Q.iii. Label the conjugate acid-base pair in the following reactions a. $\mathrm{HCl + H_2O \rightleftharpoons H_3O^+ + Cl^-}$
b. $\mathrm{CO_3^{2-} + H_2O \rightleftharpoons OH^- + HCO_3^-}$

Step 1. For each reaction, identify which species donates a proton (the acid) and which accepts one (the base), then pair each with what it becomes.

Step 2. Reaction (a): $HCl+H_2O\rightleftharpoons H_3O^++Cl^-$. HCl donates a proton to become Cl- (so HCl is acid1, Cl- is its conjugate base1). H2O accepts that proton to become H3O+ (so H2O is base2, H3O+ is its conjugate acid2).

Step 3. Reaction (b): $CO_3^{2-}+H_2O\rightleftharpoons OH^-+HCO_3^-$. Here CO3(2-) ACCEPTS a proton (from water) to become HCO3- (so CO3(2-) is base1, HCO3- is its conjugate acid1). H2O DONATES that proton to become OH- (so H2O is acid2, OH- is its conjugate base2) -- note water plays the opposite role (acid, not base) here compared with reaction (a), demonstrating its amphoteric nature (section 3.3).

Step 4. So the two conjugate pairs in each reaction are found by matching each reactant to the product it turns into by gaining or losing exactly one H+.

In each equation, track which species gains and which loses a proton, and pair each reactant with the product it becomes.

• Pairing species across the WRONG side matches (e.g. pairing HCl with H3O+ instead of with Cl-) -- a conjugate pair is always reactant-with-its-own-product, not reactant-with-the-other-species'-product.
• Missing that water plays different roles (base in reaction a, acid in reaction b) -- a direct illustration of its amphoteric nature.