Answer
$ K_{sp} (MnCO_3) = (1.8 \times 10^{-11})$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ MnCO_3(s) \lt -- \gt 1Mn^{2+}(aq) + 1C{O_3}^{2-}(aq)$
$ K_{sp} = [Mn^{2+}]^ 1[C{O_3}^{2-}]^ 1$
2. Determine the ion concentrations:
$[Mn^{2+}] = [MnCO_3] * 1 = [4.2 \times 10^{-6}] * 1 = 4.2 \times 10^{-6}$
$[C{O_3}^{2-}] = [MnCO_3] * 1 = 4.2 \times 10^{-6}$
3. Calculate the $K_{sp}$:
$ K_{sp} = (4.2 \times 10^{-6})^ 1 \times (4.2 \times 10^{-6})^ 1$
$ K_{sp} = (4.2 \times 10^{-6}) \times (4.2 \times 10^{-6})$
$ K_{sp} = (1.764 \times 10^{-11}) = (1.8 \times 10^{-11})$
