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CrEaTiVePyroScience
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Synthesis of Iodine from KI and sulfuric acid question.
Hello,
So Iodine can be synthesised from concentrated sulfuric acid and potassium iodine, since I only have 3g of potassium Iodine I do not want to waste it
so I got a question.
All sites tell that you have to use concentrated sulfuric acid and add that to the iodine solution. But how about using 32% sulfuric acid? Will adding
32% sulfuric acid in larger amounts will also work?
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DJF90
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Concentrated acid is used as it is (slightly) oxidising. You can use dilute acid and an additional oxidant, for example hydrogen peroxide. Dilute acid
on its own will do diddly-squat though.
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AndersHoveland
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If you are concerned that some of the iodide is remaining unoxidized, just add some dilute H2O2 in. I have doubts that 30 percent conc. H2SO4 will be
able to oxidize most of the iodide.
(2) I[-] + (2) H[+]aq + H2O2 --> I2 + (2) H2O
This thread probably should be sent to "beginnings"
[Edited on 15-4-2012 by AndersHoveland]
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woelen
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Indeed, 30% acid does not oxidize iodide to iodine. Using conc. H2SO4 for oxidizing iodide to iodine also is a bad idea, because that reaction is
extremely messy and dirty. If you add conc. H2SO4 to solid KI, then you get a mix of HI, H2S, S8, I2 and SO2. This mix is extremely smelly and looks
dirty brown with yellow specks like bad shit. Isolating the iodine from this dirty 'shit' is not a pleasure at all.
A much better method is to dissolve the KI in 30% H2SO4 and then add dilute H2O2 to this. Try to add a slight excess of H2O2 and do this slowly while
stirring. At a certain point you will see lots of glittering particles of iodine forming in the dark liquid. If you wait long enough, then you see
that the liquid turns lighter and iodine settles at the bottom and another part of the iodine settles near the surface. You can pour the iodine on a
filter (preferrably use a sintered glass filter and if that is not available, then use a dense paper filter). The wet iodine then can be put in a
small beaker to which about three times its own volume of concentrated sulphuric acid is added. When this beaker is heated, then you can see the
iodine melting, forming a nearly black liquid, which remains under the sulphuric acid layer. After this step and solidifying of the iodine, you can
decant the acid liquid (which contains water and left over potassium ions and a small amount of iodine) and add another small portion of conc. H2SO4.
Heating again makes the iodine really clean. After cooling down you can decant the acid again and then rinse with a lot of water. Finally, you end up
with a piece of iodine, which you can wipe dry with a small paper tissue and then break down in parts with a glass rod. You can also keep it as one
piece. The iodine, produced in this way is pure enough for all practical home chemistry experiments and syntheses.
If you want the iodine really pure, then you can heat it and let it sublime on a piece of cold glass. I myself tried this, but I had limited succes, I
stopped the process, because I lost a lot of purple vapor.
The most tricky part of the process described above is the filtering step. You end up with a dark grey mud, which stains everything. Paper becomes
black, skin becomes yellow/brown (like you sometimes see with certain tobacco users) and clothes become brown. That's why the best results are
obtained with a sintered glass filter, from which you can easily scrape the mud. The filter itself is cleaned very easily by rinsing it with a dilute
solution of sodium sulfite or bisulfite.
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MrHomeScientist
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This has worked just fine for me using ~30% battary acid grade sulfuric acid added to KI solution. I didn't need to use an oxidant, just adding the
two liquids immediately precipitated black iodine. Adding some hydrogen peroxide would certainly help to ensure you have oxidized everything fully,
though. See my video on the halogens for where I perform this: http://www.youtube.com/watch?v=OoS3q1C8zts
The iodine bit starts at about 1:00.
Edit: woelen posted this as I was posting.
| Quote: | | At a certain point you will see lots of glittering particles of iodine forming in the dark liquid. If you wait long enough, then you see that the
liquid turns lighter and iodine settles at the bottom and another part of the iodine settles near the surface. |
This is exactly how my setup behaved, without adding any peroxide. Skip to the end of the video to see how the iodine has settled on the top and
bottom. The brown color of the remaining solution is likely iodine still in solution as triiodide, I imagine, so the oxidizer would help to kick this
out. But as you can see, it didn't appear to be necessary to get most of the iodine to precipitate.
[Edited on 4-16-2012 by MrHomeScientist]
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woelen
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I hardly can believe that just adding 30% sulphuric acid to a solution of KI causes oxidation of the iodide to iodine. I have done this kind of
experiments many times and sulphuric acid at lower concentration does not oxidize iodide.
But if in your case it does oxidize the KI, then what would be the left-over product of the sulphuric acid? It clearly cannot be sulfite or sulphur
dioxide, because in the presence of water these immediately react with iodine to make sulphuric acid and hydroiodic acid. It also cannot be sulfide,
because in the presence of water this immediately reacts to form elemental sulphur and hydroiodic acid (in fact, this is an old-fashioned way of
making HI, by bubbling H2S through a suspension of iodine in water). It could be sulphur, but then you should see that as a dirty brown color, mixed
with your grey iodine. But you definately would have noticed this, it makes the appearance of the liquid totally different.
I cannot imagine a single sulphur-compound of lower than sulfate/sulphuric acid oxidation state which could be formed from the iodide/sulphuric acid
mix in aqueous solution.
So, please give a good explanation of what happened in your experiment. The only thing I now can imagine is that you did have an oxidizer in the acid,
either added intentionally (and possibly forgotten that you did), or as an impurity.
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CrEaTiVePyroScience
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@Woelen (and everyone)
So I can't get a hold on 98% sulfuric acid I have made my own acid by electrolysis of copper sulfate and its 32% (determined after tritration). I can
boil it down but I don't want to go that path, accidents with hot sulfuric acid which can happen are really scaring me. I do got 35% of hydrogen
peroxide and the potassium iodine (only 2.4g) so my question is can I make iodine from the three productes I've mentioned or do I really need the 98%
sulfuric acid which I don't got?
Really thanks for the comments!
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MrHomeScientist
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Quote: Originally posted by woelen  | | So, please give a good explanation of what happened in your experiment. The only thing I now can imagine is that you did have an oxidizer in the acid,
either added intentionally (and possibly forgotten that you did), or as an impurity. |
All your points make a lot of sense. That's certainly a possibility - I did this over a year ago so I may have forgotten some details. The video seems
pretty straightforward, though. I am sure, however, that the acid I used was battery acid, bought as a bottle of sulfuric used to refill old lead acid
batteries. It wasn't recovered from old batteries or anything, so I would hope that it should be relatively free of impurities. I think I have a small
amount of KI left, I'll try this again and see if I can replicate the results as soon as I can. I still have some of this same bottle of acid left, as
well as 'liquid fire' 98% sulfuric drain opener, so I'll try both.
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woelen
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| Quote: Originally posted by CrEaTiVePyroScience | @Woelen (and everyone)
So I can't get a hold on 98% sulfuric acid I have made my own acid by electrolysis of copper sulfate and its 32% (determined after tritration). I can
boil it down but I don't want to go that path, accidents with hot sulfuric acid which can happen are really scaring me. I do got 35% of hydrogen
peroxide and the potassium iodine (only 2.4g) so my question is can I make iodine from the three productes I've mentioned or do I really need the 98%
sulfuric acid which I don't got?
Really thanks for the comments! |
Did you read my (long) post? I said that using conc. H2SO4 is not good at
all, you need dilute H2SO4 plus an oxidizer!
@MrHomeScientist: Yes, if you can repeat what you tried, then that would be nice. Just try it at a small scale, so that you don't waste a lot of
chemicals on it. Just 2 or 3 ml of acid, a few 100's of mg of KI, dissolved in 2 ml or so of water and mixing of these solutions should do the job.
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CrEaTiVePyroScience
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I did but you said:
" The wet iodine then can be put in a small beaker to which about three times its own volume of concentrated sulphuric acid is added. When this beaker
is heated, then you can see the iodine melting, forming a nearly black liquid, which remains under the sulphuric acid layer. After this step and
solidifying of the iodine, you can decant the acid liquid (which contains water and left over potassium ions and a small amount of iodine) and add
another small portion of conc. H2SO4. Heating again makes the iodine really clean. After cooling down you can decant the acid again and then rinse
with a lot of water. Finally, you end up with a piece of iodine"
I dont got concentrated sulfuric acid..
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MrHomeScientist
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| Quote: Originally posted by woelen |
@MrHomeScientist: Yes, if you can repeat what you tried, then that would be nice. Just try it at a small scale, so that you don't waste a lot of
chemicals on it. Just 2 or 3 ml of acid, a few 100's of mg of KI, dissolved in 2 ml or so of water and mixing of these solutions should do the job.
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I just had a thought - my KI was very likely contaminated with KIO<sub>3</sub> because of the way I made it (KOH +
I<sub>2</sub>, my video on it is here). Potassium iodate is an oxidizer! I'm not sure what the reaction would be, but perhaps the iodate is enough to cause this to happen?
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nezza
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The reaction between alkali and Iodine is broadly
3I2 + 6OH- -> 5I- + IO3- + 3H2O
Adding acid to the mix of Iodide and Iodate just reverses this reaction and gives iodine
5I- + IO3- + 6H+ -> 3H2O+ 3I2
To make iodine from pure iodide and acid an oxidant is necessary. Dilute (<98%) sulphuric acid is not a strong enough oxidant.
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woelen
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The iodate thing explains it all. You don't have KI, but some mix, which on acidification instantly gives iodine. Nezza explained it well.
If no concentrated sulphuric acid is available, then making iodine still is possible, but isolating it from the wet mud is more difficult. You can
clean it by rinsing it with water and then you can add a nonpolar solvent to extract the iodine and evaporate the solvent. This, however, most likely
gives a less pure product and getting totally rid of the water is more difficult.
[Edited on 16-4-12 by woelen]
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CrEaTiVePyroScience
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Okay thank you very much Woelen.
I am new to chemistry (only 15year old but got a great ambition to learn more in chemistry and at the moment I am just studying science but next year
I will be having 8hours chemistry a week!)
So I am sorry if I asked stupid questions but things are getting quite complicated right now for me=O
Can you please correct this proces if possible.
I got 35% hydrogen peroxide, 32% sulfuric acid , and can obtain basic chemicals (Also got full glass equipment).
So first I dissolve the 2.4g in my 32% sulfuric acid , adding the acid slowly to the potassium iodine while stirring till all the potassium iodine is
dissolved.
Then I add the 35% hydrogen peroxide (how much should I add?)
Then I pour the solution in a larger beaker of lets say 400ml water and decantate it add water again to the iodide that have formed on the bottom and
just filter it, keep the residu and let it evaporate.
Then to purify I just put a beaker on a hotplate with the iodide in and a cold round flask ontop of the beaker so the vapors condens on the beaker ,
scraping off the powder from the beaker and I should have pure iodide right?
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MrHomeScientist
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Thanks woelen and nezza for the detailed explanations. I'll add an annotation to my video explaining what happened. I'm surprised no viewers of my
video caught that until now!
@CrEaTiVePyroScience:
Be careful to not confuse iodide with iodine! They mean very different things.
Iodine - I<sub>2</sub> , the element itself
Iodide - I<sup>-</sup> , part of a compound, i.e. potassium iodide
Iodate - IO<sub>3</sub><sup>-</sup> , another iodine compound mentioned above
I would think you won't need much 35% peroxide at all. In fact you may want to dilute it a bit before using it. Peroxide at that concentration can be
scary stuff - I had a reaction run away from me once because I used 30% rather than 3%. My beaker started spewing clouds of purple iodine vapor and I
lost all my product 
I think woelen would be the authority on this, though, as it was his procedure outlined above.
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woelen
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You indeed must dilute the peroxide. Dilute it with 3 to 4 times of its volume of water.
But I want to add another thing. If you just have 2.4 grams, then expect a low yield, unless you have specialized small-scale glassware. Imagine using
a filter paper and you pour your liquid with just 1.5 grams or so of iodine on it. Most of it will stick on the paper and you hardly won't be able to
recover anything. These losses are called mechanical losses and mechanical losses tend to be nearly independent of the total amount of chemicals used.
In your process the mechanical loss may be 1 gram of iodine. If you start with just 1.5 gram or so, then you loose 2/3 of your iodine. If you start
with 15 grams, then you only loose 6.7%. Other types of losses (e.g. due to solubility of compounds) frequently are percentages of used chemicals, but
mechanical losses are absolute values, depending only on the apparatus used and not on the amount of chemicals (unless your amounts are so small that
all is lost).
So, my advice is that you try to get more KI before attempting to make iodine of it. I am afraid that if you use your small sample of 2.4 grams that
you will loose nearly all of the iodine in mechanical losses.
[Edited on 17-4-12 by woelen]
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Endimion17
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| Quote: Originally posted by CrEaTiVePyroScience | Hello,
So Iodine can be synthesised from concentrated sulfuric acid and potassium iodine, since I only have 3g of potassium Iodine I do not want to waste it
so I got a question.
All sites tell that you have to use concentrated sulfuric acid and add that to the iodine solution. But how about using 32% sulfuric acid? Will adding
32% sulfuric acid in larger amounts will also work?
|
Just one useful bit of info - chemical elements can't be chemically synthesized. Compounds are synthesized, whereas elements are isolated.
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CrEaTiVePyroScience
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@Woelen Well I don't think I will have much mechanical loss because I got beakers from 5ml-800ml.
And I also don't need high yield just enough to do the magnesium/aluminium iodine test. I also got decent filter paper so it shouldn't stick too much
on the paper.
But after I added the h2so4 to the potassium iodide and the h2o2 should I then pour it in a larger container with water to clean the iodine and then
let it settle, decantate the water , add more water and then filter it? Or should I first add it to the filter and then rinse it on the filter itself?
And how much h2o2 (35% that's diluted 4times) should I use when I am using 2.4g potassium iodine
And are any other useful productes formed by the reaction that I may be able to isolate also?
(@endimion17 Makes sense=))
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woelen
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After adding the H2SO4 and H2O2 (please use capitals in chemical formulas at appropriate places) you can add the iodine to more water (e.g. take 50 ml
of water) and allow the iodine to settle at the bottom. Do not use too much water. Although iodine does not dissolve well in water, it still does
dissolve somewhat and as you have only a very limited amount you want to keep your losses at a minimum. You can repeat the rinse with another batch of
water.
I keep it as an exercise for you to compute how much H2SO4 and H2O2 you need for reacting 2.4 grams of KI. First determine the reaction equation
(assume that the H2SO4 is converted to KHSO4) and then determine how many grams of H2SO4 and H2O2 you need. Assure that you have a slight excess of
H2O2, it is much more harmful if you have too little, better take some extra, but do not use a large excess. In your computations you may assume a
density of 1 gram/l for your 35% H2O2 and you may assume a density of 1.2 for your 30% acid. It does not need to be an exact computation, just a rough
estimate and then you take an excess of 25% or so.
Please do the math, and I am willing to have a look at it when you post it over here.
There are no other useful products worth isolating. Of course, you get some KHSO4, but extracting just a gram or so of this chemical from a lot of
water which also has some iodine, unreacted H2O2 and unreacted H2SO4 is not really interesting.
[Edited on 18-4-12 by woelen]
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AJKOER
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Per my notes from an old reference (I believe Mellor around page 118), if you have a mixture of KI and KIO3, consider a reaction with dilute HCl:
5 KI + KIO3 + 6 HCl --> 6 KCl + 3I2 + 3 H2O
Interestingly, if its concentrated HCl, there are different reaction products:
2 KI + KIO3 + 6 HCl --> 3 KCl + 3ICl + 3 H2O
and:
ICl + H2O --> HOI + HCl
adding a bicarbonate should neutralize the HCl and move the reaction to the right (or, increasing the HCl concentration due to the ICl2- complex
formation ICl + Cl - = ICl2-) followed immediately by (if the hydrolysis is not stopped):
5 HIO --> 2 H2O + 2 I2 + HIO3
-----------------------------------------------------------
Note, KI + Cl2 --> KCl + ICl
so adding KI to an acidified NaClO solution (source of Cl2) similarly implies a ICl path (which is inefficient given the Iodate formation) to Iodine.
However, adding more iodide one could take advantage of the first reaction noted above:
5 KI + KIO3 + 6 HCl --> 6 KCl + 3I2 + 3 H2O
to recover all the Iodine. One probably wants to avoid Iodine monochloride as it has a choking smell like a mixture of I2 and Cl2 and strongly
attacks the nose and eyes as well as cork and rubber. ICl is yellow in water and alcohol when Iodine is brown.
-----------------------------------------------
Some old threads mention the use of slightly acidified Bleach (NaClO) to extract I2. Here is guess estimate of the chemistry:
Formation of HOCl with a weak acid (or very dilute mineral acid):
OCl (-) + H(+) --> HOCl
Reaction of the Iodide with hypochlorous acid:
KI + 3 HOCl --> KIO3 + 3 HCl
(Watt's noted the ability of Hypochlorous acid to oxidize Iodine to Iodate. See http://books.google.com/books?id=ijnPAAAAMAAJ&pg=PA18&am... )
Reaction of the Iodate and Iodide (dilute acid equation):
5 KI + KIO3 + 6 H (+) --> 6 K (+) + 3 I2 + 3 H2O
(For more on this reaction see: http://medind.nic.in/iby/t06/i4/ibyt06i4p531.pdf . I explain this reaction as follows: 3I2 + 6KOH <-->5KI + KIO3 + 3H2O, a classic
disproportionation by a halogen in a strong alkali, and upon adding a dilute acid the KOH is removed moving the reaction to the left, but with strong
HCl, a side reaction may also follow with the I2).
Therefore, on net:
6 KI + 3 HOCl + 3 H (+) --> 3 KCl + 3 K (+) + 3 I2 + 3 H2O
where a weak acid is present. I have a source that quote this very reaction with a weak acid, specifically with Acetic acid, the equation presented
is:
HOCl + 2KI + HOAc → I2 + KCl + KOAc + H2O
which is my equation above dividing through by 3. Source: Kaohsiung Medical University,
"20 Applications of Oxidation/Reduction Titrations", top of page 145. Link:
http://sjlin.dlearn.kmu.edu.tw/20.pdf
So, interestingly, one doesn't even need a strong mineral acid starting with just an Iodide, Bleach and Vinegar to liberate Iodine.
[Edited on 20-4-2012 by AJKOER]
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vmelkon
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Here are my calculation if you are going to use 3 g of KI.
For the H2O2 3 % solution, I estimated the molarity.
====================================
3 g of KI = 0.01807159 mol
32 % H2SO4 (about 5 M)
2 KI + H2SO4 => K2SO4 + 2 HI
0.01807159 mol of KI
We need 0.009035796 mol of H2SO4 which means 0.001807159 L of H2SO4
and then add H2O2 (3 %) (1.4699509 M)
2HI + H2O2 = I2 + 2H2O
0.01807159 mol of HI
We need 0.009035796 mol of H2O2 which means 0.0061470053 L
=====================================
So, you need 1.81 mL of H2SO4 32 % and 6.15 mL of H2O2 3 %
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CrEaTiVePyroScience
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Thanks alot Vmelkon if anyone can confirm those calculations I would be very pleased and will let you guys know how it went!
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AJKOER
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| Quote: Originally posted by AJKOER |
Some old threads mention the use of slightly acidified Bleach (NaClO) to extract I2. Here is guess estimate of the chemistry:
..............
Therefore, on net:
6 KI + 3 HOCl + 3 H (+) --> 3 KCl + 3 K (+) + 3 I2 + 3 H2O
where a weak acid is present. I have a source that quote this very reaction with a weak acid, specifically with Acetic acid, the equation presented
is:
HOCl + 2KI + HOAc → I2 + KCl + KOAc + H2O
......
So, interestingly, one doesn't even need a strong mineral acid starting with just an Iodide, Bleach and Vinegar to liberate Iodine.
|
An obvious but important point I will add is that one must avoid an excess of HOCl as:
I2 + H2O <---> H(+) + I(-) + HOI
and with an excess of HOCl, this equilibrium is moved to the right, that is, the free Iodine is oxidized to Iodide as:
HOCl + I(-) --> HOI + Cl(-)
followed by a slow disproportionation of HOI to molecular iodine and iodate:
5HOI ---> 2 I2 + IO3(-) + 2 H2O + H(+)
But, with even more HOCl, one has Chlorine via:
HOCl + Cl(- ) + H(+) --> Cl2 + H2O
which will consume more Iodine per the reaction:
Cl2 + I2 + 2H2O --> 2HOI + 2Cl(-) + 2H(+)
As an excellent reference on the many possible reactions, see Table I on page 3716 in "Rate Constants for Reactions between Iodine- and
Chlorine-Containing Species: A Detailed Mechanism of the Chlorine Dioxide/Chlorite-Iodide Reaction", by Istvan Lengyel, Jing Li, Kenneth Kustin and
Irving R. Epstein, J. Am. Chem. Soc. 1996, 118, 3708-3719.
Full paper link:
http://hopf.chem.brandeis.edu/pubs/pub234%20rep.pdf
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edgeofacliff
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You can dissolve the potassium iodide in water, add clorox bleach, and HCL, for example 100 grams KI, one half cup water, stir, when dissolved add
one half cup clorox bleach and shot glass of HCL the crystals will form immediately. Filter, then rinse with water, dry. You may have to add a little
more bleach, watch for purple smoke.
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woelen
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This is not the way it should be done. You have to carefully adjust the amount of bleach. Too little and you loose a lot of iodine through formation
of soluble I3(-) complex and too much and you loose a lot of iodine through formation of iodate(V) and tetrachloroiodate(III).
Bleach can be used, but you have to carefully determine the molarity of the hypochlorite and carefully adjust the amount of bleach to the amount of
iodide. The amount of HCl is not critical, just use excess amount of acid.
It is better to use hydrogen peroxide as oxidizer. If you have some excess hydrogen peroxide and the excess amount is not too concentrated, then it
does not react with iodine. So, with hydrogen peroxide as oxidizer you should try to determine the exact amount needed, based on concentration and
stoichiometry of the reaction, but then you take some extra hydrogen peroxide (e.g. 10% if you work accurately, or 20% if there is more uncertainty in
your computed amounts) in order to assure that all iodide is converted to iodine.
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