1. Sep 12, 2008

crays

Hi, i'm doing an experiment that uses iodine and sodium thiosulphate and also sodium sulphite.

First, i added 50.0 cm3 of 0.05 mol dm-3 aqueous iodine into a titration flask. Then added 25.0 cm3 of Solution that contain 24.0 g of anhydrous sodium sulphite, Na2SO3 per dm3 into it. Lastly, i added 2g of solid hydrogen carbonate into it. Then the solution is titrated with 0.1 mol dm-3 aqueous sodium thiosulphate.

From the equation given, which is
SO32- + I2 + H2O will gives you SO42- + 2HI

And
2HI + 2HCO3- will gives 2I- + 2H2O + 2CO2

So from what i know is that the aqueous iodine solution reacts with the anhydrous sodium sulphite, Na2SO3 and water will gives a SO42- (is it pronounced as sulphide ions?) and 2 Hydrogen iodide.

The hydrogen iodide then reacts with 2HCO3- (which should be the hydrogen carbonate ions) and gives out 2I- + 2H2O + 2CO2

Am i right so far? Please correct me if i'm wrong.
If i am right, here's the question.

Calculate the volume of I2, that did not react with the sulphite ions (which i think should be SO32-).

In my experiment, what i did is exactly what i mentioned above, the color was first reddish, as i added my aqueous sodium thiosulphate in, the color slowly turns pale yellow, then i added starch as an indicator when it is pale yellow, and titrate until the blue color is gone. By that, it should mean that no more Iodine/Iodide is present in the soultion, so all Iodine should have completely been reacted, how do i solve the problem above?

Thanks, sorry for the long post.

2. Sep 12, 2008

GCT

Is it asking for the theoretical amount?

Edit - You can also find the experimental yield by analyzing the titration data.

3. Sep 12, 2008

Staff: Mentor

This is relatively simple stoichiometry. You must start writing all reaction equations and balancing them. Then amount of titrant used will let you calculate amount of titrated iodine. Knowing how much iodine was left you can easily calculate how much iodine reacted (you are given initial amount) before titration. That in turn lets you calculate amount of sulfite that reacted with iodine.

4. Sep 12, 2008

crays

Thanks for the reply, from my experiment, 50.0 cm3 of aqueous iodine reqiures 31.88 cm3 of aqueous sodium thiosulphate for a complete reaction. A complete reaction, doesn't that mean there is no more iodine left?

Also, can someone tell me the difference between sulphate and sulphite? is sulphite the ions form of sulphate? or is it a totally different thing?

5. Sep 12, 2008

Staff: Mentor

Complete reaction means either that one of the reagents was consumed completely (but the other one was in excess) or that both reagents were in exactly stoichiometric amounts. It may depend on the context. In both cases at least one of the reagents is gone and reaction can't proceed further.

SO42- an SO32- - sulfate and sulfite.

6. Sep 12, 2008

crays

I've got a bit of clue on how to solve it now, but not sure if i'm on the right track.
Since iodine is in excess, so i used 0.05 mol dm3 multiply by 50/1000 (from cm to dm) to get its number of mols.

Also the number of mols for the anhydrous sodium sulphite, which is 24/126 x 25/1000 which gave me 0.00475 mol.

Am i on the right track? cause anhydrous sodium sulphite have more number of mol than iodine.

7. Sep 13, 2008

Staff: Mentor

Approach seems correct, but you are right that numbers look wrong. Are you sure about amounts and concentrations?

8. Sep 13, 2008

crays

Thanks for the reply, yes, those are the given values. But i think again, since 1 part iodine reacts with 1 part of thiosuphite ions, and 50 cm3 of iodine is used and 25 cm3 of thiosulphite is used, can i say 25 cm3 of iodine is left? Since iodine is in excess.

9. Sep 13, 2008

Staff: Mentor

According to your first post iodine reacts with excess sulfite first.

10. Sep 13, 2008

crays

That is what stated in the paper, so there is no iodine left right?

11. Sep 13, 2008

Staff: Mentor

Looks like, thus there is no need to titrate with thiosulfate. There is obviously something wrong with the description.

12. Sep 13, 2008

crays

nono, we don't titrate it with thipsulphate, the procedure is

KA1 is 0.05 mol dm-3 aqueous iodine.
KA3 is a solution containing 24.0 g of anhydrous sodium sulphite, NA2SO3, per dm3.

Pipette 50.0 cm3 of KA1 into a titration flask. Using another pipette, place 25.0 cm3 of KA3 slowly into this titration flask containing KA1 and shake. Add 2g ...

The question is calculate the volume of I2 that did not react with the sulphite ions.
It is already given in the procedure with the values, and the experiment is correct too.

13. Sep 14, 2008

Staff: Mentor

You wrote we do in your first post.

And now you repeat exactly the same information - that we start mixing iodine with EXCESS sodium sulfite (25mL/1000mL*24g = 0.6g of the sulfite). You may check amounts shown on the EBAS screenshot. Numbers displayed in red mean excess.

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14. Sep 14, 2008

crays

Sorry to confuse, but you mean sulphite and sulphate is the same thing?

15. Sep 14, 2008

Staff: Mentor

No. Sodium sulfate is Na2SO4, sulfite is Na2SO3, tiosulfate is Na2S2O3. These are three different substances. Sulfite and thiosulfate react with iodine, sulfate doesn't.

16. Sep 14, 2008

crays

@_@ what's sodium thiosulphate for then? The one i titrate my aqueous iodine with. Also what's anhydrous sodium thiosulphite added to aqueous iodine for?

17. Sep 14, 2008

Staff: Mentor

As long as you don't recognize these substances you are wasting both your and my time. It is sodium sulfite that was anhydrous and that was added to iodine.

18. Sep 15, 2008

GCT

It was sulphite and thiosulfate.

19. Sep 15, 2008

GCT

Here's the summary of what is going on in case you are not understanding it yet

- This is an experiment in back titration

- Iodine is supposed to be in excess , it is first titrated with the sulfite , then the acid Iodide that is formed is neutralized since thiosulfate is consumed by acid , I'm guessing that by " aqueous Iodine " I'm guessing that it is in the form of I3- by dissolution of a gaseous sample of I2 perhaps with KI.

- Thiosulfate titrates the excess Iodine ( in the form of I3- ) , starch is added as an indicator since the solution has become clearer , it is not added with the original solution since the solution is dark. The Thiosulfate solution should be standardized , it is a more accurate reagent than sulfite.

Edit - it is the Thiosulphate titration calculation that yields the number of I2 remaining from the Sulfite consumption.

Last edited: Sep 15, 2008
20. Sep 15, 2008

Staff: Mentor

At least that's what everyone involved (including you and me) is expecting. The only problem being - it is not, as long as the information given by OP is correct.