04-02-02 02:58
No 291174
Most methods of iodine extraction I have found on here and on the web seem to disgard the ethanol from the tincture with dilution or evapouration.
It seems a lot of potential iodine is wasted or washed away with the "orange water layer" Post 237276 (Aurelius: "newbee info", Newbee Forum), and it also seems that good old ethanol is wasted as well.
In another process (found on rhodiums site) the ethanol is evapourated on a low heat leaving iodine and potassium iodide. There must be quite a bit of iodine lost doing it this way due to sublimation when the tincture is heated.
How can this ethanol be removed without removing some iodine with it ?
Can the sublimation of iodine be reduced by evapourating the ethanol off under reduced pressure ?
Do sublimation points change like boiling points with pressure changes ?
Can we turn the ethanol into something else and seperate it from the iodide and the iodine easier ?
Could we not remove this ethanol by oxidising it to acetaldehyde and removing this with less heat than ethanol ?
Leonard Cohen cheers me up !
(Hive Bee)
04-02-02 03:15
No 291182
If the iodine in the alcohol (tincture) is converted to iodoform :
NaOCl + KI = KIO + NaCl
CH3.CO.CH3 + 3KIO = CH3.COOK + 2KOH
When Iodoform is heated strongly iodine is given off as a purple vapour.
IODOFORM REACTION
If potassium iodide solution and a little sodium hypochlorite solution are gently warmed with ethanol, pale yellow crstals of iodoform appear on standing. The same result is achieved, though rather more slowly, if ethanol is warmed with iodine and sodium hydroxide solution.
Well domestic bleach usually contains both sodium hydroxide and sodium hypochlorite. Adding this to the tincture would convert the iodine and the iodide to iodoform. The iodine can then be released from the iodoform by heating.
Leonard Cohen cheers me up !
(Chief Bee)
04-02-02 05:34
No 291249
I can't see any reason for any iodine to sublimate while the ethanolic solution is concentrated as long as it is not evaporated to dryness, but only concentrated to a high degree (which should be sufficient for subsequent manipulation).
Or - does experimental observations contradict my theoretical babble?
(Distinctive Doe)
04-02-02 06:46
No 291273
I think the iodine will definately evaporate. Iodine is not soluable in water, so you heat it to drive off ethanol and as this happens the solution is driving I2 out, Iodine has a very high vapor pressure which should rise significantly with temp. Separations are NOT purfect, if they were then you would evaporate off ethanol at the azeotrope until the ethanol is gone. I can tell you that this is not what happens.
Those who give up essential liberties for temporary safety deserve neither liberty nor safety
(Chief Bee)
04-02-02 07:31
No 291295
Yes, but iodine does not form an azeotrope with ethanol. I was under the impression that sublimation only could happen when actual crystals were subjected to heat, and not when covered with solvent (and in my suggestion, the solution wouldn't be concentrated as far as allowing the ethanol level sink below the precipitated crystalline mass.
Do you have any references on how sublimaton works in the presence of solvents?
(Distinctive Doe)
04-02-02 08:13
No 291310
I'll try to explain.
Here is what happens.
When the solution becomes supersaturated with I2 then the I2 will migrate to the edge of the solution and accumulate at the interfaces, this forms a concentrated microscopic I2 layer coating the water.
Here is why it happens.
I2 is nonpolar, air is nonpolar, water is polar.
Water doesn't like air or I2.
Air likes I2.
These enegetic effects will result in a microscopically thin film of high I2 concentration at the air/water interface. The I2 will then evaporate from there. This microscopic concentrated layer minimizes the free energy of the interface. I hope this made sence.
note: I think it is the free energy that defines interfacial effects, but I could bee mistaken.
Interfacial phenomena is a whole field of science itself. I have taken several graduate level surface chemistry classes and I have a decent feel for how things happen at interfaces.
Rhodium
I work with compounds that melt at 120C and they also evaporate fairly fast from water at 50C. And these compounds don't even have a fraction of the vapor pressure I2 has.
Those who give up essential liberties for temporary safety deserve neither liberty nor safety
(Chief Bee)
04-02-02 08:17
No 291311
Okay, I believe you. Would addition of aqueous iodide to completely convert the iodine to triiodide followed by evaporation work, or would the I2 + I- <-> I3- equilibrium screw it all up anyway?
(Distinctive Doe)
04-02-02 08:32
No 291322
If the species all remain ionic then I would assume that they would stay in solution. I don't think the equilibrium lies completely to the right, does it? I had the impression that all three species are present, but with enough iodide ions you could possible overcome that and practically eliminate I2. We need to know the charateristics of the triiodide equilibrium system.
The curious thing for me is that they use NH3I in decolorized iodine tinc. I wonder why?
Those who give up essential liberties for temporary safety deserve neither liberty nor safety
(Hive Bee)
04-02-02 12:03
No 291407
Some practical observations re: Iodine:
- I2 dissolved in acetone: no sublimation occurs at room temperature
- Saturated aq. I2 (orange water over I2 cristals; no I-): some sublimation (though not nearly as much as with dry I2).
- When oxidising a KI solution no I2 vapour is forming until crystals start falling out of solution.
I'd say it's safe to boil off ethanol as long as enough aq. I- is present.
Hansje high in proteine and fibre!
(Hive Bee)
04-02-02 20:00
No 291480
The second line of formula should be this:
CH3.CO.CH3 + 3KIO = CHI3 + CH3.COOK + 2KOH
Leonard Cohen cheers me up !
(Hive Bee)
04-02-02 22:12
No 291521
If a small amount of potassium hydroxide was added to the tincture, it would convert any free iodine into potassium iodide/iodate. This would now allow evapouration of the ethanol and water without any iodine being lost due to sublimation. The iodine would be locked up as potassium salts.
The potassium iodide and iodate remaining would then be acidified and oxidised with hydrochloric acid and hydrogen peroxide, to yield free iodine.
Just this small addition of alkali before evapouration will decrease iodine losses and also allow the extraction of ethanol of a reasonable concentration with very low iodine contamination.
Leonard Cohen cheers me up !
(Hive Addict)
10-31-02 03:55
No 374777
Here is why it happens.
I2 is nonpolar, air is nonpolar, water is polar.
Water doesn't like air or I2.
Air likes I2.
These energetic effects will result in a microscopically thin film of high I2 concentration at the air/water interface. The I2 will then evaporate from there. This microscopic concentrated layer minimizes the free energy of the interface. I hope this made sense.
note: I think it is the free energy that defines interfacial effects, but I could bee mistaken.
Interfacial phenomena is a whole field of science itself. I have taken several graduate level surface chemistry classes and I have a decent feel for how things happen at interfaces.
Having only recently become interested in iodine,
I'm snuffling after it like a rutting dog sniffs the bitch's flanks....
Hoping to get educated
in past overlooked areas of studies..
Foxy: (You duplicitous doe..)
If the solution was being stirred continuously,
under vacuum,
would the evaporation
be better accomplished
with the "plates"
being continuously dispersed?
dwarfer