If s and S are respectiveluy solubility and solubility product of a sparingly soluble binary electrolyte , then :

Options

A. `s=S^(1//2)`
B. `s=1//2S`
C. s=S
D. `s=S^(2)`

Explanation

To find the relationship between the solubility (s) and the solubility product (S) of a sparingly soluble binary electrolyte, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Dissociation of the Electrolyte**:
- Consider a sparingly soluble binary electrolyte, which we can denote as AB. When it dissolves in water, it dissociates into its constituent ions:
\[ AB \rightleftharpoons A^+ + B^- \]

2. **Defining Solubility**:
- Let the solubility of the salt AB in water be denoted as \( s \). This means that at equilibrium, the concentration of \( A^+ \) ions will be \( s \) and the concentration of \( B^- \) ions will also be \( s \).

3. **Writing the Expression for Solubility Product (S)**:
- The solubility product \( S \) is defined as the product of the molar concentrations of the ions at equilibrium:
\[ S = [A^+][B^-] \]
- Substituting the concentrations from the solubility:
\[ S = s \cdot s = s^2 \]

4. **Finding the Relationship**:
- From the equation \( S = s^2 \), we can express \( s \) in terms of \( S \):
\[ s = \sqrt{S} \]

5. **Conclusion**:
- The relationship between the solubility \( s \) and the solubility product \( S \) is given by:
\[ s = \sqrt{S} \]

### Final Answer:
The relationship between the solubility (s) and the solubility product (S) of a sparingly soluble binary electrolyte is:
\[ s = \sqrt{S} \]