Two equilibrium `ABhArrA^(+)+B^(-)` and `AB+B^(-)hArrAB_(2)^(-)` are simultaneously maintained in a solution with equilibrium constants `K_(1)` and `k_(2)`, respectively. Ratio of `[A^(+)]` to `[AB_(2)^(-)]` in the solution is

Options

A. Directly proportional to `[B^(-)]`
B. Inversely proportional to `[B^(-)]`
C. Directly proportional to `[B^(-)]^(2)`
D. Inversely proportional to `[B^(-)]^(2)`

Explanation

To solve the problem, we need to analyze the two equilibria given and derive the ratio of the concentrations of \( [A^+] \) to \( [AB_2^-] \).

### Step-by-Step Solution:

1. **Identify the Equilibria**:
- The first equilibrium is:
\[ AB \rightleftharpoons A^+ + B^- \]
with equilibrium constant \( K_1 \).
- The second equilibrium is:
\[ AB + B^- \rightleftharpoons AB_2^- \]
with equilibrium constant \( K_2 \).

2. **Write the Expressions for the Equilibrium Constants**:
- For the first equilibrium:
\[ K_1 = \frac{[A^+][B^-]}{[AB]} \]
- For the second equilibrium:
\[ K_2 = \frac{[AB_2^-]}{[AB][B^-]} \]

3. **Divide \( K_1 \) by \( K_2 \)**:
- We can express this as:
\[ \frac{K_1}{K_2} = \frac{[A^+][B^-]}{[AB]} \cdot \frac{[AB][B^-]}{[AB_2^-]} \]
- Simplifying this gives:
\[ \frac{K_1}{K_2} = \frac{[A^+][B^-]^2}{[AB_2^-]} \]

4. **Rearranging the Equation**:
- From the above expression, we can rearrange to find:
\[ [A^+] = \frac{K_1}{K_2} \cdot \frac{[AB_2^-]}{[B^-]^2} \]

5. **Finding the Ratio**:
- We need the ratio \( \frac{[A^+]}{[AB_2^-]} \):
\[ \frac{[A^+]}{[AB_2^-]} = \frac{K_1}{K_2} \cdot \frac{1}{[B^-]^2} \]
- This indicates that the ratio \( \frac{[A^+]}{[AB_2^-]} \) is inversely proportional to the square of the concentration of \( [B^-] \).

### Final Result:
Thus, the ratio of \( [A^+] \) to \( [AB_2^-] \) in the solution is:
\[ \frac{[A^+]}{[AB_2^-]} \propto \frac{1}{[B^-]^2} \]