len1
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Simple measurement of Iodide
I searched the web for hassle-free and accurate determinations of iodide. As usual lots of pages repeating one another, precious little original. A
few research papers on the myriad of possible reactions helped me esteblish a method I havent seen elsewhere. It works a treat, but small stuff like
this is unpublishable. I know hardly anyone interested in quantitative/experimental chemistry here, please dont be offended.
Simplest way to titrate I- with easily accessible chemicals is to turn it into iodine. Trouble is
a) Most oxidizing agents strong enough to do this will continue on to IO3-
b) The method has no obvious end point
H2O2 and S2O8 2- in acidic medium are the most common reagents to solve a)
H2O2 + 2H+ + 2I- -> 2H2O + I2
The way to solve b) is to titrate the I2 generated with thiosulphate, which only slowly reacts with H2O2 directly - (the links show a whole range of
oxidizers for which this is characteristic, for a reason I have not yet established)
I2 + 2S2O3 2- -> S4O6 2- + 2I-
The end point is indicated by adding clear 0.1% starch soln. (only when colour of I2 is difficult to see)
This method also solves the problem that H2O2 is unstable (and so no standard) since its used in excess.
But there are big problems
1) Thiosulphate is unstable in acidic media below about pH=2, disproportionating and generating an indeterminate amount of white/yellow cloudy
sulphur which settles to the bottom, and makes its quantity non-deterministic
S2O3 2- => SO3 2- + S
2) For low [H2O2] the half-time is of order of hours, while using higher acidities or H2O2 concentrations allows the following to take place
I2 + 5H2O2 -> 2IO3- + 4H2O + 2H+
2H2O2 -> 2H2O + O2 (I- catalyzed)
3) A basic solution of order pH ~ 10 will get the iodine to disproportionate immediately
I2 + 6OH- -> IO3- + 5I- + 3H2O
4) And the most serious problem, the thiosulphate regenerates I-, which therefore serves as a catalyst in the oxidation of thiosulphate by H2O2. The
amount of the thiosulphate in any back titration is then determined by the H2O2.
For this reason thiosulphate has generally been used to titrate I2, and any other oxidizing agents which oxidize it in the links below, but was
thought incapable of titrating iodide, for the latter purpose one generally used
i) bromine with phenol
ii) I2 separation in hydrocarbon with subsequent titration
iii) a buffer solution
These are all very cumbersome and introduce their own uncertainties.
Liebensfahl found the governing step in the oxidation to be the formation of the intermediate HIO
I- + H+ + H2O2 -> HIO + H2O
and rate proportional to
(a + b[H+])[I-][H2O2] -> a[I-][H2O2] for 6 > pH ~> 2
The precise time constant was hard (as is anything useful) to find on the net but from data available I calculated
tc = 0.6(min)/[H2O2 in mol/L]
with
[I-] = [I-]o exp(-t/tc) for 6 > pH ~>2
Over-oxidation to IO3- requires [H+]>>10-2 mol/L and even then is first oder in [I2] only and has time constant of order 20mins.
So a good procedure is to use an excess of H2O2 and conc. of the order of [H2O2] = 0.3 M, which gives tc = 2 mins. In other words after 10mins the
results are accurate to within 1% - note that falling out of I2 is not a good indication of progress of reaction, as this onsets very suddenly when
[I-] has become severly depleted at end of reaction.
The procedure has been
1) Mix 10 ml test [I-] solution with 10ml 0.6M H2O2 and 1ml 0.6M H2SO4. If later one finds [I-] >
0.1M use correspondingly more H2SO4 soln.
2) Let reaction run for 20mins.
3) Hinder reaction by diluting to 200ml with H2O and adding 0.6M NaOH till pH ~ 5 - 7. Excess NaOH obviates itself by rapid and instant
discoloration of iodine due to IO3- formation. This procedure increases characteristic reaction time (time for concentration to go down by e^-1) (for
oxidation of I-) to about 200min.
4) Titrate quickly with S2O3 2-, shaking. If titration time ~1min, error < 0.2%.
No buffers, hydrocarbon elution and evaporation, bromine water or phenol is needed!
Titration tested with known KI (0.06M) and gave repeatable expected results.
books.google.com.au/books?isbn=1406706345
books.google.com.au/books?isbn=1406736937
books.google.com.au/books?id=ldo8AAAAIAAJ
books.google.com.au/books?isbn=1406747904
books.google.com.au/books?isbn=140677314X
books.google.com.au/books?isbn=8170247179
books.google.com.au/books?isbn=1406715034
Edit by chemoleo: Title
[Edited on 19-3-2009 by chemoleo]
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Magpie
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| Quote: |
3) A basic solution of order pH ~ 10 will get the iodine to disproportionate immediately
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Does this mess up the iodine method for estimating CN- in alkalai cyanides found in Allen's book? Or does his massive dilution take care of that?
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len1
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| Quote: | Originally posted by Magpie
| Quote: |
3) A basic solution of order pH ~ 10 will get the iodine to disproportionate immediately
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Does this mess up the iodine method for estimating CN- in alkalai cyanides found in Allen's book? Or does his massive dilution take care of that?
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I dont have Allen's book, is this a simple titration of I2 vs CN-?
If that is what you are referring to this is an interesting point since the next thing I was going to do with the iodine is titrate cyanide -
obvioulys given my conclusion in the cyanide thread that the best candidate to determine CN- accurately in the presence of CNO- is I2.
Since an OH- concentration of the order of pH ~ 10 is sufficient to discolour iodine (and you can test this is so by adding a drop of NaOH to an I3-
solution) then since KCN solutions have a pH ~ 12 they should discolour the iodine just by disproportionation without any oxidation of the CN- taking
place.
We know both dispropotionation
1) I2 + 6OH- -> IO3- + 5I- + 3H2O
and cyanide oxidation
2) CN- + I2 + 2OH- -> 2I- + CNO- + H2O
are not kinetically hindered so the explanation if this method works lies in the thermodynamics.
Since reaction is aqueous, delG of formation is not appropriate and its best to use standard potentials which are easy to find for aqueous media. I
found
1) I2 + 2e- -> 2I- Vo = 0.535V
2) 2IO3- + 12H+ 10e- -> I2 + 6H2O Vo = 1.19V
3) 2IO3- + 6H2O + 10e- -> I2 + 12OH- Vo = 0.19V
So in acid ([H+] = 1M) I2 takes electrons at 0.535V to go to I-, but another atom of iodine will release them only at -1.19V to go to iodate. The sum
of cell voltages is negative hence disproportionation will not proceed if reactants mixed. On the contrary the potential for IO3- redunction in
acidic media is strong enough to drive the first reaction as an oxidation 1.19-0.535 > 0. So that is why adding KI to KIO3 solution in (strongly)
acidic media will precipitate all the iodine and is used as a coproportionation reaction and is a test for KIO3 in salt.
In basic solutions its the opposite case as the reduction voltage for iodate drops a whole volt, and themodynamics would drive half cell 1) in
reduction and 3) in oxidation. However the driving force here mainly comes from the iodine oxidation, the potential for the iodine reduction is low,
suggesting that many other reducing agents would beat it. Another was of saying this is that for the I2 concentration maintained by most reducing
agents in competition with reduction 3) at equilibrium, the corresponding iodate concentration, and therefore the contribution of the
disproportionation reaction to the total expenditure of iodine, will be fractionally small.
The oxidation potential for cyanide to cyanate in water is on the other hand very large
CN- + 2OH- -> CNO- + H2O + 2e- Vo = -1.986
so the total potential for the oxidation of CN- with I2 is
1.986 + 0.535 = 2.521V
while for the oxidation of I2 with I2 (disproportionation)
-0.19 + 0.535 = 0.516V.
Now in terms of equilibrium constant K (at 298K) we have
K(n) = 10 ^ n (Vo in volts) / 0.06
where K(n) is the equilibrium constant as writen for n moles of electrons transfered [its can obviously differ by a constant power depending on how
you transcribe the stoichiochiometry into the thermodynamic expression].
So for equilibrium 1
CN- + 2OH- + I2 -> 2I- + CNO- + H2O we have
[I-]^2 [CNO-]/([I2][CN-][OH-]^2) = 10 ^ (2 * 2.521 / 0.06) ~ 10^83 a huge number!
while for equilibrium 2
3I2 + 6OH- -> IO3- + 5I- + 3H2O
[IO3-][5I-]/([I2]^3 [OH-]^6 ) = 10 ^ (5 * 0.516 / 0.06) ~ 10^42
use of [I2] here is justified since it is assumed all solvated in the form of [I3-] for which the redox potential is about the same as for I-. If I2
reaches saturation for the amount of I- present then expression needs be changed.
Clearly while any detectable CN- concentrations are present [I2] is so small due to equilibrium one, that substitution of its value into equilibrium 2
will give tiny IO3- concentrations. The contribution of the disproportionation reaction to the whole business while any CN- is remaining is this
negligible.
A more interesting consideration which I havent checked yet, is whether the end-point is accompanied by a colour change. When all CN- is replaced by
CNO- the pH obviously drops. However cyanic acid is still a weak acid (though much stronger the HCN pKa 3.4 vs 9.5) so KCNO solutions are basic, ill
have to calculate whether this is basic enough to drive disproportionation. Better still do the the experiment.
[Edited on 20-3-2009 by len1]
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len1
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| Quote: |
while for equilibrium 2
3I2 + 6OH- -> IO3- + 5I- + 3H2O
[IO3-][5I-]/([I2]^3 [OH-]^6 ) = 10 ^ (5 * 0.516 / 0.06) ~ 10^42 |
These equations contain a typo and two errors that change everything, not that anyone noticed.
1) I- should enter to the power of its stoichiometric factor 5
2) The standard state of iodine is solid so the voltage for the iodide half cell quoted, 0.535V, would have been measured with a saturated solution
of I2 over solid iodine rather than with [I2] = 1M. Since iodine concentration is then a constant it drops out of the equilibrium constant.
3) 0.535 - 0.19 = 0.345 (not 0.516)
so
[IO3-][I-]^5/[OH-]^6 = 10^ (5 * 0.345 /.059) ~ 10^29
Assuming all iodide/iodate all come from disproportionation of I2 we have
[I-] / 5 = [IO3-]
so
[I-]^6 = 5 10^29 [OH-]^6
so
[I-] ~ 10^5 [OH-]
This gives us how much iodine is lost to disproportiation at a given pH. At pH = 7 this is 10^-2 mol/L, very small. At pH = 10 its 10 mol/L, ie
essentially no iodine left. So to see an iodine colour in iodometric titrations solutions must be acidic, especially at low iodine conentrations
(starch I presume shift equib somewhat, but ive seen not Ks for starch-I2 complex)
[Edited on 22-3-2009 by len1]
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len1
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Interesting development in this.
After the oxidation of 0.1g KI in 10ml H2O using 10ml 0.6M H2O2 at pH~2, which is essentially over in 20mins as per above, I had what must have been
about 80mg iodine precipitated and above it 20ml of almost clear water - confirming concentration of iodide remaining in solution was tiny, as else
the iodine would have dissolved.
And so this remained for about 2 weeks.
Today I was surprised to see the iodine entirely gone and the solution completely clear.
The only reaction I can think possible is further oxidation of I2 to the 5+ state KIO3 by the H2O2. But why such a sudden onset of reaction, with
then proceeds to completion? The only explanation is that its autocatalytic. However all references Ive seen mention it only first order in [I2]
rate = k[I2]
There must be a lot more to it than that.
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Magpie
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| Quote: |
There must be a lot more to it than that.
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Len1 I know you to be a very thorough and resourceful researcher so if anyone can get to the bottom of iodine chemistry it will be you. But I must
pass on my own observation. When I was working in the nuclear industry we were very concerned with the fate of iodine, because of radioactive
I129/131. So we (as engineers) hired chemists who had built much of their careers on tracking iodine. They provided much information but never
claimed they had the whole story.
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len1
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Thanks Magpie, you flatter me. I think I just persevere as far as I can, without it you get nowhere - I think Edison hinted to something like that.
Iodine in the human body though must be far more difficult than in a test tube.
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Panache
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Quote: Originally posted by len1  | | Quote: |
while for equilibrium 2
3I2 + 6OH- -> IO3- + 5I- + 3H2O
[IO3-][5I-]/([I2]^3 [OH-]^6 ) = 10 ^ (5 * 0.516 / 0.06) ~ 10^42 |
These equations contain a typo and two errors that change everything, not that anyone noticed.
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I noticed however i'm far too polite to point out what was obviously not an error in your rationale, as you're far too intelligent to make such a
trivial mistake. If you stop eating donuts your fingers will slim down and your accuracy on the keyboard will surely improve.(assuming you have
fingers)
On topic however is more in reply to basic discussion. Not to be critical but validating via KI standards, although an essential first step, only
validates that the methodology is accurate for KI samples, not more complex matrices, although it may be.
Coincidentally i had some time on my hands two weeks ago and decided to tackle 5litres of 'iodine waste' i had been accumulating on and off for
sometime.
It was a lovely day so i put the lead on my dog and walked to the local hydroponics outlet to purchase some 50%h2o2. i arrived to realise it closed
for good (it had been two years since my last visit).
Incensed and annoyed that my first effort to clear a backlog of crappy jobs had ended as such i resolved to dump the first oxidant i could lay my
hands on when i returned. It happened to be sodium permangenate and after several dilutions and two days of tedious filtering (the MnO2 was a super
fine ppt) followed by re-extraction into IPA of the I2 i had a 40L headache and my factory looked distinctly purple.
The final filtered mother liquor was still a very dark iodiney colour but i had already assessed little recoverable iodine would be gained. The
following morning the solution was colourless save for a slight slight yellow taint, however all the PP beakers and buckets i had been using were
still darkly stained still and remain so.
Even more interesting.
The only iodine waste i did not 'process' in this way was some silver iodide that was largely grey from light exposure. Recovering the iodine and
silver from this sample is always laborious i find because of the large amount of silver metal in the sample, the insolubility of the salt, the
difficulty of filtering thiosulphate solutions of that free silver, the pain cleaning fritted filters aterwards etc.
Several days later i had purchased some 50% h2o2 and for the sake of some bucket chemistry i dumped a couple of tablespoons of the Ag/AgI solid into a
500mL erlenmeyer and poured in ~200ml ~20% H2o2 solution, no I2 appeared, actually not much happened and the flask went on the 'deal with that later'
shelf.
Well i have the most interesting solution, it consists of a yellow solid (no free silver metal left) on the bottom of the flask that constantly issues
up geysers of small particles that rise to the surface through a iodine coloured solution and accululate there. On the surface a different ppt
exsists, yellow but different crystallinity and hue, which does the converse, sends solid ppt downwards.
It has been doing this now for 10days, its so pretty i transferred it to a lovely jar and it sitting on my desk. A perpetual motion machine, until it
stops of course.
Anyway, not much 'technical' to add, but iodine is kind of unpredictable.
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len1
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| Quote: | | I noticed however i'm far too polite to point out what was obviously not an error in your rationale, as you're far too intelligent to make such a
trivial mistake. If you stop eating donuts your fingers will slim down and your accuracy on the keyboard will surely improve.(assuming you have
fingers) |
Good at last someone understood - actually Ill correct that, I can make scientific errors trivial and otherwise, but they are still too intelligent
for you guys to find them. );
| Quote: | | Not to be critical but validating via KI standards, although an essential first step, only validates that the methodology is accurate for KI samples,
not more complex matrices, although it may be. |
Absolutely right! I thought of posting on this, but then why bother. The latest test with 0.06M KI but in saturated solution of various other salts
did not clear all the iodine to the bottom, some stayed disolved as evidenced by bright red iodine layer. Still titration with thiosulphate did not
fail.
The clearing of solutions overnight doesnt surprise as permanganate is too strong an oxid agent and will take it all to iodate. The other gear didnt
have the permang concentration needed.
My sample lived as I2 in conjunction with its oxidant H2O2, until after two weeks it suddenly all disappeared overnight.
I cant follow how you separate in IPA when its miscible with water?
Now with all this iodine, and your 'mood' I get this feeling you are not in this strictly for the science. I suppose you are not likely to say. I
dont expect anything else anymore, why do laborious, dirty, and 'hard' chemistry if there isnt a vested interest.
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Panache
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The clearing of solutions overnight doesnt surprise as permanganate is too strong an oxid agent and will take it all to iodate. The other gear didnt
have the permang concentration needed.
--Sorry i don't follow what 'other gear' is referring to
My sample lived as I2 in conjunction with its oxidant H2O2, until after two weeks it suddenly all disappeared overnight.
--Ha, i didn't follow that your solid I2 ppt disappeared, i assumed it was a colour of solution qualitative assesment
I cant follow how you separate in IPA when its miscible with water?
--After filtering the solid I2 and MnO2 from the matrix i redissolved the I2 into IPA and filtered off the MnO2. Not a methodology i would advocate.
Now with all this iodine, and your 'mood' I get this feeling you are not in this strictly for the science. I suppose you are not likely to say. I
dont expect anything else anymore, why do laborious, dirty, and 'hard' chemistry if there isnt a vested interest.
--I do have a terrible hoarding habit when it comes to chemicals and its frustrating because often when i would prefer to be doing personal chemistry
work i'm invariably
conflicted with the mundaities of 'housekeeping' my various miscellaneous chemical oddities. I do understand your sentiment, however its a
generalisation, evidenced by the fact that you do 'laborious, dirty, and 'hard' chemistry' without a vested interest, unless you're saying you're
unique. I know you're not claiming that, you are rare but not unique.
i know i'm eccentric in the extreme but i have little vested interest in anything and i'm resolved with it. God this is rambling.
Nothing to do with the recovered iodine as yet, unlikely to ever leave the bottle save for some i want to convert to NaI. Its a lovely element such a
shame its has the infamy it does. The AgI/h2o2 system i'm going to look at shortly there is definitely some interplay with atmospheric O2 or Co2 in
the system it's fascinating, i must video the system and upload it its rather unique.
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Panache
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| Quote: Originally posted by len1 | Interesting development in this.
After the oxidation of 0.1g KI in 10ml H2O using 10ml 0.6M H2O2 at pH~2, which is essentially over in 20mins as per above, I had what must have been
about 80mg iodine precipitated and above it 20ml of almost clear water - confirming concentration of iodide remaining in solution was tiny, as else
the iodine would have dissolved.
And so this remained for about 2 weeks.
Today I was surprised to see the iodine entirely gone and the solution completely clear.
The only reaction I can think possible is further oxidation of I2 to the 5+ state KIO3 by the H2O2. But why such a sudden onset of reaction, with
then proceeds to completion? The only explanation is that its autocatalytic. However all references Ive seen mention it only first order in [I2]
rate = k[I2]
There must be a lot more to it than that. |
Have you attempted to reduce the theorised iodate back to I2? Am i right in understanding you used a ~200 fold excess of the stoiciometric amount of
H2O2 required?
This Physical chemistry paper is interesting if you have not read it. I lazily glaze over much of the maths in such papers but i'm sure it would be
akin to a childrens book to your learned yourself.
http://www.rsc.org/ej/CP/1999/a809291e.pdf
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len1
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Hi,
Thanks the paper looks interesting, Ill go through it when I get back to iodine. I still have the solution but havent had a chance to investigate it
yet. The fact that the reaction went from almost zero to 100% overnight, after standing with nothing happening for two weeks is very surprising, and
ill have to follow it up.
I did use a large excess of peroxide as you say, reason being that the time constant for the reaction is a few minutes as is - from memory reaction
rate is first order in peroxide, so any lower concentration will take too long for a titration
[Edited on 15-5-2009 by len1]
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Jor
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I have done a titration of Cu with KI once. Cu(II) reacts quantitively with I- :
2 Cu(2+) + 4 I(-) --> 2 CuI + I2
The I2 was back titrated with thiosulfate.
So maybe you could add excess Cu(II), and then:
-filter off CuI and iodine, and perform an EDTA titration on the copper?
-Decant all liquid above the CuI (its long ago for me, so i do not know if it's heavy and decantable, but this is better than filtering, as filtering
always introduces some losses in the filter paper), washe the CuI a few times, and next heat it above 100C to drive off all water and I2. Then weight
the solid on a good balance. This should be done relatively large scale, so that mechanical losses are relatively minimal and the balance makes
smaller mistake.
Or maybe you could titrate with known Hg(II)-solution. First a red precitipate forms (just like here: http://www.amateurchemie.nl/viewtopic.php?f=20&t=155)
The solubilty is very low, 60mg per liter at 20C.
However, withe excess iodide, it redissolves forming the light yellow complex [HgI4](2-). Not sure if this works, just an idea. Beware of the
environmental toxicity of Hg, wich makes it hard to work with.
Incase you store your thiosulfate solutions and didn't know yet: they are ideal media for all kinds of bacteria wich break down the thiosulfate. This
can be solved by adding a few drops of chloroform or a trace of Hg-salt.
[Edited on 19-5-2009 by Jor]
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len1
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| Quote: Originally posted by Jor | I have done a titration of Cu with KI once. Cu(II) reacts quantitively with I- :
2 Cu(2+) + 4 I(-) --> 2 CuI + I2
The I2 was back titrated with thiosulfate.
So maybe you could add excess Cu(II), and then:
-filter off CuI and iodine, and perform an EDTA titration on the copper?
[Edited on 19-5-2009 by Jor] |
Unfortunately this procedure suffers many of the detractions which one must avoid. The Cu titration has no obvious visible end-point, and so Cu must
be used in excess as you say. Titrating with thiosulphate will then be a method for estimating the amount of Cu in solution - not I-, same as all
other methods. Filtering and weighing is a very inaccurate procedure and is not used in small-scale analysis.
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len1
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Actually my last reply might have been a bit hurried. Because in the reaction
2Cu2+ + 4I- -> 2CuI + I2
half the iodine is precipitated as the insoluble Cu(I) salt, subsequent reduction of I2 to I- by thiosulphate will not re-liberate all iodine from the
reduction, half of it will remain locked as CuI. If the Cu is used in excess, titration with thiosulphate (note this is not a back titration since
precipitation continues on the addition of thiosulphate) will go according to the following reaction:
2Cu2+ + 2(S2O3)2- + 2I- -> 2CuI + (S4O6)2-
This then is the overall reaction for the proposed titration. Its end point is then very sharply marked by the disappearance of a starch-iodine
complex.
As with anything in science, you have to do calculations to see if the idea actually can work. The main obstacles I see are two
1) The equilibrium constant for the titration might be small
2) Cu2+ may be reduced by thiosulphate directly.
To address 1) we have the (reduction) potentials
2Cu2+ + 2I- + 2e- -> 2CuI 0.86V
I2 + 2e- -> 2I- 0.535V
(S4O6)2- + 2e- -> 2(S2O3)2- 0.015V
We see that the original reduction of copper proceeds at a potential of 0.86-0.535 = 0.225V, so is thermodynamically allowed, but as the potential is
not very large (.225/.059~4) the equilibrium constant is not to large.
The reduction with thiosulphate in solution however has a much larger potential (!) of 0.86+0.015 = 0.875V and so its equilibrium constant is much
larger (.86/.059 ~ 13), or about 10^26. In this particular case we only need thermodynamics to give the green flag for it to proceed, since we know
the individual half reaction proceed kinetically. So 1) is not a concern.
As for 2), the reduction of copper has a potential of 0.338V so thermodynamically Cu can be reduced by thiosulphate. But I believe the reaction does
not actually occur in practice to any extent (mixing thiosulphate with CuSO4) in the same was as H2O2 does not react with thiosulphate directly.
So it seems we have a very simple method for titrating I-.
As an aside I find it incredible that thiosulphate is a quantitative titration with iodine (it is a standard analysis technique called iodiometry)
given that stronger oxidizing agents (chlorine bromine), oxidize it to (SO4)2- - and one would think some of it would form with I- as well, making the
reaction unpredictable. This is clearly not so - one of the unpredictabilities of science.
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Panache
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| Quote: Originally posted by Jor |
Incase you store your thiosulfate solutions and didn't know yet: they are ideal media for all kinds of bacteria wich break down the thiosulfate. This
can be solved by adding a few drops of chloroform or a trace of Hg-salt.
[Edited on 19-5-2009 by Jor] |
I was unaware of this, having always just reverted to the 'don't store thiosulphate solutions' dogma ingrained during prac classes at university.
Is this bacterial brake down the basis for the dogma, and if so can i just dope my thio solutions as described to avoid making them freshly each time?
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entropy51
Gone, but not forgotten
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I suspect that the bacterial degradation of thiosulfate solutions only occurs in the dilute solutions used for titrations.
I recently spilled about 2 mL of liquid Br2. I quickly dumped about 10 mL of my "emergency" solution of 15% thiosulfate on the spill and the Br2 and
its vapor disappeared immediately. The thiosulfate was about 10 years old. High concentrations of thiosulfate may not be good growth media for
bacteria.
I highly recommend keeping a bottle of strong thiosulfate within reach when working with Br2 in case of spills.
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unionised
International Hazard
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I guess it's too late for the OP to still care but why not titrate the I- with Ag+?
http://pubs.acs.org/doi/abs/10.1021/ac50101a004
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ChrisWhewell
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Wouldn't a potentiometric titration using NaOCl do the trick ?
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AJKOER
Radically Dubious
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Want a new imprecise and cheap visual technique that we can use to determine Iodine concentration?
From a well mixed iodine solution with unknown iodine concentration, take a drop and place on a small plate of glass. Add a drop of NH4OH. Before the
drops had evaporated completely, cover with filter paper. After drying, tap the NI3 to release (small pop) its iodine. This experiment can be
simultaneously replicated on a large surface and the median area of iodine stained recorded. Compare to a known test sample.
The power of the small explosion expands with approximately with cube of the concentration and the iodine area stained on filter paper accordingly
increases. Note, unlike other standard techniques which move linearly with concentration, this crude method actual is of order 2 or higher. Also, the
easy replication reduces measurement error and increases confident in the result.
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