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 $2\times {10}^{-13}M$

$NaOH$ is a strong base
$\alpha =1$ for $NaOH$
$NaOH\left(aq\right)\to N{a}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$
$0.1M0.1M$

$Ni(OH{)}_2\rightleftharpoons N{i}^{2+}(aq)+2O{H}^{-}(aq)$
$S(0.1+2S)$
$Ni(OH{)}_2$ is less soluble, hence $0.1>>>2S$
$\text{Ionic product}=\left[N{i}^{2+}\right][O{H}^{-}{]}^2$
$2\times {10}^{-15}=\left[N{i}^{2+}\right][O{H}^{-}{]}^2$
$2\times {10}^{-15}=\left[S\right][{10}^{-1}{]}^2$
$2\times {10}^{-13}=S$
Hence, option (d) is the correct answer.
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