If $pH = 12$
$pOH = 2$
$[O{H^ - }] = {10^{ - 2}}\,M$
${K_{sp}} = [C{d^{2 + }}]{[O{H^ - }]^2}$
$4 \times {(1.84 \times {10^{ - 5}})^3} = [C{d^{2 + }}]{[O{H^ - }]^2}$
$[C{d^{2 + }}] = \frac{{4 \times {{(1.84)}^3} \times {{10}^{ - 15}}}}{{{{10}^{ - 4}}}}$
$C{d^{2 + }} = 4 \times 6.22 \times {10^{ - 11}} = 2.49 \times {10^{ - 10}}\,M$
$A = NH_4Cl$; $ B = CH_3COONa$; $ C = NH_4OH$; $D = CH_3COOH$
[અહીં $K_b\,(NH_4OH) = 10^{-5}$ અને $log\,2 = 0.301$ ]