}+2O{H}^{-}[NaOH]=[O{H}^{-}]=0.01M$
$K_{sp}=\left[N{i}^{2+}\right][O{H}^{-}{]}^2$
$2\times {10}^{-15}=S(0.01{)}^2$
$S=2\times {10}^{-11}M$