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Author: Subject: Report on making bismuth oxyiodide
Lion850
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[*] posted on 13-10-2020 at 04:03
Report on making bismuth oxyiodide


A synthesis of bismuth oxyiodide appears in Brauer’s handbook as well on various other places online, for example on prepchem:
https://prepchem.com/synthesis-of-bismuth-oxyiodide/
I followed the same method but with different amounts:
- 28g Bi(NO3)3.5H20 in a small beaker and add 48g glacial acetic acid.
- Stir vigorously with slight heat until all dissolved in a clear solution.
- 10.3g KI and 14.5g CH3COONa added to a beaker with 570g water and stirred until all dissolved in a clear solution.
- The bismuth nitrate solution was dripped into the bigger beaker with the potassium iodide and sodium acetate solution, while stirring.
- Each drop formed a black blotch as it entered the solution, this turned yellow within a second or 2.
- The whole solution quickly turned yellow-orange after a few ml of bismuth nitrate was dropped in, see photo below.
1 Solution after start of DD.jpg - 621kB

- The color darkened as more was added, with the final color, after all the bismuth nitrate was added, being brown-red, or as it says online “brick red”. See photo.
2nFinal solution.jpg - 624kB

- Stirring was continued for another 20 minutes and then stopped. The brick red ppt settled quickly, and the supernatant liquid was decanted. The product was washed by 2 x 500ml portions of water by stirring and decantation.
- The solution was then vacuum filtered and the remainder washed with water in the funnel.
- The wet bismuth oxyiodide was dried on a steam bath. After some 3 hours it appeared dry, it was then powdered and dried for another 2 hours. At this time the weight loss reduced to near zero.
- Final recovery was 19.6g. See photo of final product in bottle below.
3 Bismuth oxyiodide.jpg - 380kB

The weight of 19.6g surprised me; I was expecting much less based on the equations on Prepchem:
1) Bi(NO3)3 + 3KI = BiI3 + 3KNO3
2) BiI3 + H2O = BiOI + 2HI
According to the above my recovery of BiOI should have been only around 7g.
I realised that if all the iodine present in the initial 10.3g of KI formed part of the BiOI (and did not proceed to also make HI) the BiOI yield would be around 20g which made much more sense. But how to write the equation for what happened? Eventually I settled on
Bi(NO3)3.5H2O + KI = BiOI + KNO3 + 2HNO3 +4H2O
The stoichiometry of this equation agrees to the weight ratios of Bi(NO3)3 and KI stated by Brauer (and prepchem) and also the weight of the recovered BiOI but whether it is the correct equation I do not know.
Another question I have is what is the purpose of the sodium acetate in the reaction?


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[*] posted on 13-10-2020 at 09:56


Thanks for the report. It's a nice colored compound. Sorry I can't help with your questions.
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[*] posted on 13-10-2020 at 14:42


If you have 100% yield, you obtain 20,31g. It is also possible, that your BiOI is hydrate.



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[*] posted on 13-10-2020 at 16:42


Nice job on this replication.

Bedlasky is right though. You should double check your yield calculations, because the equations that prep chem gives are more reasonable than your proposal, and result in a theoretical yield of 20.31 grams.




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Lion850
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[*] posted on 13-10-2020 at 18:52


Thanks for the replies. Assuming that all the iodide from the KI ended up in the BiOI, then the 28g of Bi(NO3)3.5H2O became the limiting reagent and the 19.6g I got represents a yield of 96%. A good yield should be possible because this product is easier to recover from the solution than many other salts - it is heavy and the ppt settles quickly leaving a clear supernatant solution, and also when filtered the filtrate is clear. Nevertheless a yield of 96% seems too good to be true so maybe it it a hydrate or still retains a bit of moisture.

I do think though that my results shows that all the iodine ends up in the BiOI and that no HI remains as a product.
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[*] posted on 13-10-2020 at 19:25


>> or still retains a bit of moisture.

yeah, unless I bake something at 450F for a few hours, I always assume that is the case :)
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[*] posted on 13-10-2020 at 20:20


Quote: Originally posted by Lion850  
Thanks for the replies. Assuming that all the iodide from the KI ended up in the BiOI, then the 28g of Bi(NO3)3.5H2O became the limiting reagent and the 19.6g I got represents a yield of 96%. A good yield should be possible because this product is easier to recover from the solution than many other salts - it is heavy and the ppt settles quickly leaving a clear supernatant solution, and also when filtered the filtrate is clear. Nevertheless a yield of 96% seems too good to be true so maybe it it a hydrate or still retains a bit of moisture.

I do think though that my results shows that all the iodine ends up in the BiOI and that no HI remains as a product.
I don’t think you understand my point. When you’re using 3 equivalents of KI you can’t have all of the I end up in the BiOI, because the ratio of Bi:I in your products is 1:3 while in your products it’s 1:1. The equations that Prep Chem gives are correct. You can write balanced reactions til the cows come home but it doesn’t mean they reflect reality. Please try calculating the yield again, you’ll see.



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[*] posted on 13-10-2020 at 20:54


Please see attached report on making BiOI from bismuth nitrate and potassium iodide. A slightly different method using only a small amount of water but they report the same overall equation as what I wildly guessed above.

49AC81EA-7F1B-4E7D-A6C1-687A1A3AAAC6.png - 841kB
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[*] posted on 14-10-2020 at 06:54


That’s all well and good but it’s not what you did. You used 3 equivalents of KI, and you definitely would have ended up with some HI in the end. Your conditions were totally different. My point is that just because this equation is correct for the experiment described in this report doesn’t mean that the equation for the report you initially followed isn’t valid. Please, calculate the yield from the equations on Prep Chem again. Post your calculation. I did it and it came out fine. I want you to see that it works.



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[*] posted on 14-10-2020 at 07:30


Texium: 20,31 g is calculated from this equation:

Bi(NO3)3.5H2O + KI = BiOI + KNO3 + 2HNO3 +4H2O

Why HI shouldn't react with another bismuth nitrate? HI certainly reacts with bismuth nitrate. HI is also source off iodide anions. If HI doesn't react with bismuth nitrate, you need three times more KI (so Lion should have worse yield). Equation above is overall. Prepchem equations show mechanism, but not completly, because there is missing reaction between HI and bismuth nitrate. So finally it doesn't matter which equation you choose for calculations.




If you are interested in aqueous inorganic chemistry look at https://colourchem.wordpress.com/main-page/

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[*] posted on 14-10-2020 at 08:47


Bedlasky, I made an incorrect assumption. Since the procedure that Lion followed cited an equation that shows 3 equivalents of KI, I assumed that he used 3 equivalents of KI in his reaction, and was saying that somehow all of that iodide ended up in his product. I went back and checked the amounts he used and I see that he used roughly equimolar amounts of bismuth nitrate and KI. Thus the equations presented by Prep Chem are indeed misleading.

Apologies for the misunderstanding.




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