In the analysis of inorganic substances, this branch involves the analysis of both metallic constituents as cations and non-metallic constituents as anions. In this experiment, you will analyze a known solution that contains the entire Group I cations—silver, lead, and mercury (I)—and an unknown solution to determine which ions are present and which are absent.
The chlorides of Pb2+, Hg22+ and Ag+ are all insoluble in cold water. They can be removed as a group from solution by the addition of HCl. The reactions that occur are simple precipitations and can be represented by the equations:
Ag+ (aq) + Cl- (aq) → AgCl(s) (1)
Pb2+ (aq) + 2 Cl- (aq) → PbCl2(s) (2)
Hg22+ (aq) + 2 Cl- (aq) → Hg2Cl2(s) (3)
It is important to add enough HCl to ensure complete precipitation, but not too large an excess. In concentrated HCl solution, these chlorides tend to dissolve, producing chloro-complexes such as AgCl2-.
Lead chloride is separated from the other two chlorides by heating with water. The Ksp for PbCl2 greatly increases with temperature, favouring less solid and more ions, therefore the PbCl2 dissolves in hot water:
PbCl2(s) → Pb2+ (aq) + 2 Cl-(aq) (4)
Once Pb2+ has been put into solution, we can check for its presence by adding a solution of K2CrO4. The chromate ion, CrO42-, gives a yellow precipitate with Pb2+:
Pb2+ (aq) + CrO42-(aq) → PbCrO4(s) (5)
The other two insoluble chlorides, AgCl and Hg2Cl2, can be separated by adding aqueous ammonia. Silver chloride dissolves, forming the complex ion Ag (NH3)2+:
AgCl(s) + 2 NH3 (aq) → Ag (NH3)2+(aq) + Cl-(aq) (6)
Ammonia also reacts with Hg2Cl2 via a rather unusual oxidation-reduction reaction. The products include finely divided metallic mercury, which is black, and a compound of formula HgNH2Cl, which is white:
Hg2Cl2(s) + 2 NH3 (aq) → Hg (l) + HgNH2Cl(s) + NH4+ (aq) + Cl-(aq) (7)
White black white
As this reaction...
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