To determine the acid-conjugate base pair in the given reaction, it's essential to identify which species are acting as acids and which are acting as bases according to the Bronsted-Lowry definition, which states that an acid is a proton (H⁺) donor and a base is a proton acceptor.
Let's examine the reaction:
[tex]\[ \text{HI} + \text{H}_2\text{O} \rightarrow \text{H}_3\text{O}^+ + \text{I}^- \][/tex]
1. Identify the Acids and Bases:
- HI: Hydrogen iodide (HI) donates a proton (H⁺) to water (H₂O) in this reaction. When HI loses a proton, it forms I⁻ (iodide ion). Hence, HI is an acid.
- H₂O: Water accepts the proton from HI to form H₃O⁺ (hydronium ion). Therefore, H₂O acts as a base.
2. Identify the Conjugate Pairs:
- When HI donates a proton, it becomes I⁻. Therefore, HI and I⁻ are a conjugate acid-base pair. The acid (HI) transforms into its conjugate base (I⁻) after donating a proton.
- Similarly, when H₂O gains a proton, it becomes H₃O⁺. Therefore, H₂O and H₃O⁺ are a conjugate base-acid pair.
Now, let's review the provided options and match them with our identified pairs:
1. HI and H₂O: This pair is not correct because they do not form a conjugate acid-base pair with each other directly.
2. H₂O and H₃O⁺: Although this is a valid conjugate base-acid pair, it is not asked for in the given options.
3. H₃O⁺ and I⁻: These two substances do not directly form a conjugate acid-base pair.
4. HI and I⁻: This is the correct conjugate acid-base pair. HI is the acid that donates a proton and forms I⁻, its conjugate base.
Thus, the correct answer to the question is:
[tex]\[ \text{HI and I}^- \][/tex]
Therefore, the correct option is the fourth one.
