hkparker
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Potassium iodate
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!
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Jor
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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.
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MrHomeScientist
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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!
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hkparker
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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.
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watson.fawkes
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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(-).
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blogfast25
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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!)
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mr.crow
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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
Double, double toil and trouble; Fire burn, and caldron bubble
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not_important
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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.
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MrHomeScientist
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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]
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blogfast25
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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...
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hkparker
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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
My YouTube Channel
"Nothing is too wonderful to be true if it be consistent with the laws of nature." -Michael Faraday
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Jor
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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.
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MrHomeScientist
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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.
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hkparker
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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!
My YouTube Channel
"Nothing is too wonderful to be true if it be consistent with the laws of nature." -Michael Faraday
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hkparker
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@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?
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"Nothing is too wonderful to be true if it be consistent with the laws of nature." -Michael Faraday
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woelen
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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.
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hkparker
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Thank you woelen! Ill give balancing those a try.
My YouTube Channel
"Nothing is too wonderful to be true if it be consistent with the laws of nature." -Michael Faraday
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hkparker
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...gave it a try:
chlorate + iodine --> chlorine + iodate/iodic acid in acidic conditions is pretty strait forward
2ClO<sub>3</sub><sup>-</sup> + H<sup>+</sup> + I<sub>2</sub> --> H(IO<sub>3</sub> <sub>2</sub><sup>-</sup> + Cl<sub>2</sub>
chlorate + chloride --> chlorine in acidic conditions is likewise well documented
12H<sup>+</sup> + 10Cl<sup>-</sup> + 6ClO3<sup>-</sup> --> 2Cl<sub>2</sub> +
6H<sub>2</sub>O
what I wasnt able to balance is how acidified chlorate + chloride --> chlorine dioxide or how chlorate + iodine --> chloride + iodate/iodic acid
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"Nothing is too wonderful to be true if it be consistent with the laws of nature." -Michael Faraday
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woelen
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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.
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woelen
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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-).
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hkparker
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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<sup>+</sup>...
chlorate + iodine --> chloride + iodate/iodic acid
separated into oxidation/reduction reactions:
ClO<sub>3</sub><sup>-</sup> --> Cl<sup>-</sup>
Balance oxygens:
ClO<sub>3</sub><sup>-</sup> --> Cl<sup>-</sup> + 3O<sup>2-</sup>
Added Hydrogen:
6H<sup>+</sup> + ClO<sub>3</sub><sup>-</sup> --> Cl<sup>-</sup> + 3H<sub>2</sub>O
and electrons for charge...
Final half reaction:
6e<sup>-</sup> + 6H<sup>+</sup> + ClO<sub>3</sub><sup>-</sup> --> Cl<sup>-</sup> +
3H<sub>2</sub>O
Next half reaction:
I<sub>2</sub> --> H(IO<sub>3</sub><sub>2</sub><sup>-</sup>
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 I<sub>2</sub>, and <b>I added H<sup>+</sup> to the left to balance hydrogens in the reaction
(mistake?)</b>
I<sub>2</sub> + H<sup>+</sup> --> H(IO<sub>3</sub><sub>2</sub><sup>-</sup>
I added oxygens to balance them in the iodate
I<sub>2</sub> + H<sup>+</sup> +6O<sup>2-</sup> --> H(IO<sub>3</sub><sub>2</sub><sup>-</sup>
Adding hydrogens:
I<sub>2</sub> + H<sup>+</sup> + 6H<sub>2</sub>O --> 2H(IO<sub>3</sub><sub>2</sub><sup>-</sup> + 12H<sup>+</sup>
<b>I now cancelled the hydrogen used to balance the hydrogen in the iodate with the hydrogens on the right (...mistake?)</b>
Looks like this:
6H<sub>2</sub>O + I<sub>2</sub> --> H(IO<sub>3</sub><sub>2</sub><sup>-</sup> + 11H<sup>+</sup>
Now add electrons for charge, and I get this as my final half life:
6H<sub>2</sub>O + I<sub>2</sub> --> H(IO<sub>3</sub><sub>2</sub><sup>-</sup> + 11H<sup>+</sup> + 10e<sup>-</sup>
Great, now cancel that with my first half reaction... heres what I get:
3H<sub>2</sub>O + 5ClO<sub>3</sub><sup>-</sup> + 3I<sub>2</sub> --> 5Cl<sup>-</sup> +
3H(IO<sub>3</sub><sub>2</sub><sup>-</sup> +
3H<sup>+</sup>
So... did I do things right? Or where did I go wrong. Thank you very much for working through this with me woelen!
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"Nothing is too wonderful to be true if it be consistent with the laws of nature." -Michael Faraday
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woelen
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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.
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hkparker
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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.
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"Nothing is too wonderful to be true if it be consistent with the laws of nature." -Michael Faraday
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plante1999
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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!!!
I never asked for this.
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woelen
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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.
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