Sciencemadness Discussion Board
Not logged in [Login ]
Go To Bottom

Printable Version  
 Pages:  1  2
Author: Subject: Easy synthesis of hydroiodic acid
Felab
Hazard to Self
**




Posts: 75
Registered: 9-11-2018
Location: Spain
Member Is Offline

Mood: Españó!!!

[*] posted on 7-1-2019 at 12:20


Quote: Originally posted by Boffis  
The separation of HI from sulphuric acid is not easy as at higher temperatures the reaction goes in the opposite direction and you get your SO2 and I2 back. You could of course selectively precipitate the sulphuric acid by partial neutralization with barium hydroxide but if you can't engineer a simple and safe set up to handle H2S then the partial neutralization is probably beyond you too.


The silver sunbeam shows another path where you mix iodine and calcium oxalate, which yields calcium iodide and CO2.

Add sulphuric acid, deccant the CaSO4 and get your HI.
View user's profile View All Posts By User
clearly_not_atara
International Hazard
*****




Posts: 1588
Registered: 3-11-2013
Member Is Offline

Mood: No Mood

[*] posted on 7-1-2019 at 13:34


Don't you already have an iodide salt? It's not even clear why oxalate is involved in that process at all.

Ironically, though, you could probably displace HI with oxalic acid. Potassium tetraoxalate, the 1/4 salt of K and oxalate, has a very low solubility in water. So:

KI (aq) + 2 H2C2O4 (aq) >> HI (aq) + KH3(C2O4)2 (s)

As long as [HI] is such that the pH > 1 you should do ok like this.

[Edited on 7-1-2019 by clearly_not_atara]




[Edited on 04-20-1969 by clearly_not_atara]
View user's profile View All Posts By User
Felab
Hazard to Self
**




Posts: 75
Registered: 9-11-2018
Location: Spain
Member Is Offline

Mood: Españó!!!

[*] posted on 7-1-2019 at 13:48


Quote: Originally posted by clearly_not_atara  
Don't you already have an iodide salt? It's not even clear why oxalate is involved in that process at all.

Ironically, though, you could probably displace HI with oxalic acid. Potassium tetraoxalate, the 1/4 salt of K and oxalate, has a very low solubility in water. So:

KI (aq) + 2 H2C2O4 (aq) >> HI (aq) + KH3(C2O4)2 (s)

As long as [HI] is such that the pH > 1 you should do ok like this.
[Edited on 7-1-2019 by clearly_not_atara]


Is oxalic acid a strong enough acid to generate hydroiodic acid? It is certainly much cheaper than phosphoric acid though.

I will have to buy some.

The Silver Sunbeam is a very old book so its procedures might be a bit odd.
View user's profile View All Posts By User
clearly_not_atara
International Hazard
*****




Posts: 1588
Registered: 3-11-2013
Member Is Offline

Mood: No Mood

[*] posted on 7-1-2019 at 22:11


In water, acid strength is not meaningful below pKa -1.6 or so. So while H2SO4, HCl, HBr, HClO4, and HI have a varying order of "acidity" under some definitions, in water they're all just strong acids. For tetraoxalate to form, what is necessary is that oxalic acid is deprotonated, which happens as long as the pH is higher than 1.25. So you can displace HI at about 0.1 molar, since pH <= -log10[HI]. A similar procedure has been used successfully to liberate sulfuric acid from magnesium sulfate.

The resulting solution of HI might then be concentrated by boiling and/or vacuum distillation. There is a concern that residual dissolved oxalic acid (which will not be entirely consumed) may decompose during this process. This will not affect the concentration of HI but it does generate carbon monoxide.




[Edited on 04-20-1969 by clearly_not_atara]
View user's profile View All Posts By User
Boffis
International Hazard
*****




Posts: 1196
Registered: 1-5-2011
Member Is Offline

Mood: No Mood

[*] posted on 8-1-2019 at 10:55


@Felab, if the oxalate method works why not just make cadmium oxate and treat it likewise and you will get your desired iodide directly without having to isolate HI solution.

By the way the use of cadmium iodide in the preparation of photographic emulsions is to do with the grain size of the silver iodide that results. Cadmium iodide is almost undissociated so the free iodide ion concentration is very low and so when used to precipitate silver iodide the grains grow more slowly and and larger. This make the emulsion slower (ie less light sensitive) for use in printing paper etc.
View user's profile View All Posts By User
Felab
Hazard to Self
**




Posts: 75
Registered: 9-11-2018
Location: Spain
Member Is Offline

Mood: Españó!!!

[*] posted on 8-1-2019 at 11:30


Quote: Originally posted by Boffis  
@Felab, if the oxalate method works why not just make cadmium oxate and treat it likewise and you will get your desired iodide directly without having to isolate HI solution.

By the way the use of cadmium iodide in the preparation of photographic emulsions is to do with the grain size of the silver iodide that results. Cadmium iodide is almost undissociated so the free iodide ion concentration is very low and so when used to precipitate silver iodide the grains grow more slowly and and larger. This make the emulsion slower (ie less light sensitive) for use in printing paper etc.


Collodion photography isn't really an emulsion (at least the type I want to do, cause there are other printing processes, colloido-chloride emulsions, etc) and its sensitivity depends more on the amount of silver nitrate present in the wet plate or on its acidity rather than on the grain of the silver halide particle.

Cadmium halides are normaly used in collodion because of their high solubility in ether and alcohol and because of their stability due to their high affinity for the halide.

Also, I don't like cadmium iodide/bromide either because it is a water soluble salt of cadmium, which isn't very healthy. I think lithium halides are a much safer alternative, although a bit shorter lived.

Maybe I could react lithium tetraoxalate with potassium iodide but I don't know the solubility of lithium tetraoxalate.

[Edited on 8-1-2019 by Felab]

[Edited on 8-1-2019 by Felab]
View user's profile View All Posts By User
Felab
Hazard to Self
**




Posts: 75
Registered: 9-11-2018
Location: Spain
Member Is Offline

Mood: Españó!!!

[*] posted on 9-1-2019 at 13:47


Quote: Originally posted by clearly_not_atara  
Don't you already have an iodide salt? It's not even clear why oxalate is involved in that process at all.

Ironically, though, you could probably displace HI with oxalic acid. Potassium tetraoxalate, the 1/4 salt of K and oxalate, has a very low solubility in water. So:

KI (aq) + 2 H2C2O4 (aq) >> HI (aq) + KH3(C2O4)2 (s)

As long as [HI] is such that the pH > 1 you should do ok like this.

[Edited on 7-1-2019 by clearly_not_atara]


Oxalic acid is involved to generate calcium iodide, which reacts with suphuric acid to yield CaSO4, wich precipitates out and can be easily filtered. What matters is that you get calcium or barium iodide by any procces, not only the oxalate one.

Are you sure that oxalic acid can displace iodides to form hydroiodic acid? I cannot find a lot of info online about potassium tetraoxalate. I could only find that it forms when reacction conditions are below 50 degrees in acidic conditions. I plan to buy some (it is dirt cheap, like 4 euros a kilo) but I don't know when will I get arround to that (I hope soon).

If this procces truly yields hydroiodic acid it will be very important for me so thank you.
View user's profile View All Posts By User
AJKOER
International Hazard
*****




Posts: 2826
Registered: 7-5-2011
Member Is Offline

Mood: No Mood

[*] posted on 9-1-2019 at 14:09


Quote: Originally posted by Boffis  
@Ajoker, I have used hydrogen sulphide as a reducing agent many time because it is so easily handled and the byproduct so easily removed. If you cannot use it safely then I think its time you moved to stamp collecting or a similar hobby.
.......


Your comments remind me of a thread on SM citing the many times previously uneventful preparation of an energetic compound, which unfortunately became eventful.

I hope, at least, my thread provided some detail background in the case the safe many times proves to be the one time.
--------------------------------------

Interestingly, I did start to collect coins, but as I mentioned in a thread discussing electrochemical cell dissolution of metals, I managed to dissolve one old mostly silver US dime and some commemorative US quarters as well. This, of course, resulted in a 'reduction' in my coin collection ;).
View user's profile View All Posts By User
AJKOER
International Hazard
*****




Posts: 2826
Registered: 7-5-2011
Member Is Offline

Mood: No Mood

[*] posted on 9-1-2019 at 14:09


Quote: Originally posted by Boffis  
@Ajoker, I have used hydrogen sulphide as a reducing agent many time because it is so easily handled and the byproduct so easily removed. If you cannot use it safely then I think its time you moved to stamp collecting or a similar hobby.
.......


Your comments remind me of a thread on SM by MrHomeScientist suggesting that one should always be prepared for the unexpected (see http://www.sciencemadness.org/talk/viewthread.php?tid=78201#... ).

I hope, at least, my thread provided some detail background in the case of the unexpected.
--------------------------------------

Interestingly, I did start to collect coins, but as I mentioned in a thread discussing electrochemical cell dissolution of metals, I managed to partially dissolve one old mostly silver US dime and some commemorative US quarters as well. This, of course, resulted in a 'reduction' in my coin collection ;).

[Edited on 9-1-2019 by AJKOER]
View user's profile View All Posts By User
clearly_not_atara
International Hazard
*****




Posts: 1588
Registered: 3-11-2013
Member Is Offline

Mood: No Mood

[*] posted on 9-1-2019 at 14:28


Quote: Originally posted by Felab  

Are you sure that oxalic acid can displace iodides to form hydroiodic acid? I cannot find a lot of info online about potassium tetraoxalate. I could only find that it forms when reacction conditions are below 50 degrees in acidic conditions. I plan to buy some (it is dirt cheap, like 4 euros a kilo) but I don't know when will I get arround to that (I hope soon).

If this procces truly yields hydroiodic acid it will be very important for me so thank you.

What matters isn't whether oxalic acid protonates I- -- it won't. What matters is whether oxalic acid is deprotonated enough that the solution contains plenty of [HC2O4-]. That happens because the oxalic acid protonates water molecules, not iodide. The precipitated salt is then removed mechanically, and the solution is heated, which ultimately results in protonation of I- to volatilize as HI.

I've made potassium tetraoxalate by precipitation this way. I don't have the yield at hand but I could weigh it I guess. The solubility of KH3(C2O4)2*2H2O is 25 g/L and the molar mass is 254 g so it will precipitate at around 0.1 M. Oxalic acid is soluble to 1.1M and KI to 8M so it should be easy to make a sufficiently concentrated solution. I would suggest combining a saturated solution of oxalic acid with 2M KI sol'n, then cool and filter.

The yield of HI by this method is bound to be pretty bad but you will get some. If you distill it the resulting product should be free of oxalic acid but again I am not sure how likely it is for carbon monoxide to form during distillation.




[Edited on 04-20-1969 by clearly_not_atara]
View user's profile View All Posts By User
AJKOER
International Hazard
*****




Posts: 2826
Registered: 7-5-2011
Member Is Offline

Mood: No Mood

[*] posted on 9-1-2019 at 15:01


Some research (see https://www.researchgate.net/publication/231458156_Catalysis... ), to quote:

"Oxalic acid acts as a very effective catalyst for the chromic acid oxidation of iodide. At low concentrations, the reaction is first order in oxalic acid, iodide, chromic acid, and hydrogen ions, but becomes zero order in iodide at high iodide concentrations. The proposed mechanism consists of the formation of a termolecular complex (CO2)2CrIO2- formed from oxalic acid, chromic acid, and an iodide ion followed by its decomposition into an iodine atom and a chromium(V)-oxalic acid complex, (CO2)2CrO2-. It is assumed that the catalytic activity of oxalic acid is due to its ability to stabilize chromium(V)"

suggests to me the possibility that H2C2O4 in the presence of metal impurities and I- may be driving a reaction that is not supporting HI creation.
View user's profile View All Posts By User
clearly_not_atara
International Hazard
*****




Posts: 1588
Registered: 3-11-2013
Member Is Offline

Mood: No Mood

[*] posted on 9-1-2019 at 15:32


There should not be any chromium present, so I do not understand the concern. All acids catalyze oxidations by chromate, why should the oxalic be any different?



[Edited on 04-20-1969 by clearly_not_atara]
View user's profile View All Posts By User
Felab
Hazard to Self
**




Posts: 75
Registered: 9-11-2018
Location: Spain
Member Is Offline

Mood: Españó!!!

[*] posted on 9-1-2019 at 22:32


Quote: Originally posted by AJKOER  
Some research (see https://www.researchgate.net/publication/231458156_Catalysis... ), to quote:

"Oxalic acid acts as a very effective catalyst for the chromic acid oxidation of iodide. At low concentrations, the reaction is first order in oxalic acid, iodide, chromic acid, and hydrogen ions, but becomes zero order in iodide at high iodide concentrations. The proposed mechanism consists of the formation of a termolecular complex (CO2)2CrIO2- formed from oxalic acid, chromic acid, and an iodide ion followed by its decomposition into an iodine atom and a chromium(V)-oxalic acid complex, (CO2)2CrO2-. It is assumed that the catalytic activity of oxalic acid is due to its ability to stabilize chromium(V)"

suggests to me the possibility that H2C2O4 in the presence of metal impurities and I- may be driving a reaction that is not supporting HI creation.


My potassium iodide nor oxalic acid should not contain chromium compounds.
View user's profile View All Posts By User
Doc B
Hazard to Self
**




Posts: 98
Registered: 5-4-2012
Member Is Offline

Mood: No Mood

[*] posted on 12-1-2019 at 12:40


Quote: Originally posted by Sulaiman  
Just had a read ... pp288 Brauer (good memory Boffis!)
I would like a little HI(aq) and I have diy FeS so I could follow this procedure ...
what is the benefit the over the phosphoric acid route ?

(which is mentioned under regeneration of HI(aq))


The regeneration method mentioned doesn't use phosphoric acid. It uses hypophosphorous acid (H3PO2), which is converted into phosphorus acid.
The phosphoric acid method involves it's use to oxidise an iodide salt.
View user's profile View All Posts By User
clearly_not_atara
International Hazard
*****




Posts: 1588
Registered: 3-11-2013
Member Is Offline

Mood: No Mood

[*] posted on 12-1-2019 at 20:33


HI can be produced from iodides using phosphoric acid, which is what he was referring to. It can also be used as a reduction catalyst with a sacrificial reductant like hypophosphorus acid, but that reaction is not what we were talking about originally, I think.

Incidentally, the SEP suggests that titanium (III) perchlorate could be used instead of hypophosphorous acid for this reaction. Counterions other than perchlorate or triflate will neutralize HI.

[Edited on 13-1-2019 by clearly_not_atara]




[Edited on 04-20-1969 by clearly_not_atara]
View user's profile View All Posts By User
Assured Fish
National Hazard
****




Posts: 305
Registered: 31-8-2015
Location: Noo Z Land
Member Is Offline

Mood: Misanthropic

[*] posted on 12-1-2019 at 21:07


This might be a bad idea, but how about dissolving iodine in a non polar solvent like hexane and then adding this to an addition funnel.
In a second flask add hexane and aluminium foil and attach a reflux condenser and immerse the flask in an ice bath.

Then proceed to very slowly add the iodine hexane solution to the aluminium. Hopefully what would happen is the iodine would react exothermicly with the aluminium forming aluminium iodide, the issue with this is that it will take some time for the iodine to get through the pacification layer on the aluminium.
Alternatively you may want to prepare aluminium iodide via the classic method which is messy and low yielding.
I would expect AlI3 to precipitate out in the hexane as i doubt aluminium iodide is soluble in hexane.

You would then want to add water to the aluminium iodide solution, this should hydrolyses slowly forming the hexahydrate, to this you might be able to add ammonia to get ammonium iodide and ammonium aluminate.

This might not work however as im not sure if ammonia is a strong enough base.




Sufficiently advanced science is indistinguishable from madness.
View user's profile View All Posts By User
draculic acid69
International Hazard
*****




Posts: 544
Registered: 2-8-2018
Member Is Online


[*] posted on 12-1-2019 at 21:31


Hypophosphorus acid is diluted with water 50/50, and chilled down in the freezer this is then reacted with iodine made from your iodide salt.the solution is then distilled to give Hydroiodic acid.

OR

Phosphorus acid is heated with elemental iodine made from iodide salt.the solution is then distilled to give Hydroiodic acid.

OR

phosphoric acid and iodide salt are heated to 3-400'c to distilled to give Hydroiodic
acid.

The h3po2 rxn is rather violent.the h3po3 rxn requires a little bit of heat to get started but is much gentler.
View user's profile View All Posts By User
Felab
Hazard to Self
**




Posts: 75
Registered: 9-11-2018
Location: Spain
Member Is Offline

Mood: Españó!!!

[*] posted on 13-1-2019 at 12:02


Quote: Originally posted by Assured Fish  
This might be a bad idea, but how about dissolving iodine in a non polar solvent like hexane and then adding this to an addition funnel.
In a second flask add hexane and aluminium foil and attach a reflux condenser and immerse the flask in an ice bath.

Then proceed to very slowly add the iodine hexane solution to the aluminium. Hopefully what would happen is the iodine would react exothermicly with the aluminium forming aluminium iodide, the issue with this is that it will take some time for the iodine to get through the pacification layer on the aluminium.
Alternatively you may want to prepare aluminium iodide via the classic method which is messy and low yielding.
I would expect AlI3 to precipitate out in the hexane as i doubt aluminium iodide is soluble in hexane.

You would then want to add water to the aluminium iodide solution, this should hydrolyses slowly forming the hexahydrate, to this you might be able to add ammonia to get ammonium iodide and ammonium aluminate.

This might not work however as im not sure if ammonia is a strong enough base.


You could react iron or zinc with an aqueous iodine solution and ad lithium carbonate to it. The iron/zinc carbonate precipitates and you can recover the LiI

I2 + Zn = ZnI2(aq)

ZnI2 + LiCO3 = LiI(aq) + ZnCO3(s)

I will have to give it a try.

This synthesis was also shown in The Silver Sunbeam, but I didn't notice it.


[Edited on 13-1-2019 by Felab]
View user's profile View All Posts By User
chemister2015
Harmless
*




Posts: 26
Registered: 25-2-2015
Location: Russia, Khabarovsk
Member Is Offline

Mood: No Mood

[*] posted on 1-2-2019 at 02:52


KI + H3PO4 --t--> HI(gas) + KH2PO4
I2 + turpentine ---> HI
I2 + H2S ---> 2HI + S
2I2 + N2H4 ---> 4HI + N2

[Edited on 1-2-2019 by chemister2015]
View user's profile Visit user's homepage View All Posts By User
Felab
Hazard to Self
**




Posts: 75
Registered: 9-11-2018
Location: Spain
Member Is Offline

Mood: Españó!!!

[*] posted on 2-2-2019 at 03:03


Quote: Originally posted by chemister2015  
KI + H3PO4 --t--> HI(gas) + KH2PO4
I2 + turpentine ---> HI
I2 + H2S ---> 2HI + S
2I2 + N2H4 ---> 4HI + N2

[Edited on 1-2-2019 by chemister2015]


Whats your source for the turpentine route?

Turpentine is a mixture of various turpenes which would probably react with the iodine to yield iodoalkanes or something.

Also, I tried the zinc iodide route and it succeeded in making lithium iodide.

[Edited on 2-2-2019 by Felab]
View user's profile View All Posts By User
chemister2015
Harmless
*




Posts: 26
Registered: 25-2-2015
Location: Russia, Khabarovsk
Member Is Offline

Mood: No Mood

[*] posted on 2-2-2019 at 04:14


Quote: Originally posted by Felab  


Whats your source for the turpentine route?


Patent USA US1,380,951
View user's profile Visit user's homepage View All Posts By User
 Pages:  1  2

  Go To Top