Answer
Molar solubility $Cd(OH)_2$ = $2.5 \times 10^{-4} M$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ Cd(OH)_2(s) \lt -- \gt 1Cd^{2+}(aq) + 2OH^{-}(aq)$
$5.9 \times 10^{-11} = [Cd^{2+}]^ 1[OH^{-}]^ 2$
2. Considering a pure solution: $[Cd^{2+}] = 1x$ and $[OH^{-}] = 2x$
$5.9 \times 10^{-11}= ( 1x)^ 1 \times ( 2x)^ 2$
$5.9 \times 10^{-11} = 4x^ 3$
$1.475 \times 10^{-11} = x^ 3$
$ \sqrt [ 3] {1.475 \times 10^{-11}} = x$
$2.452 \times 10^{-4} = x$
- This is the molar solubility value for this compound.