Answer
The sulfate ion concentration in this solution is equal to $1.35 \times 10^{-4}M$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ PbSO_4(s) \lt -- \gt 1Pb^{2+}(aq) + 1S{O_4}^{2-}(aq)$
$1.82 \times 10^{-8} = [Pb^{2+}]^ 1[S{O_4}^{2-}]^ 1$
$1.82 \times 10^{-8} = (1.35 \times 10^{-4})^ 1[S{O_4}^{2-}]^ 1$
2. Find the molar solubility.
$1.82 \times 10^{-8}= (1.35 \times 10^{-4})^ 1 \times ([S{O_4}^{2-}])^ 1$
$ \frac{1.82 \times 10^{-8}}{1.35 \times 10^{-4}} = [S{O_4}^{2-}]^ 1$
$1.35 \times 10^{-4} M= [S{O_4}^{2-}]$
