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Author: Subject: Synthesis of Iodine from KI and sulfuric acid question.
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[*] posted on 15-4-2012 at 11:02
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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[*] posted on 15-4-2012 at 11:47


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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[*] posted on 15-4-2012 at 15:07


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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[*] posted on 16-4-2012 at 05:28


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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[*] posted on 16-4-2012 at 05:32


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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[*] posted on 16-4-2012 at 07:03


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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[*] posted on 16-4-2012 at 07:25


@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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[*] posted on 16-4-2012 at 07:45


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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[*] posted on 16-4-2012 at 08:57


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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[*] posted on 16-4-2012 at 09:14


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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[*] posted on 16-4-2012 at 10:06


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.

I just had a thought - my KI was very likely contaminated with KIO3 because of the way I made it (KOH + I2, 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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[*] posted on 16-4-2012 at 10:38


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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[*] posted on 16-4-2012 at 11:36


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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[*] posted on 16-4-2012 at 11:53


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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[*] posted on 16-4-2012 at 12:48


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 - I2 , the element itself
Iodide - I- , part of a compound, i.e. potassium iodide
Iodate - IO3- , 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 :o
I think woelen would be the authority on this, though, as it was his procedure outlined above.
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[*] posted on 16-4-2012 at 22:28


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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[*] posted on 17-4-2012 at 05:43


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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[*] posted on 17-4-2012 at 06:43


@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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[*] posted on 17-4-2012 at 23:23


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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[*] posted on 19-4-2012 at 20:19


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&dq=HCl... )

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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[*] posted on 20-4-2012 at 07:40


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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[*] posted on 20-4-2012 at 11:26


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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[*] posted on 20-4-2012 at 21:40


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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[*] posted on 21-6-2012 at 03:34


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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[*] posted on 21-6-2012 at 04:05


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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