If S and `K_(sp)` are respectively solubility and solubility product of a sparingly soluble binary electrolyte, then

Options

A. `S = K_(sp)`
B. `S = K_(sp)^(2)`
C. `S = sqrt(K_(sp))`
D. `S = (1)/(2) K_(sp)`

Explanation

To solve the question regarding the relationship between solubility (S) and solubility product (Ksp) of a sparingly soluble binary electrolyte, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Electrolyte**:
- A binary electrolyte can be represented as AB, which dissociates into its ions in solution:
\[ AB \rightleftharpoons A^+ + B^- \]
- Here, A^+ is the cation and B^- is the anion.

2. **Defining Solubility (S)**:
- The solubility (S) of the electrolyte is defined as the maximum amount of the electrolyte that can dissolve in a given amount of solvent at equilibrium.
- For the dissociation of AB, if S moles of AB dissolve, then at equilibrium:
- The concentration of A^+ = S
- The concentration of B^- = S

3. **Writing the Expression for Ksp**:
- The solubility product (Ksp) is defined as the product of the molar concentrations of the ions, each raised to the power of their coefficients in the balanced equation:
\[ K_{sp} = [A^+][B^-] \]
- Substituting the concentrations from the solubility:
\[ K_{sp} = S \times S = S^2 \]

4. **Relating S to Ksp**:
- From the expression derived, we can express solubility in terms of Ksp:
\[ S^2 = K_{sp} \]
- Taking the square root of both sides gives:
\[ S = \sqrt{K_{sp}} \]

5. **Conclusion**:
- Therefore, the relationship between solubility (S) and solubility product (Ksp) for a sparingly soluble binary electrolyte is:
\[ S = \sqrt{K_{sp}} \]

### Final Answer:
The correct relationship is \( S = \sqrt{K_{sp}} \).