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Felab
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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.
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clearly_not_atara
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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]
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Felab
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| 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.
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clearly_not_atara
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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]
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Boffis
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@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.
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Felab
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| 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]
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Felab
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| 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.
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AJKOER
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| 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.
.......
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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
 .
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AJKOER
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| 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]
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clearly_not_atara
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| 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]
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AJKOER
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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.
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clearly_not_atara
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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]
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Felab
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| 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.
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My potassium iodide nor oxalic acid should not contain chromium compounds.
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Doc B
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| 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.
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clearly_not_atara
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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]
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Assured Fish
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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.
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draculic acid69
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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.
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Felab
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| 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]
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chemister2015
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KI + H3PO4 --t--> HI(gas) + KH2PO4
I2 + turpentine ---> HI
I2 + H2S ---> 2HI + S
2I2 + N2H4 ---> 4HI + N2
[Edited on 1-2-2019 by chemister2015]
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Felab
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| 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]
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chemister2015
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Patent USA US1,380,951
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chloric1
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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.
Fellow molecular manipulator
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