## Chemistry 9th Edition

$K_{sp} (PbBr_2) = (3.92 \times 10^{-5})$
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 $K_{sp}$: $K_{sp} = (0.0214)^ 1 \times (0.0428)^ 2$ $K_{sp} = (0.0214) \times (1.832 \times 10^{-3})$ $K_{sp} = (3.92 \times 10^{-5})$