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Unformatted text preview: Experiment: 4: Qualitative Analysis of Cations
The primary objectives of this experiment were to develop a plan for the qualitative
cations, Ag+, PB2+, and Hg2(2+). Some of the ways of carrying out this experiment includes
precipitate ion reactions to remove cations from mixture of soluble salts. We used a centrifuge in
order to remove the precipitated ions. Additionally, this experiment will help explain the policy of
solubility to figure out the variety of solubility content in salts.
As we were performing our experiment, it was necessary to add HCl to the mixture
because that is the step that helps the chlorides of the three cations separate. Too much of HCl
should be avoided because AgCl and PbCl2 have a high tendency to form soluble complexes if
you add too many drops of HCl. After you add the HCl and centrifuge it, you would add another
drop of HCl to make sure the mixture has completed its form of precipitation.
Lead (II) ion was separated from the insoluble Hg2Cl2 and AgCl salts by dissolving it in
hot water. Usually, the presence of Pb2+ ion is confirmed by addition of aqueous K2CrO4,
which forms a insoluble yellow precipitate, PbCrO4. There was a presence of Pb2+ in my
unknown when I tested for it in my second part of the experiment.
Mercury (I) ion is separated by the supernatant containing Ag(NH3)2+ + Cl- (aq) through
a process of acidification. When you neutralize the ammonia in that solution, it decomposes and
Ag+ ion forms a white AgCl precipitate. Then the mixture of Hg(l) and HgNH2Cl gives a
precipitate of grey appearance. Mercury (I) was not present in my unknown. There would be
many reasons why this occurred. The cause of this could be that there may have been Ag+
present before or maybe there was too much chlorine gas.
Silver ions were separated from Hg2Cl2 by the addition of ammonia. The silver chloride
forms a soluble complete with ammonia and dissolves, causing a white precipitate to form. The
classification if there was any presence of silver ions is the white precipitate that forms. My
unknown had a presence of silver ions because there was a formation of a white precipitate.
Some difficulties that I had with this experiment were with my unknown. For some odd
reason, I didn’t have a presence of Mercury ions which was necessary in my unknown. The
source of error for this could be that I had too much chlorine gas in my test tube before or that I
already had a presence of Ag+. Because as I added NH3, there was no black/grey precipitate
that should be formed for a statement of Hg2(2+) to be formed. ...
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- Spring '10