If the ionic product of Ni(OH)₂ is 1 .9 x 10^-15, then the molar solubility of Ni(OH)₂ in 1.0 M NaOH is

$\mathrm{NaOH}\rightleftharpoons {\mathrm{Na}}^{+}+{\mathrm{OH}}^{-}$

                      CM       CM
$\mathrm{Ni}(\mathrm{OH}{)}_2\rightleftharpoons {\mathrm{Ni}}^{2+}+2{\mathrm{OH}}^{-}$

   X                X            X
$$\begin{gathered} \text{ Total }\left[{\mathrm{OH}}^{-}\right]=\mathrm{x}+\mathrm{C} \\ \therefore {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ni}}^{2+}\right]{\left[{\mathrm{OH}}^{-}\right]}^2=\mathrm{x}(\mathrm{x}+\mathrm{C}{)}^2 \\ =\mathrm{x}\left({\mathrm{x}}^2+2\mathrm{Cx}+{\mathrm{C}}^2\right) \\ \text{ or }{\mathrm{K}}_{\mathrm{sp}}={\mathrm{xC}}^2\text{ (neglecting higher powers of }\mathrm{x}\text{ ) } \\ \mathrm{x}=\frac{{\mathrm{K}}_{\mathrm{sp}}}{{\mathrm{C}}^2}=\frac{1.9\times {10}^{-15}}{(1{)}^2}=1.9\times {10}^{-15}\mathrm{M} \end{gathered}$$