Question

Find out the solubility of $Ni(OH{)}_2$ in $0.1MNaOH$. Given that the ionic product of $Ni(OH{)}_2$ is $2\times {10}^{-15}$.

• A. $2\times {10}^{-8}M$
• B. $1\times {10}^{-13}M$
• C. $1\times {10}^{8}M$
• D. $2\times {10}^{-13}M$

Answer 1

The correct option is D $2\times {10}^{-13}M$

$\alpha =1$ for $NaOH$
$NaOH\left(aq\right)\to N{a}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$
$0.1M$ $0.1M$
$Ni\left(OH{)}_2\right(s)\leftrightharpoons N{i}^{+2}(aq)+2O{H}^{-}(aq)$
$S$ $(0.1+2S)$
$Ionicproduct=\left[{Ni}^{+2}\right]{\left[{OH}^{-}\right]}^{2}2\times {10}^{-15}=\left[{Ni}^{+2}\right]{\left[{10}^{-1}\right]}^{2}2\times {10}^{-13}=\left[{Ni}^{+2}\right]$

Option D is correct.