Answer
Since $Q_{sp} \lt K_{sp}$, there will be no precipitation
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. 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
