The two equilibrium `AB hArr A^(+) + B^(-) and AB+B^(-)hArrAB_(2)^(-)` are simultaneously maintained in a solutio with equilibrium constant `K_(1) and K_(2)` respectively. The ratio of `[A^(+)] ` to `[AB_(2)^(-)]` in the solution is

Options

A. directly proportional to the concentration of `B^(-)`
B. inversely proportional to the concentration of `B^(-)`
C. directly proporitonal to the squar eof the concentration of `B^(-)`
D. Inversely proportional to the square of the concentration of `B^(-)`.

Answer

To solve the problem, we need to analyze the two equilibria and their respective equilibrium constants.

### Step-by-Step Solution:

1. **Identify the Equilibria**:
- The first equilibrium is:
\[ AB \rightleftharpoons A^+ + B^- \]
- The second equilibrium is:
\[ AB + B^- \rightleftharpoons AB_2^- \]

2. **Write the Equilibrium Constants**:
- For the first equilibrium, the equilibrium constant \( K_1 \) is given by:
\[ K_1 = \frac{[A^+][B^-]}{[AB]} \]
- For the second equilibrium, the equilibrium constant \( K_2 \) is given by:
\[ K_2 = \frac{[AB_2^-]}{[AB][B^-]} \]

3. **Divide the Equilibrium Constants**:
- To find the ratio of \([A^+]\) to \([AB_2^-]\), we can divide \( K_1 \) by \( K_2 \):
\[ \frac{K_1}{K_2} = \frac{\frac{[A^+][B^-]}{[AB]}}{\frac{[AB_2^-]}{[AB][B^-]}} \]
- This simplifies to:
\[ \frac{K_1}{K_2} = \frac{[A^+][B^-]^2}{[AB][AB_2^-]} \]

4. **Rearranging the Equation**:
- Rearranging gives us:
\[ [A^+] = \frac{K_1}{K_2} \cdot \frac{[AB_2^-]}{[B^-]^2} \]

5. **Finding the Ratio**:
- From the above equation, we can express the ratio of \([A^+]\) to \([AB_2^-]\):
\[ \frac{[A^+]}{[AB_2^-]} = \frac{K_1}{K_2} \cdot \frac{1}{[B^-]^2} \]

### Final Result:
Thus, the ratio of \([A^+]\) to \([AB_2^-]\) in the solution is:
\[ \frac{[A^+]}{[AB_2^-]} = \frac{K_1}{K_2 \cdot [B^-]^2} \]