The molar solubility of $\mathrm{Cd}{\left(\mathrm{OH}\right)}_2$ is $1.84\times {10}^{-5} \mathrm{M}$ in water. The expected solubility of $\mathrm{Cd}{\left(\mathrm{OH}\right)}_2$ in a buffer solution of $\mathrm{pH}=12$ is:

$\mathrm{Cd}{\left(\mathrm{OH}\right)}_2\rightleftharpoons \underset{\mathrm{s}}{{\mathrm{Cd}}^{2+}}+2\underset{2\mathrm{S}}{{\mathrm{OH}}^{-}}$ At equilibrium,

${\mathrm{K}}_{\mathrm{sp}}=\mathrm{s}{\left(2\mathrm{s}\right)}^2=4{\mathrm{s}}^3$

Solubility in buffer solution having $$\begin{gathered} \mathrm{pH}=12\left[{\mathrm{OH}}^{-}\right]={10}^{-2} \\ \mathrm{Cd}{\left(\mathrm{OH}\right)}_2\rightleftharpoons \underset{{\mathrm{s}}^{\text{'}}}{{\mathrm{Cd}}^{2+}}+\underset{2\mathrm{S}+{10}^{-2}\approx {10}^{-2}}{2{\mathrm{OH}}^{-}} \\ \therefore {\mathrm{K}}_{\mathrm{sp}}=4\times {\left(1.84\times {10}^{-5}\right)}^3={\mathrm{s}}^{\text{'}}{\left({10}^{-2}\right)}^2 \\ \Rightarrow {\mathrm{s}}^{\text{'}}=\frac{24.9\times {10}^{-15}}{{10}^{-4}}=2.49\times {10}^{-10}\mathrm{M} \end{gathered}$$