Question about thiosulphate and iodine

[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
SO₃^2- + I₂ + H₂O will gives you SO₄^2- + 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 SO₄^2- (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 SO₃^2-).

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.

 

[GCT]

Is it asking for the theoretical amount?

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

 

[Borek]

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 let's you calculate amount of sulfite that reacted with iodine.

 

[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?

 

[Borek]

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.

SO₄^2- an SO₃^2- - sulfate and sulfite.

 

[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.

 

[Borek]

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

 

[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.

 

[Borek]

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

 

[crays]

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

 

[Borek]

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

 

[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.

 

[Borek]

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.

nono, we don't titrate it with thipsulphate

You wrote we do in your first post.

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 ...

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.

 

[crays]

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

 

[Borek]

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.

 

[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?

 

[Borek]

Also what's anhydrous sodium thiosulphite added to aqueous iodine for?

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.

 

[GCT]

@_@ 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?

It was sulphite and thiosulfate.

 

[GCT]

It was sulphite and thiosulfate.

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.

 

[Borek]

Iodine is supposed to be in excess

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.

 

[crays]

Thanks and sorry. Now i understand how it should be already. The sodium thiosulphate i titrate with is the extra amount that the sulphite ions didn't use.

 

[Borek]

To be precise: sodium thiosulfate you titrate with should react with the extra amount of iodine.