Sciencemadness Discussion Board - Potassium iodate

Potassium iodate

hkparker - 6-12-2010 at 16:55

Im interested in making potassium iodate from potassium iodide. Now I know that the reaction of iodine and potassium hydroxide yields potassium iodide and potassium iodate.

3I2 + 6KOH --> 5KI + KIO3 + 3H2O

So my thought was I can convert KI to iodine then react it with potassium iodide to make the KIO3, but this seems kind of like an inefficient process. Does anyone know of a one step process that could be used to make KIO3? Thanks!

Jor - 6-12-2010 at 17:46

That is a very inefficient way to prepare iodate. You can get iodates in great yield by oxidising iodine with chlorate. You will have to add a few mL of dilute HNO3, as the reaction proceeds at around pH=2 or lower. Heat until the iodine is completely dissolved. Use a reflux condenser to condense gaseous iodine (if you don't you will lose some iodine), and push it back in periodically. Next boil away excess chlorine (and some ClO2), and add KOH until alkaline (important, or you might get KH(IO3)2 ) reaction. Crystallise the KIO3 (not sure what to do incase of NaIO3) and recrystallise from water, and dry in an oven at 110C.

MrHomeScientist - 7-12-2010 at 12:34

Ah, I see you're still looking for a way to make iodate! Good to see you on these boards too - I just recently discovered this place and I'm wowed at the level of expertise found here. So much great information. If there's a way to do it, I'm sure you'll find it here.

By the way I realized I hadn't subscribed to your channel yet, whoops! Done and done. Now I've got some catching up to do

See you around!

hkparker - 7-12-2010 at 20:05

Yea, these boards are great!

Thanks for the info Jor, big help! What would the reaction look like though? So I can do the stoichiometry.

watson.fawkes - 7-12-2010 at 20:32

Quote: Originally posted by hkparker

What would the reaction look like though? So I can do the stoichiometry.

There are several by-products, and therefore several reactions. Try writing down a couple here first. For example, there's both Cl2 and ClO2 as products, and they likely don't come from the same reaction pathway. Chlorate is ClO3(-).

blogfast25 - 10-12-2010 at 08:18

I’ll ask the same question in reverse. Refluxing I2 (of which I have quite a bit – bought for an abandoned experiment) with KOH yields mostly KI and some KIO3 as indicated above. How to get rid of the KIO3 though? Recrystallising doiesn’t seem much of an option, KI being so soluble in water. Separation with different solvents?

KI is useful to me for various redox titrations but the last time I looked on eBay the only product available was an extortionate pharmacon grade… And another grade I bought was yellowish: presumably small amounts of I2 (which would necessitate titrating blanks: messy!)

mr.crow - 10-12-2010 at 08:32

How about adding reducing agents like K bisulfite? But now you have sulfate contamination. Upon adding acid you will get the reverse reaction and it will all turn into I2.

Yeah it turns yellowish over time

not_important - 10-12-2010 at 14:57

KI, although better with NaI, is soluble in MeOH, acetone, and similar solvent, while the iodates have much lower solubilities. So solution in such an organic solvent, filter, concentrate and crystalise, generally gets the iodate concentration low enough although a 2nd recrystallisation may be desired.

An alternative, used in old lab iodide recovery and industrial production, is to heat the mixed salt to dull red heat, which decomposes the iodate.

MrHomeScientist - 10-12-2010 at 14:58

blogfast25:

In my experience, getting rid of the KIO3 is actually pretty simple since its solubility in water is so different from KI's. After the reaction I immerse it in an ice bath for a bit to get as much as I can out of solution, then simply filter out the KIO3. I actually just posted a video of this process on my YouTube page last weekend

I've also read that it can be thermally decomposed to yield more KI, as not_important said. I haven't tried this myself though.

I haven't done any experiments where purity is critical though, so if there were any impurities I might not have noticed them. What's been your experience?

[Edited on 12-10-2010 by MrHomeScientist]

blogfast25 - 11-12-2010 at 06:37

Thanks not_important and MrHomeScientist!

MHS, can we see your video? I've seen a few on maling KI this way but all are gloriously silent on thhe KIO3 formed...

hkparker - 20-3-2011 at 20:32

I know this thread is ancient but I thought people might want to see a video of what Jor described. Here it is!

http://www.youtube.com/watch?v=RJBmNYKyb7s

Jor - 21-3-2011 at 05:19

Nicely done! I am also not sure what the causes the brown color to develop when you add KOH. What are the purity of your reagents and did you use clean glassware? When I did this I used pure lab-grade chemicals and did not get this color.
And how much KClO3 did you use? What was your yield?

I really liked this synthesis. It is fun (unless you have to breathe the gasses wich are evolved

) and very rewarding as you make a chemicals wich is quite expensive to buy.

MrHomeScientist - 21-3-2011 at 05:48

Here's my video on making KI, as mentioned quite a while back: (sorry for not posting the link earlier, I missed the response and forgot about the thread)

http://www.youtube.com/watch?v=SNXIY4InWFU

Interestingly, it's gotten a huge increase in traffic lately because of the disaster in Japan - as people are looking for potassium iodide anti-radiation pills.

hkparker - 21-3-2011 at 14:56

Thanks Jor! Yea that brown impurity confused me too... my iodine was ultra pure (I purified it by sublimation/recrystallization), my chlorate I got from a friend who bought it from a pyro website, so I think thats my source of impurity. The glass was clear but cleaned with tap water. Ill ask him if he has an msds sheet for the chlorate but I think it was tech grade.

I'm not sure how much chlorate I used exactly probably about 3-5g. I didnt calculate my percent yield with respect to iodine for a few reasons, one of which being I didnt mass out how much iodine I added

. I will soon upscale this and calculate it with as much accuracy as I can.

I found it to be pretty rewarding as well, it was a fun synth and I got a fairly usable product!

hkparker - 21-3-2011 at 21:44

@Jor: you once said that two seperate reactions were going on here, one producing Cl2 and the other ClO2. I tried to write some out but I'm getting stuck, can you help?

If I'm right about the one producing chlorine then its pretty strait forward:

2ClO3(-) + H(+) + I2 --> H(IO3)2(-) + Cl2

What I don't get is the one that produced chlorine dioxide. I can't seem to make those reagents form it and still have it be electrochemically balanced, so I figured maybe water in the solution is involved?

woelen - 22-3-2011 at 01:25

This reaction between chlorate and iodine is not a simple clean reaction at all. The iodine ends up as iodate (which is what you want), but the chlorate partially ends up as chlorine, an other part ends up as chloride and it is this chloride which leads to a secondary reaction in which ClO2 is formed.

So you have:
1) chlorate + iodine --> chlorine + iodate/iodic acid
2) chlorate + iodine --> chloride + iodata/iodic acid
3) chlorate + chloride --> chlorine
4) chlorate + chloride --> chlorine dioxide
All these reaction consume acid as well.

Chlorate is not an oxidizer showing a single simple reaction, its reactions strongly depend on the presence of chloride and the pH of the solution.

You could try to balance all of the above equations. The last two ones are comproportionation reactions.

hkparker - 22-3-2011 at 06:39

Thank you woelen! Ill give balancing those a try.

hkparker - 22-3-2011 at 16:00

...gave it a try:

chlorate + iodine --> chlorine + iodate/iodic acid in acidic conditions is pretty strait forward

2ClO3- + H+ + I2 --> H(IO3

2- + Cl2

chlorate + chloride --> chlorine in acidic conditions is likewise well documented

12H+ + 10Cl- + 6ClO3- --> 2Cl2 + 6H2O

what I wasnt able to balance is how acidified chlorate + chloride --> chlorine dioxide or how chlorate + iodine --> chloride + iodate/iodic acid

woelen - 23-3-2011 at 00:18

Unfortunately, only the first one is correct.

There is a methodology which you may find useful. This works with so-called half-equations, in which the elements are balanced, but charge is not balanced. Charge is compensated for with electrons on one of the two sides of the arrow, such that the total charge at the left and the right is the same for a half-equation.
If the reaction occurs at low pH and consumes acid, then oxygen atoms are compensated for with H(+) such that they appear as water molecules. Oxygen is assumed to have oxidation state -2. If a reaction occurs at high pH, then OH(-) is used for compensating.

Next, half-equations are combined to make the full equation and this is done in such a way that the number of electrons on either side of the arrow is equal and hence can be removed from the equation.

An example is the reaction of chlorate and choride to chlorine.

Chlorate is reduced to chlorine.

ClO3(-) --> Cl2

balancing of chlorine: 2ClO3(-) --> Cl2
balancing of oxygens: 2ClO3(-) --> Cl2 + 6O(2-)
compensation with 12H(+) is needed to have all oxygen end up as water:
2ClO3(-) + 12H(+) --> Cl2 + 12H(+) + 6O(2-) --> Cl2 + 6H2O

Now all elements are balanced, we have the same number of oxygens, chlorines and hydrogens on each side of the arrow. Charge is not balanced, so we need to compensate for that with electrons. At the left we have a total charge of +10, at the right we have a total charge of 0. So, we put 10 electrons at the left to make the charge equal on both sides. This completes the half-equation for conversion of chlorate to chlorine.

2ClO3(-) + 12H(+) + 10e --> Cl2 + 6H2O

Now we need to do the same for conversion of Cl(-) to Cl2.

2Cl(-) --> Cl2
This is already balanced for elements, no oxygen needs to be compensated. Only charge needs to be compensated. We have to put 2 electrons at the right to make charges equal:

2Cl(-) --> Cl2 + 2e

Now we have the two half-equations. We have to combine these in such a way that left and right we have the same number of electrons. This can be achieved by taking the first half-equation 1 time and the second half-equation 5 times. That gives

2ClO3(-) + 12H(+) + 10e + 10Cl(-) --> Cl2 + 6H2O + 5Cl2 + 10e

At the left and right we have 10 electrons. Remove these:

2ClO3(-) + 12H(+) + 10Cl(-) --> Cl2 + 6H2O + 5Cl2

Simplify:

2ClO3(-) + 12H(+) + 10Cl(-) --> 6Cl2 + 6H2O

Divide out common factor of 2 for all coefficients:

ClO3(-) + 6H(+) + 5Cl(-) --> 3Cl2 + 3H2O

This ends up the example of reaction of chloride and chlorate in acidic medium with production of Cl2.

woelen - 23-3-2011 at 00:46

I use a separate post now for clarity to do one other example, the reaction of chlorate and chloride to ClO2. This is a reaction which runs in parallel to the reaction in which chlorine is formed. Both occur simultaneously.

First the reaction of chlorate to ClO2:

ClO3(-) --> ClO2

balancing of oxygens: ClO3(-) --> ClO2 + O2(-)
compensating with H(+): ClO3(-) + 2H(+) --> ClO2 + H2O

adding electrons such that charge is equal left and right. We need to add a single electron to the left:

ClO3(-) + 2H(+) + e --> ClO2 + H2O

This completes one half-equation.

------------------------------------------------------------------

The chloride also is converted to ClO2. So we have:

Cl(-) --> ClO2
Balancing of oxygens: Cl(-) + 2O(2-) --> ClO2
Compensating with H(+): Cl(-) + 2H2O --> ClO2 + 4H(+)

At the left we have a charge equal to -1, at the right we have +4. So we need to add 5 electrons at the right in order to complete the half-equation:

Cl(-) + 2H2O --> ClO2 + 4H(+) + 5e

Now we have to combine these equations, such that the number of electrons is the same at both sides. This can be done by taking 5 times the first + 1 times the second.

5ClO3(-) + 10H(+) + 5e + Cl(-) + 2H2O --> 5ClO2 + 5H2O + ClO2 + 4H(+) + 5e

Removing the electrons:

5ClO3(-) + 10H(+) + Cl(-) + 2H2O --> 5ClO2 + 5H2O + ClO2 + 4H(+)

Removing the common water molecules and H(+) ions at left and right and combining the ClO2 molecules gives the final complete equation:

5ClO3(-) + 6H(+) + Cl(-) --> 6ClO2 + 3H2O

Just try it yourself for the last equation and then as an other exercise you can try this one:

Dichromate, Cr2O7(2-), reacts with sulphur dioxide, SO2, to form Cr(3+) and SO4(2-). In this reaction, acid is consumed. Try to derive the complete equation for this, using two half-equations (one for dichromate to Cr(3+)) and the other for SO2 to sulfate ion SO4(2-).

hkparker - 23-3-2011 at 19:56

Thanks for the refresher in redox woelen!

I haven't done this in a while, since last time I took chem in school, so it didn't occur to me to use redox here, sorry to make you type all that out.

Ok, so I tried to balance it, and I think I got it, but my reaction... produces H+...

chlorate + iodine --> chloride + iodate/iodic acid

separated into oxidation/reduction reactions:

ClO3- --> Cl-

Balance oxygens:
ClO3- --> Cl- + 3O2-

Added Hydrogen:
6H+ + ClO3- --> Cl- + 3H2O

and electrons for charge...

Final half reaction:

6e- + 6H+ + ClO3- --> Cl- + 3H2O

Next half reaction:

I2 --> H(IO3

2-

Ok, heres where it got me... I need to balance hydrogen in the reaction, not just for waters. Shouldn't be a problem right?

balance I2, and I added H+ to the left to balance hydrogens in the reaction (mistake?)

I2 + H+ --> H(IO3

2-

I added oxygens to balance them in the iodate

I2 + H+ +6O2- --> H(IO3

2-

Adding hydrogens:

I2 + H+ + 6H2O --> 2H(IO3

2- + 12H+

I now cancelled the hydrogen used to balance the hydrogen in the iodate with the hydrogens on the right (...mistake?)

Looks like this:

6H2O + I2 --> H(IO3

2- + 11H+

Now add electrons for charge, and I get this as my final half life:

6H2O + I2 --> H(IO3

2- + 11H+ + 10e-

Great, now cancel that with my first half reaction... heres what I get:

3H2O + 5ClO3- + 3I2 --> 5Cl- + 3H(IO3

2- + 3H+

So... did I do things right? Or where did I go wrong. Thank you very much for working through this with me woelen!

woelen - 23-3-2011 at 23:51

You did everything well for the reaction of chlorate and iodine to chloride and hydrogen-diiodate. I am quite confident that you now really understand it. The 'trick' with the H(+) also is exactly what I would have done, again this shows your understanding.

Now I want to let you know I have an online tool which balances equations:

http://woelen.homescience.net:18080/chemeq

No fiddling with half-reactions and that kind of things, just put at the left the ingoing compounds and at the right the products in the notation, which I used in my posts, e.g. H(IO3)2(-) for the hydrogen-diiodate ion and ClO3(-) for chlorate ion and I2 for iodine.

If you do not really know whether acid is consumed or produced in the reaction, then just add H(+) to one side and H2O to the other. The program then will come up somewhere with negative coefficients and then you know that you need to exchange reactants and products.

Please read the tutorial as well: http://woelen.homescience.net:18080/chemeq/tutor/tutor.html

I waited with posting this, because I think that understanding is important. Once you understand things, then you may use tools to make life easier.

hkparker - 24-3-2011 at 16:49

Thanks for the link! And thanks for helping me understand it first, I too think thats important and am glad thats what normally happens on SM.

plante1999 - 1-7-2012 at 13:24

I tried this methode with very pure potassium chlorate and iodine with potassium chlorate in large excess. The only diference is that I used sulphuric acid for acidification. The reaction worked very well but now I have about the same weigh if not more of chlorate in my iodate, How can I purify it.

Thanks!!!

woelen - 2-7-2012 at 01:33

You should not have used a large excess of chlorate. Now you have a problem isolating the two from each other. Both iodate and chlorate have similar solubilities in water, so that is not going to help you.

The only thing I see is adding more iodine and trying to convert that to iodate as well. Adding a reductor most likely also will result in destruction of iodate. Chlorate is a stronger oxidizer, but it reacts somewhat sluggishly in aqueous solution, while iodate reacts very fast. Any iodate which is reduced initially will be oxidized again by excess chlorate, but I'm not sure how fast it will be oxidized beyond the iodine-stage.

Actually, for this reason of mixed chlorate/iodate production, I never liked this reaction very much. I have used electrolysis to make KIO3 from KI. If you only have I2 and KOH, then you also can use electrolysis. Just add I2 to KOH, until the solution remains brown and then apply electrolysis. Details on the process are described here on sciencemadness and on my website.