Answer
$ K_{sp} (PbBr_2) = (3.92 \times 10^{-5})$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ PbBr_2(s) \lt -- \gt 1Pb^{2+}(aq) + 2{Br}^{-}(aq)$
$ K_{sp} = [Pb^{2+}]^ 1[{Br}^{-}]^ 2$
2. Determine the ions concentrations:
$[Pb^{2+}] = [0.0214] $
$[{Br}^{-}] = [Pb^{2+}] * 2 = 0.0428$
** Considering a pure solution.
3. Calculate the $K_{sp}$:
$ K_{sp} = (0.0214)^ 1 \times (0.0428)^ 2$
$ K_{sp} = (0.0214) \times (1.832 \times 10^{-3})$
$ K_{sp} = (3.92 \times 10^{-5})$
