Question:
$\text{PQ}_2$ is a sparingly soluble salt with solubility product $\text{K}_{sp} = 4 \times 10^{-12}$ in aqueous medium at some given temperature. It is observed that upon addition of a highly soluble salt RQ at the same temperature, the solubility of $\text{PQ}_2$ drops by a factor of 100. The concentration (in millimoles per litre) of added RQ in the solution is closest to:
$\text{PQ}_2$ is a sparingly soluble salt with solubility product $\text{K}_{sp} = 4 \times 10^{-12}$ in aqueous medium at some given temperature. It is observed that upon addition of a highly soluble salt RQ at the same temperature, the solubility of $\text{PQ}_2$ drops by a factor of 100. The concentration (in millimoles per litre) of added RQ in the solution is closest to:
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In common-ion effect calculations, always assume that the concentration of the common ion comes entirely from the highly soluble salt ($C \gg 2S$). This greatly simplifies the mathematics and is highly accurate.
Updated On: Jun 16, 2026
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The question asks for the concentration of a highly soluble salt RQ that must be added to a saturated solution of the sparingly soluble salt $\text{PQ}_2$ to reduce its solubility by a factor of 100 via the common-ion effect.
Step 2: Key Formula or Approach:
1. The solubility product constant ($K_{sp}$) of $\text{PQ}_2$ is:
\[ K_{sp} = [\text{P}^{2+}][\text{Q}^-]^2 \]
2. Let $S_0$ be the solubility of $\text{PQ}_2$ in pure water:
\[ K_{sp} = (S_0)(2S_0)^2 = 4S_0^3 \]
Step 3: Detailed Explanation:
• Let us first calculate the solubility ($S_0$) of $\text{PQ}_2$ in pure water:
\[ 4S_0^3 = 4 \times 10^{-12} \implies S_0^3 = 10^{-12} \implies S_0 = 10^{-4}\ \text{M} \]
• The problem states that the addition of the highly soluble salt RQ causes the solubility of $\text{PQ}_2$ to drop by a factor of $100$.
• Therefore, the new solubility ($S$) of $\text{PQ}_2$ is:
\[ S = \frac{S_0}{100} = \frac{10^{-4}}{100} = 10^{-6}\ \text{M} \]
• Let the concentration of the added soluble salt RQ be $C$. Since RQ dissociates completely:
\[ \text{RQ(aq)} \rightarrow \text{R}^+\text{(aq)} + \text{Q}^-\text{(aq)} \]
• This contributes a concentration $C$ of $\text{Q}^-$ ions to the solution.
• Now, the equilibrium concentration of ions in the solution is:
\[ [\text{P}^{2+}] = S = 10^{-6}\ \text{M} \]
\[ [\text{Q}^-] = 2S + C \approx C \quad \text{(since the solubility } S \text{ is extremely small, } C \gg 2S) \]
• Substituting these values into the $K_{sp}$ expression:
\[ K_{sp} = [\text{P}^{2+}][\text{Q}^-]^2 \]
\[ 4 \times 10^{-12} = (10^{-6})(C)^2 \]
\[ C^2 = \frac{4 \times 10^{-12}}{10^{-6}} = 4 \times 10^{-6} \]
\[ C = \sqrt{4 \times 10^{-6}} = 2 \times 10^{-3}\ \text{M} \]
• Converting the concentration from moles per litre to millimoles per litre:
\[ C = 2 \times 10^{-3}\ \text{mol/L} = 2\ \text{mmol/L} \]
Step 4: Final Answer:
Therefore, the concentration of the added RQ is closest to $2\ \text{mmol/L}$ (Option A).
The question asks for the concentration of a highly soluble salt RQ that must be added to a saturated solution of the sparingly soluble salt $\text{PQ}_2$ to reduce its solubility by a factor of 100 via the common-ion effect.
Step 2: Key Formula or Approach:
1. The solubility product constant ($K_{sp}$) of $\text{PQ}_2$ is:
\[ K_{sp} = [\text{P}^{2+}][\text{Q}^-]^2 \]
2. Let $S_0$ be the solubility of $\text{PQ}_2$ in pure water:
\[ K_{sp} = (S_0)(2S_0)^2 = 4S_0^3 \]
Step 3: Detailed Explanation:
• Let us first calculate the solubility ($S_0$) of $\text{PQ}_2$ in pure water:
\[ 4S_0^3 = 4 \times 10^{-12} \implies S_0^3 = 10^{-12} \implies S_0 = 10^{-4}\ \text{M} \]
• The problem states that the addition of the highly soluble salt RQ causes the solubility of $\text{PQ}_2$ to drop by a factor of $100$.
• Therefore, the new solubility ($S$) of $\text{PQ}_2$ is:
\[ S = \frac{S_0}{100} = \frac{10^{-4}}{100} = 10^{-6}\ \text{M} \]
• Let the concentration of the added soluble salt RQ be $C$. Since RQ dissociates completely:
\[ \text{RQ(aq)} \rightarrow \text{R}^+\text{(aq)} + \text{Q}^-\text{(aq)} \]
• This contributes a concentration $C$ of $\text{Q}^-$ ions to the solution.
• Now, the equilibrium concentration of ions in the solution is:
\[ [\text{P}^{2+}] = S = 10^{-6}\ \text{M} \]
\[ [\text{Q}^-] = 2S + C \approx C \quad \text{(since the solubility } S \text{ is extremely small, } C \gg 2S) \]
• Substituting these values into the $K_{sp}$ expression:
\[ K_{sp} = [\text{P}^{2+}][\text{Q}^-]^2 \]
\[ 4 \times 10^{-12} = (10^{-6})(C)^2 \]
\[ C^2 = \frac{4 \times 10^{-12}}{10^{-6}} = 4 \times 10^{-6} \]
\[ C = \sqrt{4 \times 10^{-6}} = 2 \times 10^{-3}\ \text{M} \]
• Converting the concentration from moles per litre to millimoles per litre:
\[ C = 2 \times 10^{-3}\ \text{mol/L} = 2\ \text{mmol/L} \]
Step 4: Final Answer:
Therefore, the concentration of the added RQ is closest to $2\ \text{mmol/L}$ (Option A).
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