Answer
$ K_{sp} (PbCl_2) = (1.7 \times 10^{-5})$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ PbCl_2(s) \lt -- \gt 1Pb^{2+}(aq) + 2Cl^{-}(aq)$
$ K_{sp} = [Pb^{2+}]^ 1[Cl^{-}]^ 2$
2. Determine the ions concentrations:
$[Pb^{2+}] = [PbCl_2] * 1 = [0.0161] * 1 = 0.0161$
$[Cl^{-}] = [PbCl_2] * 2 = 0.0323$
3. Calculate the $K_{sp}$:
$ K_{sp} = (0.0161)^ 1 \times (0.0323)^ 2$
$ K_{sp} = (0.0161) \times (1.05 \times 10^{-3})$
$ K_{sp} = (1.701 \times 10^{-5})$
