An acid-base reaction has two "halves" (a bit like redox):

• the acid changing into a base
• the base changing into an acid

The acid is HCl, and the base is O^2-. (Zn²⁺ is a spectator ion. You may think Cl^- is a spectator ion, but H-Cl is actually covalently bonded, not H⁺ and Cl^-.)

• HCl (the acid) loses a H⁺, and becomes Cl^-, which is its conjugate pair. (HCl / Cl^-)

• O^2- (the base) becomes H₂O, but it has to gain 2 H⁺ to do so.

This reaction actually consists of two acid-base reactions.

1. Acidification of O^2- into OH^-: O^2- + HCl → Cl^- + OH^-
2. Acidification of OH^- into H₂O: OH^- + HCl → Cl^- + H₂O

To confirm, adding them together gives: O^2- + HCl → Cl^- + H₂O, after cancelling the OH^- on both sides

So the conjugate pairs involved with O^2- would be:

• O^2- / OH^-
• OH^- / H₂O

Hope that explains a lot!

This reaction is a special kind of acid base reaction, but it still follows the Bronsted-Lowry theory and so you need to do is follow the movement of protons and forget about the rest of the ions present.

Colin's answer is the true explanation of the chemical workings, but aren't we looking for conjugate pairs - where one is an acid and the other is a base??

1. Find where the H+ ions are coming from. Here they come from the HCl, so this must be the acid.

2. Find where the H+ ions have joined up with. Here they have joined with an O2- ion, so that must be the base on the reactant side.

So if we are looking for conjugates:

1. Since HCl is the acid, then the conjugate is Cl-. It's only now is solution with the Zn2+ ions also. Hence the first conjugate pair is HCl / Cl-

2. O2- on the other hand was stuck with the Zn2+ to start off with, but then gains the 2H+ to form water. The second conjugate pair is O2- / H2O

As a side note, if an acid reacts with just a metal.. that isn't an acid/base reaction, that's a redox reaction