d
\({K_{sp}}\) of \(Cd{(OH)_2} = 4{s^3} = 4 \times {(1.84 \times {10^{ - 5}})^3}\)

If \(pH = 12\)

\(pOH = 2\)

\([O{H^ - }] = {10^{ - 2}}\,M\)

\({K_{sp}} = [C{d^{2 + }}]{[O{H^ - }]^2}\)

\(4 \times {(1.84 \times {10^{ - 5}})^3} = [C{d^{2 + }}]{[O{H^ - }]^2}\)

\([C{d^{2 + }}] = \frac{{4 \times {{(1.84)}^3} \times {{10}^{ - 15}}}}{{{{10}^{ - 4}}}}\)

\(C{d^{2 + }} = 4 \times 6.22 \times {10^{ - 11}} = 2.49 \times {10^{ - 10}}\,M\)