Answer
The molar solubility for $Ca(OH)_2$ is equal to $0.0105M$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ Ca(OH)_2(s) \lt -- \gt 1Ca^{2+}(aq) + 2OH^-(aq)$
$4.68 \times 10^{-6} = [Ca^{2+}]^ 1[OH^-]^ 2$
2. Considering a pure solution: $[Ca^{2+}] = 1S$ and $[OH^-] = 2S$
$4.68 \times 10^{-6}= ( 1S)^ 1 \times ( 2S)^ 2$
$4.68 \times 10^{-6} = 4S^ 3$
$1.17 \times 10^{-6} = S^ 3$
$ \sqrt [ 3] {1.17 \times 10^{-6}} = S$
$0.0105 = S$
- This is the molar solubility value for this salt.
