The solubility product of a sparingly soluble salt `AX_(2)` is `3.2xx10^(-11)`. Its solubility (in `mol L^-1`) is

Options

A. `5.6xx10^(-6)`
B. `3.1xx10^(-4)`
C. `2xx10^(-4)`
D. `4xx10^(-4)`

Explanation

To find the solubility of the sparingly soluble salt \( AX_2 \) given its solubility product \( K_{sp} = 3.2 \times 10^{-11} \), we can follow these steps:

### Step-by-Step Solution:

1. **Write the Dissociation Equation:**
The salt \( AX_2 \) dissociates in water as follows:
\[ AX_2 (s) \rightleftharpoons A^{2+} (aq) + 2X^{-} (aq) \]

2. **Define the Solubility:**
Let the solubility of \( AX_2 \) be \( S \) (in mol/L). At equilibrium:
- The concentration of \( A^{2+} \) will be \( S \).
- The concentration of \( X^{-} \) will be \( 2S \) (since there are 2 moles of \( X^{-} \) for every mole of \( AX_2 \)).

3. **Write the Expression for \( K_{sp} \):**
The solubility product \( K_{sp} \) is given by the formula:
\[ K_{sp} = [A^{2+}][X^{-}]^2 \]
Substituting the equilibrium concentrations:
\[ K_{sp} = (S)(2S)^2 \]

4. **Simplify the Expression:**
Expanding the expression gives:
\[ K_{sp} = S \cdot 4S^2 = 4S^3 \]

5. **Set Up the Equation:**
Now substitute the given \( K_{sp} \) value:
\[ 4S^3 = 3.2 \times 10^{-11} \]

6. **Solve for \( S^3 \):**
Rearranging the equation:
\[ S^3 = \frac{3.2 \times 10^{-11}}{4} \]
\[ S^3 = 0.8 \times 10^{-11} \]

7. **Calculate \( S \):**
Now take the cube root to find \( S \):
\[ S = \sqrt[3]{0.8 \times 10^{-11}} \]
\[ S \approx 2 \times 10^{-4} \text{ mol/L} \]

### Final Answer:
The solubility of the salt \( AX_2 \) is \( 2 \times 10^{-4} \) mol/L.

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