Devise a volumetric procedure to determine the percentage of Iron(II) and Iron(III) in a mixture containing both.

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M. Usman Ali        (71D)

THE ESTIMATION OF IRON(II) AND IRON(III) IN A MIXTURE CONTAING BOTH

Aim: 

To devise a volumetric procedure to determine the percentage of Iron(II) and Iron(III) in a mixture containing both.

Background/Planning: 

Iron is a transition element, and all transition elements may be found in a variety of oxidation states, for example iron exists as both, Iron(II) and Iron(III). These kinds of elements can react with both oxidising and reducing agents due to the fact that they can be converted from one oxidation state to another.

Working out the percentage composition of both of the Iron ions will require two separate titrations. One of the titrations will react with only one of the ions however the other will react with the mixture as a whole. To be able to carry out a titration in which of all the mixture will react, a preliminary reaction will need to be done. Potassium Manganate (KMnO4) is an oxidising agent and will react with Iron(II) but will not react with Iron(III). This can be therefore used to work out the percentage of Iron(II) in the solution. To work out the percentage composition of Iron(III) a separate reaction will need to be carried out first. The Iron(III) will first need to be reduced to Iron(II) by reacting the solution with Zinc. The reaction can then be titrated with Potassium (KMnO4). The result can then be used to work out the new percentage of Iron(II) present. From the results of the two titrations, the percentage composition can be found.

I am told that the solution provided contains between 1.1g and 1.3g of iron ions as a mixture of Fe2+ (aq) or Fe3+ (aq). I have therefore decided to use the average mass of 1.2g.

The approximate concentration of the iron solution is :-

No. of Moles of Iron         = Mass/RMM of iron

                                = 1.2/55.8

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                                          = 0.0215

Therefore the concentration of Iron is :-

Concentration                 = (No. of Moles x 1000)/Volume

                                  = (0.0215 x 1000)/25 cm3

                                  = 0.86 mol dm -3

The concentration of the second titration in which the Iron(III) is reduced to Iron(II)

could be calculated using the approximate concentration of iron solution and the mole ratio found in the equation. For the first titration in which the Iron(II) is reacting an approximate percentage composition ...

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