Testing for halide ions with silver nitrate

• Fluorides produce no precipitate
• Chlorides produce a white precipitate

  Ag+ (aq) + Cl- (aq)  AgCl(s)
• Bromides produce a cream precipitate

  Ag+ (aq) + Br- (aq)  AgBr(s)
• Iodides produce a pale yellow precipitate

  Ag+ (aq) + I- (aq)  AgI(s)
• The silver halide precipitates can be treated with ammonia solution to help differentiate between them if the colours look similar
• Silver chloride dissolves in dilute ammonia to form a complex ion AgCl(s) + 2NH3 (aq) [Ag(NH3 )2 ]+ (aq) + Cl- (aq) Colourless solution
• Silver bromide dissolves in concentrated ammonia to form a complex ion

  AgBr(s) + 2NH3 (aq) [Ag(NH3 )2 ]+ (aq) + Br - (aq) Colourless solution
• Silver iodide does not react with ammonia – it is too insoluble.