## Chemistry: The Molecular Science (5th Edition)

(a) Solubility = $0.016M$ (b) Solubility = $4.0 \times 10^{-4}M$
(a) 1. Write the $K_{sp}$ expression: $PbCl_2(s) \lt -- \gt 1Pb^{2+}(aq) + 2Cl^-(aq)$ $1.6 \times 10^{-5} = [Pb^{2+}]^ 1[Cl^-]^ 2$ 2. Considering a pure solution: $[Pb^{2+}] = 1S$ and $[Cl^-] = 2S$ $1.6 \times 10^{-5}= ( 1S)^ 1 \times ( 2S)^ 2$ $1.6 \times 10^{-5} = 4S^ 3$ $4 \times 10^{-6} = S^ 3$ $\sqrt [ 3] {4 \times 10^{-6}} = S$ $0.016 = S$ - This is the molar solubility value for this salt. ----- (b) 3. Write the $K_{sp}$ expression: $PbCl_2(s) \lt -- \gt 2Cl^{-}(aq) + 1Pb^{2+}(aq)$ $1.6 \times 10^{-5} = [Cl^{-}]^ 2[Pb^{2+}]^ 1$ $1.6 \times 10^{-5} = (0.2 + S)^ 2( 1S)^ 1$ 4. Find the molar solubility. Since 'S' has a very small value, we can approximate: $[Cl^{-}] = 0.2$ $1.6 \times 10^{-5}= (0.2)^ 2 \times ( 1S)^ 1$ $1.6 \times 10^{-5}= 0.04 \times ( 1S)^ 1$ $\frac{1.6 \times 10^{-5}}{0.04} = ( 1S)^ 1$ $4.0 \times 10^{-4} = S$