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conducter
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57% Hydriodic Acid Solution
For a certain reaction what if someone needed 57% HI(aq.) and could not purchase it.
But can obtain lots of Red phosphorus, Iodine, and Water.
BTW, this is not for methamphetamine so those that think just mix them all in situ thats not what is being asked.
Is there a way to mix red phosphorus/iodine/water in a ratio that would create a 57% HI solution that can filter off the red phosphorus and store?
Or if it would be easier to just use KI with H3PO4 thats simpler than id love to know.
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not_important
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Phosphorus, iodine, and water will give a mixture of phosphorous acid and hydroiodic acid. You want a slight excess of phosphorus over iodine.
I remeber the standard procedure was to place a layer of sand in a flask, then iodine and red phosphorus. Fit with a two hole stopper, one leads
through a drying tube filled with glass wool coated with moist red phosphorus. The other holds a thistle tube that leads to the bottom of the flask,
or is configured with a dropping funnel. Water is slowly added through that, the rate is controlled so that the reaction remains under control.
Hydrogen iodide is formed, the drying tube + red P removes I2 from it, then the HI is absorbed in water. Use a series of absorption bottles lets you
saturate most of them to get full strength hydroiodic acid without have the free gas escaping.
If you have good clean phosphoric acid, it might be easier to go that route.
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solo
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Why not try the hypophosphorous (50%) and iodine to make HI......solo.
It's better to die on your feet, than live on your knees....Emiliano Zapata.
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nightflight
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I like -well it´s overstated, but I succeded, so I liked the H2S reaction in water with I2.
Stinks like hell, but is joyful to watch the I2 dissapper and the purple/black solution getting to a clear, sulfur laced liquid, which is then
filtered through a funnel with a glass fibre plug. HI aq. is always very useful to have and here it seems to be more popular than HBr. Add some RP and
it will be stable or regenerated in in situ recations.
[Edited on 8-3-2007 by nightflight]
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Chris The Great
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The 57% concentration is an azeotrope, so purification to that strength is simple through distillation once you have your crude acid solutions (which
likely will not be 57% concentration).
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chemrox
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phosphorous forms
I have phosphorous in four forms:
red P
Phosphoric acid crystals (anhydrous?)
Phopsphoric acid reagent
polyphosphoric acid
Of these which would be the best to make HI from?
How?
Thanks in advance, I too could probably use some HI one of these days.
I only know the P/I2/water and water/I2/H2S bubbler methods and as i understand it the H2S method requires a rapid flow of gas
[Edited on 23-3-2007 by chemrox]
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sonogashira
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As i understand it, it can simply be distilled from a solution of an iodide salt in phosphoric acid.
Having said that, i have not actually done it. And, i confess, my primary source of information is wikipedia!!
I would be interested in hearing if anyone has made it this way, since the other ways seem so much harder in comparison- both in terms of the
difficulty in acquiring the reagents, and the practical set-up.
Is this method not as easy as it seems?
As it happens, i am not particularly interested in this substance- but i have wondered for a while why this (seemingly) simple preparation is not the
first to be suggested.
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solo
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There is always the classic synthesis by Argox from the Hive......solo
Attachment: Hydriodic Acid--Step by Step Write-Up-Argox.pdf (146kB)
This file has been downloaded 5742 times
It's better to die on your feet, than live on your knees....Emiliano Zapata.
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sonogashira
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Thanks solo.
Perhaps a stupid question, but still(!): Would Hydroiodic acid react with copper (as in a home-made copper-pipe distillation set-up)?
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solo
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I would think so ........it eats through aluminum and steel stands.....solo
It's better to die on your feet, than live on your knees....Emiliano Zapata.
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UnintentionalChaos
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But hydroiodic is not an oxidizing acid, correct, just like the other hydrohalic acids. As long as the pipe is clean and does not have any oxide on
it, I don't see why it would eat through it. Copper is below hydrogen on the electromotive series, so 0 state copper chould be immune to attack
because Hydrogen prefers to be an ion more than copper likes to become one.
[Edited on 3-24-07 by UnintentionalChaos]
Department of Redundancy Department - Now with paperwork!
'In organic synthesis, we call decomposition products "crap", however this is not a IUPAC approved nomenclature.' -Nicodem
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not_important
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Can you keep the interior of the still oxygen free? is the copper oxygen free? If not then you'll get some corrosion. And don't forget 2 HI
<=> H2 + I2 which can give a trace of free iodine even without O2.
And it's more than a question of electromotive series, as copper forms complexs with the halides.
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12AX7
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Maybe a good reducing agent in the process.....SO2 gas maybe? Of course, it contaminates your distillate... but it's that or you get green 
Tim
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conducter
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hypophosphorous acid last time i checked was damn hard to get ahold of
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Hilski
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H3PO3 in water + I2 and heat works pretty well, and the reagents aren't hard to get. And there is no red phosphorus to have to filter out.
\"They that can give up essential liberty
to obtain a little temporary safety
deserve neither liberty nor safety. \"
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guy
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How do you get phosphorous acid?
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woelen
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| Quote: | Originally posted by sonogashira
As i understand it, it can simply be distilled from a solution of an iodide salt in phosphoric acid. |
I have tried making HI from KI and H3PO4. I used 85% H3PO4, first boiled it to get rid of most of the water in it, and then added the KI. It gives HI,
when heated, but the HI is not very pure. I passed the gas through a little tube and collected it over water. The solution was not colorless, but
light brown.
A similar experiment with KBr yielded very nice colorless solutions of HBr in water, but apparently, even with H3PO4, the formation of HI is
accompanied with formation of some I2 also. This might be due to a little oxidation of the iodide by the hot H3PO4, it may also be, because of
oxidation by air, or even decomposition of HI to H2 and I2.
Anyway, getting colorless HI is something which is very hard and I did not succeed in achieving that.
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Hilski
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| Quote: | | How do you get phosphorous acid? |
Despite the fact that all the meth cooks made the stuff somewhat harder to aquire, it can still be purchased without many problems from chemical
suppliers. H3PO3 is widely used in fertilizer formulations for obvious reasons, and is also used by itself in solutions to help prevent root rot from
occurring in plants, since it is an anti-fungal. That is actually why I bought the H3PO3 that I had.
It can also be made easily from potassium phosphite, which is sold as liquid foliar spray in some garden centers.
\"They that can give up essential liberty
to obtain a little temporary safety
deserve neither liberty nor safety. \"
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guy
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| Quote: | Originally posted by woelen
| Quote: | Originally posted by sonogashira
As i understand it, it can simply be distilled from a solution of an iodide salt in phosphoric acid. |
I have tried making HI from KI and H3PO4. I used 85% H3PO4, first boiled it to get rid of most of the water in it, and then added the KI. It gives HI,
when heated, but the HI is not very pure. I passed the gas through a little tube and collected it over water. The solution was not colorless, but
light brown.
A similar experiment with KBr yielded very nice colorless solutions of HBr in water, but apparently, even with H3PO4, the formation of HI is
accompanied with formation of some I2 also. This might be due to a little oxidation of the iodide by the hot H3PO4, it may also be, because of
oxidation by air, or even decomposition of HI to H2 and I2.
Anyway, getting colorless HI is something which is very hard and I did not succeed in achieving that. |
H2 + I2 <---> 2HI
The K is slightly less than 1 so I always decomposes a bit, especially with heat.
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LSD25
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Polyphosphoric (or at least a clear-glass phase of metaphosphoric) acid will react with diluted (1:1) Povidone-Iodine to give a clear liquid with a
crappy brown precipate. I have no idea what the clear liquid is (in fact I suspect I do) but perhaps someone who is interested might want to try it
and then find out what the acid is by forming the salts of the same?
Whhhoooppps, that sure didn't work
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l0k1
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while this is somewhat an aside, talking about povidone-iodine, what would happen if you mixed activated carbon powder with the solution until it's a
nasty slurry, let it sit for a few days at room temperature, then wash the carbon to remove the PVP, dry it without heat (press flat on filter paper
and in a plastic box with dessicant) and heat inside a sublimation recrystallisation vessel to yield clean dry I2. I would think that the
electrostatic affinity of I2 to C would be far higher than the far more polar PVP, and thus a sufficient excess would shift almost all of it off given
a little time, since while in solution PVP's bonding with iodine is imperfect hence the brown colour of the solutions, the sign of suspended molecules
of I2.
more on topic to this pvp-I2 thing to going to the acid - wouldn't highly acidic conditions from polyphosphoric acid alter the solubility of the
PVP-I2 complex? my understanding is that normally phosphoric acid can only isolate HI by displacing it off a salt like KI in saturated phosphoric acid
a high enough temperature to force the suspended HI to the surface from its different density. so this brown gak sounds like a likely suspect of the
pvp-i2. i don't know if this would make it any easier at all to do anything further with it since the PVP is likely to degrade and crosslink and turn
into pure obsidian-like impenetrable gak that cannot be removed from glass except with something like an angle grinder.
I am looking at methods involving ion exchange materials and electrolysis or electrophoresis for converting KI directly into I2 or ideally, some way
of electrochemically making the respective ions separate, and then adsorb onto ion exchange materials. The way of making the I2 might be somewhat
elaborate when using chemicals to do this is pretty easy to do anyway.
The one that looks most prospective at the moment is electrolytic separation of potassium iodide in a brine over a bed of zeolite on heat set to about
113C to ensure the iodine immediately sublimes. Above the bath could be sat a cooling surface such as a hemisphere-shaped bowl with ice or dry ice. A
shroud over the hydrogen-producing electrode attached to a hose would ensure no buildup of hydrogen gas.
The direct route to an ion exchange matrix loaded with HI by some method is very appealing however. Once formed there is possibilities of directly
using it possibly in microwave synthesis that should be able to reduce the amount of HI needed without a regenerative reducing agent. Regardless of
this, one could take the dry charged ion exchange matrix and add the appropriate molarity of hydrochloric acid and the two would switch places
yielding a 36% HI solution which can then be concentrated. Alternatively, if a silica-based ion exchange is used, it could be heated to a high enough
temperature that the ions are forced out of the pores into the air and then bubbled through water or some such.
[Edited on 15-11-2012 by l0k1]
also one could mix potassium iodide and an excess of concentrated phosphoric acid and boil through a flask with an ion exchange material in it, then
swap it out by washing with phosphoric acid. the phosphoric acid won't interfere with the way the HI works for the small amount that will end up
remaining. it can also be removed by some means i'm sure. it would not come across in distillation, for example.
[Edited on 16-11-2012 by l0k1]
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MKSStal
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I've read a suggestion to put some copper in HJ solution after preparing it. So, I've got a question if it is just an oxidation reaction inhibitor or
it reacts with already formed iodine:
4 HI + O2 -> 2 I2 + 2 H2O
2 Cu + I2 -> Cu2I2
sum 4 HI + O2 + 4 Cu -> 2 Cu2I2 + 2 H2O
Both possibilities lead to no iodine in solution (making it colourless) but second one uses both copper and acid making it weaker. What do you think?
[Edited on 25-8-2016 by MKSStal]
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Melgar
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Quote: Originally posted by l0k1  | | I am looking at methods involving ion exchange materials and electrolysis or electrophoresis for converting KI directly into I2 or ideally, some way
of electrochemically making the respective ions separate, and then adsorb onto ion exchange materials. The way of making the I2 might be somewhat
elaborate when using chemicals to do this is pretty easy to do anyway. |
Considering how much thought you're putting into this, I may have some idea of what you plan on using that I2 for...
35% H2O2 is $20 a gallon, and chlorine bleach is even cheaper. Why make things more complicated than they need to be? In any case, the type of
iodine you apparently want is not ionic, so you're back at square one anyway.
I guess you could make HI using ion exchange resins, but those resins work best with dilute solutions. So scratch that.
Hell, I'll save you some effort: titrate a KI solution with peroxide until it's clear and there's a bunch of iodine precipitated. Decant the liquid
and filter and dry the iodine. Then make silicon tetraiodide the same way they make silicon tetrachloride on youtube. It's actually even easier,
because the reactants and products are solid at room temperature, so no need to use all that extensive cooling. You'll have to use CO2 to move the
vapors along though, otherwise the iodine will condense as a solid. SiI4 is 95% iodine by mass, and will react extremely vigorously with water to
form SiO2 and HI. You can get HI of any concentration you like this way.
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Cabalaba
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| Quote: Originally posted by woelen | | Quote: | Originally posted by sonogashira
As i understand it, it can simply be distilled from a solution of an iodide salt in phosphoric acid. |
I have tried making HI from KI and H3PO4. I used 85% H3PO4, first boiled it to get rid of most of the water in it, and then added the KI. It gives HI,
when heated, but the HI is not very pure. I passed the gas through a little tube and collected it over water. The solution was not colorless, but
light brown.
A similar experiment with KBr yielded very nice colorless solutions of HBr in water, but apparently, even with H3PO4, the formation of HI is
accompanied with formation of some I2 also. This might be due to a little oxidation of the iodide by the hot H3PO4, it may also be, because of
oxidation by air, or even decomposition of HI to H2 and I2.
Anyway, getting colorless HI is something which is very hard and I did not succeed in achieving that. |
ive read, IIRC, metaphosphoric acid can minimize further I2 formation. i need to double check that though.
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chemrox
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| Quote: Originally posted by not_important | Phosphorus, iodine, and water will give a mixture of phosphorous acid and hydroiodic acid. You want a slight excess of phosphorus over iodine.
I remeber the standard procedure was to place a layer of sand in a flask, then iodine and red phosphorus. Fit with a two hole stopper, one leads
through a drying tube filled with glass wool coated with moist red phosphorus. The other holds a thistle tube that leads to the bottom of the flask,
or is configured with a dropping funnel. Water is slowly added through that, the rate is controlled so that the reaction remains under control.
Hydrogen iodide is formed, the drying tube + red P removes I2 from it, then the HI is absorbed in water. Use a series of absorption bottles lets you
saturate most of them to get full strength hydroiodic acid without have the free gas escaping.
If you have good clean phosphoric acid, it might be easier to go that route. |
This sounds like an OrgSyn entry. Any chance of finding the ref?
"When you let the dumbasses vote you end up with populism followed by autocracy and getting back is a bitch." Plato (sort of)
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