Chemical Forums
Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: kemistrykid on March 02, 2006, 04:53:42 PM
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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:
2S2O32- + I2 (+NaI) ----> 2I- + S4O62-
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 S4O62-/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?
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IO3- + 6H+ + 4e------>I+ + 3H2O
Something is wrong here.
IO3- + 5I- + 6H+ -> 3I2 + 3H2O
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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+
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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.
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Writing all these equations was what I call a time-consuming process. :D
IO3- + 5I- + 6H+ -> 3I2 + 3H2O
6S2O32- + 3I2 -> 6I- + 3S4O62-
IO3- + 5I- + 6H+ + 6S2O32- -> 6I- + 3S4O62- + 3H2O
To answer your question, the ratio is 1:6
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Writing all these equations was what I call a time-consuming process.
Not with proper tools :P
IO3- + 5I- + 6H+ + 6S2O32- -> 6I- + 3S4O62- + 3H2O
Huh? It is not a correct reaction equation.
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IO3- + 6H+ + 6S2O32- -> I- + 3S4O62- + 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.
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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.
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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 (http://www.nzfsa.govt.nz/science/research-projects/iodine-fort/iodine-fort-foods.pdf)
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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 S4O62- is of no use too, as it will be not titrated by IO3- - so what is this solution for?
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The S4O62- 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
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The S4O62- 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 S4O62- against KIO3?
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I said titrate solution S (i.e. S4O62-) against the KIO3. Since I worked out the reaction ratio to be 7:1.
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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...
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What reaction do you expect to take place between S4O62- and IO3-? S4O62- 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.
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Here is the descripition of the experiment:
Oxidations using the iodate ion (IO3-) give iodine-containing products in differing ocidation states depending on the conditions. In dilute acid, IO3- and I- react in a 5:1 ratio to give I2. In the presence of a large excess of Cl-, (in the form of moderately concentrated HCl), the iodine-containing product is I+, in the form of ICl2, and the reaction ratio differs. Such titrations are know as Andrews' titrations.
Now, I have to work out that reaction ratio (I think its 7:1), and I also have to work out the half-reaction for the tetrathionate/iodate reaction and the overall equation.
It turns out that the results I got for the qualitative results were wrong. Upon adding barium chloride, I should have got a white precipitate, which is a sign that SO42- is present. In the second test, nothing happens, thus showing conclusively that SO42- is present.
I think the half-reaction is something like this:
S4O62- + 7IO3- + 26H+ -----> 4SO42- + 13H2O
I think its balanced...
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Well, if you take EBAS and play long enough you may find something like
2S4O62- + 7IO3- + 7Cl- + 2H+ -> 8SO42- + 7ICl + H2O
but whether it is the reaction you are looking for - I have no idea. And somehow I doubt, as tetrathionate seems to be stable enough to be not oxidized further in analytical solutions, that's why thiosulfate is used in many analytical methods.
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Actually I've just found the answer. The reaction is:
2S4O62- + 7IO3- + 14Cl-+ 2H+ ----> 8SO42- + 7 ICl2- + H2O
Very similar to what you suggested. The key thing here is that overall, S is being oxidised to +3, but is reduced to +3 from whatever oxidation state S is in in S4O62-. :drive1: :ph34r:
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Very similar to what you suggested.
IMHO they are identical - ICl/Cl- and ICl2- is analogous to I2/I- vs I3-.
The key thing here is that overall, S is being oxidised to +3, but is reduced to +3 from whatever oxidation state S is in in S4O62-.
We have discussed very similar problem in HS forum tonight. There are two different kinds of sulfur with different oxidation numbers in this compound. +3 is average and has nothing to do with reality (not that oxidation numbers have anything to do with reality ;) ).
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IMHO they are identical - ICl/Cl- and ICl2- is analogous to I2/I- vs I3-.We have discussed very similar problem in HS forum tonight. There are two different kinds of sulfur with different oxidation numbers in this compound. +3 is average and has nothing to do with reality (not that oxidation numbers have anything to do with reality ;) ).
lol true!