1. An unknown containing cations from Group 1 (Hg2^2+, Ag+, Pb^2+) was treated w
ID: 846091 • Letter: 1
Question
1. An unknown containing cations from Group 1 (Hg2^2+, Ag+, Pb^2+) was treated with HCL. A white precipitate formed which was completely soluble in hot water.
What ion(s) must be present (although still in need of confirmation)?
What ion(s) must be absent?
Now that you have figured out what ion(s) were in the unknown, write net ionic equations for:
(a) the formation of the precipitate
(b) the dissolving of the precipitate
2. An unknown contatining cations from Group I was treated with HCl. A white precipiate formed, which appeared to be completely insoluble in hot water. The precipiate dissolved completely in NH3 to leave a clear solution.
What ion(s) must be present (although still in need of confirmation)?
What ion(s) seem to be absent? What must be done to confirm the absence of this/these ions(s)?
Now that you have figured out what was in the unkonwn, write net ionic equations for each of the observations made during analysis.
Explanation / Answer
1]
Pb+2 is present as PbCl2 is desolved in hot water
Hg2^2+, Ag+ may be absent
(a) Pb2+(aq) + 2 Cl-(aq) ? PbCl2 (s)
(b) PbCl2 (s) ? Pb2+(aq) + 2 Cl-(aq)
2]
Ag+ is present
Pb2+ Hg22+ are absent
However, mercury(I) chloride reacts with the ammonia yielding what appears to be a gray solid
which is actually a mixture of black Hg (l) and white HgNH2Cl (s). The presence of this gray
solid is confirmation of the presence of Hg22+ hence Hg2^2+ is absent also is pb2+ as the white precipiate formed, which appeared to be completely insoluble in hot water. Hg22+and Ag+cations can be separated by adding 6 M NH3 (aq) to the solid mixture of the two chlorides. Silver chloride will dissolve since it forms a soluble complex ion with ammonia:
AgCl (s) + 2 NH3 (aq) ? Ag(NH3)2+(aq) + Cl-(aq)
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