Answer
The solubility of $CaCO_3$ is equal to $7.04 \times 10^{-5}M$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ CaCO_3(s) \lt -- \gt 1Ca^{2+}(aq) + 1C{O_3}^{2-}(aq)$
$4.96 \times 10^{-9} = [Ca^{2+}]^ 1[C{O_3}^{2-}]^ 1$
2. Considering a pure solution: $[Ca^{2+}] = 1S$ and $[C{O_3}^{2-}] = 1S$
$4.96 \times 10^{-9}= ( 1S)^ 1 \times ( 1S)^ 1$
$4.96 \times 10^{-9} = S^ 2$
$ \sqrt [ 2] {4.96 \times 10^{-9}} = S$
$7.04 \times 10^{-5} = S$
- This is the molar solubility value for this salt.