Answer
(a) Solubility = 0.013 M
(b) Solubility = $2.2 \times 10^{-4}$ M
(c) Solubility = $3.3 \times 10^{-3}$ M
Work Step by Step
(a)
1. Write the $K_{sp}$ expression:
$ PbBr_2(s) \lt -- \gt 1Pb^{2+}(aq) + 2Br^{-}(aq)$
$8.9 \times 10^{-6} = [Pb^{2+}]^ 1[Br^{-}]^ 2$
2. Considering a pure solution: $[Pb^{2+}] = 1S$ and $[Br^{-}] = 2S$
$8.9 \times 10^{-6}= ( 1S)^ 1 \times ( 2S)^ 2$
$8.9 \times 10^{-6} = 4S^ 3$
$2.225 \times 10^{-6} = S^ 3$
$ \sqrt [ 3] {2.225 \times 10^{-6}} = S$
$0.01305 = S$
- This is the molar solubility value for this salt in pure water.
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(b)
1. Write the $K_{sp}$ expression:
$ PbBr_2(s) \lt -- \gt 2Br^-(aq) + 1Pb^{2+}(aq)$
$8.9 \times 10^{-6} = [Br^-]^ 2[Pb^{2+}]^ 1$
$8.9 \times 10^{-6} = (0.2 + S)^ 2( 1S)^ 1$
2. Find the molar solubility.
Since 'S' has a very small value, we can approximate: $[Br^-] = 0.2$
$8.9 \times 10^{-6}= (0.2)^ 2 \times ( 1S)^ 1$
$8.9 \times 10^{-6}= 0.04 \times ( 1S)^ 1$
$ \frac{8.9 \times 10^{-6}}{0.04} = ( 1S)^ 1$
$2.225 \times 10^{-4} = S$
(c)
1. Write the $K_{sp}$ expression:
$ PbBr_2(s) \lt -- \gt 1Pb^{2+}(aq) + 2Br^{-}(aq)$
$8.9 \times 10^{-6} = [Pb^{2+}]^ 1[Br^{-}]^ 2$
$8.9 \times 10^{-6} = (0.2 + S)^ 1( 2S)^ 2$
2. Find the molar solubility.
Since 'S' has a very small value, we can approximate: $[Pb^{2+}] = 0.2$
$8.9 \times 10^{-6}= (0.2)^ 1 \times ( 2S)^ 2$
$ \frac{8.9 \times 10^{-6}}{0.2} = ( 2S)^ 2$
$4.45 \times 10^{-5} = ( 2S)^ 2$
$ \sqrt [ 2] {4.45 \times 10^{-5}} = 2S$
$6.671 \times 10^{-3} = 2S$
$3.335 \times 10^{-3} = S$
