Questions: 8-2 Activity Coefficients 4. The Ksp value for PbI2 is 7.9 × 10^-9. a) Neglecting activities, calculate the solubility of PbI2 in a solution that is 0.0500 M in NaI and 0.0100 M in K2SO4. (Disregard any potential soluble complexes that may form.)

Solution Steps

The dissolution of \(\text{PbI}_2\) in water can be represented by the following equation:

\[ \text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq) \]

The solubility product expression (\(K_{\text{sp}}\)) for this equilibrium is:

\[ K_{\text{sp}} = [\text{Pb}^{2+}][\text{I}^-]^2 \]

The solution already contains 0.0500 M \(\text{NaI}\), which provides a common ion, \(\text{I}^-\). Therefore, the initial concentration of \(\text{I}^-\) is 0.0500 M.

Let \(s\) be the solubility of \(\text{PbI}_2\) in mol/L. At equilibrium, the concentration of \(\text{Pb}^{2+}\) will be \(s\), and the concentration of \(\text{I}^-\) will be \(0.0500 + 2s\).

Substituting these into the \(K_{\text{sp}}\) expression:

\[ K_{\text{sp}} = s(0.0500 + 2s)^2 \]

Given \(K_{\text{sp}} = 7.9 \times 10^{-9}\), we have:

\[ 7.9 \times 10^{-9} = s(0.0500 + 2s)^2 \]

Assuming \(2s\) is much smaller than 0.0500 M, we can approximate:

\[ (0.0500 + 2s)^2 \approx (0.0500)^2 \]

Thus, the equation simplifies to:

\[ 7.9 \times 10^{-9} = s(0.0500)^2 \]

Solving for \(s\):

\[ s = \frac{7.9 \times 10^{-9}}{(0.0500)^2} \]

\[ s = \frac{7.9 \times 10^{-9}}{0.0025} \]

\[ s = 3.16 \times 10^{-6} \, \text{M} \]

Final Answer