Ammonia has the capacity to act as a Bronsted-Lowry base, due to
the availability of a lone pair on N to accept a proton, for
instance, from water solvent, thereby generating the hydroxide ion;
hence where the discriminating student decides the question
requires the involvement of the hydroxide ion (where "aqueous
ammonia" is stated"), the student should first write the equation
for the hydrolysis of the ammonia :
NH3 (aq) + H2O (l) (insert equilibrium double half arrow here) NH4+
(aq) + OH- (aq) or simply NH4OH (aq)
Subsequently, the remaining required equation(s), eg for the
formation of a blue ppt when aqueous ammonia is added to an aqueous
solution containing Cu2+ ions :
Cu2+ (aq) + OH- (aq) (insert equilibrium double half arrow here)
Cu(OH)2 (s)
and in excess aqueous ammonia :
NH3 (aq) + [Cu(H2O)6]2+ (aq) (insert equilibrium double half arrow
here) [Cu(NH3)4(H2O)2]2+ (aq)
Furthermore, there are instances, in which the student may
optionally write the equation in which ammonia is directly the
Bronsted-Lowry base, OR the equation in which ammonia first
undergoes hydrolysis to generate a separate Bronsted-Lowry base -
the hydroxide ion, which accepts the proton from the Bronsted-Lowry
acid; as an example, both of the following are acceptable at 'O'
levels and 'A' levels :
Either : NH3 (aq) + HCl (aq) (insert equation arrow here) NH4Cl
(aq) Or :
NH4OH (aq) + HCl (aq) (insert equation arrow here) NH4Cl (aq) + H2O
(l)
Technically though, the former is more accurate. The reason being,
because the hydroxide ion OH- is a stronger base than ammonia NH3
(or put in another way, NH4+ is a stronger acid than H2O), the
position of equilibrium lies strongly to the left :
NH3 (aq) + H2O (l) (insert equilibrium double half arrow here) NH4+
(aq) + OH- (aq)