LanthanumK
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Bromate/iodate reaction with hydrochloric acid
Sodium chlorate, according to Wikipedia, reacts with hydrochloric acid to produce chlorine gas, chlorine dioxide, and sodium chloride. I added
hydrochloric acid to a filter paper containing some dried bleach and it turned yellow (chlorine dioxide solution). Some chlorine was also produced, as
evidenced by smell.
By comparison, I added hydrochloric acid to some sodium bromate/bromide crystals, formed by the heating of electrolyzed sodium bromide solution. The
acid turned red immediately, and copious bubbling was observed. The smell of halogens was not extremely strong, although it was far from faint;
therefore, the large bubbles could not have been entirely chlorine.
What is the reaction for sodium bromate (or iodate) with hydrochloric acid?
hibernating...
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woelen
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I found this old post and I have done quite a few experiments in this reaction, so I think I can give a decent answer to this question.
The reaction of bromate is fairly simple.
Bromate ion is completely reduced by concentrated hydrochloric acid:
BrO3(-) + 6H(+) + 5Cl(-) ---> 2Cl2 + BrCl + 3H2O
You get a golden yellow solution and a yellowish gas mix above the liquid. The golden yellow color is due to the presence of bromine chloride in the
concentrated hydrochloric acid, the yellowish gas mix is mainly chlorine, contaminated with some BrCl.
On dilution with a lot of water, BrCl is hydrolyzed to chloride ions, bromine and bromic acid.
---------------------------------------------------------------------------------
Iodate reacts with concentrated hydrochloric acid by forming the beautiful gold/yellow polyhalogen ion ICl4(-):
IO3(-) + 6Cl(-) + 6H(+) ---> ICl4(-) + Cl2 + 3H2O
The ion ICl4(-) is in equilibrium with ICl3 and Cl(-).
In the presence of potassium ions, rubidium ions, or cesium ions you get a nice bright yellow solid: MICl4, where M is the alkali metal ion:
http://woelen.homescience.net/science/chem/exps/KIO3+HCl/ind...
http://woelen.homescience.net/science/chem/exps/rubidium_pol...
On strong dilution, the ICl4(-) ion is hydrolyzed by water with formation of chloride ions, free iodine and iodate ions. The golden yellow solution
befomes brown and turbid if a lot of water is added.
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chornedsnorkack
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The iodate reaction is interesting.
A Pourbaix diagram for iodine I found at:
http://sti.srs.gov/fulltext/tr9900270/tr9900270.html
Note that the redox potential of IO3(-)/I2 never exceeds the redox potential of O2/H2O at any pH
On the other hand, chlorine is at almost all pH a stronger oxidant than oxygen.
It would follow that the reaction
2IO3(-)+10Cl(-)+10H(+)->I2+5Cl2+5H2O
should stay on the left at any pH
But we have the formation of ICl3, to drag the reaction to the right... so IO3(-) should become a stronger oxidant when there is something available
to bind the reduction products (like Cl(-) or Ag(+)).
If you are worried about the reaction giving off gaseous chlorine, how about mixing the iodate with a stoichiometric amount of reducent, like solid
iodine or iodide, before adding hydrogen chloride?
Reactions could be like
3KIO3+I2+18HCl->3KICl4+2ICl3+9H2O
2KIO3+KI+12HCl->3KICl4+6H2O
What are the HCl concentrations where these reactions go to right, as opposed to the HCl concentrations needed to oxidize Cl2?
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bismuthate
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So I recently did some experiments with a mix of Na/KICl4. Interestingly enough when I mix it with a copper sulfate solution that has been reduced
with metabisulfite (with copper(I) in it now) it forms a brown precipitate. I'm not sure what it could be. I first thought that it could be Cu(I)ICl4
but that seems rather impossible.
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woelen
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The brown material most likely is a precipitate of CuCl, contaminated with iodine. If you use a large excess of the ICl4(-) ion, then the precipitate
will dissolve partly and become dark grey, being fairly pure iodine.
CuICl4 with copper in oxidation state +1 is not possible. ICl4(-) is a very strongly oxidizing ion and will immediately oxidize the Cu(I) to cu(II).
The ICl4(-) ion is so strongly oxidizing and reactive that it may lead to spontaneous ignition if a salt of this (e.g. KICl4) is mixed with finely
powdered Mg-metal or red P.
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MrHomeScientist
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woelen, your second link is a really beautiful experiment! I'd like to try this sometime, but I don't have most of the reagents.
Handbook of Inorganic Chemicals by Patniak lists a bunch of ways to make iodic acid:
Oxidation of iodine with conc. nitric acid:
3I2 + 10HNO3 == 6HIO3 + 10NO + 2H2O
Oxidation of iodine with hydrogen peroxide:
I2 + 5H2O2 == 2HIO3 + 4H2O
>> What concentration of peroxide?
Oxidiation of iodine with chlorine in dilute acidic solution:
I2 + 6H2O + 5Cl2 == 2HIO3 + 10HCl
>> Is this one as simple as making a solution of iodine with a pinch of iodide for solubility, bubbling chlorine into it, and reducing
volume to precipitate the solid acid?
The nitric acid route sounds like the simplest, for me at least. The other difficulty of this experiment being rubidium compounds are
expensive!
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chornedsnorkack
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Does ICl4(-) anion have any poorly soluble salts?
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woelen
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@MrHomeScientist: The only practical reaction is the first one. The second one will lead to huge losses of hydrogen peroxide and only is of interest
from an academic/theoretical point of view. The third reaction works in very dilute solutions. On concentration, you go the other way around and you
get the reaction which forms Cl2 and ICl4(-).
One reaction, which is feasible is making of periodate. Periodates give a reaction similar to iodate in conc. HCl, they also give ICl4(-) and Cl2, but
they produce two times as much of Cl2 per mole of reagent.
Try this one: http://woelen.homescience.net/science/chem/exps/KIO4_synth/i...
You can replace KIO3 with KI. You just have to bubble more Cl2 through the solution and you need somewhat more KOH besides the KI.
You can also make the sodium salt: http://woelen.homescience.net/science/chem/exps/Na2H3IO6/ind...
This one works with NaI as starting material, but also NaIO3 can be used.
@chornedsnorkack: I know of no insoluble stable salt of ICl4(-). The only stable ones I know are the potassium, rubudium and cesium salts. The
ammonium salt does not exist, because the ICl4(-) ion is incompatible with ammonium ion. The cesium salt is the least soluble in conc. HCl.
ICl4(-) also is unstable in dilute aqueous solution, it hydrolyses to HIO3, I2 and Cl(-).
An interesting experiment may be to add HIO3 to conc. HCl in which some transition metal salt is dissolved. Many transition metal salts, however, do
not have the free metal ion in solution, but strongly coordinated negatively charged ions, which do not combine with ICl4(-). E.g. copper(II) does not
exist as Cu(2+) in conc. HCl, but as CuCl4(2-). A similar thing is true for nickel, cobalt, iron(III). Many others are oxidized by the ICl4(-), e.g.
iron(II), copper(I), manganese(II), tin(II).
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bismuthate
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woelen first that is why I believe it to be impossible. Also the Cu(I) ion could not exist in the presence of a strong oxidizer so CuCl would be
impossible.Also upon analyzing the product I found that it does not react with a starch solution. This is truly curious. I don't know what could have
happened. Could a insoluble double salt be formed? Or possibly a copper halide anion?
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woelen
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I can imagine formation of CuCl in your situation, provided there also was sulfite in the mix. CuCl and I2 coexist (e.g. if you add a solution of a
copper(II) salt to a solution of an iodide, then you get insoluble CuI and I2, and if chloride is present as well, you can get CuCl with I2).
But if your brown material does not react with starch (and your starch does react with proven iodine) then you indeed have an interesting compound. I
have no idea what it can be in that case. First you should try your starch with proven iodine under the same conditions as the ones in which you
tested your brown precipitate. I once had starch, which did not react with iodine, probably it was a different kind of starch, based on different base
sugars.
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bismuthate
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I, at first, did not believe that the starch did not react so I tested it and it worked just fine multiple times. When you say "the same conditions"
what do you mean by that?
Oh also the starch is a starch solution specially made to be used for the oscillating iodine reaction, so it definitely isn't the wrong kind (I have
also used it many times).
[Edited on 20-5-2014 by bismuthate]
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chornedsnorkack
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Quote: Originally posted by woelen  |
@chornedsnorkack: I know of no insoluble stable salt of ICl4(-). The only stable ones I know are the potassium, rubudium and cesium salts. The
ammonium salt does not exist, because the ICl4(-) ion is incompatible with ammonium ion. The cesium salt is the least soluble in conc. HCl.
ICl4(-) also is unstable in dilute aqueous solution, it hydrolyses to HIO3, I2 and Cl(-).
An interesting experiment may be to add HIO3 to conc. HCl in which some transition metal salt is dissolved. Many transition metal salts, however, do
not have the free metal ion in solution, but strongly coordinated negatively charged ions, which do not combine with ICl4(-). E.g. copper(II) does not
exist as Cu(2+) in conc. HCl, but as CuCl4(2-). A similar thing is true for nickel, cobalt, iron(III). Many others are oxidized by the ICl4(-), e.g.
iron(II), copper(I), manganese(II), tin(II).
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Out of the monovalent metal cations, both that are not alkali metals have stable/insoluble halides and should therefore break down ICl4(-).
Hm - you want a big cation? Well, the biggest divalent cations are not transition metals - they are alkaline earths.
Is Ba(2+) capable of precipitating a salt with ICl4(-)? Or do the solubilities of ICl4(-) salts match those of ClO4(-) and MnO4(-) salts, so only K,
Rb and Cs salts precipitate?
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bismuthate
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So, I was thinking; could ICl4- salts of transition metals exist in the form of ammine complexes? (I don't know much about the chemistry)
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chornedsnorkack
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Not complexes with ammonia. In strongly acid environments proton is a stronger competitor for ammonia than transition metal cations, and apparently
ICl4(-) is able to carry out oxidative attack on ammonia, unlike NH4MnO4 and NH4ClO4 which explode on heating but are reasonably stable crystallized
from a cold aqueous solution.
How vulnerable is (CH3)4N(+) cation to oxidative attack, compared to NH4(+)?
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woelen
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The ion (CH3)4N(+) would indeed be a nice candidate. I have the chloride salt of this and I can try this with iodic acid and conc. HCl. I'll come back
on this, it is an easy experiment for me.
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PHILOU Zrealone
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Just a thought:
I wonder if mixing ICl3 or MeICl4 with AlCl3 would help for direct formation of iodoaromatic dichlorides like iodobenzene dichloride?
ICl3 + AlCl3 --?--> AlCl4(-) + ICl2(+)
ICl2(+) + Ar-H --> Ar-ICl2 + H(+)
Maybe "poly-iododichlorid-ation" could happen... but more likely not without an activating group like OH or NH2.
On an energetic material point of view it could be very interesting because Ar-ICl2 give acces to Ar-I=O and Ar-IO2...the later is a known
explosophoric group (iodoxy-benzene is explosive while nitrobenzene is not!)!
Phenol could give rise in one step to 2.4.6-tri-iododichloride-phenol; precursor of 2.4.6-tri-iodoso-phenol and finaly to 2.4.6-tri-iodoxy-phenol, an
explosive and sensitive compound that must be related to trinitrophenol (explosive forming picrates salts primaries) but with a very high density
owing to the presence of 3 iodine atoms (density is important in increasing the velocity of detonation of energetic materials but little is known
about that rule outside CHNO molécules...).
Triiodoxyphenates if existing could be sensitive primaries.
[Edited on 22-5-2014 by PHILOU Zrealone]
PH Z (PHILOU Zrealone)
"Physic is all what never works; Chemistry is all what stinks and explodes!"-"Life that deadly disease, sexually transmitted."(W.Allen)
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woelen
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I tried the experiment with the N(CH3)4(+) ion and the result is amazing!
Experiment 1
----------------
Add some solid N(CH3)4Cl to conc. HCl and dissolve all of the solid. This goes fairly easily.
Next, add some solid HIO3 to this solution. The HIO3 is covered by a thin bright yellow layer and the reaction stops.
Shake the test tube with the solid chunk of HIO3. Slowly, little yellow flakes are removed from the chunk of HIO3 and newly exposed HIO3 becomes
covered with the yellow solid. There also is slow release of Cl2.
After a few minutes of shaking, still only a small part of the HIO3 has dissolved and there is quite some compact yellow precipitate under a clear
colorless liquid.
Experiment2
---------------
Add a chunk of appr. 0.5 grams of HIO3 to a few ml of conc. HCl in a short test tube. As soon as the HIO3 comes in contact with the HCl, it starts
foaming vigorously and a lot of Cl2 is produced. The liquid becomes deep golden yellow/orange. The final piece of HIO3 dissolves more slowly. An
additional ml of conc. HCl was added to also quickly dissolve the last amount of HIO3.
In a separate test tube prepare a highly concentrated solution of N(CH3)4Cl, which is somewhat viscous. Pour 1 ml of this solution on the golden
yellow/orange solution. The two liquids do not mix, between the colorless solution of N(CH3)4Cl and the golden yellow solution, there is a solid very
thin yellow layer.
With a glass stick press through both liquids and thoroughly mix the two liquids. The entire mass solidifies to a bright yellow solid, with just a
small amount of yellow/orange liquid above it. The whole mass can be kneaded. With a small plastic spoon, everything is taken out of the short test
tube and put on filter paper. This solid material is pressed between filter paper and around the filter paper a lot of paper tissue. A fairly dry
solid bright yellow mass is obtained. This solid mass has a faint smell of chlorine.
The remains of the yellow solid, which cannot be scraped from the test tube are rinsed with water. The solid easily is removed from the glass and
quickly a bright yellow precipitate settles at the bottom under the colorless liquid, which is slightly acidified water.
Remove the acidified water and replace with plain tap water. The yellow solid does not change.
Again remove most of the water and add a little amount of 5% ammonia to the yellow precipitate. Immediately, the yellow precipitate becomes dark
grey.
------------------------------------------------------------------------------------
My explanation of this experiment:
- The yellow solid is N(CH3)4ICl4. This material is (nearly) insoluble in water.
- The insoluble material apparently dissolves with so great difficulty that the ICl4(-) ion does not hydrolyze in the water. This is in strong
contrast with e.g. KICl4. This immediately turns brown when added to water, due to hydrolysis.
- On addition of NH3, the ion ICl4(-) is destroyed at once, with libration of iodine, which immediately reacts to NI3.NH3 with the NH3 in the water.
------------------------------------------------------------------------------------
I now have some N(CH3)4ICl4 drying. Let's see how well it remains in contact with air and if I can isolate some of the dry solid. Further experiments
will follow.
[Edited on 22-5-14 by woelen]
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woelen
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I found the experiment with the tetramethylammonium salt of tetrachloroiodate(III) so nice, that I decided to make a webpage of it, which describes
the making of the exotic salt N(CH3)4ICl4:
http://woelen.homescience.net/science/chem/exps/tetrachloroi...
If you have the chemicals, then it really is worth the effort to try this experiment. If you have no iodic acid, but you do have potassium iodate or
sodium iodate, then you can try with smaller amounts of iodate added to 25% HCl. I'm quite sure that also works.
If you have no tetramethylammonium salts, then you could try a tetraethylammonium salt. There is some freedom of choice.
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Bezaleel
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Interesting, and - as usual - a beautiful and clear write-up. Thanks! 
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