Question

For a sparingly soluble salt \( \text{AB}_2 \), the equilibrium concentrations of \( \text{A}^{2+} \) ions and \( \text{B}^- \) ions are \( 1.2 \times 10^{-4} \, \text{M} \) and \( 0.24 \times 10^{-3} \, \text{M} \), respectively. The solubility product of \( \text{AB}_2 \) is:

• A. \( 0.069 \times 10^{-12} \)
• B. \( 6.91 \times 10^{-12} \)
• C. \( 0.276 \times 10^{-12} \)
• D. \( 27.65 \times 10^{-12} \)

Answer 1

The Correct Option is B

To find the solubility product constant (\( K_{sp} \)) for the sparingly soluble salt \( \text{AB}_2 \), we start by understanding its dissociation in water:

\[ \text{AB}_2 (s) \rightleftharpoons \text{A}^{2+} (aq) + 2\text{B}^- (aq) \]

From the dissociation, we can write the expression for the solubility product (\( K_{sp} \)):

\[ K_{sp} = [\text{A}^{2+}] [\text{B}^-]^2 \]

Given:

• Concentration of \( \text{A}^{2+} \) ions = \( 1.2 \times 10^{-4} \, \text{M} \)
• Concentration of \( \text{B}^- \) ions = \( 0.24 \times 10^{-3} \, \text{M} \)

Substitute these concentrations into the \( K_{sp} \) expression:

\[ K_{sp} = (1.2 \times 10^{-4}) \times (0.24 \times 10^{-3})^2 \]

Calculate \( (0.24 \times 10^{-3})^2 \):

\[ (0.24 \times 10^{-3})^2 = 0.0576 \times 10^{-6} \]

Next, calculate \( K_{sp} \):

\[ K_{sp} = (1.2 \times 10^{-4}) \times (0.0576 \times 10^{-6}) \]

\[ K_{sp} = 6.912 \times 10^{-12} \]

Since the options are generally rounded, the closest given option is:

• \( 6.91 \times 10^{-12} \)

Thus, the correct answer is \( 6.91 \times 10^{-12} \).

Answer 2

Let the solubility of AB$_2$ be s. The dissociation of AB$_2$ in water is given by:
AB$_2$ $\rightleftharpoons$ A$^{2+}$ + 2B$^−$.
At equilibrium, the concentration of A$^{2+}$ is 1.2 × 10$^{-4}$ M and the concentration of B$^−$ is 0.24 × 10$^{-3}$ M.
The solubility product $K_{sp}$ is given by:
$K_{sp}$ = [A$^{2+}$][B$^−$]$^2$.
Substituting the given values:
$K_{sp}$ = (1.2 × 10$^{-4}$)(0.24 × 10$^{-3}$)$^2$ = 6.91 × 10$^{-12}$.
Hence, the correct answer is (2).