Answer
The molar solubility of $MgCO_3$ in pure water is equal to $2.61 \times 10^{-3}M$.
Work Step by Step
1. Write the $K_{sp}$ expression:
$ MgCO_3(s) \lt -- \gt 1Mg^{2+}(aq) + 1CO_3^{2-}(aq)$
$6.82 \times 10^{-6} = [Mg^{2+}]^ 1[CO_3^{2-}]^ 1$
2. Considering a pure solution: $[Mg^{2+}] = 1S$ and $[CO_3^{2-}] = 1S$
$6.82 \times 10^{-6}= ( 1S)^ 1 \times ( 1S)^ 1$
$6.82 \times 10^{-6} = S^ 2$
$ \sqrt [ 2] {6.82 \times 10^{-6}} = S$
$2.61 \times 10^{-3} = S$
- This is the molar solubility value for this salt.