text: The 1st analytical group of cations consists of ions which form insoluble chlorides. As such, the group reagent to separate them is hydrochloric acid, usually used at a concentration of 1&ndash;2 M. Concentrated HCl must not be used, because it forms a soluble complex ([PbCl]) with Pb. Consequently, the Pb ion would go undetected.
  The most important cations in the 1st group are Ag, Hg, and Pb.  The chlorides of these elements cannot be distinguished from each other by their colour - they are all white solid compounds. PbCl is soluble in hot water, and can therefore be differentiated easily.  Ammonia is used as a reagent to distinguish between the other two.  While AgCl dissolves in ammonia (due to the formation of the complex ion [Ag(NH)]), HgCl gives a black precipitate consisting of a mixture of chloro-mercuric amide and elemental mercury. Furthermore, AgCl is reduced to silver under light, which gives samples a violet colour.
  PbCl is far more soluble than the chlorides of the other two ions, especially in hot water. Therefore, HCl in concentrations which completely precipitate Hg and Ag may not be sufficient to do the same to Pb. Higher concentrations of Cl cannot be used for the before mentioned reasons. Thus, a filtrate obtained after first group analysis of Pb contains an appreciable concentration of this cation, enough to give the test of the second group, viz. formation of an insoluble sulfide. For this reason, Pb is usually also included in the 2nd analytical group.
  This group can be determined by adding the salt in water and then adding dilute hydrochloric acid. A white precipitate is formed, to which ammonia is then added. If the precipitate is insoluble, then Pb is present; if the precipitate is soluble, then Ag is present, and if the white precipitate turns black, then Hg is present. 
  Confirmation test for Pb: 
  :Pb + 2 KI → PbI + 2 K
  :Pb + KCrO → PbCrO + 2 K
  Confirmation test for Ag:
  :Ag + KI → AgI +  K
  :2Ag + KCrO → AgCrO + 2 K
  Confirmation test for Hg:
  :Hg + 2 KI → HgI + 2 K
  :2 Hg + 2 NaOH → 2 HgO + 2 Na + HO