Answer
$ K_{sp} (PbBr_2) = (8.616 \times 10^{-10})$
Work Step by Step
1. Calculate the molar mass:
207.2* 1 + 79.9* 2 = 367g/mol
2. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 0.022}{ 367}$
$n(moles) = 5.995\times 10^{- 5}$
3. Find the concentration in mol/L:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 5.995\times 10^{- 5}}{ 0.1} $
$C(mol/L) = 5.995\times 10^{- 4}M$
4. Write the $K_{sp}$ expression:
$ PbBr_2(s) \lt -- \gt 1Pb^{2+}(aq) + 2Br^-(aq)$
$ K_{sp} = [Pb^{2+}]^ 1[Br^-]^ 2$
5. Determine the ions concentrations:
$[Pb^{2+}] = [PbBr_2] * 1 = [5.995 \times 10^{-4}] * 1 = 5.995 \times 10^{-4}$
$[Br^-] = [PbBr_2] * 2 = 1.199 \times 10^{-3}$
6. Calculate the $K_{sp}$:
$ K_{sp} = (5.995 \times 10^{-4})^ 1 \times (1.199 \times 10^{-3})^ 2$
$ K_{sp} = (5.995 \times 10^{-4}) \times (1.437 \times 10^{-6})$
$ K_{sp} = (8.616 \times 10^{-10})$
