[kemistrykid]

Hey guys, I'm a 2nd year chemistry student and I am writing up my lab report for an andrews' titration experiment. It goes something like this:

Make standard solution (called solution "S"): 25ml of 0.1M Sodium thiosulphate, 25ml of 0.05M I₂ in aqueous NaI (approx. 0.15M in NaI), dilute up to 250ml.

This is the reaction:
2S₂O₃^2- + I₂ (+NaI) ---->  2I^- + S₄O₆^2-

Now the titration: Take 25ml of "S", add 25ml of H₂O, 25ml of conc. HCl and 2ml of chloroform, and titrate against 25ml of 0.05M KIO₃.

This is the reaction:

IO₃^- + 6H⁺ + 4e^------>I⁺ + 3H₂O

Also, since this reaction is happening in a vast excess of hydrochloric acid, this reaction also happens:

I⁺ + 2Cl^------->ICl₂^-

OK, my question is, I have to work out the reaction ratio for S₄O₆^2-/IO₃^-. I have worked it out to be 1:7, but I can't work out the exact equation for this mechanism, namely because I am not sure how they react.

Any ideas?

[Borek]

IO₃^- + 6H⁺ + 4e^------>I⁺ + 3H₂O

Something is wrong here.

IO₃^- + 5I^- + 6H⁺ -> 3I₂ + 3H₂O

[kemistrykid]

Something is wrong here.

IO₃^- + 5I^- + 6H⁺ -> 3I₂ + 3H₂O

No, its definitey what I put. Thats the whole point of this experiment, well, at least one of them. We go from I^------>I₂------>I⁺

[Borek]

I don't get whole experiment setup. It doesn't make chemical sense.

What do you use iodine for at the very beginning?

You add thiosulfate to reduce iodine to iodides - at least that's what your first equation suggests. But you have also added iodides to the S solution. Why do you need two sources of iodides?

Usually iodideds are added to iodine solution to inxrease I₂ solubility (in form of I₃^-). But you don't add thiosulfate in such situation, as it will reduce iodine. You may use thiosulfate as a titrant, to determine amonut of iodine, but that's completely different story.

And the reaction equation I have posted is definitely the one used for determination od iodides with iodate solution in iodometry. There is no single reason for other reaction to take place in the solution you described.

[Albert]

Writing all these equations was what I call a time-consuming process.  

IO₃^- + 5I^- + 6H⁺ -> 3I₂ + 3H₂O

6S₂O₃^2- + 3I₂ ->  6I^- + 3S₄O₆^2-

IO₃^- + 5I^- + 6H⁺ + 6S₂O₃^2- -> 6I^- + 3S₄O₆^2- + 3H₂O


To answer your question, the ratio is 1:6

[Borek]

Writing all these equations was what I call a time-consuming process.

Not with proper tools  

IO₃^- + 5I^- + 6H⁺ + 6S₂O₃^2- -> 6I^- + 3S₄O₆^2- + 3H₂O

Huh? It is not a correct reaction equation.

[Albert]

IO₃^- + 6H⁺ + 6S₂O₃^2- -> I^- + 3S₄O₆^2- + 3H₂O


What do you think about it, now? However, I agree with you: the first two equations sound more correct to me and they show the ratio, while this one is maybe a faster :uhhuh: way to display the ratio.

[Borek]

Now it is correctly balanced and obey rules. The question is whether it describes real reaction that takes place in the solution. I doubt.

Besides, I still don't know what was the lab procedure, let's wait for kemistrykid.

[Albert]

I don't get whole experiment setup. It doesn't make chemical sense.

Have a look at page 36:
http://www.nzfsa.govt.nz/science/research-projects/iodine-fort/iodine-fort-foods.pdf

[Borek]

I don't see there anything explaining experiment as described by kemistrykid, besides, I was aware of the chemistry described.

Solution S doesn't make sense to me. Either it have to contain iodides (if so - what iodine and thiosulfate are used for?) or iodine (if so - what iodide and thiosulfate are used for?) or thiosulfate (if so - what iodine and iodides are used for?). Produced S₄O₆^2- is of no use too, as it will be not titrated by IO₃^- - so what is this solution for?

[kemistrykid]

The S₄O₆^2- is directly titrated against the KIO₃ solution. I've worked out the reaction ratio is 7:1 using the number of moles used. Strange thing is some people have been getting 7:2. O.o

[Albert]

The S₄O₆^2- is directly titrated against the KIO₃ solution.

This is not consistent with what you wrote in the opening post.

Could you write the chemical equation that describes the reaction behind the titration of S₄O₆^2- against KIO₃?

[kemistrykid]

I said titrate solution S (i.e. S₄O₆^2-) against the KIO₃. Since I worked out the reaction ratio to be 7:1.

[kemistrykid]

OK, the next thing is that i add 2ml of 0.5M barium chloride to one of the titrated solutions and it turns yellow and then cloudy. I do the same to a solution of 0.25ml KIO₃ and 10ml of conc. HCL, diluted to 40ml. It turns yellow, but not cloudy.

Apparently I can work out the oxidation half-reaction in the tetrathionate/iodate reaction, and the overall reaction, but I can't quite see how...

[Borek]

What reaction do you expect to take place between S₄O₆^2- and IO₃^-? S₄O₆^2- is already in oxidized form, it is product of thiosulfate oxidation with iodine.

My guess is that there is some error in your description of the experiment. Perhaps S solution was made without use of iodine, just with iodide? It may make some sense, as iodate plus iodide produces iodine which in turn reacts with thiosulfate.