Solubility product constant `K_(sp)` of salts of types MX, `MX_(2)` and `M_(3)X` at temperature ‘T’ are `4.0xx10^(-8), 3.2xx10^(-14)`, and `2.7xx10^(-15)` respectively. Which salt has maximum solubility.

Options

A. MX
B. `M_(3)X`
C. `MX_(2)`
D. can’t be predicted

Explanation

To determine which salt has the maximum solubility among the salts MX, MX₂, and M₃X, we will calculate the molar solubility for each salt using their respective solubility product constants (Ksp).

### Step 1: Calculate the solubility for MX
1. The dissociation of MX can be represented as:
\[ MX \rightleftharpoons M^+ + X^- \]
If the solubility of MX is \( S \), then at equilibrium:
- \([M^+] = S\)
- \([X^-] = S\)

2. The expression for the solubility product \( K_{sp} \) is:
\[ K_{sp} = [M^+][X^-] = S \cdot S = S^2 \]

3. Given \( K_{sp} = 4.0 \times 10^{-8} \):
\[ S^2 = 4.0 \times 10^{-8} \]
\[ S = \sqrt{4.0 \times 10^{-8}} = 2.0 \times 10^{-4} \, \text{M} \]

### Step 2: Calculate the solubility for MX₂
1. The dissociation of MX₂ can be represented as:
\[ MX_2 \rightleftharpoons M^{2+} + 2X^- \]
If the solubility of MX₂ is \( S \), then at equilibrium:
- \([M^{2+}] = S\)
- \([X^-] = 2S\)

2. The expression for the solubility product \( K_{sp} \) is:
\[ K_{sp} = [M^{2+}][X^-]^2 = S \cdot (2S)^2 = S \cdot 4S^2 = 4S^3 \]

3. Given \( K_{sp} = 3.2 \times 10^{-14} \):
\[ 4S^3 = 3.2 \times 10^{-14} \]
\[ S^3 = \frac{3.2 \times 10^{-14}}{4} = 8.0 \times 10^{-15} \]
\[ S = \sqrt[3]{8.0 \times 10^{-15}} \approx 2.0 \times 10^{-5} \, \text{M} \]

### Step 3: Calculate the solubility for M₃X
1. The dissociation of M₃X can be represented as:
\[ M_3X \rightleftharpoons 3M^{2+} + X^{3-} \]
If the solubility of M₃X is \( S \), then at equilibrium:
- \([M^{3+}] = 3S\)
- \([X^{3-}] = S\)

2. The expression for the solubility product \( K_{sp} \) is:
\[ K_{sp} = [M^{3+}]^3[X^{3-}] = (3S)^3 \cdot S = 27S^4 \]

3. Given \( K_{sp} = 2.7 \times 10^{-15} \):
\[ 27S^4 = 2.7 \times 10^{-15} \]
\[ S^4 = \frac{2.7 \times 10^{-15}}{27} = 1.0 \times 10^{-16} \]
\[ S = \sqrt[4]{1.0 \times 10^{-16}} = 1.0 \times 10^{-4} \, \text{M} \]

### Step 4: Compare the solubilities
- Solubility of MX: \( 2.0 \times 10^{-4} \, \text{M} \)
- Solubility of MX₂: \( 2.0 \times 10^{-5} \, \text{M} \)
- Solubility of M₃X: \( 1.0 \times 10^{-4} \, \text{M} \)

### Conclusion
The salt with the maximum solubility is **MX** with a solubility of \( 2.0 \times 10^{-4} \, \text{M} \).

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