Question

A solution containing a mixture of metal cations was treated as outlined. 1. Dilute HCl was added and a precipitate formed. The precipitate was filtered off. 2. H,S was bubbled through the acidic solution. Again, a precipitate formed and was filtered off. 3. The pH was raised to about 9 and H,S was again bubbled through the solution. No precipitate formed. 4. Finally, sodium carbonate was added. A precipitate formed and was filtered off. What can be said about the presence of each of these groups of cations in the original solution? Cation group Description Present in the original solution? Ag+, Pb form insoluble chlorides | form acid-insoluble sulfides Bi3+, Cd2+, Cu2+, Hg2+, Pb2+, Sb3+, Sn2+, Sn4+ Al3+, Co2+, Cr3+, Fe2+, Fe3+, Ni2+, Mn2+, Zn2+ Ba2+, Ca2+, Mg2+, Sr2+ form base-insoluble sulfides or hydroxides form insoluble carbonates Li+, Na+, K+, NH completely soluble

Answer 1

The information obtained from each of the steps are:

1) a precipitate was obtained upon addition of dilute HCl. This means that the cation forms insoluble chlorides. Hence, atleast one of the ions from group 1 is present.

So, corresponding to group I ( Ag+ , Pb​​​​​2+ , Hg​​​​​2​​​​2+ ) the answer will be Atleast one of these ions was present.

We can't be sure if all of the ions from this group are present or not without carrying out specific tests for each cation on the chloride precipitate.

M​​​​​​n+ + nHCl = MCl​​​​n + nH+

(General reaction between cation and HCl to form chloride)

2) Addition of HCl to the solution had decreased the pH (making it acidic) . Now, H2S is passed. So the remaining cations will form sulphides in acidic medium. Since a precipitate is formed, this means that atleast one cation is present which forms acid insoluble sulphides. This means that the answer corresponding to Group II (BI​3+ , Cd​​​​2+ , Cu​​​​​2+ ...etc) is atleast one of these ions present. Again, without any information on colour of precipitate , or carrying out further analysis on precipitate, we can't say which and how many such ions are present.

3) H2S gas is passed in alkaline medium (increased pH). This means that sulphides are formed in alkaline medium. Had a precipitate been formed, then it would have indicated the presence or Group III ( Al​​​​​3+ , Co​​​​​2+ ...etc) cations. Since no precipitate is formed, then it means that no cation which forms base insoluble sulphides or hydroxides (hydroxides may be formed due to addition of base to increase pH) is present.

Hence, the answer corresponding to this group is None of these ions were present.

4) Na​​​​​2​​​CO​​3  (sodium carbonate) is added. This means that the remaining cations in the solution will form carbonates. Since a precipitate is formed, this indicates the presence of atleast one cation which forms insoluble carbonates. So, atleast one cation from group IV ( Ba​​​​2+ , Sr​​​​​2+ , Ca​​​​​2+ , Mg​2+ ) is present. Hence, the answer for this group is Atleast one of these ions were present.

(Similar to previous steps, we can determine the identity and number of the group IV cations only by further analysis of the precipitate.)

Finally, the cations from the last group ( Li​​​​​+ , Na+ ...) may or may not have been present in the solution. Since they donot form precipitates with any of the reagents added, we can't be sure about their presence. Hence , the answer for this group is unknown.

(If you want to detect the presence of ions from the last group, then there are certain cation specific steps which are carried out from the solution left after removal of precipitate in step 4.)