In this investigation I have to find the percentage of Iron II and Iron III in a solution containing both and do so by a volumetric method.

Away to turn iron III into iron II is to add zinc powder to the iron II and iron III solution this reduces the iron III by donating an electron to it (chemistry 2, OCR). To test the zinc has reduced the iron III you can add a few drops of the iron solution and add them to some potassium thiocynate. If any iron III is left the thiocynate should turn blood red.

To workout the concentration of potassium manganate VII needed I have to workout the maximum and minimum amount of iron II and iron III in the solution. The minimum amount of iron in the solution is 30% of 1.1g and the maximum is 70% of 1.3g.

1.1_ = moles of iron = 0.0197moles   0.0197   x 30 = min moles of iron = 5.91x10-3

9. 100

The maximum amount of iron possible is: -

1.3  = moles of iron = 0.0233moles.  0.0233 x 70 = max moles of iron = 0.0163

9. 100

Both of the above equations show me that 0.01 moles is between the two amounts of iron so I will use 0.01 M potassium manganate solution to titrate with iron to workout the amount of iron in the solution.

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Diagram

Method

1. Set up the equipment as in the above diagram.
2. Take a 25cm3 sample of the iron II and iron III solution and put it in a conical flask.
3. Titrate the potassium manganate solution with the iron solution until the solution turns a pale pink.
4. Take a note of the amount of potassium manganate solution used and then repeat steps 1 to 3 until you have four results.
5. After you have four results add 1 gram (approx) of zinc powder and rest for 2 minutes.
6. After 2 minutes take a few drops of the solution and add them to 1cm3 of potassium thiocynate if the solution turns blood red wait another 2mins and repeat until no colour change occurs.
7. If no colour change occurs filter the zinc powder from the iron solution.
8. Next take a 25cm3 sample of the filtered iron solution and titrate this with the potassium manganate solution when the solution turns a pale pink take note of the amount of potassium manganate solution used.
9. Repeat step 8 until you have a further four results (eight in total).

Results

MnO4- + 5Fe2+ + 8H+                5Fe3+ + Mn2+ + 4H2O

Moles   x 1000 = Molarity

   Volume

Therefore for Fe2+ : -

Moles x 1000 = 0.01                =                   0.01 x 20.1 = Moles of MnO4- = 2.01 x 10-4

1. 1000

1 mole of MnO4-  makes 5 moles of Fe3+

Then

(2.01 x 10-4) x 5 = Moles of Fe2+ = 0.001005

So

0.001005 x 55.9 = Mass of Fe2+ = 0.0562g in 25cm3 

0.0562 x 8 = Mass Fe2+ in 200cm3 = 0.4496g

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For all the Fe : -

Moles x 1000 = 0.01                =                  0.01 x 34.6 = Moles of MnO4- = 3.46 x 10-4

6. 1000

1 mole of MnO4-  makes 5 moles of Fe3+

Then

(3.46 x 10-4) x 5 = Total Moles of Fe = 0.00173

Therefore

0.00173 x 55.9 = Mass of Fe in 25cm3 = 0.0967g in 25cm3

0.0967 x 8 = Mass of Fe in 200cm3 = 0.7736g in 200cm3

So the amount of Fe3+ in the solution will be : -

Total Amount of Fe – Amount of Fe2+ = 0.7736 – 0.4496 = 0.324g of Fe3+ in the iron solution.

The percentage of each of the Fe2+ and Fe3+ are : -

Fe2+= 0.4496 x 100 = 58%  Therefore Fe3+ = 42%

          0.7736