Ki3

[budullewraagh]

a year ago i made KI3 by failing to add enough Cl2 to a KI solution. my mistake was that the tube leading to the KI soln was forced above the surface because of pressure.

 

so i've had this 60mL bottle of potassium triiodide in water for about a year now and the label on the bottle has begun to show stains, which is weird. i kind of want to finish the oxidation get iodine crystals.

 

last time i used HCl+NaOCl after fractionally crystallizing the hypochlorite from bleach. i know it sounds pretty crazy, but i kept the water of crystallization and it didnt explode. but come to think of it, i should be able to crystallize again, then add these crystals to the KI3 solution to yield KCl and I2 because aqueous NaOCl is effectively NaOH+Cl2 in this scenario and as the Cl2 reacts, the equilibrium will be pushed towards creating more Cl2.

 

come to think of it i could add the rest of my KI and i could just convert it all to I2 at once.

 

so, my problem is finding the best possible way to collect the iodine crystals and dry them.

 

i have a 500mL erlenmeyer flask, two 400mL beakers, a glass u tube, a glass funnel, a 500mL florence flask, a 100mL graduated cylinder and a crucible.

 

and no rubber stoppers.

 

so i was thinking of possibly performing the initial reaction in the florence flask then sort of draining the excess water off, then heat to, say, 75 celsius and allow the iodine to sublime upwards into the u tube and down into a collecting flask containing anhydrous MgSO4.

 

my problem is that the u tube won't fit perfectly into the top of the florence flask, so i'll need to seal whatever gaps with duct tape. or would duct tape and/or the adhesive thereon be attacked by the iodine? luckily, on the other end, i could put my funnel upside-down with the stem inside the u tube and the funnel leading into the collecting flask.

 

next, i'd use a similar apparatus and heat the iodine at, say, 50 celsius to have it sublime and go to a final collecting flask where it will (hopefully) be dry.

 

what say you? is there an easier way?

[akcapr]

I rember we had this dilema before. With what you have I would try draining the idone as dry as possible, then getting ur florence flask vry cold on the neck with dry ice perhaps. That should keep most of the iodine from escaping and it should condense on the neck. From there u can just scrape it off

[jdurg]

I would take the funnel you have and insert it into the open end of a U tube. (So the outelt of the flask you have would be connected to a tube leading into the open side of a U tube, then the funnel would be on the other opening with the narrow portion pointing downwards). At this point in time, take a watch glass and place it over the opening of the funnell. Put some ice on the watch glass and the I2 vapor will condense on the bottom. From that point, just scrape off the purified iodine. If you want to purify it some more, take your scrapings and repeat the sublimation/condensation procedure.

[budullewraagh]

actually i just thought of something moderately weird but kind of interesting.

 

consider the reaction between KI3 and NaOCl:

(because NaOCl is effectively Cl2+NaOH and NaOH is really irrelevant in the reaction, we can call NaOCl Cl2, just multiply Cl2 coefficient by 2 to find NaOCl equivalents)

2KI3+Cl2-->2KCl+3I2

or is it the following?

2KI3+Cl2-->2KClI2+I2

 

consider the triiodide anion:

I-I-I each with a 1/3 negative charge. i kind of think that the ClI2- anion will be formed, which would be cool. woelen, given your experience with polyhalide salt formation, perhaps you could shed light on this situation?

[woelen]

In the I3(-) ion, the three iodine atoms are not all equivalent, in fact, one of them can be regarded as having oxidation state -1, while the others have oxidation state 0. This is quite different from azide, N3(-), where indeed all N-atoms are similar and each N atom truly has oxidation state -⅓.

 

If you have a solution of KI and Cl2 is bubbled through, then you first get I3(-) ions. Continued bubbling yields ICl2(-) ions (orange) and even further bubbling yields yellow ICl4(-) ions. The latter ions are much more stable and it is very difficult to isolate ICl2(-) ions, e.g. as KICl2.

 

The precise compound formed also depends on concentration and pH. At high pH, these polyhalide ions are unstable and instead, iodate ion is formed, IO3(-).

 

At intermediate or low pH, at low concentration, I2 and Cl(-) are formed. When more chlorine is bubbled in, IO3(-), H(+) and Cl(-) are formed.

 

Only at very high concentration, the polyhalide ions ICl2(-) and ICl4(-) are formed.

 

In fact, the chemistry of these polyhalide ions is very subtle and sensitive to the reaction conditions.

[jdurg]

In the I3(-) ion' date=' the three iodine atoms are not all equivalent, in fact, one of them can be regarded as having oxidation state -1, while the others have oxidation state 0. This is quite different from azide, N3(-), where indeed all N-atoms are similar and each N atom truly has oxidation state -⅓.

 

[/quote']

 

That's a very good point, and in fact, you can almost show that chemically in that any reaction involving iodine will occur in pretty much the same manner whether you have I3(-) or I2 there. It somewhat shows that perhaps the I3(-) ion is better written as I2-(I-) where it's simply an iodide ion with an attached iodine molecule.

[budullewraagh]

the attached image is what i have in my room. the pic is pretty old, but the bottle still looks like it did back then, except my label looks like something spilled on it then dried. and i never exposed it to water. and my other wet samples dont have that sort of thing on their labels, so it's a bit odd. an important note is that the glass is clear glass, not amber, as the color of the contents above the liquid may suggest. interestingly, today, that area doesn't seem so foggy.

 

so i take it that i have a bunch of ICl2-?

[woelen]

@budullewraagh: If I look at that bottle, I think you have a solution of I3(-) ions. Such a solution gives off iodine vapor very slowly and this vapor probably escapes through the cap slowly, causing stains on all kinds of things nearby. I myself also have a bottle of a solution of KI3, made by dissolving I2 in concentrated solution of KI. This bottle looks very similar. The liquid has a strong smell of iodine.

 

@Jdurg: I would write [i.I2](-) for the triiodide ion and [Cl.ICl3](-) for the ICl4(-) ion, and [Cl.ICl](-) for the ICl2(-) ion. These formulas best represent the structure of these ions.

[budullewraagh]

i really dont feel like making more free chlorine gas. i dont have particularly good equipment and it's stressful work, so i want to use hypochlorite. problem is, of course, that hypochlorite solutions are pretty alkaline and i sort of want to play around with polyhalides or just make I2. i'd imagine that pH doesn't matter as much as amount of available hydroxide, so diluting bleach even further wouldn't help me, right? or is there a way around this?

[woelen]

If you want to make free iodine, I would go for adding a dilute solution of HCl+H2O2. This oxidizes all still existing I(-) to I2 and I3(-) also is fully converted to I2. If you use excess H2O2+HCl then all iodine will precipitate

out of solution. By filtering you can isolate the I2.

 

Don't mess around with hypochlorite/bleach solutions. This will destroy all your iodide/iodine. Because of its alkalinity, the iodine will be converted to iodate and you'll have even a harder time to isolate the material.

[budullewraagh]

and the H2O2 wouldn't oxidize the HCl to Cl2? i've heard of H2O2/HCl mixtures being used in a number of reactions, but it sounds like Cl2 will be formed if there is any excess of H2O2/HCl; of course the initial oxidation is of the iodide/triiodide, but then wouldn't the chloride be next?

 

also, i sort of want to play around with interhalogen compounds. any suggestions for making things like ICl2- and ICl4-?

[jdurg]

Any Cl2 that forms would be a very miniscule amount and would most likely go on to oxidize any I- in solution to I2, so in a sense it would actually kind of help you out there.

[woelen]

H2O2 + HCl does not give Cl2 when diluted. Mixing 30% H2O2 with 30% HCl indeed does give Cl2, but mixing 3% H2O2 with 10% HCl does not give Cl2. The H2O2 directly oxidizes the I(-) in that case and when there is excess H2O2, it remains in solution unchanged and it does not oxidize any I2 further. Working with Cl2 is more difficult, because that DOES oxidize I2 further. In dilute aqueous solution, this will lead to formation of iodate and chloride. At high concentration this will lead to formation of ICl4(-) and chloride.

 

I have quite some experience with oxidizing I(-) to I2 and to my experience the dilute H2O2/HCl is a very sure method, with hardly any loss of I2. Because of the low solubility of I2 in water, virtually all of it precipitates from solution (because no I(-) is left for forming I3(-)) and that makes isolation easy. The precipitated slurry then can be heated and the crystals of iodine can be collected on a cold piece of glass. A second heat/collect glass may be needed if you want it really dry.

 

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If you want to play around with ICl2(-) and ICl4(-), then you need to do one of the following:

- add potassium iodate to conc. HCl

- or add potassium periodate to conc. HCl

- or bubble a large excess of Cl2 through a concentrated solution of KI.

On my website, I describe the first experiment.

 

The compound KICl2 hardly can be formed, it is too unstable. KICl4 can be made fairly easily, it nicely crystallizes from aqueous solution and is not hygroscopic. It has interesting properties:

 

http://woelen.scheikunde.net/science/chem/exps/exppatt.cgi?compound=potassium%20tetrachloro%20iodate%20(III)

[jdurg]

Woelen, are you certain that the presence of the metal ions and/or the iodide ions won't catalyze the decomposition of the H2O2?

[woelen]

Which metal ions? Only K(+) is present. That does not catalyze decomposition of H2O2.

I2 also does not have any effect, at least not at the low pH involved. I have done this reaction many times and it works like a charm. No evolution of oxygen, 100% oxidation of I(-) and 99+% of all I2 precipitates as a dense and easy to separate solid mass. Really neat and easy. This is one of the best preps I know of I2 from KI or NaI, using only simple OTC chemicals, like 3% H2O2 and 10% HCl. As I stated before, you should not use concentrated acid and H2O2, using that will result in a lot of side reactions.

[jdurg]

I just recall in high school chemistry class dropping crystals of KI into some H2O2 and watching it rapidly decompose while little bits of iodine are formed.

[woelen]

Yes, you're totally right. But this is without acid. Repeat the same experiment with H2O2 + dilute strong acid (any strong acid will do). Then you'll not get bubbling, but the solid iodine is formed instead.

 

In the experiment you describe you get catalysis through the formation of hypoiodite:

 

I(-) + H2O2 ---> IO(-) + H2O

 

IO(-) oxidizes H2O2 (oxygen goes from -1 to 0 oxidation state):

 

IO(-) + H2O2 --> I(-) + H2O + O2

 

In acidic solutions, however, the hypoiodite ion cannot exist and the acid HIO is so extremely unstable that it in practice cannot be formed. The only alternative then is formation of iodine and there is no catalytic effect.

[jdurg]

Yes' date=' you're totally right. But this is [i']without [/i]acid. Repeat the same experiment with H2O2 + dilute strong acid (any strong acid will do). Then you'll not get bubbling, but the solid iodine is formed instead.

 

In the experiment you describe you get catalysis through the formation of hypoiodite:

 

I(-) + H2O2 ---> IO(-) + H2O

 

IO(-) oxidizes H2O2 (oxygen goes from -1 to 0 oxidation state):

 

IO(-) + H2O2 --> I(-) + H2O + O2

 

In acidic solutions, however, the hypoiodite ion cannot exist and the acid HIO is so extremely unstable that it in practice cannot be formed. The only alternative then is formation of iodine and there is no catalytic effect.

 

Ahhhhh. Thanks for the explanation. I never really looked into it much further other than "Add KI to H2O2 and it decomposes". Interesting explanation.