Equilibrium - Result Question 64
####67. Find out the solubility of $Ni(OH)_2$ in $0.1 M$ $NaOH$. Given that the ionic product of $Ni(OH)_2$ is $2 \times 10^{-15}$
[2020]
(a) $2 \times 10^{-8} M$
(b) $1 \times 10^{-13} M$
(c) $1 \times 10^{8} M$
(d) $2 \times 10^{-13} M$
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Solution:
- (d) $Ni(OH)^{2} \rightarrow \underset{s}{Ni^{2+}}+\underset{2 s}{2 OH^{-}}$
$ NaOH \longrightarrow Na+OH^{-} $
Total $[OH^{-}]=2 s+0.1 \approx 0.1$
Ionic product $=[Ni]^{2+}[OH]^{2}$
$2 \times 10^{-15}=s(0.1)^{2}$
$s=2 \times 10^{-13}$
Solubility of $Ni(OH)_2=2 \times 10^{-13} M$
