Question

At $30^\circ\text{C}$ the solubility of $\text{PbI}_2$ salt in $2\text{ M KI}$ solution will be X, if the solubility product of $\text{PbI}_2$ at $30^\circ\text{C}$ is $4\times10^{-9}$. Identify the value of X.

• A. $3.0\times10^{-8}\text{ M}$
• B. $2.4\times10^{-8}\text{ M}$
• C. $4.8\times10^{-7}\text{ M}$
• D. $6.0\times10^{-7}\text{ M}$

Hint:

The common ion effect simplifies the math significantly! By approximating the total common ion concentration to just the value from the strong electrolyte source ($0.2\text{ M}$), you avoid a cubic equation. The calculation reduces to a simple division: $X = \frac{K_{\text{sp}}}{(0.2)^2} = \frac{2.4 \times 10^{-9}}{0.04} = 6.0 \times 10^{-7}\text{ M}$.

Answer 1

The Correct Option is D

Concept: The dissolution of lead(II) iodide establishes the following solubility equilibrium: \[ \text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq) \quad K_{\text{sp}} = [\text{Pb}^{2+}][\text{I}^-]^2 \] In the presence of a strong electrolyte like potassium iodide ($\text{KI}$), the common ion effect suppresses the solubility of the salt. The total concentration of the common ion ($\text{I}^-$) is the sum of the contributions from both sources.

Step 1: Set up concentration expressions incorporating the common ion source.
Let the molar solubility of $\text{PbI}_2$ in the solution be $X\text{ mol L}^{-1}$.
• $[\text{Pb}^{2+}] = X$
• $[\text{I}^-]$ from $\text{PbI}_2 = 2X$
• $[\text{I}^-]$ from $0.2\text{ M KI} = 0.2\text{ M}$ The total equilibrium concentration of iodide ions is $[\text{I}^-] = (0.2 + 2X)$. Since the solubility is highly suppressed by the common ion effect, $2X \ll 0.2$, allowing us to approximate $[\text{I}^-] \approx 0.2\text{ M}$.

Step 2: Substitute values into the solubility product expression ($K_{\text{sp}}$) and solve for X.
\[ K_{\text{sp}} = [\text{Pb}^{2+}][\text{I}^-]^2 \implies 2.4\times10^{-9} = X \cdot (0.2)^2 \] \[ 2.4\times10^{-9} = X \cdot 0.04 \] \[ X = \frac{2.4\times10^{-9}}{0.04} = \frac{2.4}{4} \times 10^{-7} = 0.6 \times 10^{-7} = 6.0\times10^{-7}\text{ M} \]