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

Concept: The solubility product constant \((K_{sp})\) relates the concentration of ions present in a saturated solution of a sparingly soluble salt. For a salt: \[ A_xB_y \rightleftharpoons xA^{y+} + yB^{x-} \] If solubility is \(s\), then ionic concentrations are determined from stoichiometry. For \(PbCl_2\): \[ PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq) \]

Step 1: Writing ion concentrations in terms of solubility.}
Let solubility of \(PbCl_2\) be: \[ s\ \text{mol dm}^{-3} \] Then, \[ [Pb^{2+}] = s \] \[ [Cl^-] = 2s \]

Step 2: Writing the \(K_{sp}\) expression.}
\[ K_{sp} = [Pb^{2+}][Cl^-]^2 \] Substituting concentrations: \[ K_{sp} = s(2s)^2 \] \[ K_{sp} = 4s^3 \] Given: \[ K_{sp} = 3.2 \times 10^{-5} \] Therefore, \[ 4s^3 = 3.2 \times 10^{-5} \]

Step 3: Calculating solubility \(s\).}
\[ s^3 = \frac{3.2 \times 10^{-5}}{4} \] \[ s^3 = 0.8 \times 10^{-5} \] \[ s^3 = 8 \times 10^{-6} \] Taking cube root: \[ s = \sqrt[3]{8 \times 10^{-6}} \] \[ s = 2 \times 10^{-2}\ \text{mol dm}^{-3} \]

Step 4: Final conclusion.}
Hence, the solubility of \(PbCl_2\) is: \[ \boxed{2 \times 10^{-2}\ \text{mol dm}^{-3}} \] Therefore, the correct option is: \[ \boxed{(2)\ 2 \times 10^{-2}} \]