Author Topic: Iodine Tincture: Osmium flames are welcome.  (Read 3436 times)

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interested

  • Guest
Iodine Tincture: Osmium flames are welcome.
« on: April 22, 2004, 09:47:00 AM »
After reading many posts regarding the extraction of Iodine crystals from tincture, this is what swim has come up with.

Most tincture's contain Potassium Iodide (KI) and Iodine salt (I).  KI has to be reacted with HCl acid and an oxidiser (Hydrogen Peroxide (H2O2)) to obtain Iodine. 

2 KI + 2 HCl -> 2 HI + 2 KCl
2 HI + H2O2 -> I2 + 2 H2O

The Iodine salt whithin the tincture has to be oxidised to obtain I2

This equation is a stab in the dark.  Is this the reaction?  Could somebody correct me?
2 I +  4 H2O2 -> I2 + 2 H2O + O2

If the contents of a 1L bottle of tincture contained:

25mg/mL Iodine
25mg/mL Potassium Iodide
Ethanol

I assume the following parameters would have to be met.

mol = weight of sample (g) / molar weight (g/mol)
mol = 25g KI contained in bottle / 166.02g/mol
mol = 0.151mol KI


KI + HCl -> HI + KCl
0.151mol KI + 0.151mol HCl -> 0.151mol HI + 0.151mol KCl
25g KI + 4.14mg HCl -> 1.18mg HI + 2.03mg KCl

When performing this reaction adequate H2O should be present as to dissolve the HI as it is formed.

2 HI + H2O2 -> I2 + 2 H2O
0.151mol HI + 0.0755mol H2O2 -> 0.0755mol I2 + 0.151mol H2O
1.18mg HI + 2.22mg H2O2 -> 0.595mg I2 + 8.39mg H2O

Once these reactions have completed the I2 should fall out of the solution because iodine is immiscible in water.

Swim has left the last equation as it is probably incorrect.  But there is still another 25g of iodine salt contained in the tincture.

Questions.

Wouldn't the iodine salt in the tincture also be liberated once the iodine from the KI has been obtained?

Swim asks this because of this statement.

"Iodine is soluble in a solution of potassium iodide due to the formation of the soluble potassium tri-iodide."

Found at

http://www.ucc.ie/ucc/depts/chem/dolchem/html/elem/elem053.html



Is the oxidation of I into I2 necessary as my understanding of elemental Iodine is that it is already in I2 form? (diatomic? ::) )

Would the ethanol in the tincture change the reaction(s) in any way?

This whole post is up for scrutiny.  Swim just wishes to understand what the hell is going on.  Feel free to flame away or provide some helpful insight. :)


fierceness

  • Guest
Hmm
« Reply #1 on: April 22, 2004, 10:49:00 AM »
Just as a side note, what would happen if you refluxed KI/NaI, HCl, and pseudo?  If (some) of the pseudo was converted, would you end up with a mixture of KI/NaI, HI, I2, HCl, (some) unconverted pseudo, and meth?  ( and maybe some iodometh?)

Rhodium

  • Guest
No good
« Reply #2 on: April 22, 2004, 11:12:00 AM »
Just as a side note, what would happen if you refluxed KI/NaI, HCl, and pseudo?

Nothing of interest. You need a concentration of HI exceeding 50% for the reduction to happen, and you can't attain that in a solution of KI/NaI in 37% HCl. The ionic strength of the dissolved salts will push HCl and HI gas out of the solution.


elfspice

  • Guest
bugger
« Reply #3 on: April 22, 2004, 07:23:00 PM »
ffs (*slaps self around a bit*) I don't know what i thought i was replying to... sorrry...  :-[

interested

  • Guest
Thanks for the responses guys.
« Reply #4 on: April 22, 2004, 11:01:00 PM »
Thanks for the responses guys.

Swim has no formal chemistry education and has only aquired knowledge from what he has read here at the hive and the rest of the WWW.  Swim's post above was his first attempt at making any calculations.  He had no way to verify whether he was correct so he hoped that fellow bee's could point him in the right direction.

Also, are swim's questions sensible, or are they so ridiculous that no one has bothered to answer?

Swim just wants to take a moment to say that the hive is fuckin' awesome and would like to thank everyone that has had a positive contribution to it. :)

Rhodium

  • Guest
Some comments on the original post
« Reply #5 on: April 24, 2004, 07:44:00 AM »
Most tincture's contain Potassium Iodide (KI) and Iodine salt (I)

No, it contains a mixture of Potassium Iodide (KI) and Elemental Iodine (I2).

"Iodine is soluble in a solution of potassium iodide due to the formation of the soluble potassium tri-iodide."

Is the oxidation of I into I2 necessary as my understanding of elemental Iodine is that it is already in I2 form? (diatomic?)


It isn't, as you seldom have free I atoms. Iodine is almost always in the form of the brown-black I2, or as the colorless iodide ion (I-) in solution (in solution, KI is better written as K+ + I- and HI as H+ + I-).

In aqueous solution, the following takes place: I2(s) + I-(aq) -> I3-(aq) (iodine and iodide ions form the soluble triiodide ion).

Wouldn't the iodine salt in the tincture also be liberated once the iodine from the KI has been obtained?

Correct. When you have oxidized the iodide ions to iodine (I2), no triiodide ions can form, and all iodine (I2) will precipitate.

Would the ethanol in the tincture change the reaction(s) in any way?

Yes. If you do not remove the ethanol, it too will react with the peroxide and waste your oxidizer (it isn't a good idea to compensate for this by adding a large excess of peroxide, as the iodine can be oxidized further to soluble iodate salts).

Also, iodine (I2) is soluble in ethanol (but almost insoluble in water) so any remaining ethanol will prevent some of the iodine from precipitating.


interested

  • Guest
Thanks, Rhodium. Seeing as the search engine...
« Reply #6 on: April 24, 2004, 09:53:00 AM »
Thanks, Rhodium.

Seeing as the search engine has been inaccessable, swim failed to read the following posts written by elfspice and Osmium.  These posts demonstrate the correct calculations and clear up any questions regarding the presence of ethanol within the tincture.

Post 496756

(elfspice: "stochiometry!", Stimulants)

Post 496815

(Osmium: "> I think the best way to proceed would be...", Stimulants)


I was in the ball park with my calculations, I just made a few mathematical errors (ie. dividing instead of multiplying).  At least swim has learnt something about stoichiometry and extracting iodine from tincture. 

Regards.

Organikum

  • Guest
Rhodium wrote: Re: If you do not remove the...
« Reply #7 on: April 24, 2004, 12:19:00 PM »
Rhodium wrote:

If you do not remove the ethanol, it too will react with the peroxide



I dont think so.
According to "The reactions of Ethylalcohol" from Herbert E. Morris, 1931, EtOH does not react with H2O2, neither in alkaline, acidic nor neutral medium.  An iron-salt is needed for to catalyze the reaction.
Ferrous and ferric acetate is told to catalyze the reaction to acetic acid alone.
Ferrous sulfate as catalyst is said to produce acetic acid and acetaldehyde.
Ferrous oxalate catalyzes to mainly acetic acid.
Diluting the alcohol seems to favor the oxidation, preoxidation of the ferrous sulfate to the ferric sulfate seems to reduce the activity of the catalyst.

If the author not plain forgot to mention iodine and its salts I would tend to believe that these compounds have no catalytic properties here and a reaction of ethylalcohol with hydrogen wont take place.

Also by Rhodium:


The ionic strength of the dissolved salts will push HCl and HI gas out of the solution


I was by now not able to push HCl out of an aqueous solution by adding NaCl, but perhaps I made something wrong. It would be an interesting way to produce HCl-gas if it works.  :)

No disrespect intended.
ORG




Rhodium

  • Guest
Good point
« Reply #8 on: April 24, 2004, 02:58:00 PM »
Thanks for pointing out that hydrogen peroxide is unable to oxidize ethanol. I really thought that was possible (seeing that tBuOH is easily oxidized to tBuOOH), but maybe the reaction only progresses in an effective manner on tertiary alcohols. The article you are referring to is Chem. Rev. 31, 465-506 (1931):



Regarding your experiment with salt in hydrochloric acid - no gas evolved at room temp because very little of your sodium chloride dissolved, as the solution is already saturated in chloride. If you heat the solution, HCl gas will evaporate and sodium chloride will dissolve in its place, but the opposite will not happen in an open system (HCl gas dissolving in the solution, forcing NaCl to crystallize). 

Therefore the net change will be evaporation of HCl gas and dissolution of sodium chloride.

NaCl(s) -dissolution-> Na+(aq) + Cl-(aq) + H+(aq) -evaporation-> HCl(g)


lugh

  • Guest
More References
« Reply #9 on: April 30, 2004, 05:12:00 AM »
The following articles cited in Morris's paper:

Ber. 34. 3579 (1901)



Ber. 36. 2003 (1903)



J.Am.Chem.Soc.48,2083 (1926)



:)


Organikum

  • Guest
Thanks a lot lugh! :-D ORG
« Reply #10 on: April 30, 2004, 10:46:00 AM »
Thanks a lot lugh!

;D  ORG