3) It is a type of redox titration in which one substance undergo oxidation and

Question

3) It is a type of redox titration in which one substance undergo oxidation and other one gets reduced.

In present case , titration occurs between KMnO4 and oxalic acid, this titration takes place in acidic medium. The reaction involved is as follows

2 KMnO4 + 5 H2C2O4.2H2O + 3 H2SO4 ----->

2 MnSO4 + K2SO4 + 10 CO2 + 18H2O

From this reaction it can be seen 2 moles of KMnO4 reacts with.5moles of oxalic acid.

So 1mol of KMnO4 will react with 2.5mol of oxalic acid.

4) KMnO4 is self indicator in this titration, to avoid side reactions i..e to prevent its conversion into MnO2 which is brown in color interferes with the detection of end point and To observe color change carefully ,it has to be added slowly.

5) KMnO4 is dark purple in color , it becomes colorless as it changes to MnSO4 by oxalic acid .  As all the oxalic acid gets oxidised further addition of KMnO4 does not change color , solution becomes pink.That point is end point of this titration.

Explanation / Answer

It's my third time to post this and ask you guys to type your answers, not to write on paper!

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Please type all answers. I need you to make your own answers

these are general questions, before the experiment.

3. For each mole of KMnOs that reacts in this experiment, how many moles of H C,0 2H2O react? 4. Why should the KMn04 be added very slowly at the beginning of the titration? 5. What is the endpoint of this titration?

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