Answer
$ K_{sp} (BaCrO_4) = (1.17 \times 10^{-10})$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ BaCrO_4(s) \lt -- \gt 1Ba^{2+}(aq) + 1Cr{O_4}^{2-}(aq)$
$ K_{sp} = [Ba^{2+}]^ 1[Cr{O_4}^{2-}]^ 1$
2. Determine the ions concentrations:
$[Ba^{2+}] = [BaCrO_4] * 1 = [1.08 \times 10^{-5}] * 1 = 1.08 \times 10^{-5}$
$[Cr{O_4}^{2-}] = [BaCrO_4] * 1 = 1.08 \times 10^{-5}$
3. Calculate the $K_{sp}$:
$ K_{sp} = (1.08 \times 10^{-5})^ 1 \times (1.08 \times 10^{-5})^ 1$
$ K_{sp} = (1.08 \times 10^{-5}) \times (1.08 \times 10^{-5})$
$ K_{sp} = (1.17 \times 10^{-10})$
