[samoyan]

When we added NH3 to the AgCl/Hg2Cl2 mixture, the AgCl dissolved and the Hg2Cl2 turned into black mercury metal and the white insoluble solid HgClNH2.

The net reaction for Hg2Cl2 is:

Hg2Cl2 (s) + 2 NH3(aq) → HgClNH2(s) + Hg(0)(s) + NH4Cl(aq)

Why we can see only the black precipitate as the result of this reaction? What happens with white precipitate in the form of HgClNH2?

[AWK]

If you do this test in a test tube, you can see a black precypitate and a milky suspension after mixing

[samoyan]

Thank you for the help.

Let me add more details about this procedure. In this experiment, we heated the tube containing PbCl₂, Hg₂Cl₂ and AgCl - the precipitate that formed when we added HCl to the solution containing these 3 cations. Then, we heated the tube to separate Pb cations. We did this procedure by pouring the solution with precipitate into the funnel containing filter paper. So, we can assume that residue that was left on the filter paper was containing Hg₂Cl₂ and AgCl. To dissolve AgCl and to test for Hg cations we poured NH₃ to the residue. AgCl dissolved, and Hg₂Cl₂ formed black precipitate that we could see on the filter paper. However, according to the net ionic equation we had to notice the white precipitate also, in the form of HgClNH₂, but we saw only black precipitate (Hg⁰). Do you know any reasons why we were not able to see the white solid HgNH₂Cl on the filter paper?

[AWK]

HgNH2Cl is slightly soluble in ammonia+water. It can be sometimes visible in the filtrate (milky filtate). HgNH2Cl is soluble in HNO3, hence it does not disturb analysis of silver in the filtrate.