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Standardization of H2O2 Solution

  1. Dec 10, 2004 #1
    Standardization of H2O2 Solution
    H2O2 (aq) + 2 I-(aq) + 2H+(aq)------> I2 (aq) + 2 H2O(l)
    2 S2O32-(aq) + I2 (aq) à 2 I-(aq) + S4O62-(aq)

    i wanna know
    1.why the volume of S2O32- used in trial 2 is more than trial 1?
    2.What is the use of Na2S2O3 (aq) in Part I?
  2. jcsd
  3. Dec 10, 2004 #2


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    Well, the reaction is a classical one, with sufficient precision for most purposes. Hydrogen peroxide is the oxidant here, oxidizing iodide to iodine, which is very easy to observe from disappearance of brownish red color of iodine. In the second reaction, a radicalic coupling of two thiosulfate anions is present.

    If you give more info about what trial 1 and 2, and also Part I mean, maybe we can be more helpful to you.
  4. Dec 11, 2004 #3
    for part 1
    Rinse a burette with deionized water and then with 0.05 M Na2S2O3 solution.
    Record the initial burette reading in Table 1.
    Pipette 1.00 cm3 of the ~0.8 M H2O2 solution into a clean 125 cm3 conical flask.
    Measure 25 cm3 of deionized water with a 50 cm3 measuring cylinder. Pour it into the conical flask.
    Measure 10 cm3 of 2.0 M sulphuric acid with a 10 cm3 clean measuring cylinder. Pour it into the conical flask.
    Add 1 g of solid KI (record the exact mass) and 3 drops of ammonium molybdate catalyst into the conical flask.
    Swirl the solution mixture until the KI dissolves.
    Titrate the reaction mixture in the conical flask with the sodium thiosulphate solution until
    it just turns pale yellow.
    Add 3 drops of freshly prepared starch solution to the conical flask.
    Continue to titrate the reaction mixture until it just changes from dark blue to colorless.
    Record the final reading in Table 1.

    trial 1 i got 31.2 cm3
    trial 2 i got 33.4 cm3
    i wanna know why the reading of trial 2 is larger?
  5. Dec 11, 2004 #4


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    your difference in titrating seems to come from here; in trial 1, you may have determined the color change just on time, but in the second, you may have missed the point. It is generally recommended that you do a third final titration and calculate their average will give you the sufficient titer, okay for most purposes.
  6. Dec 11, 2004 #5
    but all of my classmates got the same result...so i think the increase of the reading of trial 2 maybe not related to the problem you mentioned
    um....will it related to H2O2???
  7. Dec 11, 2004 #6


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    Do you rinse the conical flack (erlenmeyer as we call it) after each trial? Any remaining iodine will cause a positive error. I recommend washing the flask with deionized water+thiosulfate+deionized water and drying prior to titration.
  8. Dec 11, 2004 #7
    but....i used another conical flask in trial 2
  9. Dec 11, 2004 #8
    i also want to know What is the use of Na2S2O3 (aq)???????
  10. Dec 12, 2004 #9


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    Thiosulfate here is the unique reactant for iodine; with this, we indirectly find the moles of hydrogen peroxide. You have written the equations already. If you go back a bit, you'll see that all of them are inter-related.
  11. Feb 6, 2008 #10

    I have some questions on this matter too.. The labscript said that I dont need to calculate the concentration of H2O2 in mol/dm, but just use the volume of thiosulphate that i used in the titration.

    Let say , i used 40ml of thiosulphate, and from the equation 1 mole of I2 react with 2 moles of thiosulphate. DOes it mean the concnetration of H2O2 is twice as the volume of thiosulphate or half of the volume of thiosulphate?thanks.
  12. Feb 6, 2008 #11


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    Neither. It means that you will need two moles of thiosulfate to react with one mole of I2. (moles=molarity*volume(L))
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