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[Chemistry] Determine the concentration of h2o2 by redox titration!

  1. Mar 24, 2009 #1

    You are provided with a solution of hydrogen peroxide which is known to be approximately ‘100-volume’. Plan two experiments that would enable you to determine the exact concentration of H2O2 in mol dm-3, in this solution. One of your experiments should involve a redox titration with potassium manganate (VII). The other experiment should not involve a titration.

    2. Relevant equations

    2KMnO4 (aq) + 5H2O2 (aq) + 3H2SO4 (aq) --> 2MnSO4 (aq) + K2SO4 (aq) +5O2 (g) + 8H2O (l).

    2H2O2 --> 2H2O + O2

    3. My questions

    Why is KMnO4 it's own indicator?

    What is meant by 100-Volume and how can i put it into context?

    What catalyses the disproportionation of h2o2?

    4. The attempt at a solution

    The end point of this reaction is noted by a pale pink colour. No indicator is required as the solution goes pale pink once all of the H2O2 has reacted. I know this has something to do with the redox chemistry but im not exactly sure.

    100-Volume one i believe this means one volume of h2o2 realeses 100 volumes of oxygen when catalytically decomposed. what does it mean by volume? And how can i use this to determine conc of h2o2 as the second experiment.

    Last edited: Mar 24, 2009
  2. jcsd
  3. Mar 26, 2009 #2


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    Staff: Mentor


    No special chemistry involved - once you add excess of permanganate, its strong color is well visible in the solution.

    Not sure what what you are asking about. 1 mL of solution gives 100 mL of gas.

    Start with known volume of solution, measure volume of gas evolved. MnO2 is a good catalyst, drop of blood will do the trick too, and I think cigarette ash is quite efficient at decomposing hydrogen peroxide as well, not too sure about the last one.
  4. Apr 1, 2009 #3
    Hi Borek, I was wondering if you could help me. I'm doing the titration of H2O2 with KMnO4 in order to determine the concentration of H2O2. I understand that H2SO4 is required in the reaction, now if I dissolve the KMnO4 in H2SO4 does that count? Also, if I dissolve 2.5g of solid KMnO4 in 47.5cm3 of H2SO4. Will this be the resulting concentration?
    2.5/ 50 = 0.05 mol dm-3.

    Also, if I dissolve the KMnO4 in the acid, the number of moles of H2SO4 as required in the reaction (3) will not exist will it because the KMnO4 and the H2SO4 have been combined?

    I appreciate any help! thank you!
  5. Apr 1, 2009 #4


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    Staff: Mentor

    No idea what you mean. If you dissolve permanganate in sulfuric acid you get strong oxidizing mixture.

    Why do you think final volume will be 50 mL? I am almost sure it won't.

    Same problem as above - no idea what you mean. You have mixed these things, that's OK. Why do you think they don't exist any more?
  6. Apr 1, 2009 #5
    I'm sorry, I'm really confused! How would I make a solution of KMnO4 of concentration 0.05 mol dm-3 by dissolving the manganate in H2SO4? How much H2SO4 would I need?
  7. Apr 2, 2009 #6


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    Staff: Mentor

    Take correct amount of permanganate, dissolve in smaller volume of acid, once it is dissolved - fill up to 50 mL. Voumes are not additive and it is very difficult to predict what voume of acid you will need. You can do it sometimes having density tables.

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