experiment to find the concentration of the ethanedioic acid

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A2 Chemistry

Experiment 1 – Redox Titration

As this is a redox reaction there are two half equations:

5C2O42-  -10e- ➔ 10CO2              and                     2MnO4- + 16H+ + 10e- ➔ 2Mn2+ + 8H2O

Combining these 2 half equations we get:

5C2O42-  +  2MnO4-   +   16H+ ➔ 10CO2 + 2Mn2+ + 8H2O

Therefore to find the concentration of the ethanedioic acid I can titrate with Potassium Permanganate which acts as an oxidising agent, with Ethanedioic acid acting as a reducing agent. The sulphuric acid will act as an acid catalyst, providing extra H+. H2SO4 IN EXCESS

The  Sulphuric acid is in excess to ensure all the ethanedioic acid will react with the KMnO4 until the equivalence point is reached. I will use a 0.05mol dm-3 concentration of KMnO4.

Method:

  • Set up titration apparatus, and fill the burette with 0.05M Potassium Permanganate, fill up to zero on the burette, use a funnel and place apparatus on a stool if necessary.
  • Ensure that the acid mixture is properly mixed by shaking. Then fill a graduated pipette with 25cm3 of the acid mixture and pipette this into a 250cm3 conical flask.
  • You will not need to add any indicator as KMnO4 is self indicating; the end point has been reached when a permanent pale pink colour appears.
  • This reaction will only take place above 60 degrees Celsius, therefore heat the conical flask before titrating to a little over 60 degrees, using a Bunsen burner. Use an insulator to hold the conical flask when heated.
  • Carry out the titration as normal, ensuring that temperature is above 60 degrees using a thermometer. If needed stop titration and heat until over 60 degrees again and then continue. Repeat enough times until you have at least 2 consistent results within 0.1cm3 of each other, which strengthens reliability of results. Ensure you go drop wise as you approach the end point.
  • Record your results in a suitable table as shown below.
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Work out concentration of the ethanedioic acid

Now that I Know the volume of KMnO4 that was required for the redox reaction, I can work out the number of moles of KMnO4 

Concentration x Volume = Moles, therefore 0.05 x average titre=moles of KMnO4

From the equation I know that the Molar ratio between KMnO4 and H2C2O4  is:

5C2O42-    2MnO4-    , therefore 5:2 ratio. So if I divide moles of KMnO4 by 2 and then multiply by 5 this gives the moles of H2C2O4. And finally to get the concentration of H2C2O4 simply do moles/volume, so moles of ...

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