Question

Can you please write a written detailed procedure (like paragraph not a chart please) for the two experiments:

1- qualitative analysis of cation group 1

2- qualitative analysis of cation group 2

Answer 1

For the Group 1 Cations:

Add the solid compound into the 2N HCl solution, you will get a white precipitate. this confirms the presence of Group 1 Cations. Only three species comes in this category- Ag(I), Hg(I), and Pb(II). Now seperate the precipitate and wash with water which is having some HCl. HCl will remove the impurities from the precipitate. This precipitate will have the AgCl, Hg2Cl2 and PbCl2. Treating the mixture of chlorides with hot water will dissolve the PbCl2 in it and only Silver and mercury chloride will left in the mixture. Presence of PbCl2 dissolved in the hot water can be confirmed by adding the AgNO3 in it. It will give the white precipitate of AgCl.

Now, the left over precipitate which may contain AgCl and HgCl will be treated with NH4OH. If the mixture gives a precipitate on treatment with NH4OH it confirms the presence of Hg. if Ag is present it will not make a precipitate, it will form a solution.

For Group 2 Cations:

After the addition of HCl in the mixture, the precipitate is removed and the filtrate is taken for testing the presence of Group 2 Cations. Passing the H2S gas in the boiling test tube of filtrate from the first test, will give the colored precipitate if group 2 cations are present. The colored will be as follows:

Black precipitate: Mercury(II) sulphide HgS, lead(II) sulphide PbS, copper(II) sulphide CuS.

Brown precipitate: Bismuth(III) sulphide Bi₂S₃, tin(II) sulphide SnS.

Yellow precipitate: Cadmium(II) sulphide CdS, arsenic(III) sulphide As₂S₃, arsenic(V) sulphide As₂S₅, tin(IV) sulphide SnS₂.

Orange precipitate: Antimony(III) sulphide Sb₂S₃, antimony(V) sulphide Sb₂S₅