Question

The solubility product of $\text{PbF}_2$ is $3.2 \times 10^{-8}$; Calculate the solubility of $\text{PbF}_2$ in mole $\text{L}^{-1}$}

• A. $2\times10^{-3}$
• B. 0.49
• C. $3 \times 10^{-2}$
• D. 0.2

Hint:

For any sparingly soluble salt of type $\text{AB}_2$ (like $\text{PbF}_2$ or $\text{CaF}_2$):
$K_{sp} = 4S^3$.
Rearrange the equation to isolate $S^3$, and express the exponent as a multiple of 3 to simplify finding the cube root.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The solubility product constant ($K_{sp}$) is an equilibrium constant that describes the dissolution of a sparingly soluble ionic compound in water.
The relationship between $K_{sp}$ and molar solubility ($S$) is determined by the stoichiometry of the dissociation reaction.
Key Formula or Approach:
For a salt of the form $\text{AB}_2$, the dissociation equation is:
\[ \text{AB}_2(s) \rightleftharpoons \text{A}^{2+}(aq) + 2\text{B}^-(aq) \]
If the molar solubility of $\text{AB}_2$ is $S$, then:
\[ [\text{A}^{2+}] = S \]
\[ [\text{B}^-] = 2S \]
The solubility product expression is:
\[ K_{sp} = [\text{A}^{2+}][\text{B}^-]^2 = (S)(2S)^2 = 4S^3 \]

Step 2: Detailed Explanation:
For lead fluoride ($\text{PbF}_2$):
\[ \text{PbF}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{F}^-(aq) \]
Given $K_{sp} = 3.2 \times 10^{-8}$.
Using the relation $K_{sp} = 4S^3$, we solve for the molar solubility $S$:
\[ 4S^3 = 3.2 \times 10^{-8} \]
\[ S^3 = \frac{3.2 \times 10^{-8}}{4} \]
\[ S^3 = 0.8 \times 10^{-8} \]
To simplify the cube root calculation, we can rewrite the decimal value:
\[ S^3 = 8.0 \times 10^{-9} \]
Taking the cube root of both sides:
\[ S = \sqrt[3]{8.0 \times 10^{-9}} \]
\[ S = \sqrt[3]{8} \times \sqrt[3]{10^{-9}} \]
\[ S = 2 \times 10^{-3}\text{ mol L}^{-1} \]

Step 3: Final Answer:
The molar solubility of $\text{PbF}_2$ is $2 \times 10^{-3}\text{ mole L}^{-1}$, which corresponds to Option (A).