Answer
The solubility of $PbCl_2$ in this solution is equal to :$6.4 \times 10^{-5}$
Work Step by Step
- $NaCl$: 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$
$6.4 \times 10^{-5} = S$
