Solubility of M(OH)₂ in 0.1 M KOH is  $\left({\mathrm{K}}_{\mathrm{sp}} \mathrm{of} \mathrm{M}{\left(\mathrm{OH}\right)}_2=9\times {10}^{-18} {\mathrm{M}}^3\right)$

$$\begin{gathered} \mathrm{M}{\left(\mathrm{OH}\right)}_2\leftrightharpoons \underset{\mathrm{S}}{{\mathrm{M}}^{2+}}+\underset{2\mathrm{S}}{2{\mathrm{OH}}^{-}} \\ \mathrm{KOH}\to \underset{0.1 \mathrm{M}}{{\mathrm{K}}^{+}}+\underset{0.1 \mathrm{M}}{{\mathrm{OH}}^{-}} \\ {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{M}}^{2+}\right]{\left[{\mathrm{OH}}^{-}\right]}^2 \\ {\mathrm{K}}_{\mathrm{sp}}=\mathrm{S}{\left(2\mathrm{S}+0.1\right)}^2; \\ \text{'}\mathrm{S}\text{'} \mathrm{is} \mathrm{too} \mathrm{small} \mathrm{compared} \mathrm{to} 0.1\mathrm{M}, \mathrm{thus} 2\mathrm{S}+0.1 \simeq 0.1 ; \\ \mathrm{S}=\frac{{\mathrm{K}}_{\mathrm{sp}}}{{\left({10}^{-1}\right)}^2}=\frac{9\times {10}^{-18}}{{10}^{-2}}=9\times {10}^{-16} \mathrm{M} \end{gathered}$$