12-12-01 09:13
No 246818
Notes on Experiments with OTC 57% HI(aq)
Feeling a little stupid at constantly re-inventing the wheel, when so many members know exactly how to do this (because they actually studied chemistry in school), SWIA investigated hydriodic acid.
Checking the criminal code, SWIA did not find HI to be regulated in this country. In the land of the maple leaf peso, hydriodic is not illegal, at least according to a cursory check of the on-line criminal code. (Any information to the contrary would be greatly appreciated. Neither SWIA nor I desire to break any laws.)
SWIA used TFSE and checked Rhodium’s chemistry page for information and methods. TFSE revealed mass confusion. Several members declared the Rhodium Chemistry write-up to be bullshit (but for all the wrong reasons). One bee, SuperAssman, posted a full page of his best chemistry on phosphates declaring that the write-up on the Rhodium Chemistry page was not physically possible--he also stated that terbium, of all people, needed a chem lesson. Then SuperAssman Post 171036 (SuperAssman: "Re: Hydriodic Acid", Stimulants) boastfully concluded by betting both his gonads that HI could not be made from NaI or KI with H3PO4. (Boy, is he gonna feel funny walking around without any balls).
In the face of such disparate views on the Hive about a seemingly simple inorganic reaction, SWIA went to the library. A review of the Kirk-Othmer and chem texts showed that the assertions of several other bees that HI could be made from reacting either HCl or H2SO4 with either KI or NaI to be absolutely wrong. HI cannot be made by reacting KI or NaI with either HCl or H2SO4, which proves Rhodium correct as usual, since he has posted this fact several times, patiently trying to fend off the flood waters of misinformation.
There are only five practical ways to make hydriodic acid, as far as SWIA could determine:
1. React H2 and I2 under pressure and over catalyst at high temperature. (Complicated and expensive engineering. I2 not exactly OTC in these perilous times.)
2. React I2, red P, and water. (Definitely not OTC. Sort of defeats the purpose. However, this is how it’s done in sunny California by cooks in multiple 22L flasks and on a large scale, according to review of news articles.)
3. Pass a gas current of H2S through an aqueous slurry of I2. Filter elemental S. Drive off residual H2S with heat. Distill hydriodic acid. (Second runner up as a possibility, but I2 not OTC and H2S quite deadly. Hydrogen sulfide gas numbs the senses after a minute or so. Then a person no longer smells it. It is just as deadly as hydrogen cyanide(HCN) or arsine (H3As). The H2S generator would have to be perfectly engineered and monitors installed to detect leaks.)
4. React BaI with H2SO4. Forms BaSO4 and HI. (This would be the easiest reaction, except for fact that BaI is rare, expensive, and poisonous.)
5. React KI or NaI with H3PO4, per Rhodium chemistry page. (OTC ingredients, simply reaction, seems perfect, why do so many bees insist it won’t work?)
What to do in the face of such confusion on Hive and such uncertain choices?
SWIA began experimenting.
After a few days the situation became clear(er). The write-up on Rhodium’s page is more-or-less correct inasmuch as HI can be made from KI and H3PO4, but it leaves a lot out, which makes SWIA wonder whether the guy actually did it.
Here’s how to make 57% HI(aq) from OTC ingredients with at least 55% yield mol/mol. Improvements on yield are expected as experience grows. Improvements will be posted.
KI (potassium iodide) can be purchased OTC. It is a heavy white powder. It is currently being sold online as anti--radiation prophylaxis at exorbitant prices. Check it out, but don’t buy it as pills--too expensive. Use the yellow pages and you will find it in abundant supply. SWIA obtained (USP) KI for $us 36.75/kg retail, which isn’t too much of a mark-up over its world market price of $us 25/kg. Don’t PM me for sources, I personally don’t know anything about this, just what SWIA tells me.
H3PO4 (Ortho-phosphoric acid) can be purchased as “pH Down” from hydroponics stores. Technical grade is 75%. A clear syrupy liquid. Below is a chart showing b.p. of H3PO4 at various grades. A simple way to determine the concentration of your H3PO4 is to measure the boiling point. SWIA obtained 75% H3PO4 for $15/gallon, which is high retail. Bulk will be less. The guys at the store had no clue as to the concentration of the “pH Down”, so SWIA boiled it and discovered it was 75%. DO NOT use the tile cleaner containing phosphoric acid as was recommended in several posts--this tile cleaner is only 20% H3PO4 and 25% surfactants (soap) and other additives--it will NOT work.
Per Cent Boiling Point
H3PO4 ºC
10 100.2
20 100.8
30 101.8
50 108.0
75 135.0
85 158.0
105 **1
115 **2
**1 --A true boiling point can not be given, since, in the liquid state, 105% phosphoric acid reverts to a mixture of ortho-phosphoric acid and condensed phosphoric acids. At 325ºC, the vapor pressure is 760mm.
**2 --At 532ºC the vapor pressure is 760mm.
The write-up on Rhodium’s chemistry page infers the following reaction mechanism:
KI(s) + H3PO4(s) => HI(g) + KH2PO4(s)
However, this does not actually happen in SWIA’s experience. SWIA does not know what the reaction mechanism is, perhaps the formal chemists could help out. But the one inferred on the Rhodium page is definitely not it. Besides, crystalline H3PO4 is not easy to find. And if liquid H3PO4 was used by the person writing the synth, then his (her) write-up is dead wrong on several key points.
As best SWIA can fathom, the initial reaction mechanism is like this:
KI(s) + 2H3PO4(aq) + delta t(65ºC) => HI(aq) + KH2PO4•H3PO4 (aq)
Some kind of reaction is readily observed. Upon heating, the mixture quickly turns the characteristic brown color of hydriodic acid and this acid will boil over at 125-127ºC and can be condensed and recovered. According to the above reaction (and subsequent reactions--see next paragraph), from 1 to 2 moles of H3PO4 will react with 1 mole of KI to produce 1 mole of HI. However, this does not occur--at least SWIA has not worked out the ratios and conditions to make it occur. There must be other reactions unknown to SWIA. The initial brown acid is not even close to being a stoichiometric amount. Once the initial brown acid has been boiled off, other reactions must take place. Hydriodic acid continues to be generated at higher temperature in the flask, although it still comes over at 127ºC, even though the temperature of the clear goo bubbling in the bottom of the flask is over 200ºC. The further formation of acid may be due to dehydration of ortho-phosphoric acid and formation of tri-potassium phosphate, according to this equation which SWIA cannot balance because the waters of hydration of tri-potassium phosphate vary from 2 to 12:
KI +(?)KH2PO4•2H2O+ delta t => HI•H2O + (?)K3PO4•(2 to 12?)H2O
(I know this equation is not balanced--don’t flame it--this is just SWIA’s idea of what might be going on)
Metaphosphates probably form--these are polymers--long chain or ring molecules--also known as condensed phosphates. Phosphate reactions can be complex and much of what happens to ortho-phosphoric acid at higher temperature as it dehydrates and polymerizes is a few steps beyond SWIA’s level of understanding.
Why SWIA cannot achieve 100% recovery of HI is a mystery, since there are H+ ions in abundance with H3PO4. The first H+ is highly active. The second less, the third even less, but they all should react with KI to form HI. Unless iodine is attaching itself to a phosphate molecule somehow, it should all come over as HI, either as hydriodic acid or as HI gas. When SWIA figures it out, he’ll let everybody know. Of course, if a wise bee would tell him, much better, less work.
Contrary to what is stated in the write-up on Rhodium’s page, HI gas is not produced in abundance by reacting H3PO4 with KI. Even reacting KI with 105% phosphoric acid yields little gas. This was proven by experimentation. 105% H3PO4 was obtained by boiling 75% acid to 186ºC and calculating weight loss as water loss (close enough). Ortho-phosphoric acid cannot be boiled to dryness. It polymerizes... Anyway--only about 10% of the iodine comes over as hydrogen iodide--the bulk is hydriodic acid.
In the write-up on Rhodium’s page you are instructed to connect a hose to a flask containing the ingredients, heat up the ingredients to a balmy 60ºC and dip the other end of the hose (connected to an aquarium bubbler) into a flask filled with water and prepare to recover HI gas in the water, weighing the solution until a 57% HI solution (hydriodic acid) has been formed. SWIA finds this hard to believe, given that liquid phosphoric acid (even boiled to 105%) gives mostly hydriodic acid, not hydrogen iodide. Anhydrous phosphoric acid (crystalline powder) is not easy to find, whereas liquid phosphoric acid is plentiful and OTC. SWIA wonders what the conversion rate of KI to HI was using the Rhodium page write-up method, which we have to assume used powder H3PO4.
SWIA’s conducted several experiments--they go like this (qualitative description, not quantitative--that will come later):
KI powder and 75% H3PO4(aq) were combined in a erhlenmeyer flask placed on a hotplate. A still head and condenser were attached to the flask. A receiver was attached to the condenser. After the receiver, a trap was connected; after the trap, a second receiver was attached, which was filled with dH2O to capture any HI gas. All under a fume hood.
The contents of the flask were heated. At 50ºC there was an obvious and immediate reaction. HI could be seen to form, which reacted with the water portion of the 75% ortho-phosphoric acid to form hydriodic acid. Upon further heating, the contents of the flask boiled. The distillate that came over at 125-127ºC was collected. This was 57% hydriodic acid. No HI gas came over while the brown acid was boiling.
[Hydriodic acid is formed from the combination of hydrogen iodide--HI(g)--and water. At a concentration of 57% it forms a constant boiling mixture that will boil at 125-127ºC at sea level. Dilute hydriodic acid boils at lower temperature. A small portion of dilute acid will come over until the temperature at the still head stabilizes at 127ºC. The dilute acid can be further concentrated by reboiling, but there will always be a fraction that is less than 57% and which comes over at a lower temperature. SWIA imagines that the dilute acid could be added to subsequent reactions of KI and H3PO4. Recycled so to speak, and thus no HI(aq) would be lost. Anyway, not much dilute acid is formed. Pretty quickly the temperature at the still head will reach 127º and stay there. The fraction that comes over at 127ºC can be collected as 57% hydriodic acid. Density 1.7. A dark brown fuming liquid.]
Now the curious part begins. Continuing heating, all the brown liquid will boil over and be collected. Nowhere near the theoretical yield of hydriodic acid. What remains in the flask is a viscous clear syrup that continues boiling. Amazingly, from that clear syrup, yellow-brown hydriodic acid will continue to condense and come over. “Amazingly”, because supposedly all the H2O has been boiled off, so in fact the water portion of the hydriodic acid must be coming from the dehydration of H3PO4. A certain amount of HI gas is formed, but not a significant amount. However, there is a tendency toward suck-back, so the stop-cock and trap are important.
Even after the contents of the flask are boiled to near “dryness”, the total amount of hydriodic acid collected (condensed 57% HI•H20 and dilute HI(aq) from the bubbler) is about 60% of theoretical. (The contents of the flask are never solid while heated, but remain syrupy. Upon cooling, a hard, dense white cake forms in the bottom of the flask. This cake is extremely hard and difficult to remove from the flask. It is not very soluble in water.) K3PO4? Or a polymerized metaphosphate combined with the tri-potassium phosphate in a complex lattice? Anybody know?
If the contents of the flask are not boiled to “dryness” a different type of residue forms, more waxy, almost greasy. Both of these residues coincide with descriptions of different types of condensed phosphates, according to the degree of dehydration.
Here’s the million dollar question: Where is the rest of the iodine? Not in the receivers, where SWIA would like it to be, that’s for sure. Does it combine with phosphate? K2HPO4 (perhaps formed by the continued reaction of KI and KH2PO4) is basic, does HI combine with it?
When SWIA solves the mystery of the missing iodine and gets 95%+ yield of HI, a detailed write-up will follow. But for now, even though yield is not at maximum, SWIA can affirm that OTC HI(aq) can be produced from KI and H3PO4 with at least 60% conversion through simple heating and condensing.
SuperAssman owes terbium an apology along with those two gonads. Ouch, that’s gonna hurt. The gonads, I mean.
SWIA would appreciate the comments of informed members. Where is the remaining iodine?
Regards
Argox
(Distinctive Doe)
12-12-01 10:44
No 246835
"from 1 to 2 moles of H3PO4 will react with 1 mole of KI to produce 1 mole of HI."
I don't think so. Only the first Hydrogen will react, not the second, so you need to have enough H3PO4 for at least a one to one molar ratio, but I think excess H3PO4 is what you want. I would try the reaction agin with 1.5 moles of H3PO4 per mole of KI.
The excess Iodine is sitting at the bottom of your flask as KI!!! No amount of ordinary heating will change that.
"SWIA obtained (USP) KI for $us 36.75/kg retail"
That sounds damn cheap to me!! Iodide salts are expensive.
This was a nice write-up.
Do Your Part To Win The War
(Old P2P Cook)
12-12-01 19:42
No 246966
Your problem is likely just not enough water to keep your reaction mix fluid. You would need 200 grams of 70% phosphoric acid just to provide the amount of water contained in 135 grams of 57% HI (from 100 grams of KI) but this does not allow for any water in the reaction pot to keep the reaction mix fluid.
I would guess that just a little over 1 mole of H3PO4 (from 70% phosphoric acid) per mole of KI would be fine but then you should supply an additional amount of water so that you have the amount needed to form 57% aqueous HI plus keep the KH2PO4 dissolved or at least fluid.
So I would try:
100 grams KI (0.6 mole)
100 grams 70% phosphoric acid (0.7 mole)
Now the question is how much extra water you need.
The above is equivalent to 77 grams anhydrous HI, 82 grams KH2PO4, and 30 grams water. Now to form 57% HI from the 77 grams anhydrous HI you will need 60 grams of water and to dissolve the 82 grams of KH2PO4 you would need about 350 grams of water (at room temp). So, it seems that to 100 grams of KI and 100 grams of 70% phosphoric acid you will need to add anywhere from 30 to 400 grams of additional water.
(Newbee)
12-12-01 23:27
No 247037
Terbium: SWIA already thought of that and experimented right from the git-go. A 1.1:1 molar ratio of H3PO4 to KI was put in the flask along with enough additional water (taking into account the water in the 75% H3PO4) to form the exact amount of 57% HI. Guess what happened? A whole flask of dilute acid. The acid boiled over at 101 to 115. Only a small part came over at 127. No, this is not the answer. The reaction mixture is fluid enough at all stages for the ingredients to react with each other. Extra water doesn't seem to help.
Something else SWIA did was add extra water to the mix and then boil it under reflux (thinking that it was just a matter of giving the acid enough time to react with the KI. This did not work either--in fact there was a reversal of the reaction. The brown hydriodic acid that first formed dissapeared. Upon normal condensation, little came over. The usual white salt in the bottom turned into large green crystals.
And SWIA also tried 1.5:1 molar ratio of H3PO4 to KI. No improvement in yield, just more dilute acid.
Has anybody been producing quantities of HI and mind sharing their secret? (As long as it's not rp and I2 or H2S and I2--SWIA knows how to do both of those.)
PM me if you don't want the word to get out. SWIA's a grateful guy.
As regards the price being too cheap to be true. Not really. SWIA never make things up (unlike a lot of poseurs in these forums) or lies (except about his identity, or to the ex-wife). $us36.75/kg ($cnd 58.80/kg) is what SWIA paid, including PST and GST (Canadian taxes). A check on line shows that KI (USP) sells for about $25/kg in bulk. Yeah.. it's cheap. That's why SWIA is interested in figuring out this reaction.
(Old P2P Cook)
12-13-01 00:08
No 247040
Terbium: SWIA already thought of that and experimented right from the git-go. A 1.1:1 molar ratio of H3PO4 to KI was put in the flask along with enough additional water (taking into account the water in the 75% H3PO4) to form the exact amount of 57% HI. Guess what happened? A whole flask of dilute acid. The acid boiled over at 101 to 115.
Yeah, it seems that the HI/water "constant boiling" property is the problem. If you have extra water and distill then just water comes over first until there is only just enough water left to form the 57% material.
(Stoni's sexual toy)
12-13-01 03:57
No 247096
Dry adding the extra water either dropwise during the distillation, or add it in 30-50ml amounts after the distillation ceased and the flask contents have cooled down somewhat. Swirl around, maybe even reflux for a bit, and then continue distilling.
(Newbee)
12-13-01 18:54
No 247344
Osmium: have you ever tried this? Or is this just an idea? SWIA thought about doing that--having an addition funnel on top of the reactor to add water at intervals, but held off because it would fill up with noxious gas. Maybe he'll give it a try.
All of this casting about begs the question, if SWIA has found the reaction to be complicated, in the sense that yield is hard to maximize, why does it sound so easy in the write-up on Rhodium's page? Did the guy actually do the write-up he claims? I wonder...
(Chief Bee)
12-13-01 18:58
No 247346
The writeup is originally from Post 167625 (not existing), but I woul welcome any updates on the method.
(Newbee)
12-13-01 19:52
No 247363
Oh Christ!! I just read the original post. I should have known it was bullshit, but I took it for gospel, since it was on your chemistry page. Damn. Damn. Damn. 81% phosphoric, what a moron! Damn, I really thought it was for real.
OK. SWIA will really dig in and solve this thing. I mean, it is inorganic chemisty, right? Nice and simple inorganic. SWIA was holding off, thinking somebody would point him in the right direction. Maybe there is hope still, I'll tell SWIA to hold off for another day or so, before he accepts only 60% yield, just in case a wise bee helps us out.
(Master Whacker)
12-13-01 21:29
No 247400
Why not try a reduction of eph with in situ. produced HI from H3PO4 and KI? This is a standard reaction in org. syn. and is detailed on rhodiums page. Use excess glacial acetic acid as the solvent and enough H3PO4/KI to produce about 3 moles of HI per mol of eph. Add some red phosphorous and reflux for 12 hours and you should be set!
(Newbee)
12-14-01 17:26
No 247704
Ritter: Thanks for your reply. Believe it or not, SWIA is not making methamphetamine, just hydriodic acid. What other numb-nuts do with the acid is not his concern.
Meth is illegal and ingredients definitely not OTC. Hydriodic acid is legal (in Canada--AFAIK) and ingredients OTC. Profit to be made in both. But reward/risk ratio much better with hydriodic. Does this make sense?
Ritter, a chemist of your calibre surely must know how to maximise the production of HI from H3PO4 and KI. Any suggestions?
(irritable and cranky)
12-14-01 23:21
No 247788
Besides, adding red phosphorus would defeat the purpose entirely.
Bored...
(Hive Bee)
12-15-01 03:09
No 247813
Dry adding the extra water either dropwise during the distillation,
What Uemura wants to contribute is his observation, when he wanted to prepare HBr from KBr and 100% H3PO4 nothing happened. So he decided to add dropwise water and ...
With each drop of water into the hot KBr/H3PO4 mix a push of a white cloud of equous HBr went thru the condenser and Uemura got in the end an almost saturated HBr acid. He never succeeded to get waterfree HBr out of the rxn mix. This should work for HI as well he thinks.
Carpe Diem