MrDoctor
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Seperating a 50-50 mixture of potassium iodide and chloride
randomly ill sometimes get mixtures of chloride and iodide when i try buying iodide salts online. based on yields, the mixture is almost equal parts
chloride.
I was wondering what would be a nice neat way to seperate the two when its the potassium salt, my best idea so far has just been to reflux in acetone
in the soxhlet extractor.
And if that didnt work out, then i would try using dilute phosphoric acid, which would hopefully simply not react with the chloride at all. Its a bit
tricky searching for reactions that dont happen, as opposed to ones that do.
Most of the time sodium iodide is not as readily available at reasonable prices, but when it is i would go for that since simply washing some warm
acetone through it would quickly demonstrate if its fake iodide salt or not.
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Radiums Lab
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You can try repeated crystallization, but that's tedious with accurate amount of deionized or distilled water you can get a pretty pure product.
Remember temperature too plays a role on its solubility. Even using a seed crystal of pure KI can help getting it out of saturated solution.
Water is dangerous if you don't know how to handle it, elemental fluorine (F₂) on the other hand is pretty tame if you know what you are doing.
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MrDoctor
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i dont know how i missed that they play so differently in water, i must have read one in g/100ml and the other in g/L, since they are 10x different
and would then look the same.
my understanding though is one of the major driving forces is the formation of crystals, for getting a pure product, but the product i want, is KI,
which shouldnt really be crystalizing until it is the only thing left, assuming no even MORE soluble fillers.
How thoroughly would the chloride be removed by precipitation once the KI is saturated to the point of also precipitating?
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teodor
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It's usually hard to clean separate halogen salts. You can check Brauer for the methods, I remeber there is a method for purifying KBr. But I
personally think adding CuSO4 and distilling I2 from this would allow clean separation of anions. Some part of the rest can be precipitated as
insoluble CuI. But hardly you can get that CuI completely free from Cl-, so that part will require some additional treatment.
[Edited on 12-8-2025 by teodor]
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woelen
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I would take a mix of some acid (dilute HCl and dilute H2SO4 both do the job), mixed with H2O2 and add a slight excess of that to a solution of your
KI/KCl mix. Iodine will precipitate from the solution, and if you use slight excess of H2O2 and assure that there is enough acid, then hardly any I2
remains in solution. Filter the I2 and add the slurry to concentrated H2SO4. From this, you can boil off the I2 and distill it, forming crystals on a
cold glass surface. In this way you recovered the iodine.
If you don't have concentrated H2SO4, then you can also distill I2, but it is more difficult to separate it from water. You then could store the
distilled, but humid I2 for a while in a container, in which you also put some dry CaCl2. You take a bigger container, put in a container with the I2
and a container with the CaCl2). Keep the bigger cotainer as small as possible to minimize loss of iodine in the form of vapor. The CaCl2 absorbs
water and makes your I2 dry. But it is tedious and takes a long time.
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clearly_not_atara
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I don't think you want to use acetone. KCl is already mostly insoluble in ethanol, while KI should be highly soluble.
KI has some solubility in acetone but significantly less than NaI, which is why the latter is preferred for the Finkelstein reaction.
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Texium
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Do you have any more conclusive proof that you have a mixture of iodide and chloride besides bad yields? Proof of Cl2/ICl generation when oxidized,
for instance? If not, I’m highly skeptical of the claim.
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teodor
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Quote: Originally posted by woelen  | | I would take a mix of some acid (dilute HCl and dilute H2SO4 both do the job), mixed with H2O2 and add a slight excess of that to a solution of your
KI/KCl mix. Iodine will precipitate from the solution, and if you use slight excess of H2O2 and assure that there is enough acid, then hardly any I2
remains in solution. |
Is it possible that by this process ICl will be also formed?
But I think Texium is right. We are answering a wrong question. I didn't read the initial post carefully.
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RU_KLO
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I have also tried to separate iodine from "iodized" salts.
the true is that its impossible (at least for me) because the iodide in the "iodized" salt was near 0.
So I will first suggest a simple experiment. take some salt, add some acid till acidic and add some oxidating agent. if it change colors - red/brown,
it probably has iodine. you could check the intensity to more or less find how much.
Also, not tried, but maybe some could try it:
mix some starch with a solution of your iodized salt.
From WIKI:
The iodine–starch test is a chemical reaction that is used to test for the presence of starch or for iodine. The combination of starch and iodine is
intensely blue-black.[1][2] The interaction between starch and the triiodide anion (I−3) is the basis for iodometry.
after re reading the post, here is some info you could use:
Separation of Iodide, Bromide, and Chloride From One
Another and Their Subsequent Determination
Thomas J. Murphy, W. Stanley Clabaugh, and Raleigh Gilchrist
https://nvlpubs.nist.gov/nistpubs/jres/53/jresv53n1p13_a1b.p...
[Edited on 13-8-2025 by RU_KLO]
Go SAFE, because stupidity and bad Luck exist.
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teodor
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| Quote: Originally posted by RU_KLO | I have also tried to separate iodine from "iodized" salts.
the true is that its impossible (at least for me) because the iodide in the "iodized" salt was near 0.
|
What you are trying is a really interesting chemical challenge.
The problem with the oxidation method as I can imagine is that Cl2 is generated in amount capable of oxidising any I2 further to IO3- ion. See, for
example, possible routes of reaction ICl with water in wikipedia.
The CuSO4 method is based on a peculiar property of instability of CuI2 in water solution, so it doesn't make any chlorine free, but you have a
mixture of I- and I2 which coexists until you heat the solution. In this case the solution demonstrates interesting physical phenomenon that I2 having
formally a bigger boiling temperature than H2O leaves the solution first and can be collected on a cold surface in a form of I2 crystalls.
But I am not aware whether this method is capable to detect I2 in minute quantities like your example with a table salt.
Also, I would not trust any table salt, if I would like to perform this experiment I will mix KI and KCl myself.
P.S. Because 2 CuI2 -> 2CuI + I2 and not so for CuCl2 I believe you can detect a minute quantity of iodine by a volumetric method for determination
of free iodine (oxidation/reduction).
[Edited on 13-8-2025 by teodor]
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MrDoctor
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| Quote: Originally posted by Texium | | Do you have any more conclusive proof that you have a mixture of iodide and chloride besides bad yields? Proof of Cl2/ICl generation when oxidized,
for instance? If not, I’m highly skeptical of the claim. |
Yesterday i tried just crystalizing from water. this turned out to be quite foolish.
From 100g i obtained 11g iodine in the end, extracting from different fractions of the crystalization, with the very end ones producing no precipitate
whatsoever.
the very first test, i mixed 3g of the iodide with 10ml of water then added 6g of oxone dissolved in 10ml of water. the result, was a flask full of
deep deep red gas somewhat reminiscent of nitrogen dioxide, it smelled very clearly of chlorine. afterwards i did extractions with peroxide instead. i
would reduce it clear with bisulfite, and start over, and get an additional quantity indicating that iodate had formed. I think oxone specifically
makes this worse though so i stopped using it. In the past i have usually always used peroxide as well but.
My assumption was, as long as most of it can be concentrated, the ICI losses should be minimal, but this batch turned out to be around only 10% not
50%
i mixed 100g of the salt with 50ml of water and heated past 100C, only a third seemed to dissolve, the resulting liquid worked well compared to
solids, but still yielded poorly. in hindsight i should indeed have used the soxhlet after verifying the presence of chloride.
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woelen
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The H2O2 oxidation method works quite cleanly. No ICl is formed. H2O2 is only marginally capable of oxidizing chloride to chlorine, that only works in
conc. HCl. In dilute acid, this does not happen. All iodide is oxidized though, to iodine. If no iodide is left, then hardly any iodine remains in
solution. For this reason you should add a slight excess amount of H2O2 and you must be sure to have added enough acid.
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