Core practical 11 - Find the amount of iron in an iron tablet using redox titration

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8 Terms

1
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Combine the two half-equations given below to write the equation for the reaction:

Fe2+(aq) → Fe3+ + e−

MnO4− (aq) + 8H+ (aq) + 5e− → Mn2+(aq) + 4H2O(l)

5Fe2+ + MnO4– + 8H+ → 5Fe3+ + Mn2+ + 4H2O

2
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Use your average titre to calculate the number of moles of manganate(VII) ions that were used in the titration.

Average titre = 21.40 cm3

moles = concentration × volume

1000 moles = 0.005 × 21.40/1000 = 0.000107

moles of MnO4– = 0.000107 moles

3
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Use the equation to calculate the number of moles of iron(II) ions in the 25 cm3 sample of iron(II) sulfate from the iron tablet.

Moles of iron = 0.000535

4
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Calculate the number of moles of iron(II) ions in the 250 cm3 graduated flask at the start of the experiment.

Moles of iron in the 250cm3 graduated flask = 0.00535

5
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Calculate the mass of iron in the original five iron tablets, and hence the mass of iron in one iron tablet. Mr Fe = 55.8 g mol–1

0.00535 moles = 0.3g (or 300mg in five tablets or 60mg in one tablet)

6
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Compare your value for the mass of iron with the information from the supplier about the composition of each iron tablet.

The answer to this question will depend on your results but generally the published result for an iron tablet is 65mg of iron per tablet.

7
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Make a list of any procedural errors. Suggest ways in which these errors can be avoided.

Procedural errors:

● Stirring may not be sufficient to ensure that all the iron dissolves – warming the solution may help.

● Transfer of the solution and filtering – ensure that the beaker and the filter paper are rinsed with water.

● The solution may not be mixed – invert the volumetric flask several times to ensure thorough mixing.

● Glassware measurements may not be read accurately – read glassware marks from the bottom of the meniscus.

● The end point may not be clear – use a white tile so that you can see the end point clearly

8
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Calculate the percentage measurement uncertainty for the burette.

0.05/21.40 × 100 = 0.23%