dawsonsuen
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Questions about red lead iodide...
Quote from Wikipedia:
"Lead(II) iodide (PbI2) or plumbous iodide is a bright yellow solid at room temperature, that reversibly becomes brick red by heating."
Can anyone please explain what it means by "reversibly becomes red"?
Does this imply that it could be turned back to its original golden colour?
I have made a saturated solution out of lead iodide at 100°C.
I then continued heating the solution until it turned brick red.
But upon cooling, no precipitate was formed and the solution remained a clear brick red colour.
So what happened to the lead iodide?
Did it decompose? If it did, what did it decompose into?
[Edited on 6-8-2015 by dawsonsuen]
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diggafromdover
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Looks like water with dissolved Iodine. See if starch induces a color change to deep purple, which would confirm the free iodine. Consider a process
by which the lead iodide would decompose to iodine and ???.
Enjoying second childhood with REAL chemistry set.
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dawsonsuen
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I've just added a few drops of the solution into a starch solution and it didn't turn blue black...
So I'm guessing it's not aqueous iodine...
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Pumukli
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Lead iodide has a minimum solubility when no excess iodide ions are present. In case there were free iodide ions in the solution then they likely
formed complexes with PbI2 and effectively dissolved it, preventing it from precipitating again.
This phenomenon was described long ago:
The Solubility of Lead Iodide in Solutions of Potassium Iodide-Complex Lead Iodide Ions
Oscar E. Lanford , Samuel J. Kiehl
J. Am. Chem. Soc., 1941, 63 (3), pp 667–669
DOI: 10.1021/ja01848a010
Publication Date: March 1941
Try to repeat the process with thoroughly washed (distilled water) lead iodide and see what happens.
[Edited on 5-8-2015 by Pumukli]
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dawsonsuen
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So in order to precipitate the lead iodide, I have to reduce the amount of iodide ions in the solution?
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Pumukli
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Yes.
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Volanschemia
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I have done this reaction with a purposeful excess (~5-10%) of Iodide ions in solution, and it still worked beautifully. What kind of an excess did
you have?
"The chemists are a strange class of mortals, impelled by an almost insane impulse to seek their pleasures amid smoke and
vapor, soot and flame, poisons and poverty; yet among all these evils I seem to live so sweetly that may I die if I were to change places with the
Persian king" - Johann Joachim Becher, 1635 to 1682.
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dawsonsuen
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I have added a few drops of concentrated sulfuric acid to the "red PbI2" solution and I ended up with a very fine white precipitate.
But upon adding an excess of concentrated sulfuric acid, no more of the while precipitate was formed; the solution also remained a brick red colour.
I am not convinced that the fine white precipitate was the PbI2 I made in the first place because only a very small amount of white precipitate was
formed and the red colour of the solution did not dissipate at all. The solution remained a deep brick red colour throughout.
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Volanschemia
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The white precipitate is probably PbSO4, and if you added an excess of Sulphuric Acid, some of has probably converted to Lead(II) Hydrogen Sulphate.
I honestly don't know what the red colour is and I look forward to someone figuring it out!
[Edit] Actually, does starch turn blue on reaction with aqueous Iodine? I thought it only reacted with Triiodide
([I3<sup>-</sup>]). Maybe try taking some of the red solution and add a small amount of soluble iodide salt (eg. KI) and then
try adding starch.
(I could be totally wrong here).
[Edited on 6-8-2015 by TheAustralianScientist]
"The chemists are a strange class of mortals, impelled by an almost insane impulse to seek their pleasures amid smoke and
vapor, soot and flame, poisons and poverty; yet among all these evils I seem to live so sweetly that may I die if I were to change places with the
Persian king" - Johann Joachim Becher, 1635 to 1682.
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diddi
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the red colour is reversed by addition of Na2S2O5, in my experience
Beginning construction of periodic table display
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Volanschemia
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If that's the case, then reduction is what regenerates the yellow Lead(II) Iodide. So it follows that the red colour is an oxidized form of Lead(II)
Iodide. Does Lead(III) Iodide, or a higher oxidization state exist?
Or it could be some kind of complex that is destroyed by reduction.
[Edited on 6-8-2015 by TheAustralianScientist]
"The chemists are a strange class of mortals, impelled by an almost insane impulse to seek their pleasures amid smoke and
vapor, soot and flame, poisons and poverty; yet among all these evils I seem to live so sweetly that may I die if I were to change places with the
Persian king" - Johann Joachim Becher, 1635 to 1682.
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woelen
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dawsonsuen: Could you precisely specify what you have done? I also would like to know the source of the chemicals. If you made something yourself,
also specify exactly how you made it.
I ask this, because I have the impression that your lead(II)iodide contains some impurity. I have done a similar experiment with PbI2 and cannot
reproduce it. Only molten PbI2 turns red at a much higher temperature than 100 C. In solution it remains colorless, also at 100 C. On cooling down,
glittering yellow crystals are formed if the concentration is sufficiently high.
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dawsonsuen
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I first reacted 2.07g of Pb with some 68% HNO3 to form a solution of Pb(NO3)2.
With stirring, I brought the Pb(NO3)2 solution to a gentle boil until the no further NO2 was released.
At this point, the Pb(NO3)2 solution was a bit cloudy, so I added a few drops of nitric acid to try to acidify the solution but the cloudiness did not
dissipate.
Next, I topped up the final volume of the Pb(NO3)2 to about 300ml and filtered the solution.
I then prepared the second solution with 1.66g of KI and 200ml of water.
In the end, I simply mixed the two solutions together and a beautifier golden precipitate was formed.
Since I actually wanted to perform the golden rain experiment, I heated up the solution to try to get as much of the PbI2 dissolved in the water.
But unfortunately, I think I might have overheated the solution...
As the solution clears up, it also went from an orange solution to a red solution...
PS: I got the Pb metal off the internet and it states that it is 99.999% pure.
Thats basically what I have done 
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gdflp
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How much conc. nitric acid did you use? It sounds as though you had a considerable excess, which quite easily oxidized the KI to
I<sub>2</sub>. I would check the acidity of the solution with litmus paper. If it is acidic, warm it up again and slowly add a solution
of sodium hydroxide dropwise until it reaches a neutral pH. This will remove any excess acid and regenerate iodide ions from the iodine,
unfortunately however, 1/6th of the iodine will be iodate. Finally, allow the solution to cool, this should yield crystals of a mixture of lead
iodide and lead iodate.
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dawsonsuen
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Thank you gdflp!
I've used about 10-15ml of 15M nitric acid and that might have been the problem!
I will check the pH of the lead nitrate and the lead iodide solution tomorrow!
Thanks
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diddi
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in you OP you did not mention HNO3 at all. If this was the result of you simply dissolving Pb in HNO3 and adding KI without crystallising the Pb(NO3)2
first our answers would have arrived here much quicker. The I- ions are in an oxidative environment and hance the iodine colour.
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Pumukli
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In the 3rd comment he (she) said starch did not turn blue from the red solution. So I doubt it was iodine.
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gdflp
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If there was a large excess of nitric acid though, no color is to be expected with starch solution. The blue-black color is a complex between amylose
and triiodide ions, no reaction is observed with pure iodine when iodide ions are not present.
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Pumukli
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If OP had a few drops of non water-miscible organic solvent then he could extract the iodine and verify its presence by the purple color of the
organic phase.
Edit: purple in SOME solvents and brown in others.
[Edited on 8-8-2015 by Pumukli]
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