Final answer:
The reaction between CH3COOH and NaOH is a neutralization reaction resulting in sodium acetate and water, with Na+ as a spectator ion. Acetic acid is monoprotic, donating a proton to the hydroxide ion, and such reactions favor the production of weaker acids and bases.
Step-by-step explanation:
The reaction between CH3COOH (acetic acid) and NaOH (sodium hydroxide) is an example of an acid-base neutralization reaction. When acetic acid reacts with NaOH, the acetic acid donates a proton (H+) to NaOH, thereby forming sodium acetate (CH3COO- Na+) and water (H2O). This process can be better understood by looking at the equilibrium reaction:
CH3COOH(aq) + NaOH(aq) ↔ H2O(l) + CH3COONa (aq)
In this reaction, the Na+ ion is a spectator ion, which means it does not participate actively in the reaction and is usually not shown in the equation. Acetic acid is a monoprotic acid, which means it can donate one proton, and the OH- from NaOH is the base that accepts the proton, resulting in a water molecule.
Ammonia reacts with water in a similar manner, where ammonia (NH3) acts as a base accepting a proton from water:
NH3(aq) + H2O(l) ↔ NH4+(aq) + OH-(aq)
Such acid-base reactions always favor the formation of the weaker acid and base.