If the salts `M_2X,QY_2 and PZ_3` have the same solubilities (but `lt0.1M`) their , `K_(sp)` values are related as

Options

A. `K_(sp)(M_2X)=K_(sp)(QY_2)gtK_(sp)(PZ_3)`
B. `K_(sp)(M_2X)gtK_(sp)(QY_2)=K_(sp)(PZ_3)`
C. `K_(sp)(M_2X)=K_(sp)(QY_2)ltK_(sp)(PZ_3)`
D. `K_(sp)(M_2X)gtK_(sp)(QY_2)gtK_(sp)(PZ_3)`

Answer

To solve the problem, we need to determine the relationship between the solubility product constants (Ksp) of the salts \( M_2X \), \( QY_2 \), and \( PZ_3 \) given that they have the same solubility (denoted as \( S \)).

### Step-by-Step Solution:

1. **Dissociation of Salts**:
- For \( M_2X \):
\[ M_2X \rightleftharpoons 2M^+ + X^{2-} \]
The stoichiometric coefficients are:
- \( M^+ \): 2
- \( X^{2-} \): 1

- For \( QY_2 \):
\[ QY_2 \rightleftharpoons Q^{2+} + 2Y^- \]
The stoichiometric coefficients are:
- \( Q^{2+} \): 1
- \( Y^- \): 2

- For \( PZ_3 \):
\[ PZ_3 \rightleftharpoons P^{3+} + 3Z^- \]
The stoichiometric coefficients are:
- \( P^{3+} \): 1
- \( Z^- \): 3

2. **Solubility Expression**:
- Let the solubility \( S \) be the same for all three salts.
- For \( M_2X \):
\[ K_{sp} = [M^+]^2[X^{2-}] = (2S)^2(S) = 4S^3 \]

- For \( QY_2 \):
\[ K_{sp} = [Q^{2+}][Y^-]^2 = (S)(2S)^2 = S(4S^2) = 4S^3 \]

- For \( PZ_3 \):
\[ K_{sp} = [P^{3+}][Z^-]^3 = (S)(3S)^3 = S(27S^3) = 27S^4 \]

3. **Comparing Ksp Values**:
- We have:
- \( K_{sp}(M_2X) = 4S^3 \)
- \( K_{sp}(QY_2) = 4S^3 \)
- \( K_{sp}(PZ_3) = 27S^4 \)

4. **Analyzing the Relationships**:
- Since \( S < 0.1 \, M \), we can analyze the powers:
- \( 4S^3 \) and \( 4S^3 \) are equal for \( M_2X \) and \( QY_2 \).
- \( 27S^4 \) will be less than \( 4S^3 \) because as \( S \) increases, \( S^4 \) grows slower than \( S^3 \) when \( S < 1 \).

5. **Final Relationship**:
- Therefore, we conclude:
\[ K_{sp}(M_2X) = K_{sp}(QY_2) > K_{sp}(PZ_3) \]

### Conclusion:
The Ksp values are related as follows:
\[ K_{sp}(M_2X) = K_{sp}(QY_2) > K_{sp}(PZ_3) \]