For which of the following sparingly soluble salt the solubility (s) and solubility product `(K_(sp))` are related by the expression `s=(K_(sp)//4)^(1//3)`

Options

A. `BaSO_(4)`
B. `Ca_(3)(PO_(4))_(2)`
C. `Hg_(2)Cl_(2)`
D. `Ag_(3)PO_(4)`

Explanation

To solve the problem of determining which sparingly soluble salt has its solubility (s) and solubility product (Ksp) related by the expression \( s = \left(\frac{K_{sp}}{4}\right)^{\frac{1}{3}} \), we will analyze each salt provided in the options step by step.

### Step 1: Analyze the first salt, \( \text{BaSO}_4 \)

1. **Dissociation Reaction**:
\[ \text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]
- From this reaction, we see that 1 mole of BaSO4 produces 1 mole of Ba²⁺ and 1 mole of SO₄²⁻.

2. **Solubility (s)**:
- Let the solubility of BaSO₄ be \( s \). Therefore, the concentration of Ba²⁺ and SO₄²⁻ at equilibrium will both be \( s \).

3. **Ksp Expression**:
\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = s \cdot s = s^2 \]
- Thus, \( K_{sp} = s^2 \).

4. **Relation**:
- Rearranging gives \( s = \sqrt{K_{sp}} \), which does not match the required expression.

### Step 2: Analyze the second salt, \( \text{Ca}_3(\text{PO}_4)_2 \)

1. **Dissociation Reaction**:
\[ \text{Ca}_3(\text{PO}_4)_2 (s) \rightleftharpoons 3 \text{Ca}^{2+} (aq) + 2 \text{PO}_4^{3-} (aq) \]
- Here, 1 mole of Ca₃(PO₄)₂ produces 3 moles of Ca²⁺ and 2 moles of PO₄³⁻.

2. **Solubility (s)**:
- Let the solubility be \( s \). Therefore, the concentrations will be \( 3s \) for Ca²⁺ and \( 2s \) for PO₄³⁻.

3. **Ksp Expression**:
\[ K_{sp} = [\text{Ca}^{2+}]^3[\text{PO}_4^{3-}]^2 = (3s)^3(2s)^2 = 27s^3 \cdot 4s^2 = 108s^5 \]

4. **Relation**:
- Rearranging gives \( s^5 = \frac{K_{sp}}{108} \) or \( s = \left(\frac{K_{sp}}{108}\right)^{\frac{1}{5}} \), which does not match the required expression.

### Step 3: Analyze the third salt, \( \text{Hg}_2\text{Cl}_2 \)

1. **Dissociation Reaction**:
\[ \text{Hg}_2\text{Cl}_2 (s) \rightleftharpoons \text{Hg}_2^{2+} (aq) + 2 \text{Cl}^- (aq) \]
- Here, 1 mole of Hg₂Cl₂ produces 1 mole of Hg₂²⁺ and 2 moles of Cl⁻.

2. **Solubility (s)**:
- Let the solubility be \( s \). Therefore, the concentrations will be \( s \) for Hg₂²⁺ and \( 2s \) for Cl⁻.

3. **Ksp Expression**:
\[ K_{sp} = [\text{Hg}_2^{2+}][\text{Cl}^-]^2 = s \cdot (2s)^2 = s \cdot 4s^2 = 4s^3 \]

4. **Relation**:
- Rearranging gives \( s^3 = \frac{K_{sp}}{4} \) or \( s = \left(\frac{K_{sp}}{4}\right)^{\frac{1}{3}} \), which matches the required expression.

### Step 4: Analyze the fourth salt, \( \text{Ag}_3\text{PO}_4 \)

1. **Dissociation Reaction**:
\[ \text{Ag}_3\text{PO}_4 (s) \rightleftharpoons 3 \text{Ag}^+ (aq) + \text{PO}_4^{3-} (aq) \]
- Here, 1 mole of Ag₃PO₄ produces 3 moles of Ag⁺ and 1 mole of PO₄³⁻.

2. **Solubility (s)**:
- Let the solubility be \( s \). Therefore, the concentrations will be \( 3s \) for Ag⁺ and \( s \) for PO₄³⁻.

3. **Ksp Expression**:
\[ K_{sp} = [\text{Ag}^+]^3[\text{PO}_4^{3-}] = (3s)^3 \cdot s = 27s^3 \cdot s = 27s^4 \]

4. **Relation**:
- Rearranging gives \( s^4 = \frac{K_{sp}}{27} \) or \( s = \left(\frac{K_{sp}}{27}\right)^{\frac{1}{4}} \), which does not match the required expression.

### Conclusion

The only salt that satisfies the condition \( s = \left(\frac{K_{sp}}{4}\right)^{\frac{1}{3}} \) is **Hg₂Cl₂**.