Answer
The molar solubility of $PbS$ in pure water is equal to $9.51 \times 10^{-15}M$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ PbS(s) \lt -- \gt 1Pb^{2+}(aq) + 1S^{2-}(aq)$
$9.04 \times 10^{-29} = [Pb^{2+}]^ 1[S^{2-}]^ 1$
2. Considering a pure solution: $[Pb^{2+}] = 1S$ and $[S^{2-}] = 1S$
$9.04 \times 10^{-29}= ( 1S)^ 1 \times ( 1S)^ 1$
$9.04 \times 10^{-29} = S^ 2$
$ \sqrt [ 2] {9.04 \times 10^{-29}} = S$
$9.51 \times 10^{-15} = S$
- This is the molar solubility value for this salt.
** Check Table 15.2 for $K_{sp}$ values.
