Answer
1 liter of this solution has $1.342 \times 10^{-7} moles$ of $Pb^{2+}$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ PbCrO_4(s) \lt -- \gt 1Cr{O_4}^{2-}(aq) + 1Pb^{2+}(aq)$
$1.8 \times 10^{-14} = [Cr{O_4}^{2-}]^ 1[Pb^{2+}]^ 1$
2. Considering a pure solution: $[Cr{O_4}^{2-}] = 1x$ and $[Pb^{2+}] = 1x$
$1.8 \times 10^{-14}= ( 1x)^ 1 \times ( 1x)^ 1$
$1.8 \times 10^{-14} = 1x^ 2$
$1.8 \times 10^{-14} = x^ 2$
$ \sqrt [ 2] {1.8 \times 10^{-14}} = x$
$1.342 \times 10^{-7} = x$
- This is the molar solubility value for this salt.
- Therefore, one saturated solution of this salt has $1.342 \times 10^{-7}M PbCrO_4$.
Which means: 1 Liter of solution = $1.342 \times 10^{-7} moles$ of $Pb^{2+}$
