Sunday, November 21, 2010

Semi-micro analysis of cations in an inorganic substance

To identify cations the solid salt needs to be dissolved in a soluble solvent.To find the soluble solvent it is tried to solve 5mg samples in the following solvents. As soon as a solvent that the substance can be dissolved is found further steps are not carried out.

Firstly the small sample is dissolved in distilled water.If the substance dissolves in water there is no need to use any acids. Thus chemicals are saved so is the money used for them and the chemical waste is reduced.

If it cannot be dissolved in water then 5mg samples are tried to dissolve in 4 mol/dm^3 / dil. HCl ,4 mol/dm^3 / dil.HNO3 , Conc. HCl, Conc. HNO3 in the respective order. Dil. HCl is added because some salts that are not soluble in water are soluble in acidic medium. But the Chlorides in the Silver group are not soluble and if Silver group cations are present they may not dissolve. Dil. HNO3 is much stronger acid than dil.HCl thus it may dissolve the salts that HCl doesn't dissolve. In Conc. HCl the salts that do not dissolve in dil. HCl forms complexes with the excess of Cl- anions present, and dissolve. Conc. HNO3 is much stronger acid than conc. HCl and may dissolve more salts.Apart from being more acidic both dil. and conc. HNO3 are oxidizing agents. Thus they may oxidize salts and dissolve them too. And both HCl and HNO3 removes hydroxo and ammonia complexes from the inorganic substance.HNO3 in particular dissolves insoluble sulfides by oxidizing sulfide ion to sulphate ions , thus dissolving the given compound.First it is tried to dissolve in the cold solvent and then heated and tried to dissolve.

When the appropriate solvent is decided 30 mg of the substance is dissolved in a portion of 3 cm^3 of the solvent.

Silver group and Calcium group cations are tested using the general scheme of cation group separation as below.

A portion of 1 cm^3 of the above prepared solution is taken and a few drops of 4 mol/dm^3 is added.THe Silver group cation chlorides are precipitated in this step.

AgCl, PbCl2 , Hg2Cl2

More HCl is added from the same solution until no more precipitate is formed. The precipitate is filtered and the filtrate is kept for further analysis.

Silver group analysis (Ag+ , Pb2+ , Hg2 2+)

The precipitate is washed with 10 drops of  2 mol/dm^3 HCl solution so that the precipitate is formed well and does not go away from the solid system. No further HCl is not added because the silver group anions forms complexes in excess Cl- such as [AgCl2]- and [PbCl4]2- which are soluble and thus we won't be able to keep Silver group cations in the precipitate.

A portion of 2 cm^3 of water is added to it boiled , stirred and centrifuged and separated while hot.

AgCl  1.6 x 10−10  PbCl2 1.17 x 10−5  Hg2Cl2 1.1 x 10−18

When boiled the solubility product of the each of these precipitates increases. But the Ksp value of PbCl2 is much higher than the other two(about 10 5 times! ) Thus Pb2+ ions dissolves and when centrifuged they are in the solution while Ag+ and Hg2 2+ ion s are still in the residue.

Confirmation test for Lead

To the hot solution 1 drop of 2% of KI solution is added.
Pb2+ (aq) + 2I- (aq) --------> PbI2 (s)

The yellow precipitate of PbI2 (s) , crystalline in silky plates indicates the presence of Pb2+ cations in the initial substance.

If more I- is present [PbI4]2-(aq) is formed and PbI2  may dissolve 

The residue with the rest of the silver group cations (IF present!) is washed with hot water so that if more PbCl2 are left in the precipitate it dissolves in the solution.And the wash is tested with K2Cr2O7 solution until no Yellow precipitate is formed. Here the precipitate formed is PbCrO4 (s) 

Cr2O7 2- (aq)  + H2O (l)   <-------------> 2CrO4 2- (aq) + 2H+(aq)

CrO4 2- (aq) + PbCl2 (s) -------------->  PbCrO4 (s)  + 2Cl-(aq)

So it is made sure that Pb2+ ions are removed from the solution so it may not interfere with further tests. 
Then a portion of 1 cm^3 of 4 mold/dm^3 of NH4OH is added. It is warmed, stirred and then centrifuged.

NH4OH is added so that AgCl(s) is dissolved as [Ag(NH3)]+ (aq) and Hg2Cl2(s) is precipitated as HgNH2Cl (s) white precipitate and Hg(s) silver precipitate which looks black because it is not clean.

AgCl(s) + NH3(aq) ---------------> [Ag(NH3)] + (aq) + Cl- (aq)

Confirmation test for Hg2 2+

The above black precipitate

Hg2Cl2(s) + 2NH3(aq) --------------> HGNH2Cl (s) + Hg(s) _NH4(aq) + Cl- (aq)

Confirmation test for Silver

To a part of the solution 1 drop of 2% KI is added. A yellow precipitate indicates the presence of Ag in the initial substance.

I- (aq) + [Ag(NH3)]+ (aq) ------------> AgI(s) + NH3(aq) 

Calcium group analysis (Ca2+ , Sr2+ , Ba 2+ ) and Pb2+

1 comment:

  1. Shenya, Hope you are doing fine and do your studies well. I (try to)read your posts. :)They are far away from me since I have very little knowledge on science subjects.
    Could not write anything in my blog due to some difficulties. But will start writing again.
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    Best of luck.