Sciencemadness Discussion Board

Making Sodium iodide and separating it from Sodium iodate

rstar - 26-9-2011 at 04:55

Hi friends,

I recently tried to make Sodium iodide from NaOH and Iodine, but ended unsuccessfully having excess NaOH.

However, second time I tried the same reaction but using Na2CO3 in place of NaOH, and ended successfully. But I am having some problems while separating Sodium iodide from its iodate, formed during the reaction.

Thanx in advance for any help :)

Chemistry Alchemist - 26-9-2011 at 05:05

I made a video of it recently, i used acetone to clean the iodide, but im kinda unsure about the results a bit, Sodium iodide is soluble in Acetone but when i tried to dissolve it in acetone, i couldn't get any dissolved, but it may of dissolved the iodate instead...

AndersHoveland - 26-9-2011 at 06:51

Elemental iodine can be reduced to iodide by reaction with thiosulfate. The thiosulfate is reduced to tetrathionate, S4O6[-2]. Or if you prefer, sodium bisulfite (not bisulfate) can reduce iodate to iodide.

Thiosulfate can be prepared simply by reacting sulfites with sulfur. Thiosulfate is also available at many aquarium stores.

Since elemental iodine can be reduced to iodide so easily, one wonders why meth labs go to so much trouble to obtain phosphorous to do the reaction.

Na2SO3 + S --> Na2S2O3
(2)Na2S2O3 + I2 --> Na2S2O6 + (2)NaI

(6)NaOH + (3)I2 --> NaIO3 + (5)NaI + (3)H2O

(3)I2 + (2)P --> PI3 (no water)
PI3 + (3)H2O --> P(OH)3 + (3)HI





[Edited on 26-9-2011 by AndersHoveland]

blogfast25 - 26-9-2011 at 07:53

Quote: Originally posted by rstar  
However, second time I tried the same reaction but using Na2CO3 in place of NaOH, and ended successfully. But I am having some problems while separating Sodium iodide from its iodate, formed during the reaction.

Thanx in advance for any help :)


You sure you made NaI from Na2CO3 + I2??? Where does the carbon go???

Iodate is removed by heating strongly: NaIO3 === > NaI + 3/2 O2. Like heating chlorate...

AndersHoveland - 26-9-2011 at 08:01

Quote: Originally posted by blogfast25  

You sure you made NaI from Na2CO3 + I2??? Where does the carbon go???


Sodium carbonate is also basic, like sodium hydroxide. The only difference is that carbon dioxide comes out.

(3)Na2CO3 + (3)I2 --> NaIO3 + (5)NaI + (3)CO2
Quote: Originally posted by blogfast25  

Iodate is removed by heating strongly


That would indeed be one way of doing it.

[Edited on 26-9-2011 by AndersHoveland]

blogfast25 - 26-9-2011 at 08:25

Quote: Originally posted by AndersHoveland  
That would indeed be one way of doing it.

[Edited on 26-9-2011 by AndersHoveland]


It's the best way. Recrystallisation is tedious. Better to get rid of the iodate thermally, the recrystallise the whole batch to obtain nice, 100 % water soluble product, iodine free...

rstar - 26-9-2011 at 09:08

Quote: Originally posted by blogfast25  

Iodate is removed by heating strongly


That's the easiest way, and it requires 425 C

2NaIO<sub>3</sub> = 2NaI + 3O<sub>2</sub>

Can MnO<sub>2</sub> catalyze the reaction, as that of Chlorates ? ?

Traveller - 29-9-2012 at 18:30

Very interesting thread. I believe this is the forum I have been looking for.

AJKOER - 30-9-2012 at 09:13

Another path is to use the produced Sodium iodide (just leave it in the solution) and a dilute acid to recover Iodine from the NaIO3. Then, repeat your original reaction by adding Na2CO3.

The dilute acid reaction equation is given by:

5 NaI + NaIO3 + 6 H(+) --> 6 Na(+) + 3 I2 (s) + 3 H2O

Reference: see, as one source, the Wikipedia Iodine Clock Reaction at http://en.wikipedia.org/wiki/Iodine_clock_reaction

I would use a dilute solution of HI (from distilling aqueous NaHSO4 with a new batch of NaI perhaps from a prior thermal decomposition of NaI/NaIO3) to avoid a salt separation issue.

Bottom line, you have recovered 1/6 of the original Iodine locked in as Iodate per the original reaction:

3 Na2CO3 + 3 I2 --> NaIO3 + 5 NaI + 3 CO2

At this point after the 1st recovery round, you have converted 5/6 of the recovered 1/6 (or, have a loss of 1/6 of 1/6) for a total theoretical NaI yield of 35/36th. For those working with significant quantities, repeat my recommended procedure again for a total theoretical recovery of (1 -(1/6)^3) or 99.5%. This, of course, would require having additional HI on hand.

In summary, my recommendation requires a batch planning approach to achieve a targeted total amount of Sodium iodide. Note, one can plan on increasing batch sizes as the total produced NaI grows. No salt separation is required and the only external inputs are I2, NaHSO4, NaHCO3 and an initial NaI batch.


[Edited on 30-9-2012 by AJKOER]