The Estimation Of Iron(II) And Iron(III) In A Mixture Containing Both

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Claire Freer

A2 Assessed Chemistry Practical

January 2003

The Estimation Of Iron(II) And Iron(III) In A Mixture Containing Both

Iron is a transition element; a transition element is an element that forms at least one ion with a partially filled d subshell.  All transition elements may be found in a variety of oxidation states, for example iron exists as both as Iron(II) and Iron(III).  Transition elements can react with both oxidising and reducing agents.  An oxidising agent will convert Iron(II) into Iron(III), the iron will lose an electron.  A reducing agent will convert Iron(III) into Iron(II), the iron will gain an electron. #

Working out the percentage composition of the iron ions will require two separate titrations.  One of the titrations will react with only one of the ions and the other to react with the whole mixture.  To be able to carry out a titration in which all of the mixture will react, a preliminary reaction will need to be done.

From previous As practical work I know that Potassium manganate is an oxidising agent and will therefore react with Iron(II) but will not react with Iron(III).  This can therefore be used to work out the percentage of Iron(II) present 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).  This can be done by reacting the solution with Zinc. This mixture can then be titrated with Potassium Manganate.  The results 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.

We are told that the mass of iron present in the solution is between 1.1g and 1.3g.  I have therefore decided to use the average mass of 1.2g.

The approximate concentration of the iron solution is:

  • Moles of Iron                     = Mass

                                         Mr

                            = 1.2_

                                 55.8

= 0.0215

  • Concentration of Iron                 = Moles x 1000

                   Volume

= 0.0215 x 1000

                       200

= 0.1075 mol dm-3

I have also decided that for this experiment I wish to use the same concentration of Potassium Manganate for each of the titrations.  The concentration of the second titration in which the Iron(III) is reduced to Iron(II) can be easily calculated by using the approximate concentration of the iron solution and the mole ratio found in the equations.  For the first titration in which only the Iron(II) is reacting an approximate percentage composition will need to be found which will allow the same concentration of Potassium Manganate to be used

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Concentration of MnO4- required for Second Titration:

  • 1/5 concentration of Iron solution:                0.1075

(Using mole ratio from equation, pg3)             5

                                =        0.0215 mol dm-3

From rough calculations I know that a 30 – 70% percentage composition, where 70% is Iron(II) will give rise to a concentration of Potassium Manganate which is similar to the concentration calculated for the second titration.

Concentration of MnO4- required for First Titration:

  • 70% of total Iron solution concentration:        0.1075 x 0.7

=        0.07525

  • Concentration of MnO4:                        0.07525

      5

                                        =        0.0151 mol dm-3

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