Solution:
$$
\begin{aligned}
&\mathrm{Ni}(\mathrm{OH})_{2} \rightleftharpoons \underset{(S)}{\mathrm{Ni}^{2+}}+\underset{(2 S)}{2 \mathrm{OH}^{-}} S=\text { molar conc. of } \mathrm{Ni}(\mathrm{OH})_{2} \\
&\mathrm{NaOH} \longrightarrow \mathrm{Na}^{+}+\underset{(0.1 \mathrm{M}}{\mathrm{OH}}^{-} \\
&K_{S P}=\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2} \\
&=(S)(2 S+0.1)^{2} \\
&=(S)\left(4 S^{2}+0.01+0.4 S\right) \\
&=4 S^{3}+0.01 S+0.4 S^{2}
\end{aligned}
$$
Neglecting higher power of \(S\),
$$
\begin{aligned}
&2 \times 10^{-15}=0.01 S \\
&S=2 \times 10^{-13} M
\end{aligned}
$$
