In the given reaction:CH3COOH + H2O CH3COO- + H3O+We can identify the conjugate acid-base pairs as follows:1. CH3COOH acetic acid donates a proton H+ to H2O, forming CH3COO- acetate ion . So, CH3COOH is an acid and CH3COO- is its conjugate base.2. H2O water accepts a proton H+ from CH3COOH, forming H3O+ hydronium ion . So, H2O is a base and H3O+ is its conjugate acid.To determine which side of the reaction is favored at equilibrium, we can look at the acid dissociation constant Ka of the acid involved in the reaction. The Ka value for acetic acid CH3COOH is approximately 1.8 x 10^-5. Since this value is small, it indicates that the reaction does not proceed very far to the right, and the equilibrium lies towards the reactants' side.Therefore, at equilibrium, the reaction is favored towards the left side the side with CH3COOH and H2O , meaning that only a small amount of CH3COO- and H3O+ are formed, and the majority of the acetic acid remains undissociated. This is because acetic acid is a weak acid and does not ionize completely in water.