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Homework Help: Equilibrium reactions with iodine

  1. Aug 4, 2008 #1
    1. The problem statement, all variables and given/known data

    this question is contained within the equilibruim area of study.

    in a practical activity, we were supposed to 1/3 fill a vessel with iodine, and then add NaOH(aq) - when this was done, the yellowy colour of the iodine completely diappeared.

    we then added the same amount of H2SO4(aq) - i think it was three drops or something - to the same vessel, and the iodine colour returned.

    now, we were not given equations for these reactions - the point is that we were supposed to find the equation.
    so I must be present as I2(l), correct?
    and i know the formulae of the other reactants, but the equations i am not so sure about.

    is the I2 species actually involved in a reaction with NaOH/H2SO4?? i would have thought so, but word around the class is that it is somehow acting as an indicator instead.
    is this right?

    2. Relevant equations

    i don't really know what the equation should be, but i can't think of how it could be written with I2 as an indicator.
    so i was thinking maybe this, if it's even possible...

    4I2(l) + 2NaOH(aq) ---> 2I3-(aq) + 2NaI(aq) + 2OH-(aq) + H2SO4(aq) ---> Na2SO4(aq) + 2H2O(l) + 4I2(l)

    (obviously there should be double-sided arrows in place of the normal arrows)

    is this even remotely right?
    i would be really grateful if you could help me out, i'm really confused with this prac.

    3. The attempt at a solution
  2. jcsd
  3. Aug 4, 2008 #2


    User Avatar

    Staff: Mentor

    NaOH and H2SO4 used, so obviously reaction is pH dependent.

    Do you know how hypochlorites are made? (and it is no mistake that I ask about chlorine compounds; remember that to some extent chemistry of all halogens is similar). Chlorine bleach?
  4. Aug 5, 2008 #3
    um, no i don't really know much about hypochlorites...:(
    do you mean in this case something like IOH being formed??

    so then the equation could be:
    2I2(l) + 2NaOH(aq) ---> 2NaI(aq) + 2IOH(aq)
    2NaI(aq) + 2IOH(aq) + H2SO4(aq) ---> Na2SO4(aq) + 2H2O(l) + 2I2(l) ???

    oh and i don't really understand what you mean by "the reaction is pH dependent"?
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