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Author: Subject: Sodium Iodide from Iodine Povidone
Infamous_Reddy
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[*] posted on 4-5-2021 at 19:38
Sodium Iodide from Iodine Povidone


Where I live, the only available OTC sources of Iodine are either Iodine-Povidone or Iodine tincture. I chose Iodine Povidone as it is readily available in my home.
I want to turn it into Sodium Iodide which is required for a synthesis.

I started by adding NaOH to precipitate the Povidone polymer and obtain NaI and NaIO3. I then filtered the polymer and boiled down the solution to precipitate more of it and concentrate my solution. Now, how do I go on to separate the Iodide from both the Iodate and the excess Sodium Hydroxide? Also, a weird syrupy liquid was formed in the process and this was separating from the rest of the solution. Any help would be greatly appreciated (Nilered made a video on this).

Many Thanks.


[Edited on 5-5-2021 by Infamous_Reddy]




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paulll
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[*] posted on 5-5-2021 at 11:15


I think your best bet is to liberate the Iodine with HCl and peroxide, clean it up by sublimation, and reform the NaI with NaOH.
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Infamous_Reddy
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[*] posted on 5-5-2021 at 20:54


But then, if NaOH (cold and dil.) is reacted with I2, it goes on to form iodide and hypoiodite whereas if NaOH (hot and conc.) is used, we get iodide and iodate. Either way, I am sure that irrespective of the temperature and concentration of the NaOH sol. it is bound to form both NaIO and NaIO3 albeit in small quantities. Now, how do I separate the iodate, hypoiodite and possibly excess NaOH from the iodide?

I don't think this method is viable because we just added an extra factor to separate with the added steps worsening the already low yield obtained from 5% iodine solution because of the sublimation of I2. I live near the equator where temperatures regularly cross 45C - 50C and I don't think the I2 will not sublime in such an environment.
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[*] posted on 5-5-2021 at 21:13


It's been a while since I did this. IIRC, strong base is used to break apart the polymer. I believe it is more effective than acid hydrolysis.
I do recall getting some students to do redox titrations to determine yield and getting exactly 2/3, which seemed very suspicious. Evidently we were not getting all of the available iodine. But I forget exactly the procedure used.
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paulll
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[*] posted on 6-5-2021 at 00:11


Quote: Originally posted by Infamous_Reddy  

I don't think this method is viable because we just added an extra factor to separate with the added steps worsening the already low yield obtained from 5% iodine solution because of the sublimation of I2. I live near the equator where temperatures regularly cross 45C - 50C and I don't think the I2 will not sublime in such an environment.

That's how it tends to work, though; If you want greater purity you need to accept lower yield because thermodynamics just plain says so.
Taking it back to elemental Iodine and working from there... I mean it's not perfect and you will get some side-product contamination because halogens are bastards like that(it's why we love 'em, after all), and it's certainly lossy, but compared to the morass that you'd crys out of your initial Povidone extraction it's the way ahead.
I suspect you're not going to have much luck finding a practical procedure to separate iodides from iodates.
You're gonna need some ice;)
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[*] posted on 6-5-2021 at 02:22


Hypoiodite eventually decompose to iodate, it isn't stable.

You can seperate iodide from iodate by dissolving in ethanol. NaI is well soluble in EtOH, NaIO3 not.




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Infamous_Reddy
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[*] posted on 6-5-2021 at 04:15


Quote: Originally posted by paulll  

That's how it tends to work, though; If you want greater purity you need to accept lower yield because thermodynamics just plain says so.

I think you misunderstood me; I meant that I2, in general, is a pain to work with because it sublimes readily. This is especially a problem because if I didn't filter the sol. quick enough, most of the product would sublime and all I have on me are filter papers which take quite a bit of time to filter through.

Quote: Originally posted by Bedlasky  

NaI is well soluble in EtOH, NaIO3 not.

Thanks Bedlasky, I'll use EtOH to precipitate out the iodate and filter it. Then, I'll add HCl to convert the excess NaOH into NaCl until the solution is slightly basic. I don't think the HCl will react with the NaI until almost all of the NaOH is converted to NaCl. Then, I'll add acetone to precipitate out the NaCl while the NaI still remains dissolved. After filtration, I'll be left with almost pure NaI with slight NaOH contamination but that's okay for the application I intend to use it for.

I would also like to know what the oil formed during the filtration of NaI & NaIO3 sol. is. It is yellow and has the consistency of castor oil. It is denser than the solution and blocks the filtration from proceeding. I think this information is probably not enough to solve this mystery so, I'll be posting a few images of the oil as well. It also forms a emulsion with water.


[Edited on 7-5-2021 by Infamous_Reddy]
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[*] posted on 6-5-2021 at 09:58


You also could try heating the mixture up to melting, so that the iodate decomposes.



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Infamous_Reddy
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[*] posted on 8-5-2021 at 23:41


This is an image of the oil. It is a pale yellow liquid below the turbid water layer.

IMG_20210507_113331__01.jpg - 1.2MB


I am surprised as to how this oil is even formed in the first place. It seems that this oil is removed during filtration but, the filtrate is still turbid. In Nilered's 'Making Iodine' video, he raised this problem where his filtrate was interacting with the with filtering sol. in a weird way. I think this effect is caused due to the above-mentioned oil and the Marangoni effect. Any insight as to how this oil is formed would be helpful.
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