# Analysis of group 2 cations

#### Copper subgroup analysis

We have seen that the elements of copper arsenic subgroup are soluble in an excess of base, in particular (NH4)2S, while not surprisingly the sulfides of the copper subgroup are soluble in acid.

The precipitate is treated therefore with 2-3 mL of HNO3 2N, heating and stirring with the glass rod. The nitric acid is an oxidizing acid, and oxidizes the sulfide ions, S2-, to elemental sulfur, S. The only sulfide that is not solubilized after the treatment is, if present, the mercury sulfide, HgS. This compound is known to be very slightly soluble. Anyway, its solubility increases proportionally to the acid concentration and the temperature. For this reason it preferable to use diluted HNO3 (2N).

Here are the oxidation reactions that sulphides of copper subgroup face:

3CuS + 2NO3- + 8H3O+ $\dpi{120}&space;\rightleftharpoons$ 3Cu2+ + 3S° + 2NO + 12H2O

3CdS+ 2NO3- + 8H 3O+ $\dpi{120}&space;\rightleftharpoons$ 3Cd2+ + 3S° + 2NO + 12H2O

3PbS + 2NO3- + 8H3O+ $\dpi{120}&space;\rightleftharpoons$ 3Pb2+ + 3S° + 2NO + 12H2O

Bi2S3 + 2NO3- + 8H3O+ $\dpi{120}&space;\rightleftharpoons$ 2BI3+ + 3S° + 2NO + 12H2O

HgS however, is not attached (if present) and constitutes an undissolved residue. This residue should appear black, but may also appear reddish because of the presence of Hg(NO3)2.

Elemental sulfur forms a colloidal yellowish precipitate but more probably dark, almost black, very light, which lies at the level of the surface of the solution and may be removed easily with the glass rod. Obviously, the precipitate of HgS must first be centrifuged.

Here is the copper subgroup analysis pattern:

The analysis of this subgroup is relatively simple. We have already seen what are the effects of treatment with HNO3 2N. Now let's see the treatments with NH3 and NaOH.

All the sulfides or this subgroup except mercury sulfide (then sulfides of Pb 2+, Bi 3+, Cu 2+, Cd 2+) have been brought back in solution by treatment with 2N nitric acid. Now, we break into two parts this little group by treatment with 6N NH4OH. The solution is acid because of HNO3, a strong acid, so it is preferable to work with NH4OH concentrated, in order to save time and work with smaller volumes (it is always a good practice). If until now we have done well, thoroughly washing the precipitate, we should not worry too much about bringing the solution to low values of pH.

This treatment has a dual effect, and it is a beautiful demonstration of knowledge of the chemistry of the elements.

The raising of the pH, indeed, causes the precipitation of lead and bismuth hydroxides (if present), leaving in solution Cu2+ and Cd2+ (if present), as amino complexes (evidently, lead and bismuth do not form complexes with ammonia) .Moreover, copper amino complexes (if they were present), would color in blue the solution, and this represents itself a solid proof of the presence of Cu2+ in unknown substance.

Although lead has amphoteric properties, ammonia does not have sufficient strength as base to bring it into solution as plumbite ion.

Cu2+ + 4NH3 $\dpi{120}&space;\rightleftharpoons$ Cu(NH3)42+ amino complexes blue staining

Cd2+ + 4NH3 $\dpi{120}&space;\rightleftharpoons$ Cd(NH3)42+

The ammonia leads to an excess of OH- in solution:

NH3 + H2O $\dpi{120}&space;\rightleftharpoons$ NH4+ + OH -

Increasing [OH-] the Ksp value is exceeded for lead and bismuth hydroxides and as a consequence:

Pb2+ + 2OH- $\dpi{120}&space;\rightleftharpoons$ Pb(OH)2 white precipitate

Bi3+ + 3OH- $\dpi{120}&space;\rightleftharpoons$ Bi(OH)3 white precipitate

Pb(OH)2 + 2OH- $\dpi{120}&space;\rightleftharpoons$ Pb(OH)42- $\dpi{120}&space;\rightleftharpoons$ PbO22- + 2H2O