MY and `NY_(3)` two nearly insoluble salts, have the same `K_(sp)` values of `6.2xx10^(-13)` at room temperature. Which statement would be true in rearged to MY and `NY_(3)` ?

Options

A. The addition of the salt of KY to solution of MY and `NY_(3)` will have no effect on their solubilities.
B. The molar soulbities of MY and `NY_(3)` in water are identical.
C. The molar solubility of MY in water is less than that of `NY_(3)` .
D. The salts MY and `NY_(3)` are more solble in `0.5` M KY than in pure water.

Explanation

To solve the problem regarding the solubility of the salts MY and NY₃, we need to analyze their solubility products (Ksp) and how they relate to their molar solubility.

### Step-by-Step Solution:

1. **Understanding Ksp**:
- Both salts MY and NY₃ have the same Ksp value of \(6.2 \times 10^{-13}\).
- Ksp is defined as the product of the molar concentrations of the ions in a saturated solution of the salt.

2. **Dissociation of MY**:
- MY dissociates into \(M^+\) and \(Y^-\):
\[ MY \rightleftharpoons M^+ + Y^- \]
- If the solubility of MY is \(S\), then at equilibrium:
- \([M^+] = S\)
- \([Y^-] = S\)
- Therefore, the Ksp expression for MY is:
\[ Ksp = [M^+][Y^-] = S \times S = S^2 \]
- Substituting the Ksp value:
\[ S^2 = 6.2 \times 10^{-13} \]
- Solving for \(S\):
\[ S = \sqrt{6.2 \times 10^{-13}} \approx 7.88 \times 10^{-7} \text{ M} \]

3. **Dissociation of NY₃**:
- NY₃ dissociates into \(N^{3+}\) and \(3Y^-\):
\[ NY₃ \rightleftharpoons N^{3+} + 3Y^- \]
- If the solubility of NY₃ is \(S'\), then at equilibrium:
- \([N^{3+}] = S'\)
- \([Y^-] = 3S'\)
- Therefore, the Ksp expression for NY₃ is:
\[ Ksp = [N^{3+}][Y^-]^3 = S' \times (3S')^3 = S' \times 27(S')^3 = 27(S')^4 \]
- Substituting the Ksp value:
\[ 27(S')^4 = 6.2 \times 10^{-13} \]
- Solving for \(S'\):
\[ (S')^4 = \frac{6.2 \times 10^{-13}}{27} \]
\[ S' = \left(\frac{6.2 \times 10^{-13}}{27}\right)^{1/4} \approx 0.4 \times 10^{-4} \text{ M} \]

4. **Comparison of Solubilities**:
- We have calculated:
- Solubility of MY: \(S \approx 7.88 \times 10^{-7} \text{ M}\)
- Solubility of NY₃: \(S' \approx 0.4 \times 10^{-4} \text{ M} = 4.0 \times 10^{-5} \text{ M}\)
- Comparing these values:
- \(7.88 \times 10^{-7} \text{ M} < 4.0 \times 10^{-5} \text{ M}\)
- Thus, the molar solubility of MY is less than that of NY₃.

5. **Conclusion**:
- The correct statement is that the molar solubility of MY is less than that of NY₃.

### Final Answer:
The molar solubility of MY is less than that of NY₃.