The sliver group consists of Ag1+, Pb2+ and Hg22+ (Mercury I) ions.
They each react with Cl1- to form a precipitate.
All other chlorides are soluble.
This lab starts out with an unknown solution provided by the instructor.
1. Separating the Silver Group
We make a small sample from our unknown and we add some HCl to it.
A precipitate forms and we separate that precipitate from the rest of the solution by decantation.
This precipitate must contain one or more of the following compounds:
AgCl2 (s), PbCl2 (s), HgCl (s)
Of these compunds PbCl2 (s) is soluble in hot water. The others are not.
2. Separating the Lead
We mix the precipitate with hot water and separate the precipitation from the rest of the solution by decantation.
The precipitate must now contain one or more of the following 2 compounds:
AgCl2 (s), HgCl (s)
3. Detecting Lead in the Solution
The solution may or may not contain lead. There are two methods of detecting lead.
a. Add a drop of 0.2 M K2CrO4 solution and if a yellow precipitate forms (PbCrO4) then we have proved the presence of lead.
b. Add a drop of 2M H2SO4 to the solution and if a white precipitate forms (PbSO4) then we have also proved the presence of lead.
4. Separate the Silver and Mercury (I)
We mix the precipitate with NH3 and separate the precipitate from the rest of the solution by decantation.
If the precipitate is black this proves the presence of mercury (I) (Hg + HgNH2Cl).
We check for silver in the solution (decantate) by mixing it with HNO3 and if a white precipitate forms then we have proven the presence of silver AgCl2 (s).
Sources:
http://physics.gallaudet.edu/classes/GQUAL1.HTML as it appeared (8-16-2005)
Chapter 9 of Introduction to Semimicro Qualitative Analysis (8th Edition) by J.J. Lagowski and C.H. Sorum