Answer
Pure water: $1.0 \times 10^{-6}M = Solubility$
$0.010M NaSCN:$ $1.0 \times 10^{-10}M = Solubility$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ AgSCN(s) \lt -- \gt 1Ag^{+}(aq) + 1SCN^{-}(aq)$
$1.0 \times 10^{-12} = [Ag^{+}]^ 1[SCN^{-}]^ 1$
2. Considering a pure solution: $[Ag^{+}] = 1S$ and $[SCN^{-}] = 1S$
$1.0 \times 10^{-12}= ( 1S)^ 1 \times ( 1S)^ 1$
$1.0 \times 10^{-12} = 1S^ 2$
$1.0 \times 10^{-12} = S^ 2$
$ \sqrt [ 2] {1.0 \times 10^{-12}} = S$
$1.0 \times 10^{-6} = S$
- This is the molar solubility value for this salt.
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3. Write the $K_{sp}$ expression:
$ AgSCN(s) \lt -- \gt 1SCN^{-}(aq) + 1Ag^{+}(aq)$
$1.0 \times 10^{-12} = [SCN^{-}]^ 1[Ag^{+}]^ 1$
$1.0 \times 10^{-12} = (0.01 + S)^ 1( 1S)^ 1$
4. Find the molar solubility.
Since 'S' has a very small value, we can approximate: $[SCN^{-}] = 0.01$
$1.0 \times 10^{-12}= (0.01)^ 1 \times ( 1S)^ 1$
$ \frac{1.0 \times 10^{-12}}{0.01} = ( 1S)^ 1$
$1.0 \times 10^{-10} = S$
