Answer
(a) Solubility = $0.016M$
(b) Solubility = $4.0 \times 10^{-4}M$
Work Step by Step
(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.
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(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$
$4.0 \times 10^{-4} = S$
