Sciencemadness Discussion Board

Question about iodine tincture + lye

simba - 30-9-2011 at 14:56

I wanted some sodium iodide, and decided to get some from iodine tincture. I bought 10% iodine tincture and added lye to it, until the solution became clear.

The problem is that the result wasn't exactly what I expected. What I got was a yellow precipitate which I don't know what it is, since sodium iodide is a white solid which is soluble in water, alcohol and acetone, so I shouldn't get a yellow insoluble solid.

The lye I used contains aluminium hydroxide, sodium chloride and sodium carbonate besides sodium hydroxide, which I think none explain the yellow stuff I got.

Does anyone know what is this yellow powder?

simba - 30-9-2011 at 15:07

Ok I think I found out already, the yellow solid is iodoform.

Chemistry Alchemist - 30-9-2011 at 20:04

Did you get any Sodium Iodide?

ThatchemistKid - 30-9-2011 at 21:58



[Edited on 1-10-2011 by ThatchemistKid]

simba - 1-10-2011 at 07:23

Quote: Originally posted by Chemistry Alchemist  
Did you get any Sodium Iodide?


Probably some iodide was formed yes, but I didn't bother to evaporate the solution since judging by the amount of iodoform I got, not much iodide would be recovered.

The problem was that iodine + NaOH will give sodium iodide, but iodine + NaOH + EtOH will give iodoform...my tincture used EtOH as solvent so it messed up the reaction.

Chemistry Alchemist - 1-10-2011 at 07:43

try evaporate the alcohol a bit but dont let the iodine to sublime...

Mixell - 1-10-2011 at 07:52

Remember, for every hypoiodite anion that is formed (and consumed to make iodoform) an iodide anion is also formed.
I2+ NaOH --> Na+ (aq) +I-(aq) +IO-(aq) +H+(aq).

[Edited on 1-10-2011 by Mixell]

simba - 8-10-2011 at 13:23

Quote: Originally posted by Mixell  
Remember, for every hypoiodite anion that is formed (and consumed to make iodoform) an iodide anion is also formed.
I2+ NaOH --> Na+ (aq) +I-(aq) +IO-(aq) +H+(aq).

[Edited on 1-10-2011 by Mixell]


I did the experiment again (now without alcohol), and this time I got a light blueish color solution, more than likely because of formed triiodide ions, and seeing your equation I suppose hypoiodite ions are the responsivel for this, converting iodides to triiodides.

Anyway, I was wondering if other bases like sodium (bi)carbonate will work in place of NaOH, since this would make a cleanup much easier. Anyone knows?

[Edited on 8-10-2011 by shivas]

simba - 12-10-2011 at 08:56

Ok, so I just tried the carbonate variation of this reaction, and it works good. The only difference is that it does not occur expontaneously like it does with NaOH, and needs some heat to happen. Maybe NaOH also needs it, but since it releases heat by itself when dissolved in water, no external heat needs to be applied.

I mixed iodine crystals (extracted from tincture), with sodium carbonate and then placed it in boiling water. In a few seconds the solution turned a light-blueish color, then boiled down and powdered sodium iodide was obtained. Unlike NaOH, Na2CO3 can be cleaned away by simply mixing the product with ethanol.




Chemistry Alchemist - 21-12-2011 at 08:09

Is there a way to convert Sodium Iodide to Iodoform with out converting to to Iodine first?

AJKOER - 26-12-2011 at 08:05

Quote: Originally posted by shivas  

I mixed iodine crystals (extracted from tincture), with sodium carbonate and then placed it in boiling water. In a few seconds the solution turned a light-blueish color, then boiled down and powdered sodium iodide was obtained.


Actually, characteristic of Iodine, there is a fast disproportionation so in addition to sodium Iodide there will be some Sodium Iodate.

Ephoton - 26-12-2011 at 16:27

when adding NaOH too KI I dont think you will get a displacement.
but the I2 you should get a mixture of NaI and NaOI

to make NaI from Iodine Tinc which is actualy NaI3 or NaI and I2 mixture you should first
drop the iodine with an oxidising agent like peroxide or ammonium persulfate (any persulfate will work
and usually gives larger and easier crystals to filter than a peroxide.)

then put the newly formed iodine in NaOH solution that is of equal molar content.

evap water then heat NaI and NaOI powder to 200 plus you will have to check the brake down point of NaI
as it will brake down at a certian temp.

this will expell the oxygen from the NaOI and make NaI.

Al in your caustic give up now.

Ephoton - 26-12-2011 at 16:31

@ chemistry Alchemist

I would think peroxide mixed with NaOH solution would do the trick on acetone
when mixed with NaI. but then it is making the I2 insitue then and then forming sodium hypoiodide to
complete haloform reaction.

What can be done that is funky with iodoform ive never used it ?
that is other than its boiling point.