A solution containing a mixture of metal cations was treated as follows. 1.Dilut

Question

A solution containing a mixture of metal cations was treated as follows. 1.Dilute HCI was added and a precipitate formed. The precipitate was filtered off. 2.H_2S was bubbled through the acidic solution. No precipitate formed. 3.The pH was raised to about 9 and H_2S was again bubbled through the solution. A precipitate formed and was filtered off. 4.Finally, sodium carbonate was added to the filtered solution. 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?

Explanation / Answer

1) Formation of ppt with HCl (due to formation of chlorides of cations)

Ag+ : It forms white ppt, soluble in ammonia

Pb+2: it forms white ppt, soluble in hot water

Hg2+2 : forms white ppt,

So at least one of these ions are present

2) H2S bubbled through acidic solution, no ppt

The sulphide ions concentration is minimum in acidic medium, so here 2nd Group cations are absent

Cd2+, Bi3+, Cu2+,Sb3+, Sn2+, Sn4+ and Hg2+. Pb2+ : None of these were present

3) The pH rasied to 9 and H2S passed ppt forms

Fe2+, Fe3+, Al3+, and Cr3 etc may be present , Group 3 cations.

The Group 3 cations form different colours of ppt with H2S in basic medium (NH4Cl + NH4OH)

So at least one of these cations is present

4) Na2CO3 added to filtered solution: ppt forms

so Ba+2, Ca+2, Mg+2 and Sr+2: at least one of the ions was present

5) for L+, K+ , Na+ and NH4+ : unkown.

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