Identify conjugate acid and conjugate base in the following reaction.
                          H₂PO₄⁻ (aq) + H₂O (l) ⇌ HPO₄²⁻ (aq) + H₃O⁺ (aq)

• A. HPO₄²⁻ (conjugate acid), H₃O⁺ (conjugate base)
• B. HPO₄²⁻ (conjugate base), H₃O⁺ (conjugate acid)
• C. H₂PO₄⁻ (conjugate acid), H₂O (conjugate base)
• D. H₂PO₄⁻ (conjugate base), H₂O (conjugate acid)

Answer: (B)

Verified step by step guidance

1. Identify the acid and base on the reactant side by looking at which species donates a proton (H\(\u\)207a) and which accepts it. In the reaction, H\(\u\)2082PO\(\u\)2084\(\u\)207b (dihydrogen phosphate ion) can donate a proton, and H\(\u\)2082O (water) can accept a proton.
2. Write the general definitions: an acid is a proton donor, and its conjugate base is what remains after it loses a proton; a base is a proton acceptor, and its conjugate acid is what forms after it gains a proton.
3. Determine the conjugate base of H\(\u\)2082PO\(\u\)2084\(\u\)207b by removing one proton (H\(\u\)207a), resulting in HPO\(\u\)2084\(\u\)207b\(\u\)207b (hydrogen phosphate ion). This species is on the product side.
4. Determine the conjugate acid of H\(\u\)2082O by adding one proton (H\(\u\)207a), resulting in H\(\u\)2083O\(\u\)207a\(\u\)207a (hydronium ion), which is also on the product side.
5. Match the conjugate acid-base pairs: H\(\u\)2082PO\(\u\)2084\(\u\)207b (acid) and HPO\(\u\)2084\(\u\)207b\(\u\)207b (conjugate base), and H\(\u\)2082O (base) and H\(\u\)2083O\(\u\)207a\(\u\)207a (conjugate acid).