If the solubility product `K_("sp")` of a sparingly soluble salt `MX_(2)" at "25^(@) C " is " 1.0 xx 10^(-11)` , then the solubility of the salt in mole `"litre"^(-1)` at this temperature will be

Options

A. `2.46 xx 10^(14)`
B. `1.36xx 10^(-4)`
C. `2.60 xx 10^(-7)`
D. `1.20 xx 10^(-10)`

Answer

To find the solubility of the sparingly soluble salt \( MX_2 \) at 25°C given that the solubility product \( K_{sp} \) is \( 1.0 \times 10^{-11} \), we can follow these steps:

### Step-by-Step Solution:

1. **Write the Dissociation Equation:**
The salt \( MX_2 \) dissociates in water as follows:
\[ MX_2 (s) \rightleftharpoons M^{2+} (aq) + 2X^{-} (aq) \]

2. **Define the Solubility:**
Let the solubility of \( MX_2 \) be \( S \) moles per liter. This means:
- The concentration of \( M^{2+} \) ions will be \( S \).
- The concentration of \( X^{-} \) ions will be \( 2S \) (since there are 2 moles of \( X^{-} \) for every mole of \( MX_2 \)).

3. **Write the Expression for \( K_{sp} \):**
The solubility product \( K_{sp} \) is given by the expression:
\[ K_{sp} = [M^{2+}][X^{-}]^2 \]
Substituting the concentrations:
\[ K_{sp} = (S)(2S)^2 = S \cdot 4S^2 = 4S^3 \]

4. **Set Up the Equation:**
Given that \( K_{sp} = 1.0 \times 10^{-11} \), we can set up the equation:
\[ 4S^3 = 1.0 \times 10^{-11} \]

5. **Solve for \( S^3 \):**
Rearranging gives:
\[ S^3 = \frac{1.0 \times 10^{-11}}{4} = 2.5 \times 10^{-12} \]

6. **Calculate \( S \):**
Now, take the cube root to find \( S \):
\[ S = \sqrt[3]{2.5 \times 10^{-12}} \approx 1.36 \times 10^{-4} \text{ moles per liter} \]

### Final Answer:
The solubility of the salt \( MX_2 \) at 25°C is approximately \( 1.36 \times 10^{-4} \) moles per liter.