Sciencemadness Discussion Board - Easy synthesis of hydroiodic acid

Easy synthesis of hydroiodic acid

Felab - 30-12-2018 at 01:15

Hi! I´m in need of some hydroiodic acid not because I wanna make meth but to synthesise lithium, cadmium and amonium iodide (for photographic use).

All the synthesis that where provided by the silver sunbeam book where either out of my reach or extremely dangerous.

This morning, I was thinking about a way to synthesise it and I camed up with an idea: reducing elemental iodine by the means of ascorbic acid. Sure enough, I tried ading some ascorbic acid to iodine in a beaker and the solution turned deep red at first but then it cleared up as the ascorbic acid had time to work. The final solution was a very faintly yellow liquid (almost colorless) with a hidrochloric acid like odor (although slightly different).

Now, should I try this at a larger scale? Am I generating hydroiodic acid or some other weird complex? Is there any other method to produce hydriodic acid without dangerous or dificcult to get chemicals?

Tsjerk - 30-12-2018 at 01:52

Maybe reduce with sodium dithionite and seperate NaI by crystalization of sodium sulfate, followed by acidifying with sodium hydrogen sulfate or sulfuric acid, (another crystallization of sodium sulfate in the case of sodium hydrogen sulfate) and then distillation?

Felab - 30-12-2018 at 02:19

I already have potassium iodide and the problem is that the sulphuric acid oxidises it. I- is very easily oxidised.

Tsjerk - 30-12-2018 at 03:23

Then what about phosphoric acid?

wg48temp9 - 30-12-2018 at 03:24

Quote: Originally posted by Felab

I already have potassium iodide and the problem is that the sulphuric acid oxidises it. I- is very easily oxidised.

What precisely do you need the iodide for that potassium iodide will not work?

Felab - 30-12-2018 at 03:56

Quote: Originally posted by wg48temp9

Quote: Originally posted by Felab

I already have potassium iodide and the problem is that the sulphuric acid oxidises it. I- is very easily oxidised.

What precisely do you need the iodide for that potassium iodide will not work?

The problem is that potassium iodide is almost insoluble in absolute ethanol and ether. Also, collodion made with it is quite unsensitive to light.

Felab - 30-12-2018 at 03:58

Quote: Originally posted by Tsjerk

Then what about phosphoric acid?

It's what's normaly used but I don't have any. I could buy but only by the liter so I want an alternative with chemicals that I already own.

draculic acid69 - 30-12-2018 at 05:12

HI from h3po4 requires 300+'c to achieve any result.

WGTR - 30-12-2018 at 08:38

Since you want to make ammonium iodide anyway, if you can source (or make) some ammonium phosphate (fertilizer), you may be able to sublimate ammonium iodide by heating a mixture of the phosphate and potassium iodide. There would probably be some loss of ammonia from this, with the resulting production of some HI directly. You could also use ammonium sulfate, but as you mentioned before, may end up oxidizing the iodide somewhat.

Attachment: chaiken1962.pdf (726kB)
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Felab - 30-12-2018 at 13:01

Quote: Originally posted by WGTR

Since you want to make ammonium iodide anyway, if you can source (or make) some ammonium phosphate (fertilizer), you may be able to sublimate ammonium iodide by heating a mixture of the phosphate and potassium iodide. There would probably be some loss of ammonia from this, with the resulting production of some HI directly. You could also use ammonium sulfate, but as you mentioned before, may end up oxidizing the iodide somewhat.

The thing is that ammonium iodide yields a very sensitive to light collodion, but very unstable, decaing only in a few weaks. To arrest this effect, cadmium iodide is usualy added, but the cadmium iodide is an extremely water soluble cadmium salt, so therefore it is very toxic. Also, you need hydroiodic to make it. I would prefer to make lithium iodide and bromide, which yields a very good, almost non toxic collodion.

AJKOER - 1-1-2019 at 15:31

My old prior thread on the prep of HI may be of interest at http://www.sciencemadness.org/talk/viewthread.php?tid=18302&... .

Especially:

Quote: Originally posted by simba

Quote: Originally posted by blogfast25

Same with the 'bisulphite method': if it was that easy we'd be using it NOW. Often these ideas work only on paper and for very weak solutions...

Indeed, it sounds too good to be true. I don't know if it is a good method for producing HI because I haven't tried it, but it does generate HI afterall.

I2 + NaHSO3 + H2O → 2 HI + NaHSO4

where sulfite salts are available from home brewing sites.
-------------------------------------------------------------------

A path to test (see Patent https://patents.google.com/patent/US5520793A/en ) would be via the action of the reducing hydrogen atom radical. For example, with I2 and H2O, I would describe the system by the reactions:

I2 + H2O = H+ + I- + HOI

HOI + .H = H2O + .I (pH acidic dependent products)

.I + .I = I2

........(other products)

where, for example, a path to the hydrogen atom radical is:

H+ + e- = .H

and the e- is sourced from electrolysis per patent above.

Interestingly, one could apply the same argument to the so called 'bleach battery' employing Aluminum metal and HOCl (from weakly acidified bleach) where the products are Cl2 and H2O:

Al --> Al3+ + 3 e-

e- + H+ = .H

HOCl + .H = H2O + .Cl (non-alkaline conditions, see note below)

.Cl + .Cl = Cl2
...

where the e- are sourced in an electrochemical (or battery) cell and not from an external source of electricity as above.

Note, a more accurate radical path is likely:

HOCl + e- = .ClOH- (see Table 1.2a: Chlorine at https://www.bnl.gov/isd/documents/92710.pdf and Supplement Table S1, link below, Reaction [1], and [116] for Iodine)

.ClOH- + H+ = H2O + .Cl (pH < 5, k = 2.1×10^10 Source: Supplement Table S1 at: http://www.mdpi.com/1420-3049/22/10/1684 and click on Supplement F1 to download)

[Edited on 2-1-2019 by AJKOER]

Felab - 2-1-2019 at 00:12

The bisulfite method might work but in a very diluted solution because hydroiodic acid converts the bisulfite into sulfurous acid, which imediatly disproporcionates into water and sulfur dioxide, consuming the reagent and stopping the wanted reacction.

HI + NaHSO3 = NaI + H2SO3

H2SO3 = H2O + SO2(g)

If you get a yield of hydroiodic acid it will probably be terrible.

AJKOER - 2-1-2019 at 06:06

I view the reaction more like:

I2 + H2O = H+ + I- + HOI

HOI + HSO3- = H+ + I- + HSO4-
------------------------------------------------------
Net: I2 + H2O + HSO3- = 2 H+ + 2 I- + HSO4-

The actual reaction is more complex, as I have ignore the disproportionation reaction of HOI to HIO3. To quote a reference:

"Under acidic conditions Iodate reacts with Iodide to make Iodine, then immediately the Iodine reacts with the Sodium Sulfite until there is no more Sodium Sulfite. Then the Iodine reacts with the water and the Soluble Starch and that is what forms the blue complex."

Link: http://cssf.usc.edu/History/2002/Projects/J0512.pdf

[Edited on 2-1-2019 by AJKOER]

Ogannessionn - 3-1-2019 at 00:01

H2 + I2 = HI

Felab - 3-1-2019 at 00:49

Quote: Originally posted by Ogannessionn

H2 + I2 = HI

* H2 + I2 = 2HI*

I would't try having a controled hydrogen fire.

lordcookies24 - 3-1-2019 at 11:07

then I would use phosphoric acid

edit: I said something dumb that is wrong

[Edited on 3-1-2019 by lordcookies24]

clearly_not_atara - 3-1-2019 at 13:35

Quote: Originally posted by Felab

[Phosphoric acid is] what's normaly used but I don't have any. I could buy but only by the liter so I want an alternative with chemicals that I already own.

Well, if you don't mind doing extra work,

- you can make some p-toluenesulfonic acid by heating sulfuric acid and toluene. That will liberate HI without any oxidation.

- Alternatively if magnesium sulfate (Epsom salt) is available you may be able to perform some salt metathesis with potassium iodide, precipitating K2SO4, then boil down aqueous MgI2 which will eventually liberate HI. This method will have a poor yield, but it should give you some.

- Or if ammonium phosphate is available you can simply heat it to drive off ammonia and leave behind phosphoric acid. Beware of a very bad smell. It is present in fire extinguishers but usually as a mixture with the sulfate.

- If sulfamic acid (rust remover) is available, that might work.

- If zinc is available, you can try Zn + I2 >> ZnI2, and then dry distillation of the hydrate should liberate HI to leave behind a hydroxyiodide.

Overall I'd buy a liter of H3PO4. Sounds much easier than any of this crap.

Felab - 4-1-2019 at 13:52

I think I will try the ascorbic acid methode.

Also, I want high yeld since I don't have a lot of iodine.

If I wanted to make more than 40 ml I would probably buy the phosphoric acid, but photographic applications don't require much iodide.

[Edited on 4-1-2019 by Felab]

Boffis - 6-1-2019 at 08:17

What is wrong with the traditional method of using Hydrogen sulphide and iodine. The hydrogen sulphide can be generated from iron sulphide by the usual process and passed into a suspension of iodine in water. Providing the rate of hydrogen sulphide generation is not too rapid and an inverted funnel is used to do the gassing then practically not H2S escapes until almost all of the iodine has gone. The milky suspension is then boiled to expel any excess H2S and also to coagulate the sulphur, to make it easier to filter, and cooled. The solution is normally filtered and then distilled but you could probably just filter it to remove the sulphur.

This process is described in detail somewhere though I can't remember exactly where, try Brauer though (SM library)

Sulaiman - 6-1-2019 at 09:25

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))

AJKOER - 7-1-2019 at 05:21

The marginal difference in efficiency of the reactions:

HOI + HSO3- = H+ + I- + HSO4-

HOI + H2S = H+ + I- + H2O + S

does not warrant, especially in a home lab setting, of employing H2S over H2SO3 (aqueous SO2)!

Some possible scenarios when using H2S:

1. Your home (and you) smells like rotten eggs, and you hope nobody notices.

2. You dump the residue H2S rich solution and it flows into your neighbor's yard, who then calls the county to investigate a sewage leak.

3. You have investigators knocking on your door and the first point applies.

4. Law enforcement arrives and there don't even bother to knock....

5. H2S is quite toxic (see https://www.ncbi.nlm.nih.gov/books/NBK208170/ ), in the range of 10 to 20 times the lethal dose of HCN (see https://www.cdc.gov/niosh/idlh/74908.html )!

6. At higher exposure levels (which includes both inhalation and possible skin exposure, see http://www.sciencemadness.org/talk/viewthread.php?tid=10491&... ), you may no longer be even able to smell it (making it more deadly), but others can and you are unaware of the ongoing danger!

7. Because you are still apparently alive, does not mean you have NOT received a fatal dose (a case of the walking dead, with eventual organ failure progressing)!

8. For those poisoned and surviving, here is an interesting abstract (fom https://www.ncbi.nlm.nih.gov/pubmed/1772008 ):

"A shipyard worker was poisoned by hydrogen sulfide (H2S), and rescued after 15-20 min. He regained consciousness after 2 days. Three days later his condition deteriorated, and he was more or less comatose for a month. When he woke up, he was amnesic, nearly blind, had reduced hearing, and had a moderate spastic tetraparesis and ataxia. Two months after the accident, he had greatly improved. Audiograms showed hearing loss with maximum at 2000 Hz and significantly poorer speech discrimination. EEG showed generalized dysrhythmia. At follow-up 5 years later he had not been able to resume his work, and had slight motor, memory and visual symptoms. CT and MRI showed slight cerebral atrophy. EEG and evoked responses were normal."

While with SO2:

1. A transient smell due to some home brewing, your new legal hobby.

[Edited on 7-1-2019 by AJKOER]

Felab - 7-1-2019 at 06:16

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 silver sunbeam details this procedure but I only want to try it as a last resource because H2S is a pain to work with and it is also incredibly toxic.

Ascorbic acid methode went wrong

Felab - 7-1-2019 at 06:23

This morning I went to try the ascorbic acid methode I described.

Everithing was going well: a slightly yellow liquid came over and got redder and redder as the distilation proceeded.

When I stopped the distilation I smelled the liquid and it smelled like toluene? I was stuned by that and then I tested acidity and it wasn't acidic . I have no idea of what this thing is.

If someone has, please tell me.

Boffis - 7-1-2019 at 11:47

@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.

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.

Clearly ascorbic acid is not a great route either.

clearly_not_atara - 7-1-2019 at 12:10

HI reacts with ascorbic acid to make iodoalkanes. That's probably what you smelled.

I do not recommend working with H2S as, based on your other posts here, you are not skilled enough to handle a substance with an LC50 below 1000 ppm. Inappropriate technique or poor confinement when generating H2S may cause a severe health hazard to both you and your neighbors.

Felab - 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.

clearly_not_atara - 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]

Felab - 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.

clearly_not_atara - 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.

Boffis - 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.

Felab - 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]

Felab - 9-1-2019 at 13:47

Quote: Originally posted 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.

AJKOER - 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

AJKOER - 9-1-2019 at 14:09

Quote: Originally posted by Boffis

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]

clearly_not_atara - 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.

AJKOER - 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.

clearly_not_atara - 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?

Felab - 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.

Doc B - 12-1-2019 at 12:40

Quote: Originally posted by Sulaiman

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.

clearly_not_atara - 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]

Assured Fish - 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.

draculic acid69 - 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.

Felab - 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]

chemister2015 - 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]

Felab - 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]

chemister2015 - 2-2-2019 at 04:14

Quote: Originally posted by Felab

Whats your source for the turpentine route?

Patent USA US1,380,951

chloric1 - 14-12-2022 at 20:35

I have an elegant suggestion. Barium iodate is relatively insoluble. Barium hydroxide is easily made from Barium chloride and potassium hydroxide. Barium hydroxide is quite soluble in near boiling water so you could just iodize hot or boiling saturation barium hydroxide and 5/6 of it will be barium iodide. The iodate should settle. Add an equimolar amount of elemental iodine to the cooled and filtered barium iodide solution to get the barium triiodide solution. Prepare a gas generator bottle with sodium metabisulfite solid and add 30% sulfuric or hydrochloric acid to pipe in sulfur dioxide into the barium triiodide. Do so until all barium is precipitated as sulfate. You should have a diluted hydriodic acid solution quite capable of making iodides with. The 57% acid is really for organic chemistry.