## Chemistry and Chemical Reactivity (9th Edition)

(a) $9.2 \times 10^{-9}M$ (b) $2.2 \times 10^{-6}g/L$
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 = $2.2 \times 10^{-6}$