In the preceeding problem, if`[A^(+)]"and"[AB_(2)^(-)] "are " y " and " x` respectively, under equilibrium produced by adding the substance `AB` to the solvents, then `K_(1)/K_(2) "is equal to"`

Options

A. `(y)/(x)(y-x)^(2)`
B. `(y^(2)(x+y))/(x)`
C. `(y^(2)(x+y))/(x)`
D. `(y)/(x)(x-y)`

Answer

To solve the problem, we need to find the ratio \( \frac{K_1}{K_2} \) given the concentrations of \( [A^+] \) and \( [AB_2^-] \) at equilibrium.

### Step-by-Step Solution:

1. **Identify Given Concentrations:**
- Let \( [A^+] = Y \)
- Let \( [AB_2^-] = X \)

2. **Calculate Concentration of \( [B^-] \):**
- The concentration of \( [B^-] \) can be calculated using the relationship:
\[ [B^-] = [A^+] - [AB_2^-] = Y - X \]

3. **Write the Equilibrium Reactions:**
- The first step of the reaction can be represented as:
\[ AB \rightleftharpoons A^+ + B^- \]
The equilibrium constant for this reaction is \( K_1 \).
- The second step can be represented as:
\[ B^- + AB \rightleftharpoons AB_2^- \]
The equilibrium constant for this reaction is \( K_2 \).

4. **Write the Expressions for \( K_1 \) and \( K_2 \):**
- For \( K_1 \):
\[ K_1 = \frac{[A^+][B^-]}{[AB]} \]
- For \( K_2 \):
\[ K_2 = \frac{[AB_2^-]}{[B^-][AB]} \]

5. **Calculate \( \frac{K_1}{K_2} \):**
- To find \( \frac{K_1}{K_2} \), we divide the expressions:
\[ \frac{K_1}{K_2} = \frac{[A^+][B^-]}{[AB]} \cdot \frac{[B^-][AB]}{[AB_2^-]} \]
- Simplifying this gives:
\[ \frac{K_1}{K_2} = \frac{[A^+][B^-]^2}{[AB_2^-]} \]

6. **Substitute the Known Values:**
- Substitute \( [A^+] = Y \), \( [B^-] = Y - X \), and \( [AB_2^-] = X \):
\[ \frac{K_1}{K_2} = \frac{Y \cdot (Y - X)^2}{X} \]

7. **Final Result:**
- Therefore, the final expression for \( \frac{K_1}{K_2} \) is:
\[ \frac{K_1}{K_2} = \frac{Y \cdot (Y - X)^2}{X} \]