Answer
The concentration of lead (II) cations is equal to 0.0143M
Work Step by Step
1. Write the $K_{sp}$ expression:
$ PbCl_2(s) \lt -- \gt 1Pb^{2+}(aq) + 2Cl^{-}(aq)$
$1.17 \times 10^{-5} = [Pb^{2+}]^ 1[Cl^{-}]^ 2$
$1.17 \times 10^{-5} = [Pb^{2+}]^1 \times (2.86 \times 10^{-2})^ 2$
$1.17 \times 10^{-5} = [Pb^{2+}] \times 8.18 \times 10^{-4}$
$\frac{1.17 \times 10^{-5}}{8.18 \times 10^{-4}} = [Pb^{2+}]$
$[Pb^{2+}] = 0.0143M$
