Is Iodine Acting as an Indicator in Equilibrium Reactions with NaOH and H2SO4?

AI Thread Summary
The discussion focuses on the role of iodine in equilibrium reactions with NaOH and H2SO4. Participants observed that iodine's yellow color disappeared upon adding NaOH and returned with H2SO4, prompting questions about iodine's involvement in the reactions. There is uncertainty about whether iodine acts as a reactant or simply as an indicator. Proposed reaction equations suggest iodine may form iodide and hypoiodite species, but clarity on the correct chemical equations is lacking. The conversation highlights the pH dependency of the reactions and the need for further understanding of the chemistry involved.
jason1989
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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

 
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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?
 
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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