Three sparigly soluble salts `M_(2)X, MX`,and `MX_(3)` have the same solubility product. Their solubilities will be in the order

Options

A. `MX_(3) gt MX gt M_(2)X`
B. `MX_(3) gt M_(2)X gt MX`
C. `MX gt MX_(3) gt M_(2)X`
D. `MX gt M_(2)X gt MX_(3)`

Explanation

To determine the order of solubility of the salts \( M_2X \), \( MX \), and \( MX_3 \) given that they all have the same solubility product (\( K_{sp} \)), we will analyze the dissociation of each salt and derive their solubility expressions.

### Step-by-Step Solution:

1. **Identify the dissociation of each salt:**
- For \( MX \):
\[ MX \rightleftharpoons M^+ + X^- \]
If the solubility of \( MX \) is \( s \), then at equilibrium:
- Concentration of \( M^+ = s \)
- Concentration of \( X^- = s \)

2. **Write the expression for \( K_{sp} \) for \( MX \):**
\[ K_{sp} = [M^+][X^-] = s \cdot s = s^2 \]
Thus, the solubility \( s \) can be expressed as:
\[ s = \sqrt{K_{sp}} \]

3. **For \( M_2X \):**
\[ M_2X \rightleftharpoons 2M^+ + X^{2-} \]
If the solubility of \( M_2X \) is \( s' \), then at equilibrium:
- Concentration of \( M^+ = 2s' \)
- Concentration of \( X^{2-} = s' \)

4. **Write the expression for \( K_{sp} \) for \( M_2X \):**
\[ K_{sp} = [M^+]^2[X^{2-}] = (2s')^2 \cdot s' = 4s'^3 \]
Rearranging gives:
\[ s' = \left( \frac{K_{sp}}{4} \right)^{1/3} \]

5. **For \( MX_3 \):**
\[ MX_3 \rightleftharpoons M^{3+} + 3X^- \]
If the solubility of \( MX_3 \) is \( s'' \), then at equilibrium:
- Concentration of \( M^{3+} = s'' \)
- Concentration of \( X^- = 3s'' \)

6. **Write the expression for \( K_{sp} \) for \( MX_3 \):**
\[ K_{sp} = [M^{3+}][X^-]^3 = s'' \cdot (3s'')^3 = 27s''^4 \]
Rearranging gives:
\[ s'' = \left( \frac{K_{sp}}{27} \right)^{1/4} \]

7. **Compare the solubilities:**
- \( s = \sqrt{K_{sp}} \)
- \( s' = \left( \frac{K_{sp}}{4} \right)^{1/3} \)
- \( s'' = \left( \frac{K_{sp}}{27} \right)^{1/4} \)

8. **Determine the order of solubility:**
To find the order, we can compare the expressions:
- \( s \) is proportional to \( K_{sp}^{1/2} \)
- \( s' \) is proportional to \( K_{sp}^{1/3} \)
- \( s'' \) is proportional to \( K_{sp}^{1/4} \)

Since \( K_{sp} \) is constant for all three salts:
\[ \sqrt{K_{sp}} > \left( \frac{K_{sp}}{4} \right)^{1/3} > \left( \frac{K_{sp}}{27} \right)^{1/4} \]

Therefore, the order of solubility is:
\[ MX > M_2X > MX_3 \]

### Final Answer:
The solubility order is \( MX > M_2X > MX_3 \).