Answer
(a) $9.2 \times 10^{-9}M$
(b) $2.2 \times 10^{-6}g/L$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ AgI(s) \lt -- \gt 1Ag^{+}(aq) + 1I^-(aq)$
$8.5 \times 10^{-17} = [Ag^{+}]^ 1[I^-]^ 1$
2. Considering a pure solution: $[Ag^{+}] = 1S$ and $[I^-] = 1S$
$8.5 \times 10^{-17}= ( 1S)^ 1 \times ( 1S)^ 1$
$8.5 \times 10^{-17} = 1S^ 2$
$8.5 \times 10^{-17} = S^ 2$
$ \sqrt [ 2] {8.5 \times 10^{-17}} = S$
$9.22 \times 10^{-9} = S$
- This is the molar solubility value for this salt.
3. Determine the molar mass of this compound (AgI):
107.87* 1 + 126.9* 1 = 234.77g/mol
4. Calculate the mass
$mm(g/mol) = \frac{mass(g)}{n(mol)}$
$mm(g/mol) * n(mol) = mass(g)$
$ 234.77 * 9.22 \times 10^{-9} = mass(g)$
$2.2 \times 10^{-6} = mass(g)$
Therefore, the concentration in g/L = $2.2 \times 10^{-6} $
