A solution containing a mixture of metal cations was treated with dilute HCl and no precipitate formed. Next, H2S was bubbled through the acidic solution. A precipitate formed and was filtered off. Then, the pH was raised to about 8 and H2S was again bubbled through the solution. A precipitate again formed and was filtered off. Finally, the solution was treated with a sodium carbonate solution, which resulted in no precipitation. Which metal ions were definately present, which were definitely absent, and which may or may not have been present in the original mixture?
Ag+, Ba2+, Mg2+, Hg2+, Pb2+, Hg2^2+, Cu2+, Zn2+, Ni2+, Co2+, Sn2+, Li+, Sb3+
I understand how to find out which ones POSSIBLY are there, but how can one find out which ones are DEFINITELY there?
Well, its definitely too late to help you, but I'll post this for the benefit of others. None of the listed atoms are definitely present, as there is no way to tell from the limited information. Likewise, we know from the chart of the seperation of cations that Ba2+ and Ca2+ have no chance of being there, since they form precipitates when carbonate is added. Ag+, Pb2+, and Hg2^2+ have no chance of being there either, since they form precipitates when dilute HCl is first added. The rest on the list are all possibly there.
SHOTS FIRED
You're incorrect.
To determine which metal ions are definitely present in the original mixture, we need to analyze the experimental observations and reactions.
1. No precipitate formed when the solution was treated with dilute HCl:
This indicates that no metal ions capable of forming insoluble chlorides were present in the mixture. Therefore, we can conclude that Li+ is definitely absent from the original mixture since lithium chloride is highly soluble.
2. A precipitate formed when H2S was bubbled through the acidic solution:
H2S (hydrogen sulfide gas) reacts with metal cations to form insoluble metal sulfides. Since a precipitate formed in this step, we can conclude that at least one or more of the metal cations in the original mixture can form insoluble sulfides. Therefore, we cannot definitively say which metal ions are present at this stage.
3. A precipitate formed again when H2S was bubbled through the solution with a pH raised to about 8:
Increasing the pH to about 8 helps to precipitate certain metal sulfides that are less soluble at a more basic pH. By observing the second precipitation with H2S, we can conclude that the metal ions responsible for this precipitate were not filtered off during the first filtration step. So, we can eliminate the possibility of having Pb2+ and Hg2^2+ since their sulfides would be filtered off in the first step.
4. No precipitation occurred when the solution was treated with a sodium carbonate solution:
Sodium carbonate reacts with certain metal cations to form insoluble carbonates. Since no precipitate formed during this reaction, we can conclude that metals capable of forming insoluble carbonates such as Ba2+ and Pb2+ are not present in the original mixture.
Based on the observations and reactions described, we can conclude that Li+, Pb2+, Hg2^2+, Ba2+ are definitely absent from the original mixture. The remaining metal ions, Ag+, Mg2+, Hg2+, Cu2+, Zn2+, Ni2+, Co2+, Sn2+, Sb3+, may or may not have been present in the original mixture.