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AndersHoveland
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I think the reaction, in the presence of water, is actually initially:
2 KI + 3 NO2 + H2O --> 2 KNO3 + 2 HI(aq) + NO
However, when the concentration of the hydroiodic acid in solution starts to get more concentrated, the equilibrium will shift, and iodine will begin
to be formed.
2 KI + 2 NO2 --> 2 KNO2 + I2
Normally, I would tend to think, nitrogen dioxide would preferentially oxidize nitrite to nitrate before it would displace iodide from its salts, but
in this situation things are different. All that concentrated hydroiodic acid in solution shifts the equilibrium, by making the environment acidic and
reducing.
While iodine will oxidize nitrite to nitrate under alkaline conditions, under acidic conditions nitric oxide can reduce nitrate to nitrogen dioxide.
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“Chlorine, bromine, and iodine also oxidize nitrous acid solutions to nitrate. The reaction between aqueous iodine and nitrite ion is measurably
slow, and the rate in buffered solutions (pH = 6 to 7) has been studied by Durrant, Griffith, and McKeown [Trans. Faraday Soc., 32, 999 (1936)]. The
net reaction is
NO2- + I2 + H2O = NO3- + 2I- + 2H+
Hydroiodic acid acts as a reducing agent towards nitrous acid solutions, the reaction products being iodine and nitric oxide.
RATE CONSTANTS FOR THE OXIDATION OF NITRITE ION BY IODINE for Systematic Inorganic Chemistry (1946)
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KNO2 + NO2 --> KNO3 + NO
KNO2 + I2 + H2O --> 2 HI(aq) + KNO3
NO + I2 + H2O --> 2 HI(aq) + NO2
3 KNO2 + 2 HI(aq) --> KNO3 + KI + H2O + 2 NO
4 KNO2 + I2 --> 2 KNO3 + 2 KI + 2 NO
I think the best equation to describe the reaction, after hydriodic acid has already accumulated and made the solution acidic, may be:
4 KI + 4 NO2 --> 2 KNO3 + 2 KI + I2 + 2 NO
Some of you may be wondering why the iodine does not just oxidize the nitric oxide to nitric acid. Here are my thoughts on that. Nitric acid is a
strong oxidizing agent, but generally only at high concentrations (>60%). However, even 20% conc. nitric acid is still an oxidizing agent. It is
just that its reaction rate as an oxidizer is very very slow. Nevertheless, this slow reaction rate would still tend to make the equilibrium favor
nitric acid, since the rate of reaction forming nitric acid would be much faster than the reverse action. The big factor here is probably all those
nitric oxides floating around, which act as catalysts to reduce back the nitric acid. While dilute nitric acid is essentially not an oxidizer, nitric
oxide which is a reactive free radical can reduce it to lower oxides of nitrogen which can act as much more reactive oxidizing agents. Again, although
dilute nitric acid is a more "powerful" oxidizing agent than nitrous acid, it is just not a reactive one.
[Edited on 9-11-2012 by AndersHoveland]
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AJKOER
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The only concern I have with this Chlorine basis method is the possible formation of ICl and ICl3 with reduced yield if it happens that excess Cl2 is
employed or Iodine crystals are exposed locally to Cl2 gas. Basis, see Wiki (http://en.wikipedia.org/wiki/Iodine_monochloride ), to quote:
"Preparation of iodine monochloride entails simply combining the halogens in a 1:1 molar ratio, according to the equation
I2 + Cl2 → 2 ICl
When chlorine gas is passed through iodine crystals, one observes the brown vapor of iodine monochloride. Dark brown iodine monochloride liquid is
collected. Excess chlorine converts iodine monochloride into iodine trichloride in a reversible reaction:
ICl + Cl2 <--> ICl3 "
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White Yeti
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That's why iodide is used in slight excess. It's easier to deal with triiodide than with assorted halogen halides. The reaction is conducted in
aqueous solution; you may have overlooked the fact that ICl decomposes in water.
"Ja, Kalzium, das ist alles!" -Otto Loewi
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AJKOER
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Actually, ICl rapidly reacts as follows:
ICI + H2O —> Cl– + OI– + 2H+
And, under acidic conditions, the final products (through a series of intermediaries) are:
Acid:
5 ICl (aq) + 3 H2O <--> HIO3 + 2 I2 + 5 HCl
Base:
3 ICl(aq) + 6 NaOH <--> NaIO3 + 2 NaI + 3 NaCl + 3 H2O
Source: "Kinetics of hydrolysis of iodine monochloride measured by the pulsed-accelerated-flow method" by Yi Lai Wang, Julius C. Nagy, Dale W.
Margerum, published in
J. Am. Chem. Soc., 1989, 111 (20), pp 7838–7844. Online link
http://pubs.acs.org/doi/abs/10.1021/ja00202a026
So the formation of ICl in an aqueous solution can lead to some Iodine loss via the formation of Iodate. Hence, my prior comment "formation of ICl and
ICl3 with reduced yield".
[Edited on 21-11-2012 by AJKOER]
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White Yeti
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If you're going to start splitting hairs again, I'm gonna piss off. Before I do, I'd like to say that the disproportionation reaction in base is
unfavourable unless there is a large excess of hydroxide and a huge amount of iodine chloride is present. This would not be the case the reagents are
used in stoichiometric amounts. The loss of iodine as iodate is as insignificant as considering the formation of chlorate when dissolving chlorine in
water.
"Ja, Kalzium, das ist alles!" -Otto Loewi
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woelen
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The loss of iodine to iodate only occurs when there is excess chlorine. As long as there is excess iodide no iodate/iodic acid is formed at all (if
any were formed, then in the acidic solution, usually used to make iodine from iodide and bleach it would be converted to iodine at once).
So, in practice the chlorine/bleach method works fine as long as no excess amount of bleach is used and one assures that the solution remains acidic.
If the solution becomes alkaline, then the reaction does not work anymore. In that case you get iodate as final product.
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AJKOER
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OK, I apology for the seemingly hair splitting, but my concern lies with my original comment "reduced yield if it happens that excess Cl2 is employed
or Iodine crystals are exposed locally to Cl2 gas." As no one has challenged the local concentration issue, I have some theoretical concerns on how it
may impact Iodine yield (and sometimes my theory are how things can go wrong are all too accurate).
Assume no stirring permitting local Cl2 concentration issues and then, with even perfect stoichiometric amounts, some ICl and then Iodate is formed
(or formed by another route). Then via the reaction (see http://paws.wcu.edu/bacon/vitamin%20c.pdf ):
IO3- + 6 H+ + 8 I- = 3 I3- + 3 H2O
we have a significant conversion of the Iodide to the tri-iodide and not Iodine.
The comments presented do not appear, however, to suggest that this is a significant issue observed in practice.
[EDIT] Per wikipedia (http://en.wikipedia.org/wiki/Iodine_clock_reaction ):
IO3− + 6 H+ + 5 I− → 3 I2 + 3 H2O
so as long as we avoid a stoichiometric excess of Iodide, loss due to tri-iodide (from iodate and excess iodide) may not be an issue after all.
[Edited on 22-11-2012 by AJKOER]
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woelen
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AJKOER, what you describe in your last post certainly will occur, also in practice, but it is not an issue. Suppose, due to some locally concentrated
spot with lots of Cl2 that ICl and/or HIO3 are formed, then this is not a true loss. Somewhere else there will be insufficient chlorine to convert all
iodide to iodine. So at one spot, iodide remains present (either as free iodide, or bound to iodine as I3(-)) and at other spots excess Cl2 leads to
formation of HIO3. As soon as the mix is stirred the iodine will be formed anyway. HIO3 and I(-) immediately react with each other to iodine. What I
write is true under the following conditions:
- total amount of chlorine is not in excess, any local excess is compensated with other spots where there is too little chlorine.
- solution is acidic, otherwise iodate and iodide do not react to form iodine.
A good way to make iodine from iodide and bleach is first to titrate some of the bleach such that its concentration is known and then add a
stoichiometric amount of potassium iodide (heated in an oven to 100 C for some time to drive off any water from the crystals). When all iodide is
dissolved, then acid is added in sufficiently large amounts.
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vmelkon
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This reaction got me interested since it produces KNO2. I'm interested in having a nitrite salt since some organic chemistry experiments call for it.
Is the only way to separate the nitrite and nitrate is by taking advantage of solubility differences?
Another question : when you melt KNO3, it decomposes to KNO2 and O2, right?
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