Answer
$ K_{sp} (Ag_2CO_3) = (1.3 \times 10^{-4})$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ Ag_2CO_3(s) \lt -- \gt 2Ag^{+}(aq) + 1C{O_3}^{2-}(aq)$
$ K_{sp} = [Ag^{+}]^ 2[C{O_3}^{2-}]^ 1$
2. Determine the ion concentrations:
$[Ag^{+}] = [Ag_2CO_3] * 2 = [0.032] * 2 = 0.064$
$[C{O_3}^{2-}] = [Ag_2CO_3] * 1 = 0.032$
3. Calculate the $K_{sp}$:
$ K_{sp} = (0.064)^ 2 \times (0.032)^ 1$
$ K_{sp} = (4.2 \times 10^{-3}) \times (0.032)$
$ K_{sp} = (1.3 \times 10^{-4})$
