\(\text{CH}_3\text{COOH}_{(aq)} + \text{OH}^-_{(aq)} \; \rightleftharpoons \; \text{CH}_3\text{COO}^-_{(aq)} + \text{H}_2\text{O}_{(l)}\)

In the reaction above, \(\ce{CH3COO^-_{(aq)}}\) is

The reaction is:

\(\text{CH}_3\text{COOH}_{(aq)} + \text{OH}^-_{(aq)} \; \rightleftharpoons \; \text{CH}_3\text{COO}^-_{(aq)} + \text{H}_2\text{O}_{(l)}\)

Step-by-step reasoning:

• Identify the acid and base on the left side: Acetic acid (\(\text{CH}_3\text{COOH}\)) donates a proton (\(H^+\)), so it acts as an acid. Hydroxide (\(\text{OH}^-\)) accepts that proton, so it acts as a base.
• Look at the products: When \(\text{CH}_3\text{COOH}\) loses a proton, it forms \(\text{CH}_3\text{COO}^-\). This species is the conjugate base of acetic acid.
• Definition reminder:
  • The product that results from an acid after donating a proton is called its conjugate base.
  • The product that results from a base after accepting a proton is called its conjugate acid.

Answer: \(\text{CH}_3\text{COO}^-_{(aq)}\) is the conjugate base.

Correct option: conjugate base