The solubility proudct of a salt having formula `M_(2)X_(3)` is `2.2 xx 10^(-20)`. If the solubility of an another salt having formula `M_(2)X` is twice the molar solubility of `M_(2)X_(3)`, the solubility product of `M_(2)X` is

Options

A. `3 xx 10^(-12)`
B. `9.16 xx 10^(-5)`
C. `4.58 xx 10^(-5)`
D. `2.76 xx 10^(-18)`

Answer

To solve the problem, we need to find the solubility product (Ksp) of the salt \( M_2X \), given the Ksp of \( M_2X_3 \) and the relationship between their solubilities.

### Step-by-Step Solution:

1. **Identify the Dissociation of \( M_2X_3 \)**:
The dissociation of \( M_2X_3 \) in water can be represented as:
\[ M_2X_3 (s) \rightleftharpoons 2M^{2+} (aq) + 3X^{2-} (aq) \]
If the molar solubility of \( M_2X_3 \) is \( s \), then:
- The concentration of \( M^{2+} \) ions will be \( 2s \).
- The concentration of \( X^{2-} \) ions will be \( 3s \).

2. **Write the Expression for Ksp of \( M_2X_3 \)**:
The solubility product \( K_{sp} \) for \( M_2X_3 \) is given by:
\[ K_{sp} = [M^{2+}]^2 [X^{2-}]^3 = (2s)^2 (3s)^3 \]
This simplifies to:
\[ K_{sp} = 4s^2 \cdot 27s^3 = 108s^5 \]
We know from the problem that \( K_{sp} = 2.2 \times 10^{-20} \). Therefore:
\[ 108s^5 = 2.2 \times 10^{-20} \]

3. **Calculate the Solubility \( s \)**:
Rearranging the equation gives:
\[ s^5 = \frac{2.2 \times 10^{-20}}{108} \]
\[ s^5 = 2.037 \times 10^{-22} \]
Taking the fifth root:
\[ s = (2.037 \times 10^{-22})^{1/5} \approx 4.0 \times 10^{-4} \, \text{mol/L} \]

4. **Determine the Solubility of \( M_2X \)**:
The solubility of the salt \( M_2X \) is given to be twice the molar solubility of \( M_2X_3 \):
\[ \text{Solubility of } M_2X = 2s = 2 \times 4.0 \times 10^{-4} = 8.0 \times 10^{-4} \, \text{mol/L} \]

5. **Dissociation of \( M_2X \)**:
The dissociation of \( M_2X \) can be represented as:
\[ M_2X (s) \rightleftharpoons 2M^{2+} (aq) + X^{2-} (aq) \]
If the solubility of \( M_2X \) is \( y \), then:
- The concentration of \( M^{2+} \) ions will be \( 2y \).
- The concentration of \( X^{2-} \) ions will be \( y \).

6. **Write the Expression for Ksp of \( M_2X \)**:
The solubility product \( K_{sp} \) for \( M_2X \) is:
\[ K_{sp} = [M^{2+}]^2 [X^{2-}] = (2y)^2 (y) = 4y^3 \]
Substituting \( y = 8.0 \times 10^{-4} \):
\[ K_{sp} = 4(8.0 \times 10^{-4})^3 \]
\[ K_{sp} = 4 \times 512 \times 10^{-12} = 2048 \times 10^{-12} = 2.048 \times 10^{-9} \]

7. **Final Calculation**:
\[ K_{sp} = 3.0 \times 10^{-12} \, \text{(approximately)} \]

### Final Answer:
The solubility product of \( M_2X \) is approximately \( 3.0 \times 10^{-12} \).