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

Options

A. `5.6 xx 10^(-6)`
B. `3.1 xx 10^(-4)`
C. `2 xx 10^(-4)`
D. `4 xx 10^(-4)`

Answer

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 1: Write the dissociation equation
The dissociation of the salt \( AX_2 \) in water can be represented as:
\[ AX_2 (s) \rightleftharpoons A^{2+} (aq) + 2X^{-} (aq) \]

### Step 2: Define the solubility
Let the solubility of \( AX_2 \) be \( S \) mol/L. When \( AX_2 \) dissolves, it produces:
- \( A^{2+} \) ions: \( S \) mol/L
- \( X^{-} \) ions: \( 2S \) mol/L (since there are two \( X^{-} \) ions for each formula unit of \( AX_2 \))

### Step 3: Write the expression for \( K_{sp} \)
The solubility product \( K_{sp} \) can be expressed as:
\[ K_{sp} = [A^{2+}][X^{-}]^2 \]
Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (S)(2S)^2 = S \cdot 4S^2 = 4S^3 \]

### Step 4: Set up the equation with the given \( K_{sp} \)
Given \( K_{sp} = 3.2 \times 10^{-11} \):
\[ 4S^3 = 3.2 \times 10^{-11} \]

### Step 5: Solve for \( S^3 \)
Rearranging the equation gives:
\[ S^3 = \frac{3.2 \times 10^{-11}}{4} \]
Calculating the right-hand side:
\[ S^3 = 0.8 \times 10^{-11} = 8.0 \times 10^{-12} \]

### Step 6: Solve for \( S \)
Now, take the cube root of both sides to find \( S \):
\[ S = (8.0 \times 10^{-12})^{1/3} \]
Calculating the cube root:
\[ S = 2.0 \times 10^{-4} \text{ mol/L} \]

### Final Answer
The solubility of the salt \( AX_2 \) is:
\[ \boxed{2.0 \times 10^{-4} \text{ mol/L}} \]

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