teodor
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KI + HCl + sample + Starch -> yellow colour, why?
I have some samples of sec-butanol which give a yellow color of KI + HCl solution, so I suspect they can contain little peroxides. But when I try to
repeat the test with starch it gives not blue but a more intense yellow color than if I do the same test without starch.
If I add then a few drops of H2O2 (old samples with unknown concentration) it intensifies the yellow color until almost orange and at some
concentration, it develops blue color which was expected by me from the beginning.
Is it something special with my starch or is it normal and means a low level of peroxides?
I use "Aardappelzetmeel HONIG". I use 1.5g per 300ml of boiling water.
For peroxide testing, I use Vogel's procedure which is similar to any other procedure published, so:
1. 0.5 ml of the alcohol
2. 1ml of 10% KI in water
3. 0.2ml of 6M HCl
After mixing this and waiting a bit I start to add starch solution, drop by drop, but I see a yellow-orange color as a result. Also tried to mix KI +
HCl +starch first and then alcohol, with the same result.
So, what could be wrong? Why it can give yellow color but not blue? How sensitive could be this test (when it gives blue) in terms of H2O2 amount
which I can add as a reference value?
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macckone
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You have to have enough free iodine to give the blue reaction.
You need to titrate against a known sample.
Starch differs so you calibrate with a known amount of iodine
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Maurice VD 37
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Are you sure of the quality of your starch ? Because the word HONIG means honey in German. And honey does not contain starch.
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woelen
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I also had bad experiences with "aardappelzetmeel" from a supermarket (the word is the dutch word for "potato starch"). It simply does not work with
iodine for making the blue complex. I don't know why. Now I have starch from a chemical supplier and that works like a charm. You may have luck with
another brand, apparently starch and starch are not the same thing in all cases.
HONIG is just a brand name in the Netherlands, it has nothing to do with honey in this case. HONIG has many sauces, soups, mixes and spices for
preparing food.
[Edited on 16-9-21 by woelen]
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teodor
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Woelen, I have tried also "Maizena KoopMans", it is corn starch from a supermarket and the effect is the same. Probably I will also buy starch from a
chemical supplier for my iodometry exercises.
But my observation is slightly different. It can detect iodine. For example, if I mix KI, HCl, and the starch solution it turns slightly blue but this
color disappears when I add alcohol. It looks like alcohol binds iodide or iodine somehow which prevents the iodine-starch complex formation, at least
with these 2 types of starch. The effect of 2-butanol and n-butanol is the same. In the case of n-butanol which is not miscible with water, only the
organic layer gets the yellow color. With 2-butanol it is more interesting, because, I noticed, while it is not quite miscible with water without HCl,
the addition of HCl makes 2-butanol fully miscible with water which is remarkable.
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Maurice VD 37
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Using KI + HCl is not sufficient to produce iodine I2. The yield is miserable. It is also necessary to add an oxidizing agent, like hydrogen peroxide,
or diluted bleach, to the solution HCl + KI. But only with diluted bleach. With ordinary bleach, you will produce chlorine Cl2 which is toxic and
corrosive.
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unionised
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The effect of many or most solvents on the blue starch / iodine (strictly, pentaiodide I think) complex is fairly well known.
It doesn't work if you have solvents present.
Incidentally, have you considered starting with a potato?
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teodor
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"A potato method of peroxide testing". Sounds good. "Soak potato in alcohol and fry. No cracks = no peroxides. Serve with vegetables"
But dripping KI+HCl on different stuff to get blue color first and decolorizing it with a "solvent" is probably a good educational experiment because
it is not dependent on a chemical vendor. Unfortunately, this is a different thing from checking how safe is to fractionate my 2-butanol sample.
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macckone
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You have a bunch of competing reactions.
After looking at the other responses, HI + R-OH -> RI + H2O
So you are not getting the level of free iodine needed to yield a reaction.
You might want to work it the other way and just destroy any peroxides and redistill the alcohol.
This has a procedure but it isn't super effective with secondary or tertiary alcohols because of the iodination.
https://ehs.mst.edu/labsafetyprogramtoc/peroxidetestprotocol...
One other note is that too much HCl interferes with the reaction.
Diluting with a primary alcohol reduces the pH which reduces the iodination as well.
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teodor
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The point is that with n-butanol the result is the same as with 2-butanol. But I plan to do more tests soon to understand what is going on. I got
similar results when tried to check diethyl ether with this method in the past. I don't think ether reacts with HI.
I suspect really, as said unionised, that the iodine method could be used for many organic solvents only without starch, but many organic manuals say
the opposite. (As a variant: only with proper, magically brewed starch)
Quite interesting: https://wikieducator.org/Chemistry/Testing_for_Peroxides:
METHOD C
Prepare a sodium ferrothiocyanate reagent by dissolving 9 g of FeSO4•7H2O in 50 mL of 18% HCl. Add 0.5 to 1.0 g of granulated zinc followed by 5 g
of sodium thiocyanate. When the transient red colour fades, add 12 g of sodium thiocyanate and decant the liquid from the unused zinc into a clean
stoppered bottle. Add 1 drop of this solution to 1 drop of the compound to be tested in a clear glass vial.
Barely discernable pink colour = 0.001% peroxide as H2O2
Pink to cherry red = 0.002%
Red (*) = 0.008%
Deep red (*) = 0.04 %
* A percentage of 0.01% or more is not safe.
[Edited on 23-9-2021 by teodor]
[Edited on 23-9-2021 by teodor]
[Edited on 23-9-2021 by teodor]
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teodor
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I must say this: "I love METHOD C". Ferrous-thiocyanate solution is extremely sensitive and handy. Don't bother with the iodide method for testing
organic peroxides if you can prepare the ferrous-thiocyanate indicator and stay alive.
For that, you need at least a fume hood or a gas mask, good outside, and a good fan. The addition of Zn causes unmistakable H2S emission but that
means that SCN- is also split and you may ask "what happens with lovely CN- ions in this highly acidic (6M by HCl) solution". I have only 4 possible
explanations: they combine with Fe, they stay dissolved as HCN, they leave without even saying "goodbye" and they form something else. I tried to find
evindence of cases 1 and 2 by neutralizing HCl with NaHCO3 solution hoping to get Prussian blue but nothing. I hope they are gone but still, I would
prefer to wear gloves when working with this indicator.
Some comments about the receipt. Obviously, KSCN works as well as NaSCN, just put 6/5.
The idea is to create Fe2+ / SCN- solution without Fe3+ ion which is not easy at all especially if you have only granulated zinc. The solution is
readily oxidized by air. I was able to manage Fe3+ elimination only using zinc powder. Be careful, add small portions preferably with stirring.
When it becomes colorless add the rest of KSCN and if after that it becomes pinkish add a bit more Zn. It must be colorless. Then you probably have to
filter it but it will become pink upon contact with air/paper. So, the hint here is to put a tiny amount of Zn dust on a filter paper before
filtering.
It is so sensitive that when you open the bottle and allow oxygen to enter it will slightly develop pink color. I will try to find some immiscible
liquid that I can use as a top layer to protect the solution from oxidation. But it could be always filtered through a filter paper containing a bit
of Zn to become colorless again.
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