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

The solubility of $PbCl_2$ in this solution is equal to :$6.4 \times 10^{-5}$
- $NaCl$: Totally dissociated $[Cl^-] = [NaCl] = 0.50M$ 1. Write the $K_{sp}$ expression: $PbCl_2(s) \lt -- \gt 2Cl^{-}(aq) + 1Pb^{2+}(aq)$ $1.6 \times 10^{-5} = [Cl^{-}]^ 2[Pb^{2+}]^ 1$ $1.6 \times 10^{-5} = (0.50 + S)^ 2( 1S)^ 1$ 2. Find the molar solubility. Since 'S' has a very small value, we can approximate: $[Cl^{-}] = 0.5$ $1.6 \times 10^{-5}= (0.50)^ 2 \times ( 1S)^ 1$ $1.6 \times 10^{-5}= 0.25 \times ( 1S)^ 1$ $\frac{1.6 \times 10^{-5}}{0.25} = ( 1S)^ 1$ $6.4 \times 10^{-5} = S$