## Chemistry and Chemical Reactivity (9th Edition)

Since $Q_{sp} \lt K_{sp}$, there will be no precipitation
1. Write the $K_{sp}$ expression: $PbCl_2(s) \lt -- \gt 1Pb^{2+}(aq) + 2Cl^-(aq)$ $K_{sp} = [Pb^{2+}]^ 1[Cl^-]^ 2$ 2. Find the $Q_{sp}$ value $Q_{sp} = (1.2 \times 10^{-3})^ 1 \times (0.01)^ 2$ $Q_{sp} = (1.2 \times 10^{-3})^ 1 \times (0.01)^ 2$ $Q_{sp} = 1.2 \times 10^{-3} \times (1 \times 10^{-4}$) $Q_{sp} = 1.2 \times 10^{-7}$ $K_{sp} (PbCl_2) = 1.7 \times 10^{-5}$ Since $Q_{sp} \lt K_{sp}$, there will be no precipitation