Answer
(a) 0.012 M
(b) 4.4 g/L
Work Step by Step
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)$
$4.4 = mass(g)$
