Question

The solubility product of \(PbCl_2\) at \(298\ \text{K}\) is \(3.2 \times 10^{-5}\). What is its solubility in \(\text{mol dm}^{-3}\)?

• A. \(8 \times 10^{-6}\)
• B. \(2 \times 10^{-2}\)
• C. \(5.6 \times 10^{-3}\)
• D. \(5 \times 10^{-2}\)

Hint:

For salts of the type: \[ MX_2 \rightleftharpoons M^{2+} + 2X^- \] Always remember: \[ K_{sp} = 4s^3 \] Examples:
• \(PbCl_2\)
• \(CaF_2\)
• \(BaI_2\) This shortcut helps solve numerical questions very quickly.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Lead(II) chloride (\( PbCl_2 \)) is a 1:2 electrolyte salt. Its dissociation in water can be represented by the following equilibrium equation:
\[ PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq) \]
The solubility product constant (\( K_{sp} \)) relates the concentrations of the product ions in a saturated solution.
Step 2: Key Formula or Approach:
Let the molar solubility of \( PbCl_2 \) be \( s \text{ mol/L} \).
According to the stoichiometry:
\[ [Pb^{2+}] = s \]
\[ [Cl^-] = 2s \]
The expression for \( K_{sp} \) is:
\[ K_{sp} = [Pb^{2+}][Cl^-]^2 \]
\[ K_{sp} = (s)(2s)^2 = 4s^3 \]
Step 3: Detailed Explanation:
Given \( K_{sp} = 3.2 \times 10^{-5} \).
Substituting the value into the derived formula:
\[ 4s^3 = 3.2 \times 10^{-5} \]
Divide by 4:
\[ s^3 = \frac{3.2 \times 10^{-5}}{4} \]
\[ s^3 = 0.8 \times 10^{-5} \]
To take the cube root easily, adjust the exponent to a multiple of 3:
\[ s^3 = 8 \times 10^{-6} \]
Taking the cube root of both sides:
\[ s = \sqrt[3]{8 \times 10^{-6}} \]
\[ s = \sqrt[3]{8} \times \sqrt[3]{10^{-6}} \]
\[ s = 2 \times 10^{-2} \text{ mol dm}^{-3} \]
Step 4: Final Answer:
The solubility of \( PbCl_2 \) is \( 2 \times 10^{-2} \text{ mol dm}^{-3} \).