Equations of Neutralisation

Case 1 : strong acid + strong alkali
For example, consider the neutralization between hydrochloric acid and sodium hydroxide solution. HCl is completely ionized in water to give H+(aq) ions, and NaOH is completely dissociated in water to give OH-(aq) ions. The neutralization only involves the reaction between H+(aq) and OH-(aq) ions.
H+(aq) + OH-(aq) → H2O(l)

Case 2 : weak acid + strong alkali
For example, consider the neutralization between ethanoic acid and sodium hydroxide solution. CH3COOH is partly ionized in water to give H+(aq) ions, and NaOH is completely dissociated in water to give OH-(aq) ions. The neutralization also involves the reaction between H+(aq) and OH-(aq) ions. However, due to the removal of H+(aq) ions in neutralization, CH3COOH is further ionized to give more H+(aq) ions.
H+(aq) + OH-(aq) → H2O(l)
CH3COOH(aq) = CH3COO-(aq) + H+(aq)
Adding the above two equations together, the equation becomes
CH3COOH(aq) + OH-(aq) → CH3­COO-(aq) + H2O(l)

Case 3 : strong acid + weak alkali
For example, consider the neutralization between hydrochloric acid and ammonia solution. HCl is completely ionized in water to give H+(aq) ions, and NH3 is partly ionized in water to give OH-(aq) ions. The neutralization also involves the reaction between H+(aq) and OH-(aq) ions. However, due to the removal of OH-(aq) ions in neutralization, NH3 is further ionized to give more OH-(aq) ions.
H+(aq) + OH-(aq) → H2O(l)
NH3(aq) + H2O(l) = NH4+(aq) + OH-(aq)
Adding the above two equation together, the equation becomes
NH3(aq) + H+(aq) → NH4+(aq)

Case 4 : weak acid + weak alkali
For example, consider the neutralization between ethanoic acid and ammonia. CH3COOH is partly ionized in water to give H+(aq) ions, and NH3 is partly ionized in water to give OH-(aq) ions. The neutralization also involves the reaction between H+(aq) and OH-(aq) ions. However, due to the removal of H+(aq) and OH-(aq) ions in neutralization, CH3COOH and NH3 are further ionized to give more H+(aq) and OH-(aq) ions respectively.
H+(aq) + OH-(aq) → H2O(l)
CH3COOH(aq) = CH3COO-(aq) + H+(aq)
NH3(aq) + H2O(l) = NH4+(aq) + OH-(aq)
Adding the above three equation together, the equation becomes
CH3COOH(aq) + NH3(aq) → CH3COONH4(aq)