Analysis of group 2 cations

Antimony recognition

Antimony is searched from the solution that we have separated from the precipitate containing arsenic sulfides. It had been solubilized as chloo-complex (SbCl6- ,SbCl4-).

The antimony research is hampered by tin (interfering).

1) Assay with oxalic acid and Na2S

Oxalic acid forms a very stable complex with the tin, so that it does not precipitate subsequently the addition of sodium sulfide , Na2S. Therefore, oxalic acid acts as masking agent.

Sn4+ + 3C2O42- Sn(C2O4)32-

The precipitation reactions of sulfides are those that we have already seen at the start of group 2 analysis. If antimony is present, it precipitates as orange antimony sulfide.

2) Assay with Sn

Tin is able to behave as a reductant towards different elements, and among these right antimony. Elemental antimony is colored in black.

2Sb + + 3Sn  2Sb + 3Sn2+

2Sb 5+ + 5Sn  2Sb + 5Sn2+

This, as mentioned, is confirmed by their respective reduction potential std.:

Sn2+/Sn = - 0,14 V                                               Sb3+/Sb = 0,16 V                                                         Sb5+/Sb3+  = 0.6 V

2Sb 3+ +   3Sn  2Sb ° + 3Sn2+

  2Sb 5+ + 5Sn  2Sb ° + 5Sn2+

Tin recognition

As the tin is annoying for antimony research so antimony is annoying for tin research. We have to get rid of antimony exaclty as we did to identify cadmium in presence of copper (treatment with dithionite).

First, however, the solution must surely be "cleaned" from sulfide ions S2-. Then, previously, we heat the test tube in a boiling bain-marie, to facilitate the removal of the gas H2S. Also in this case its' performed a control with a piece of paper soaked with lead acetate. As long as H2S comes out from the test tube, the paper will turn black because of the formation of PbS.

The sulfide could indeed be easily oxidized to elemental sulfur, and that's why it is easily removed

  • Treatment with iron filings and HgCl2

Iron has actually a double utility. It reduces the antimony (+3 or +5) to metallic antimony and tin (IV) to tin (II). It's rightn Sn2+ that we use to carry out the specific recognition essay.

 E°Sn+4/Sn2+ = 0.15 V                                      E°Fe2+/Fe° E ° = -0.44V                                   E°Sb3+/Sb = 0.16 V                            E°Sb5+/Sb3+  = 0.6 V

Therefore treating with iron filing, heating and stirring:

Sn4+ + Fe  Sn2+ + Fe2+

It reduces all the Sn4+ to Sn2+. (Theoretically, the iron may reduce the tin(IV)to tin metal, but this does not happen because we are in concentrated hydrochloric solution).

2Sb3+ + 3Fe 2Sb°↓ + 3Fe2+

Antimony, if present, gives a black precipitate, which is separated by centrifugation. Very quickly we add then mercuric chloride, HgCl2 (tin (II) is a very sensitive species, which is easily oxidized to tin (IV) by atmospheric oxygen)

Sn2+ + 2HgCl2 Sn4+ + Hg2Cl2  + 2Cl-

Sn2+ + Hg2Cl2 Sn4+ + 2Hg↓ + 2Cl-

The reaction, as you can see, is based on the reducing properties of Sn2+. The mercuric chloride has to be the limiting reagent, in defect. Thus, we are sure that the reduction wil continue up to metallic mercury, black.

In any case, if got just a white precipitate, we could verify whether it is or not the mercurous chloride. The reaction is very well known, because it is applied to identify sublimated mercorous chloride. We isolate the precipitate, and we treat it with few drops of NH3, induces the dismutation of the chloride in mercuric amidochloride and elemental mercury, black.

* The tin in solution on whichwe perform the essays should be all tin (IV). This is because the treatment with ammonium polysulphide (carried out before) oxidizes tin (II) to tin (IV):

SnS + (NH4)2S2(I) SnS32-+ 2NH4+