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 of the salt \( M_2X \) given the solubility product of the salt \( M_2X_3 \) and the relationship between their solubilities.

### Step-by-Step Solution:

1. **Identify the solubility product expression for \( M_2X_3 \)**:
The formula for the salt \( M_2X_3 \) indicates that it dissociates in the following manner:
\[ M_2X_3 (s) \rightleftharpoons 2M^{2+} (aq) + 3X^{3-} (aq) \]
The solubility product \( K_{sp} \) for this salt can be expressed as:
\[ K_{sp} = [M^{2+}]^2 [X^{3-}]^3 \]
Let the molar solubility of \( M_2X_3 \) be \( s \). Then:
\[ [M^{2+}] = 2s \quad \text{and} \quad [X^{3-}] = 3s \]
Substituting these into the solubility product expression gives:
\[ K_{sp} = (2s)^2 (3s)^3 = 4s^2 \cdot 27s^3 = 108s^5 \]

2. **Set the known value of \( K_{sp} \)**:
We know from the problem that \( K_{sp} \) for \( M_2X_3 \) is \( 2.2 \times 10^{-20} \):
\[ 108s^5 = 2.2 \times 10^{-20} \]
Solving for \( s \):
\[ s^5 = \frac{2.2 \times 10^{-20}}{108} \approx 2.037 \times 10^{-22} \]
\[ s = (2.037 \times 10^{-22})^{1/5} \]

3. **Calculate the solubility \( s \)**:
Using a calculator:
\[ s \approx 1.58 \times 10^{-5} \, \text{mol/L} \]

4. **Determine the solubility of \( M_2X \)**:
The solubility of the salt \( M_2X \) is given to be twice that of \( M_2X_3 \):
\[ \text{Solubility of } M_2X = 2s = 2 \times 1.58 \times 10^{-5} \approx 3.16 \times 10^{-5} \, \text{mol/L} \]

5. **Write the dissociation equation for \( M_2X \)**:
The dissociation of \( M_2X \) can be represented as:
\[ M_2X (s) \rightleftharpoons 2M^{2+} (aq) + X^{2-} (aq) \]
The solubility product \( K_{sp} \) for this salt is:
\[ K_{sp} = [M^{2+}]^2 [X^{2-}] \]
With the solubility of \( M_2X \) being \( 3.16 \times 10^{-5} \):
\[ [M^{2+}] = 2(3.16 \times 10^{-5}) = 6.32 \times 10^{-5} \]
\[ [X^{2-}] = 3.16 \times 10^{-5} \]

6. **Substituting into the \( K_{sp} \) expression**:
\[ K_{sp} = (6.32 \times 10^{-5})^2 (3.16 \times 10^{-5}) = 4.00 \times 10^{-9} \times 3.16 \times 10^{-5} \approx 1.26 \times 10^{-13} \]

### Final Answer:
The solubility product of \( M_2X \) is approximately \( 1.26 \times 10^{-13} \).