Answer
$(a) Solubility = 1.4 \times 10^{-5}M$
$(b) Solubility = 1.4 \times 10^{-7}M$
Work Step by Step
(a)
1. Write the $K_{sp}$ expression:
$ BaCrO_4(s) \lt -- \gt 1Ba^{2+}(aq) + 1CrO{_4}^{2-}(aq)$
$2.1 \times 10^{-10} = [Ba^{2+}]^ 1[CrO{_4}^{2-}]^ 1$
2. Considering a pure solution: $[Ba^{2+}] = 1S$ and $[CrO{_4}^{2-}] = 1S$
$2.1 \times 10^{-10}= ( 1S)^ 1 \times ( 1S)^ 1$
$2.1 \times 10^{-10} = 1S^ 2$
$ \sqrt [ 2] {2.1 \times 10^{-10}} = S$
$1.4 \times 10^{-5} = S$
- This is the molar solubility value for this salt.
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(b)
3. Write the $K_{sp}$ expression:
$ BaCrO_4(s) \lt -- \gt 1Cr{O_4}^{2-}(aq) + 1Ba^{2+}(aq)$
$2.1 \times 10^{-10} = [Cr{O_4}^{2-}]^ 1[Ba^{2+}]^ 1$
$2.1 \times 10^{-10} = (1.5 \times 10^{-3} + S)^ 1( 1S)^ 1$
4. Find the molar solubility.
Since 'S' has a very small value, we can approximate: $[Cr{O_4}^{2-}] = 1.5 \times 10^{-3}M$
$2.1 \times 10^{-10}= (1.5 \times 10^{-3})^ 1 \times ( 1S)^ 1$
$2.1 \times 10^{-10}= (1.5 \times 10^{-3})^ 1 \times ( 1S)^ 1$
$2.1 \times 10^{-10}= 1.5 \times 10^{-3} \times ( 1S)^ 1$
$ \frac{2.1 \times 10^{-10}}{1.5 \times 10^{-3}} = ( 1S)^ 1$
$1.4 \times 10^{-7} = ( 1S)^ 1$
$1.4 \times 10^{-7} = S$
