Answer
$ K_{sp} (ZnC_2O_4) = (6.2 \times 10^{-5})$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ ZnC_2O_4(s) \lt -- \gt 1Zn^{2+}(aq) + 1C_2O{_4}^{2-}(aq)$
$ K_{sp} = [Zn^{2+}]^ 1[C_2O{_4}^{2-}]^ 1$
2. Determine the ion concentrations:
$[Zn^{2+}] = [ZnC_2O_4] * 1 = [7.9 \times 10^{-3}] * 1 = 7.9 \times 10^{-3}$
$[C_2O{_4}^{2-}] = [ZnC_2O_4] * 1 = 7.9 \times 10^{-3}$
3. Calculate the $K_{sp}$:
$ K_{sp} = (7.9 \times 10^{-3})^ 1 \times (7.9 \times 10^{-3})^ 1$
$ K_{sp} = (7.9 \times 10^{-3}) \times (7.9 \times 10^{-3})$
$ K_{sp} = (6.2 \times 10^{-5})$
