Redox Titration Experiment
Experiment to Determine the Concentration of Iron (II) by Manganate (VII) in an Acidic Solution
Oxidation is the loss of electrons, while reduction is the gain of electrons by an element. Redox titration is a laboratory technique used to determine the concentration of an analyte by reacting it with a titrant, exchanging electrons between the two species.
Potassium manganate is a strong oxidizing agent, changing from deep purple (manganate (VII)) to colorless (manganese (II) ion) when it oxidizes another substance.
Iron (II) ions are pale green in color and, when oxidized by manganate (VII) ions, change to iron (III), which is pale yellow. The end point of the titration is marked by the color change from purple to colorless.
Chemical Reaction:
MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
Procedure
- Weigh out an iron sample and place it in a beaker.
- Add 100 mL of deionized water and heat the solution to dissolve it.
- Allow the solution to cool, then transfer it to a 250 mL volumetric flask and fill it up to the mark.
- Prepare 1000 mL of 0.01 M potassium manganate (VII) solution, dissolve it in 10 mL of 0.2M sulfuric acid.
- Pipette out 25 mL of potassium manganate solution into a conical flask.
- Use a burette to add the iron (II) solution into the flask.
- Record the reading when the purple color of the solution is decolorized.
- Repeat the process until concordant readings are obtained.
Observations and Calculations
The following data should be recorded during the experiment:
| Sr. # | Initial Volume (Vi) (mL) | Final Volume (Vf) (mL) | Volume (Vf - Vi) (mL) |
|---|---|---|---|
| 1 | |||
| 2 | |||
| 3 |
Mean Volume V = ________
Moles of potassium manganate (VII) used: N1 = M2 × V1 = ________
Moles of iron (II): N2 = 5 × N1 = ________
In 250 mL: N = (250 / V) × N2 = ________
Mass of iron in 250 mL: 56 × N = ________
Result:
The mass of iron in the sample = ________
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