The solubility product of $Ni{\left(OH\right)}_2$  at 298 K is $4\times {10}^{-15}$ $mo{l}^{3}d{m}^{-9}$ . The $pH$ value of, its aqueous and saturated solution is (round off to the nearest integer)

Solution:  $Ni{\left(OH\right)}_2\rightleftharpoons N{i}_s^{2+}+2O{H}_{2s}^{-}$
 $\because K_{sp}=\left[N{i}^{2+}\right]{\left[O{H}^{-}\right]}^2$
 $$\begin{gathered} =S\times {\left(2S\right)}^2=4S^3 \\ 2\times {10}^{-15}{(mol/L)}^3=4S^3 \\ 2\times {10}^{-15}mo{l}^{3}d{m}^{-9}=4S^3 \end{gathered}$$
  Or     $S={(\frac{2}{4}\times {10}^{-15})}^{1/3}\simeq 8\times {10}^{-6}molL$
 ${\left[OH\right]}^{-}=2S=2\times 8\times {10}^{-6}molL$
 $$\begin{gathered} pOH=-{\log }_{10}\left[O{H}^{-}\right] \\ pOH=-{\log }_{10}[16\times {10}^{-6}] \\ pOH=6-\log 16=4.983\simeq 5 \\ pH=14-pOH=14-5=9 \end{gathered}$$