Question

Calculate the molar solubility of $Ni{\left(OH\right)}_2$ in $0.10M$ solution. The ionic product of $Ni{\left(OH\right)}_2$ is $2.0\times {10}^{-15}$.

Answer 1

$Ksp=\left[{Ni}^{+2}\right]{\left[{OH}^{-}\right]}^{2}2\times {10}^{-15}=\left[{Ni}^{+2}\right]{\left[{10}^{-1}\right]}^{2}2\times {10}^{-3}=\left[{Ni}^{+2}\right]$

Molar Solubility of $\left[{Ni}^{+2}\right]=2\times {10}^{-13}M$