Determination of iodine value - Hanus method - reaction problem

[tomplo93]

Hello,

Please help me solve the puzzle related to determining the iodine value using the Hanus method. No matter how I try, it always seems to me that something is not working.The method I used:

1. I added 10mL of Hanus reagent (0.1M IBr) to 10 mL of chloroform
2. I mixed everything and put it in a dark place for 30 minutes - after that time I added 10mL of 10% KI solution and 50mL of water.
3. I titrated everything with 0.1M sodium thiosulfate.
This was a blank test - and at this stage a problem arose; it seemed to me that I had used too much titrant - 44mL.

After the analysis, I started to exclude individual elements - I started by checking the concentrations of the reagents to exclude the manufacturer's error.

1. I checked the titer of thiosulfate - I did it in 2 ways to test each reaction environment; a. Titration with potassium chromate
The first stage of the reaction is the conversion of chromate to dichromate, and then the reaction of dichromate with potassium iodide:
2K₂CrO₄ + 2HCl -> K₂Cr₂O₇ + H₂O + 2KCl
K₂Cr₂O₇ + 6KI + 14HCl -> 3I₂ + 2CrCl₃ + 7H₂O + 6KCl

In 200mL of water I dissolved 2.4587g of K₂CrO₄ – a 0.0615M solution was created
I took 10mL of it and added 20mL of 0.5M HCl (a large excess of HCl guarantees the transformation of all chromate into dichromate)
After discoloration (transition to chromate) I added 20mL of KI solution (also a large excess guaranteeing the complete reaction of dichromate)
At this stage in the sample should be about 0.0009225 mol I₂.
After 30 minutes in the dark I titrated the released I₂ with thiosulfate Na₂S₂O₃ according to the reaction:

I₂ + 2Na₂S₂O₃ -> 2NaI + Na₂S₄O₆

I lost about 19 mL of thiosulfate which gives 0.0019 mol of thiosulfate. From stoichiometry for 0.0009225 mol I₂ there is 0.001845 mol Na₂S₂O₃ – so the results matched; the titer of thiosulfate in this method was confirmed.

2. Titration with pure I₂ – I did a separate reaction with I₂ alone (we have 0.1M expired r-r)

I took 5mL of 0.1M I₂ and titrated with thiosulfate

I₂ + 2Na₂S₂O₃ -> 2NaI + Na₂S₄O₆

From the stoichiometry it appears that 10mL of thiosulfate should be used – I used 10.2 mL and 10.3 mL – the results match. The titer of thiosulfate in this method was confirmed. Additionally, I saw a “naked” reaction as it looks in practice without additional elements.


3. Hanus reaction – according to the recipe, you should mix Hanus reagent with chloroform – I decided to give up chloroform (it is for dissolving oils) to get rid of any potential interference. Of course, I didn't add any oil either, to check how the pure reaction works.
I took 5mL of Hanus reagent, which they declare to be 0.1M IBr.
The Hanus reaction is:

IBr + 2KI + KBr + KI + I₂

So in 5mL of IBr there should be 0.0005 moles of IBr – which gives 0.0005 moles of I₂.
I added an excess of KI (approx. 10mL of 10% KI solution) until the IBr reacted completely.
The released iodine is titrated by thiosulfate:

I₂ + 2Na₂S₂O₃ -> 2NaI + Na₂S₄O₆

So for 0.0005 mol of I₂ we should use 0.001 mol of Na₂S₂O₃. With a r-r of 0.1M this is a volume of 10mL.
During the titration I used 22.7 mL and 22.9 mL of thiosulfate – which gives a consumption of 0.00228 mol of Na₂S₂O₃, which translates to 0.00114 mol of I₂ in the sample.
This concentration of I₂ corresponds to 0.00114 mol of IBr in 5mL, which translates to the actual concentration of IBr being 0.228M – more than twice as much as on the label.

4. I used another bottle of Hanus reagent - from a different batch. The result was the same. I have no more ideas what could be wrong. The glass I am using is clean, the reagents are checked, but the stoichiometry does not match. Additionally, using starch itself, I checked whether there was no iodine in the Hanus reagent, which would increase the concentration - the starch did not turn color.

Please help me with this topic.

[ChenBeier]

If I start with the first chapter.

0.1 M IBr should correspond to 0,1 M I₂ according IBr + KI => I₂ + KBr
and this should be 0,2 M Thiosulfate.

I₂ + 2 Na₂S₂O₃ => 2 NaI + Na₂S₄O₆

So if you titrate 10 ml 0,1 M IBr the consumption should be about 20 ml of 0,1 M Thiosulfat or 10 ml of 0,2 M Thiosulfate. But you got 44 ml.

Question mark here. The cleaness of the added KI solution, does it contain already iodine?
Suggestion add fresh solid KI or NaI. Not a solution which is old.
The concentration of IBr you verified already.
The Thiosulfate seems also OK.

[tomplo93]

Thanks for your quick response. I just checked the KI solution (I use a reagent with the purity: pure for analysis). I added a few drops of starch solution to the KI solution - no signs of color.

[ChenBeier]

Did you add also some acid. Hanus solution contains acetic acid.

If still OK then it can only the IBr solution itself it's not 0.1 M.
This you already confirmed you found 0.228 M

I found this 0,1 M IBr is 0,2 N , means mathematical problem. Molarity and difference to normality. Divide by 2 everything looks fine.

https://www.itwreagents.com/rest-of-wor ... sis/281572

[tomplo93]

Hi,
Thanks for your answer!
I have purchased Hanus reagent - I don't make it myself. According to the quality certificate, it is 0.1M IBr (no mention of N) and contains glacial acetic acid and IBr.

I performed the tests on two different bottles from different batches - I thought the first one was defective, but it turned out that the second one gave identical results.

I also thought about the normality of the solution - but unfortunately I can't put it together logically, why 0.1M IBr is 0.2N.

[ChenBeier]

This is similar to iodine itself.

0,1 m I₂ is 0,2 N I₂ . It can transfer 2 electrons, each of one iodine atom.

Iodine and also iodine bromide is used to determine the iodine number for fats and oils.
The iodine I₂ or the IBr will be added to the double bonds. So also bromine get a reaction to the carbon chain.

The IBr is transferred to iodine by adding of the KI for analysis.

So 1 m IBr is 1m I₂. But in this case one iodine is coming from the KI.

So the normality is twice of the molarity, for I₂ and IBr.

The titration of the iodine obtained from IBr gives a result for two Iodine atoms.

So the 20 ml of 0.1 m Na₂S₂O₃ is equal 20 ml 0,05 m I₂ But this 0,1 N I₂

More simple 1 IBr is 0,5 I₂, the ofter 0,5 I₂ is obtained from Potassium iodide KI.

[tomplo93]

Ok, now I understand what's going on with M and N but either I'm missing something or it still doesn't explain why in an otherwise relatively simple reaction the concentration is double the ratio described on the bottle.

[ChenBeier]

No it is not. The concentration is the same.
Other example
1 mol H₂SO₄ is 2 N, because it has two H+.
For titration you need twice the amount of 1 m NaOH.

The concentration is still 1 mol/l for sulfuric acid.

For IBr the Br is exchanged to Iodine so we also have twice of iodine.

The analysis gives a result of 2 N. But it reflect to 1 m IBr. As already told the half iodine came from KI.

Other thinking. 1 m IBr is containing 0.5 m I₂ if we don't exchange the bromine.

For determining the I₂ the double amount of thiosulfate is needed. This is 1 m.
So 1 m IBr correspond to 1 m Thiosulfate.

The lable is correct.

[tomplo93]

I think I understand this, but it still doesn't add up in my calculations. IBr is a substance that limits the amount of iodine generated:

IBr + 2KI --> KBr + KI + I₂

I took 5mL of 0.1M IBr - that's 0.0005 mol
according to stoichiometry it should also give 0.0005 mol of I₂ - because for me the amount of IBr limits the amount of I₂

Then I have a reaction with thiosulfate:
I₂ + 2Na₂S₂O₃ --> 2NaI + Na₂S₄O₆

so 0.0005 mol of I₂ should react with 0.001 mol of Na₂S₂O₃
I have 0.1M Na₂S₂O₃ - from here it should take about 10 mL
and it takes twice as much

[ChenBeier]

I took 5mL of 0.1M IBr - that's 0.0005 mol
according to stoichiometry it should also give 0.0005 mol of I₂ - because for me the amount of IBr limits the amount of I₂

But the 0.0005 mol I₂ is made of 0,0005 mol IBr and same amount of KI.
So the consumption will be twice of thiosulfate.

[tomplo93]

OK, I think I fully understand it now - thank you so much for your help!