In order to understand the nature of each species in the buffer solution, let's analyze the given equilibrium equation:
[tex]\[ \text{CH}_3\text{CO}_2\text{H}_{(aq)} \rightleftharpoons \text{H}_3\text{O}_{(aq)}^+ + \text{CH}_3\text{CO}_{2(\text{aq})}^- \][/tex]
To identify the acids and bases and their conjugate pairs, we need to review the concepts of acids, bases, and their conjugates according to the Brønsted–Lowry theory:
- A Brønsted–Lowry acid is a proton (H⁺) donor.
- A Brønsted–Lowry base is a proton (H⁺) acceptor.
- The conjugate base of an acid is what remains after the acid has donated a proton.
- The conjugate acid of a base is what is formed after the base has accepted a proton.
Let's break it down:
1. CH[tex]\(_3\)[/tex]CO[tex]\(_2\)[/tex]H (Acetic acid) can donate a proton (H⁺) to become CH[tex]\(_3\)[/tex]CO[tex]\(_2^−\)[/tex] (Acetate ion).
2. H[tex]\(_3\)[/tex]O⁺ (Hydronium ion) is formed when water accepts a proton.
From this, we can identify the following:
- CH[tex]\(_3\)[/tex]CO[tex]\(_2\)[/tex]H is the acid because it donates a proton to form CH[tex]\(_3\)[/tex]CO[tex]\(_2^−\)[/tex].
- CH[tex]\(_3\)[/tex]CO[tex]\(_2^−\)[/tex] is the conjugate base of CH[tex]\(_3\)[/tex]CO[tex]\(_2\)[/tex]H.
- H[tex]\(_3\)[/tex]O⁺ is the conjugate acid that is produced by the donation of a proton.
Among the given choices:
- (A) is correct: CH[tex]\(_3\)[/tex]CO[tex]\(_2\)[/tex]H is an acid, and CH[tex]\(_3\)[/tex]CO[tex]\(_2^−\)[/tex] is its conjugate base.
- (B) is incorrect as it states that CH[tex]\(_3\)[/tex]CO[tex]\(_2\)[/tex]H is a base, which it is not.
- (C) is incorrect since it states H[tex]\(_3\)[/tex]O⁺ forms from CH[tex]\(_3\)[/tex]CO[tex]\(_2^−\)[/tex], which is not the case.
- (D) is incorrect as it states that CH[tex]\(_3\)[/tex]CO[tex]\(_2\)[/tex]H is the conjugate base of H[tex]\(_3\)[/tex]O⁺, which it is not.
Thus, the correct answer is [tex]\( \text{A)} \)[/tex].
