Answer
The molar solubility of this compound in pure water is equal to: $3.7 \times 10^{-6}M$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ ZnCO_3(s) \lt -- \gt 1Zn^{2+}(aq) + 1C{O_3}^{2-}(aq)$
$1.4 \times 10^{-11} = [Zn^{2+}]^ 1[C{O_3}^{2-}]^ 1$
2. Considering a pure solution: $[Zn^{2+}] = 1x$ and $[C{O_3}^{2-}] = 1x$
$1.4 \times 10^{-11}= ( 1x)^ 1 \times ( 1x)^ 1$
$1.4 \times 10^{-11} = 1x^ 2$
$1.4 \times 10^{-11} = x^ 2$
$ \sqrt [ 2] {1.4 \times 10^{-11}} = x$
$3.742 \times 10^{-6}M = x$
- This is the molar solubility value for this salt.
