Question

Solubility product $(K_{sp})$ of saturated PbCl$_2$ in water is $1.8 \times 10^{-4}\ \text{mol}^3\text{dm}^{-9}$. What is the concentration of Pb$^{2+}$ in the solution?}

• A. $(0.45 \times 10^{-4})^{1/3}$ mol dm$^{-3}$
• B. $(1.8 \times 10^{-4})^{1/3}$ mol dm$^{-3}$
• C. $(0.9 \times 10^{-4})^{1/3}$ mol dm$^{-3}$
• D. $(2.0 \times 10^{-4})^{1/3}$ mol dm$^{-3}$
• E. $(2.45 \times 10^{-4})^{1/3}$ mol dm$^{-3}$

Hint:

For salt MX$_2$: always remember $K_{sp} = 4s^3$.

Answer 1

The Correct Option is A

Concept:
For sparingly soluble salts: \[ K_{sp} = \text{product of ionic concentrations raised to their stoichiometric powers} \] For PbCl$_2$: \[ \text{PbCl}_2 \rightleftharpoons \text{Pb}^{2+} + 2\text{Cl}^- \]

Step 1: Assume solubility = $s$.
\[ [\text{Pb}^{2+}] = s,\quad [\text{Cl}^-] = 2s \]

Step 2: Write $K_{sp}$ expression.
\[ K_{sp} = s(2s)^2 = 4s^3 \]

Step 3: Substitute given value.
\[ 1.8 \times 10^{-4} = 4s^3 \] \[ s^3 = \frac{1.8 \times 10^{-4}}{4} = 0.45 \times 10^{-4} \]

Step 4: Find $s$.
\[ s = (0.45 \times 10^{-4})^{1/3} \] Thus concentration of Pb$^{2+}$ is $s$.