For a sparingly soluble salt ` A_pB_q,` the relationship of its solubility product `(K_s) ` with its solubility (S) is

Options

A. `K_s = S^(p+q) . P^(p) .q^(q)`
B. ` K_s = S^(p+q) . P^(q) .q^(p)`
C. ` K_s =S^(pq) .p^(p) .q^(q)`
D. ` K_s= S^(pq) , (pq) ^(p+q) `

Explanation

To solve the problem regarding the relationship between the solubility product (K_s) and the solubility (S) of a sparingly soluble salt \( A_pB_q \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Dissociation of the Salt**:
The salt \( A_pB_q \) dissociates in water according to the following reaction:
\[ A_pB_q (s) \rightleftharpoons p A^{+} (aq) + q B^{-} (aq) \]
Here, \( p \) is the number of moles of cation \( A^{+} \) produced, and \( q \) is the number of moles of anion \( B^{-} \) produced.

2. **Define Solubility (S)**:
Let the solubility of the salt \( A_pB_q \) be \( S \) moles per liter. This means that when the salt dissolves, the concentration of \( A^{+} \) ions will be \( pS \) and the concentration of \( B^{-} \) ions will be \( qS \).

3. **Write the Expression for the Solubility Product (K_s)**:
The solubility product \( K_s \) is defined as:
\[ K_s = [A^{+}]^p \cdot [B^{-}]^q \]
Substituting the concentrations in terms of solubility \( S \):
\[ K_s = (pS)^p \cdot (qS)^q \]

4. **Simplify the Expression**:
Expanding the expression gives:
\[ K_s = p^p \cdot S^p \cdot q^q \cdot S^q \]
Combining the \( S \) terms:
\[ K_s = p^p \cdot q^q \cdot S^{p+q} \]

5. **Final Expression**:
Thus, the relationship between the solubility product \( K_s \) and the solubility \( S \) is:
\[ K_s = S^{p+q} \cdot p^p \cdot q^q \]

### Conclusion:
From the derived expression, we can conclude that the correct option that matches our result is:
\[ K_s = S^{p+q} \cdot p^p \cdot q^q \]
This corresponds to the first option given in the question.