## Chemistry 10th Edition

1 liter of this solution has $1.342 \times 10^{-7} moles$ of $Pb^{2+}$
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 $Pb^{2+}$