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Redox Titration, Iodine

  1. Mar 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Bleaching powder, Ca(OCl)Cl, reacts with iodide in acidic medium according to the equation:

    OCl– + 2 I– + 2 H3O– ---> I2 + Cl– + 3 H2O

    How many millilitres of 0.0645 M Na2S2O3 are required to titrate the iodine liberated from a 0.6028 g sample of bleaching powder containing 10.50 % (w/w) Cl ?

    2. Relevant equations

    I2 + 2S203 --> 2I- + S4O6

    3. The attempt at a solution

    I have already submitted an answer for this assignment, and this was one of only 2 questions I got wrong, we can do a second submission so I went to figure this one out.

    I think this is where I went wrong.

    I Figured out the Cl- moles by: g Cl/0.6028 g = 0.01050

    0.063294 g Cl / 35.453 g/mol = 1.78529 mmol

    I think I may have done that wrong, but from that.
    From mol Cl- I used 1:1 ratio of original reaction to get mol I2

    Mol I2 in second reaction are 1/2 mol S203

    mol S2O3 = 3.57058641 mmol

    Then calculated mL needed.

    V = 3.57058641 mmol/0.0645 M
    = 55.4 mL

    Any suggestions?
     
  2. jcsd
  3. Mar 19, 2009 #2

    Borek

    User Avatar

    Staff: Mentor

    None :wink:

    I got 55.3 mL, close enough.
     
  4. Mar 19, 2009 #3
    Hm.. that's odd then. The computer usually gives part marks when we're close.
     
  5. Mar 19, 2009 #4

    Borek

    User Avatar

    Staff: Mentor

    Hm, could be we are both wrong. It is 10.5% of Ca(OCl)Cl. Part of the chlorine is not taking part in the iodine oxidation. I have concentrated on your solution, instead of reading the question :grumpy:
     
  6. Mar 21, 2009 #5
    Would I just divide it by 2? So only 1/2(1.785 mmol Cl) take place in the reaction?
     
  7. Mar 24, 2009 #6
    Ah, possibly I'm going about it the wrong way, can I instead just avoid the w/w of Cl?

    0.6028 g / MW Cl-/ MW Ca(OCl)Cl = mol Cl-

    Does anyone know, I'm really lost in the solution.
     
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