Question

The solubility of \( \text{PbI}_2 \) in 0.10 M KI(aq) is ........... \( \times 10^{-7} \) M (rounded up to two decimal places). 
\(\text{[The solubility product, \( K_{\text{sp}} = 7.1 \times 10^{-9} \)]}\)

Hint:

Use the solubility product expression to calculate the solubility of a salt in a solution containing a common ion.

Answer 1

Correct Answer: 7 - 7.2

To find the solubility of \( \text{PbI}_2 \) in 0.10 M KI(aq), we start with the dissolution equation:

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

The solubility product (\( K_{\text{sp}} \)) is given as \(7.1 \times 10^{-9}\). When \( \text{PbI}_2 \) dissolves, let the solubility be \( s \). In a 0.10 M KI solution, the concentration of \(\text{I}^-\) is increased by the dissociation of KI to 0.10 M. Therefore, the equilibrium concentrations are \([\text{Pb}^{2+}] = s\) and \([\text{I}^-] = 0.10 + 2s \approx 0.10\) since \(s\) is small.

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

\( K_{\text{sp}} = [\text{Pb}^{2+}][\text{I}^-]^2 = s \times (0.10)^2\)

\(7.1 \times 10^{-9} = s \times 0.01\)

Solving for \(s\):

\(s = \frac{7.1 \times 10^{-9}}{0.01} = 7.1 \times 10^{-7} \, \text{M}\)

Answer: \(7.1 \times 10^{-7}\, \text{M}\)