EXPERIMENT
Experiment to determine the concentration of iron(II) by manganate (VII) in an acidic solution
Introduction:
Oxidation is defined as the loss of electrons while reduction is defined as the gain of electrons by an element. A laboratory technique called redox titration is used to determine the analyte concentration by carrying out a reaction between the analyte and the titrant. It depends on the exchange of electrons between two reacting species.
Potassium manganate is a good oxidizing agent, it turns from a deep purple ( manganate (VII)) to colourless (manganese (II) ion).
Ions of iron (II) are pale green in colour. When they are oxidized by manganate (VII) ions from iron (II) to iron (III), it turns to pale yellow colour. When the potassium permanganate solution is titrated against the iron (II) the change in colour of potassium permanganate from purple to colourless shows the endpoint of redox titration.
Reaction:
\[MnO_{4}^{-} \; + \; 8H^{+} \; + \; 5Fe^{+2} \; \rightarrow \; Mn^{+2} \; + \; 4H_{2}O \; + \; 5Fe^{+3}\]
Procedure:
- Weight out an iron sample and put it in a beaker
- Add 100mL of deionized water and heat the solution to dissolve it.
- Allow to cool the solution and transfer the solution to 250mL volumetric flask and make it up to mark.
- Prepare 1000mL of 0.01 M potassium manganate (VII) solution and dissolve it in 10mL of 0.2M sulphuric acid.
- Pipette out 25mL of potassium manganate solution into a conical flask.
- Add iron (II) solution into a flask from a burette.
- Record the reading when the purple colour of the solution is decolourized.
- Repeat the process until concordant readings are obtained.
Observations and calculations:
Mass of potassium manganate (VII) used = M1 = ________
Concentration of standard potassium manganate solution = M2 = ( M1 / 158.032) = ________
Volume of potassium manganate (VII) used for titration = V1 = ________
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Sr # |
Initial Volume (Vi) |
Final Volume (Vf) |
Volume=(Vf-Vi) |
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1 |
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2 |
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3 |
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Mean Volume V = ________
Moles of potassium manganate (VII) used = N1 = M2 X V1 = _______
Moles of iron (II) = N2 = 5 X N1 = ________
This is in V=_______, therefore in 250mL = N = (250 / V ) X N2 = ________
Therefore the mass of iron in 250mL = 56 X N = ________
Result:
The mass of iron in the sample = ________