## Chemistry and Chemical Reactivity (9th Edition)

The molar solubility for $PbSO_4$ is equal to $1.6 \times 10^{-4}$ Correct answer: $(b)$
1. Write the $K_{sp}$ expression: $PbSO_4(s) \lt -- \gt 1Pb^{2+}(aq) + 1S{O_4}^{2-}(aq)$ $2.5 \times 10^{-8} = [Pb^{2+}]^ 1[S{O_4}^{2-}]^ 1$ 2. Considering a pure solution: $[Pb^{2+}] = 1S$ and $[S{O_4}^{2-}] = 1S$ $2.5 \times 10^{-8}= ( 1S)^ 1 \times ( 1S)^ 1$ $2.5 \times 10^{-8} = S^ 2$ $\sqrt [ 2] {2.5 \times 10^{-8}} = S$ $1.6 \times 10^{-4} = S$ - This is the molar solubility value for this salt.