Sciencemadness Discussion Board - Separating Iodide and Iodate?

Separating Iodide and Iodate?

K12Chemistry - 10-4-2013 at 02:26

I am trying to make KI by reacting KOH with pure crystal iodine. But when this happens, it reacts to form potassium iodide and iodate. They are both in solution so does anyone know how i can separate the two.

woelen - 10-4-2013 at 03:32

Use slight excess of iodine, such that the solution remains pale brown instead of colorless after all iodine has dissolved. Now you have 5 parts KI and 1 part KIO3 with slight excess of I2.

Next, evaporate all of the material. This will give a light brown solid. Heat this solid fairly strongly. The KIO3 will decompose, giving KI and O2. Excess I2 will be driven off as well (purple vapor). What remains is fairly pure KI.

Dissolve the fairly pure KI in as little as possible water and allow to evaporate such that only a small amount of solution remains over a white crystalline mass. Decant the liquid as much as possible and put the mass on a filter, which is put on a pile of paper tissues. Much of the adhering liquid is absorbed by the filter and paper tissue. Rinse the solid with some pure and clean acetone or diethyl ether and let this liquid evaporate. This treatment will give you quite pure KI. Store in a tightly closed vessel, KI is fairly hygroscopic.

blogfast25 - 10-4-2013 at 04:27

Quote: Originally posted by K12Chemistry

I am trying to make KI by reacting KOH with pure crystal iodine. But when this happens, it reacts to form potassium iodide and iodate. They are both in solution so does anyone know how i can separate the two.

Please use the Search Facility: this must be thread #15 on this boring subject!

Pyro - 10-4-2013 at 05:37

or crystallize it. KIO3 isn't nearly as soluble as KI, but MUCH more expensive.

K12Chemistry - 10-4-2013 at 05:40

KIO3 dissolves in KI solution.

can someone give me an idea as to what temperature i need to heat this to decompose the KIO3 into KI and Oxygen.

Also why would you use an excess of iodine

MrHomeScientist - 10-4-2013 at 05:41

Woelen's method is a great way to make a high purity product. I made a video about making KI from KOH and I2, but was unaware how much KIO3 contamination there was in it. This actually caused one of my other videos to be inaccurate, when I used my product to make iodine using acid but no extra oxidizer! The KIO3 impurity was enough to create iodine with just the addition of acid, which shouldn't happen. I recently ordered some pure KI, so once that arrives I need to post a correction video showing all this.

K12Chemistry - 10-4-2013 at 07:03

oh,

I've seen so many of your videos. Big fan!

I didn't know you were on sciencemadness.

Ral123 - 10-4-2013 at 07:15

I had no idea iodine can displace oxygen and produce iodate like chlorine does. I though iodine is with reactivity similar to sulphur.

K12Chemistry - 11-4-2013 at 04:26

What about using acetone. Iodate is soluble but Iodide isn't. Would that work?

woelen - 11-4-2013 at 06:26

@Ral123: Iodine does not displace oxygen, the reaction described here is a simple disproportionation reaction. First, iodine is disproportionating to iodide and hypoiodite. Next, the hypoiodide in turn disproportionates to iodide and iodate. The iodate is decomposed by heat, loosing its oxygen.

Reactivity of iodine is larger than reactivity of sulphur, but in some respect they are similar. Sulphur also can disproportionate in alkalies (it requires heating and the reaction is slower), but it forms sulfide and thiosulfate as the initial step of disproportionation.

@K12Chemistry: Is iodate soluble in acetone? I doubt this. KI is very slightly soluble, but this solubility is so low that there is no need to fear great losses when rinsing KI with acetone to get rid of water and impurities.

K12Chemistry - 11-4-2013 at 10:58

oh sorry i meant it the opposite.

KI dissolves in acetone but not KIO3

K12Chemistry - 11-4-2013 at 11:02

Woelen in your method you say to use an excess of iodine, why? couldn't you just heat it.

Also what temperature would it take to decompose the KIO3 to KI. Would a basic hotplate work? Or a spirit burner?

Is there any solution (More or less readily available) that KIO3 is soluble in but not KI, or the other way around?

K12Chemistry - 13-4-2013 at 04:57

I think I have found the answer,

1g of potassium iodide dissolves in 8ml of methanol (pretty soluble), potassium iodate however, is completely insoluble in methanol. You add methanol to the mixture to dissolve the iodide and filter it. You then boil down the methanol to get a solid crystalline product of pure potassium iodide

unionised - 13-4-2013 at 06:30

If you want KI and not KIO3 then the classic method is to mix the I2 with iron filings and water to get a solution of FeI2
then add KOH to ppt Fe(OH)2 and leave a solution of KI which you can evaporate the water from.

blogfast25 - 13-4-2013 at 07:09

Quote: Originally posted by unionised

If you want KI and not KIO3 then the classic method is to mix the I2 with iron filings and water to get a solution of FeI2
then add KOH to ppt Fe(OH)2 and leave a solution of KI which you can evaporate the water from.

Interesting tip. Live and learn. Must try that once...

woelen - 13-4-2013 at 10:48

Quote: Originally posted by K12Chemistry

Woelen in your method you say to use an excess of iodine, why? couldn't you just heat it.

If you use excess of I2, then no KOH remains left. Excess of I2 can be driven off by heating. If you have excess of KOH, then you get hard to remove impurities which cannot be driven off.

K12Chemistry - 14-4-2013 at 01:50

Quote:

If you want KI and not KIO3 then the classic method is to mix the I2 with iron filings and water to get a solution of FeI2 then add KOH to ppt Fe(OH)2 and leave a solution of KI which you can evaporate the water from.

Sorry guys, I guess I wasn't clear enough in the starting question where I said I wanted to separate the two. What I actually was trying to do was get pure potassium iodide and probably discard of the KIO₃.

Anyway would the equation for the FeI2 method be:

I2 + Fe -> FeI2
FeI2 + 2KOH -> 2KI + Fe(OH)2

Is this balanced?

Sorry I'm a noob so it's probably wrong

blogfast25 - 14-4-2013 at 04:56

Quote: Originally posted by K12Chemistry

Anyway would the equation for the FeI2 method be:

I2 + Fe -> FeI2
FeI2 + 2KOH -> 2KI + Fe(OH)2

Is this balanced?

Sorry I'm a noob so it's probably wrong

No, you happen to be right on this one. But I'd like to see practical implementation of this idea, not just balancing equations.

[Edited on 14-4-2013 by blogfast25]

K12Chemistry - 14-4-2013 at 05:08

Yeah I'm going to stoichiometrically calculate how much of each reagent I use, but first I need to find the solubility of FeI2.

Does anyone know the solubility of FeI2 in water?

K12Chemistry - 14-4-2013 at 05:19

To make Iron Iodide:

I2 = 254 g/mol
Fe = 56 g/mol.

So I add about 127g of iodine to 28g of Fe, obviously not on such a great scale.

To make KI:

FeI2 = 310 g/mol
KOH = 56 g/mol

So I add approximately 155g of Iron Iodide to 56g of KOH

Is this right?

If it is then it's actually the first time I've done this all on my own

blogfast25 - 14-4-2013 at 05:32

Yes, the math is correct but you would add the KOH to the FeI₂.

Presumably one would mix the requisite amount of I₂ with an excess of iron filings or steel wool. The excess is needed to complete the reaction in a reasonable amount of time. Mild heating and good stirring will also speed things up considerably. When the solution contains apparently no more free iodine, it would be filtered to separate out the excess of iron. The filtrate would contain dissolved FeI₂, in the amount calculated.

Then a calculated amount of KOH (or NaOH, MW = 40 g/mol) solution would be added to the filtrate, to achieve a pH of about 7 – 9. This will precipitate the Fe²⁺ as Fe(OH)₂ and that is filtered off. The process is known as a 'displacement reaction'.

The filtrate would contain relatively pure KI (or NaI).

Re. the solubility of ferrous iodide, iodides are generally very soluble and so are ferrous compounds. Chances are that ferrous iodide is very water soluble.

[Edited on 14-4-2013 by blogfast25]

K12Chemistry - 14-4-2013 at 11:43

I thought steel wool wasn't pure iron

K12Chemistry - 14-4-2013 at 11:44

oops just searched it, about 98% iron.

blogfast25 - 14-4-2013 at 13:09

Quote: Originally posted by K12Chemistry

oops just searched it, about 98% iron.

No, it's not pure but it doesn't really have to be.

AndersHoveland - 14-4-2013 at 21:20

Quote: Originally posted by Ral123

I had no idea iodine can displace oxygen and produce iodate like chlorine does. I though iodine is with reactivity similar to sulphur.

Iodine seems to act more like the other halogens in the presence of water.

But much like sulfur, the covalent bond between iodine and hydrogen is very weak.

[Edited on 15-4-2013 by AndersHoveland]

woelen - 14-4-2013 at 22:55

This FeI2/KOH method does not sound like a pleasant method to me. In practice it will be very hard to make reasonably pure KI with this, much harder than with I2 + KOH. I also would use excess iron to be sure that all I2 reacts and then rinse all FeI2 from the excess of iron. I2 can be measured reasonably accurately. KOH, on the other hand, is very bad when it comes to precise measurements. It usually contains some water and this may be anything between a few percent up to 15%. If you want really good results then you first need to titrate the FeI2 solution and the KOH solution for concentration and then add the right amounts of liquid to each other. Both excess FeI2 and excess KOH are not easily removed from KI, so you need to work very precisely and you definitely need one or mayeb even two recrystallizations of KI.

blogfast25 - 15-4-2013 at 08:58

I don't think there's a real need for titration here. I use KOH to precisely neutralise things all the time. Calculate the precise amount of KOH needed, in the knowledge that it contains at least 10 % water. Add this, then bit by bit add more until pH 7 - 9. Not very hard.

If you start from KOH + I2, you basically have the same problem: you'll always be a bit off on the actual amount of KOH used and you'll need to recrystallise your KI/KIO3 too.

I think the Fe/I2/KOH method is worth a shot, if only to satisfy curiosity.

K12Chemistry - 15-4-2013 at 10:25

Ok I have £2.20,

I need to save up to buy iodine, koh and iron (unless i find steel wool)

AndersHoveland - 15-4-2013 at 15:11

What if you added a copper salt? That would react with the iodide to precipitate out insoluble CuI and iodine in elemental form (which is not very soluble in water so long as all the iodide is used up).

2 Cu⁺² + 4 I^- --> 2 CuI + I₂

woelen - 15-4-2013 at 22:48

What do you want to reach with this? The goal is pure KI and I do not see how this relates to your suggestion of adding a copper salt.

blogfast25 - 16-4-2013 at 04:34

Hoveland: do READ the thread before spouting irrelevant nonsense, it's about MAKING iodide, not destroying half one's stockpile of I2 as CuI.