Answer
$ K_{sp} (HgI_2) = (3.2 \times 10^{-29})$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ HgI_2(s) \lt -- \gt 1Hg^{2+}(aq) + 2I^{-}(aq)$
$ K_{sp} = [Hg^{2+}]^ 1[I^{-}]^ 2$
2. Determine the ions concentrations:
$[Hg^{2+}] = [HgI_2] * 1 = [2 \times 10^{-10}] * 1 = 2 \times 10^{-10}M$
$[I^{-}] = [HgI_2] * 2 = 4 \times 10^{-10}M$
3. Calculate the $K_{sp}$:
$ K_{sp} = (2 \times 10^{-10})^ 1 \times (4 \times 10^{-10})^ 2$
$ K_{sp} = (2 \times 10^{-10}) \times (1.6 \times 10^{-19})$
$ K_{sp} = (3.2 \times 10^{-29})$
