Author Topic: HI pre-prepared out of a tincure...  (Read 4727 times)

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tpower9s2003

  • Guest
HI pre-prepared out of a tincure...
« on: September 04, 2003, 11:18:00 PM »
While looking around UTFSE, SWIT noticed a possibility for iodine tinctures. It seems that nobody has made much mention, probly not the most economical if using 30mL tincs, nonetheless...

The available tincture used in this example has 2%I2, 2.4%NaI, 47% ethyl, rest H20, no extra or hidden gaaks. First NaOH with around the theoretical amount required to convert the I2 to NaI. With the given amount, the tincs eventually turns clear and is heated until dry. Each 30mLs yield totals a couple grams of NaI, possibly some iodate, but that doesn't seem as though it would affect the cleanliness later on, esp since it is being used immediately vs leaving it for hours or days etc. 

Standard muriatic acid (HCL 33%) would be added to the NaI... NaI+HCl > HI+NaCl. A dilute transparent yellow aqueous HI layer forms above a solid bottum salt layer. When done and cool from reacting and layers now well separated, upper layer can be decanted and filtered off and would be fairly clean HI. Boiling this to 127C in some kind of receiver/condensor based distiller should concentrate whats boiling to 57%. In the event that it does happen to be too dirty for use, a reciever could collect the HI once 127C is met instead.  

This would be easier and seems as though it would get more grams worth to rXn use vs. having to convert it all back to I2, esp if H202 is the oxidizer, and then possibly having to recryst that as well to get it clean (resulting in another yield hit). The chances of screwing up concentrating the acid provided reasonable equip appears minimal. It would also significantly reduce the RP needed.

With another method seen, phosphoric acid and KI can get HI, but that method seems far more risky and difficult (ie more prone to failiure for a newbee). The quality of the chemicals necessary also incur more difficult availability for some ppl, inc SWIT.

Comments. Is this a reasonable method?


isopropylcumshot

  • Guest
WOW
« Reply #1 on: September 04, 2003, 11:42:00 PM »
If this works, ISO wants to buy you sex flowers and candy.
I can't wait to hear other peoples feed back.

ISO

Osmium

  • Guest
> Standard muriatic acid (HCL 33%) would be
« Reply #2 on: September 05, 2003, 01:00:00 AM »
> Standard muriatic acid (HCL 33%) would be added to the
> NaI... NaI+HCl > HI+NaCl. A dilute transparent yellow
> aqueous HI layer forms above a solid bottum salt layer

What happens if you add HI to NaCl? Would that produce HCl?  ;)


SHORTY

  • Guest
Actually....
« Reply #3 on: September 05, 2003, 01:06:00 AM »
I would rather make it with phosphoric acid and potassium iodide.  Its really easy to do and potassium iodide is much easier to get than iodine isn't it?

I recycle my iodine and i no longer resublime it with no problems so far. 

If this does work it would bee more work in my opinion.

Also, don't forget that you will still need to have either rp, hypo, or phosphorous acid.


tpower9s2003

  • Guest
SWIT hasn't actually tried this for HI so...
« Reply #4 on: September 05, 2003, 08:20:00 PM »
SWIT hasn't actually tried this for HI so he doesn't know for sure if this is effective. Hes gone all the way to I2, obviously beats trying from a 2%...Back in the I2 faq, this was just a theory based upon what was said by the chief.

A bee mentions "Reactions shemes:
 With H2O2: 2KI+2HCl+H2O2-->I2+2KCl+2H2O"
Chief responds "Your first scheme (with peroxide) is a bit off, you can divide all the figures with 2, and also separating the reaction in two steps, first the acidification and next the oxidation makes it easier to follow:
2 KI + 2 HCl -> 2 HI + 2 KCl
2 HI + H2O2 -> I2 + 2 H2O"


If this is correct then NaI would work the same way as with KI. If chief is right, adding HCL to NaI (Assuming its totally dried without appreciable amounts of additive to NaI hence no doubling theoretical values, etc...) would result in equimolar amounts of HI and NaCl.

Os: If your point was even relevant to this topic, then the products of the reaction would have to be able to react BACKWARDs as quickly as it does fowards (or HI could still be easily removed with minimal loss). This luckily is wrong, and if it wasn't would devalidate the use of HCl in any I2 tinc extraction should the tinc contain any NaI or KI whatsoever (and conveiniently most seem to).  

In SWIT's situation, this would be preferred to H2PO4, KI genre, that if it is even a potential way to HI. More feedback? SWITs gotta go kerplunk ;)


hypo

  • Guest
learn basic chemistry.
« Reply #5 on: September 06, 2003, 08:18:00 AM »
first of all: HI is a stronger acid than HCl.
(generally: Ka(HI) > Ka(HBr) > Ka(HCl) > Ka(HF))

> This luckily is wrong

says who? you?

> and if it wasn't would devalidate the use of HCl in any I2 tinc extraction
> should the tinc contain any NaI or KI whatsoever (and conveiniently most seem to).

learn the concept of equilibrum, pretty please.


tpower9s2003

  • Guest
dammit....mistaken
« Reply #6 on: September 09, 2003, 02:37:00 PM »
"2 KI + 2 HCl -> 2 HI + 2 KCl"

SWIT is surprised that Rhodium was not accurate. More accurate: KI+HCl <-> HI+KCl.
Hypo is right about the equilibrium favoring the weaker acid, a diluted HCl. What Os said was relevant since this be the case. They are both considered strong acids though only a very insignificant amount of HI would be produced. Surely it looks like this method for HI would be a lost cause :(  oh well.

At least evaporating the tinctures after converting I2 to an iodide (and a little iodate), then using HCl followed by the oxidizer will still work MUCH better than any wet tinc extraction. Just got some 7%-473mL tinc totalling over 33g I2 and 23g KI...same price as 6 x 2%-30mL I2/NaI tincs   ;)


Osmium

  • Guest
When you dissolve KI and HCl in water you have
« Reply #7 on: September 10, 2003, 02:00:00 AM »
When you dissolve KI and HCl in water you have a solution containing K+, H+, Cl- and I- IONS. Exactly the same as if you dissolved KCl and HI, there is no difference in the two solutions whatsoever!
There is no "HI" in solution. Neither does the H2O2 reaction require HI to be present, what you need is the presence of I- and a sufficiently low pH for this reaction to take place.


ADDkid

  • Guest
It has been a long time since I mess with...
« Reply #8 on: September 10, 2003, 04:21:00 AM »
It has been a long time since I mess with inorganic, but I belive that the bond with I2 is much stronger then Na+  diatomic bonds are stronger then most people to relize. I once tried to react pure Na with Iodine and failed, if any suggestion please let me know.  Second of all  the equilium does not favor the reaction, if any very minor.  The biggest mistake though is the fact that it seems that you belive that this reaction is entergically favorable.  The release of enery is a postive number I think.  When would me that it would be endothermic.  But I could be wrong. ADDkid

Rhodium

  • Guest
Behavior of ions in solution
« Reply #9 on: September 15, 2003, 12:21:00 PM »
I'm sorry for confusing people around here with what I said in

Post 323629 (missing)

(Rhodium: "Good start!", Stimulants)


What I said in that post is not for general usage, iodide salts will only be converted to hydriodic acid completely and  spontaneously as long as the second part of the equation { 2 HI(aq) + H2O2(aq) -> I2(s) + 2 H2O(l) } is also happening in the same flask (i.e. the iodide ions constantly being removed from the equilibrium by oxidation and subsequent precipitation).

If that isn't happening, then it all results in an ion mixture as described above: KI(aq) + HCl(aq) -> H+(aq) + I-(aq) + K+(aq) + Cl-(aq)


Question to all: How do you keep explanations of reactions involving equilibriums simple enough for people to easily understand them, but precise enough for people to not generalize what's written too far under other circumstances?

halfkast

  • Guest
Bee as general as is possible.
« Reply #10 on: September 15, 2003, 02:31:00 PM »
Bee as general as is possible.

I would say supply a set of conditions after the generalisation to maintain the integrity of it in all cases that you think are relevant.

Everyone makes mistakes...even you  ;)

Rhodium

  • Guest
Hydrogen Peroxide and the Iodine-Iodide Couple
« Reply #11 on: September 20, 2003, 09:50:00 AM »
Here is an article which might be helpful to people who discuss the chemistry involved in the oxidation of iodine tincture (KI/I2) with hydrogen peroxide:

The Catalytic Decomposition of Hydrogen Peroxide by the Iodine-Iodide Couple, Part 2 and 3.
The Rate of Oxidation in Neutral, and in Acid, Solution of Hydrogen Peroxide by Iodine

J. Am. Chem. Soc. 54, 3409-3508 (1932)

(https://www.thevespiary.org/rhodium/Rhodium/pdf/h2o2-i2.pdf)