## Chemistry and Chemical Reactivity (9th Edition)

1. Write the $K_{sp}$ expression: $PbBr_2(s) \lt -- \gt 1Pb^{2+}(aq) + 2Br^-(aq)$ $6.6 \times 10^{-6} = [Pb^{2+}]^ 1[Br^-]^ 2$ 2. Considering a pure solution: $[Pb^{2+}] = 1S$ and $[Br^-] = 2S$ $6.6 \times 10^{-6}= ( 1S)^ 1 \times ( 2S)^ 2$ $6.6 \times 10^{-6} = 4S^ 3$ $1.7 \times 10^{-6} = S^ 3$ $\sqrt [ 3] {1.7 \times 10^{-6}} = S$ $0.012 = S$ - This is the molar solubility value for this salt. 4. Determine the molar mass of this compound (PbBr_2): 207.2* 1 + 79.9* 2 = 367g/mol 4. Calculate the mass $mm(g/mol) = \frac{mass(g)}{n(mol)}$ $mm(g/mol) * n(mol) = mass(g)$ $367 * 0.012 = mass(g)$ $4.4 = mass(g)$