Test Tube 7: 0.1 M SnCl 4 in 3 M HCl: When I mixed samples from tube 3 (6 M NH 3 ) with
samples from test tube 7, a white precipitate formed and heat was produced. Tube 7 could con-
tain Al3+or Sn4+. Since heat was produced, an acid must have been present. Between the two
choices, only SnCl 4 was originally mixed with acid, therefore I concluded that test tube 7 con-
tained SnCl 4. When I mixed samples of test tube 7 with samples from test tube 2 that I believed
contained OH–, a white precipitate formed, consistent with Sn(OH) 4. Upon adding more NaOH,
the precipitate dissolved resulting from a shift in equilibrium.
Test Tube 8: 0.1 Cu(NO 3 ) 2 : I was able to identify this tube by observation, knowing that the
Cu2+ion is blue. No other solution in the list should have a blue color. I confirmed this by form-
ing Cu(NH 3 ) 4 2+(dark blue solution) with samples from test tube 2 that I had initially identified
as NH 3.
Test Tube 9: 0.1 M Ca(NO 3 ) 2 : When I mixed samples from test tube 9 with samples from test
tube 2 that I had already identified as OH–, a white precipitate was produced consistent with
Ca(OH) 2.
Test Tube 10: 0.1 M AgNO 3 : When I mixed samples from test tube 10 with samples from test
tube 2 that I had already identified as OH–, a brownish-gray precipitate consistent with the
color of AgOH. Since I had already ruled out test tube 9 and this was the last tube that needed
identification, I can conclude that test tube 10 contained Ag+. Finally, when I mixed the sam-
ples of test tube 10, which I believed to contain Ag+, with samples from test tube 1 which con-
tained what I believed to be HCl, a white precipitate formed consistent with the color of AgCl.
Laboratory Experiments