Equilibrium reactions with iodine

[jason1989]

Homework Statement

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?

Homework 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.
:)

The Attempt at a Solution

 

[Borek]

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?

 

[jason1989]

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"?
:):)