In the equilibrium
`CH_(3)COOH+HF Leftrightarrow CH_(3)COOH_(2)^(+)+F^(-)`

Options

A. `F^(-)` is the conjugate acid of `CH_(3)COOH`
B. `F^(-)` is the conjugate base of HF
C. `CH_(3)COOH` is the conjugate acid of `CH_(3)COOH_(2)^(+)`
D. `CH_(3)COOH_(2)^(+)` is the conjugate base of `CH_(3)COOH`

Answer

To solve the problem, we need to identify the conjugate acids and bases in the given equilibrium reaction:

**Reaction:**
\[ \text{CH}_3\text{COOH} + \text{HF} \leftrightarrow \text{CH}_3\text{COOH}_2^+ + \text{F}^- \]

### Step-by-Step Solution:

1. **Identify the Acid and Base:**
- In the reaction, we have acetic acid (\( \text{CH}_3\text{COOH} \)) and hydrofluoric acid (\( \text{HF} \)).
- Acids are substances that can donate protons (\( \text{H}^+ \)), while bases are substances that can accept protons.

2. **Determine the Proton Transfer:**
- Here, \( \text{HF} \) donates a proton (\( \text{H}^+ \)) to \( \text{CH}_3\text{COOH} \).
- As a result, \( \text{CH}_3\text{COOH} \) becomes \( \text{CH}_3\text{COOH}_2^+ \) (the conjugate acid), and \( \text{HF} \) becomes \( \text{F}^- \) (the conjugate base).

3. **Identify Conjugate Acids and Bases:**
- The conjugate acid of \( \text{CH}_3\text{COOH} \) is \( \text{CH}_3\text{COOH}_2^+ \) because it is formed by the addition of a proton.
- The conjugate base of \( \text{HF} \) is \( \text{F}^- \) because it is formed by the removal of a proton.

4. **Conclusion:**
- In this equilibrium, \( \text{F}^- \) is the conjugate base of \( \text{HF} \), and \( \text{CH}_3\text{COOH}_2^+ \) is the conjugate acid of \( \text{CH}_3\text{COOH} \).

### Final Answer:
- **Conjugate Base:** \( \text{F}^- \) (from \( \text{HF} \))
- **Conjugate Acid:** \( \text{CH}_3\text{COOH}_2^+ \) (from \( \text{CH}_3\text{COOH} \))