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Author: Subject: Make Potassium (from versuchschemie.de)
blogfast25
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[*] posted on 18-12-2010 at 13:35


Thanks also to Nicodem for his contribution above. This is more or less the reaction mechanism I proposed.

[Edited on 18-12-2010 by blogfast25]
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[*] posted on 18-12-2010 at 13:38


I'm more of a browser than a poster, but I just wanted to say that I fucking love you sciencemadness. What a great thread with a interesting discussion. Gotta add this to the list of reactions to undertake. peace!
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[*] posted on 18-12-2010 at 14:05


Yes, this really is a victory. I now have 3 balls of potassium, one has a diameter of just below 1 cm and the other 2 have diameters of well above 1 cm. Precise measurements are hard, due to the grey liquid in which the balls reside. Besides the 3 big balls I have many smaller ones, stuck in grey gunk. Given the density of potassium which is appr. 0.89 grams/ml I estimate to have around 3 grams which can be isolated. The tiny balls of less than 1 mm diameter I will not isolate, simply too much of a hassle.

This night I will let all of it cool down in a sealed flask and tomorrow I'll transfer the balls of potassium to some clean mineral oil.




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[*] posted on 18-12-2010 at 14:10


The difference in what nicodem wrote from the reaction I and blogfast had been proposing is that MgOH2 is formed instead of MgO.

No Im not sure that I got MgO instead of MgOH2. I know the dry hydroxide decomposes at higher temperatures - I wrote that in the post 3 years ago - but my feeling is that the hydroxide wont survive at 200C with t-BuOK around, that stuff is a desicator, and I presume the thermodynamics will favour dehydration, but maybe not the kinetics. The test of course is very simple and Ill do it.

The other thing is that there IS a constant gas evolution in the course of the reaction. I suppose one could maintain its residual water from the KOH reacting with the magnesium. 5.8gm K corresponds to about 1.8L hydrogen, or about 10ml per minute for a 3 hour reaction. I had been observing about that. On the other hand MgOH2 is known to form crusts, I havent heard of MgO doing that. It could be a combination of both. Ill check.

Well done Wilco!

[Edited on 18-12-2010 by len1]
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[*] posted on 18-12-2010 at 14:13


Quote: Originally posted by len1  
The other thing is that there IS a constant gas evolution in the course of the reaction. I suppose one could maintain its residual water from the KOH reacting with the magnesium. 5.8gm K corresponds to about 1.8L hydrogen, or about 10ml per minute for a 3 hour reaction. I had been observing about that. On the other hand MgOH2 is known to form crusts, I havent heard of MgO doing that. It could be a combination of both. Ill check.


You sure about that? I mean with all that boiling, how do you distinguish between vapour bubbles and other gas?

It really would be worth separating the 'slag' and checking for MgO v. Mg(OH)2, IMHO...



[Edited on 18-12-2010 by blogfast25]
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[*] posted on 18-12-2010 at 14:27


Remember I had a 30cm condenser connected, so that not much condensable matter would have made it past that. And anything that did would have condensed in the bubbler.

Yet a had a steady stream of bubbles throughout the experiment. So there is no doubt hydrogen is evolved. One could mainatin its all from Mg reacting with the KOH water. Or MgO and MgOH2 being formed simultaneously.

This is such an unexpected result, I thinking only testing can show.
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[*] posted on 18-12-2010 at 14:28


I've notified Theodore Gray (from periodictable.com (in)famy). He might want to put this forum on the map via PopSci...
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[*] posted on 18-12-2010 at 14:30


Quote: Originally posted by len1  
Remember I had a 30cm condenser connected, so that not much condensable matter would have made it past that. And anything that did would have condensed in the bubbler.

Yet a had a steady stream of bubbles throughout the experiment. So there is no doubt hydrogen is evolved. One could mainatin its all from Mg reacting with the KOH water. Or MgO and MgOH2 being formed simultaneously.

This is such an unexpected result, I thinking only testing can show.


Hmmm...my money's on H2 from water + Mg. Intuitively speaking...
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[*] posted on 18-12-2010 at 14:42


Many small K balls could possibly get unified by adding a few drops of isopropanol to the inert fluid in which the molten potassium balls are. Also oxidized potassium can be cleaned by this method.

http://translate.google.de/translate?u=http%3A%2F%2Fwww.vers...

See description of photo 1 - 4 there.

[Edited on 18-12-2010 by Pok]
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[*] posted on 18-12-2010 at 14:50


I'm about to place an order for 3L of Shellsol D70, but just had the idea of trying some of the heating oil that our central heating runs on as a solvent. We have like 3000L of this in our basement.
This is a simple crude oil distillate with a boiling range of 150- 390°C to which a marker substance (furfural) and red dye has been added to allow the detection of its diversion towards use as a diesel fuel, which is illegal due to the reduced tax on heating oil and the high tax on all automotive fuels.

I would have to vacuum distill this to get rid of the furfural in the first fraction and obtain a cut with (atmospheric) boiling range of 200-300°C. I could then try this out as a reaction medium. Shellsol D70 boils from 195-245°C.

However, I've read that potassium cannot be stored under aromatics like toluene since it can abstract an alpha hydrogen, forming potassium benzyl which undergoes further reactions, slowly consuming the potassium already at room temperature.
Heating oil contains aromatics, while Shellsol D70 is specified as aromatic-free.
Does anyone here know anything about the true nature and extent of the incompatibility of K with aromatics?

[Edited on 18-12-2010 by garage chemist]




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[*] posted on 18-12-2010 at 15:22


Quote: Originally posted by Pok  
Many small K balls could possibly get unified by adding a few drops of isopropanol to the inert fluid in which the molten potassium balls are. Also oxidized potassium can be cleaned by this method.

http://translate.google.de/translate?u=http%3A%2F%2Fwww.vers...

See description of photo 1 - 4 there.

[Edited on 18-12-2010 by Pok]



I can confirm this works.

I reccomend though keeping the temperature around 80 celsius. Too high and you risk spontaneous combustion if a tiny piece of potassium floats up and hits some stray moisture.
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[*] posted on 18-12-2010 at 15:30


This was my work. I'm Stefan at Versuchschemie.
I made Na/K alloy from the metals, cleaned the oxide off the alloy by adding IPA to the paraffin oil that it was melted in, and put the metal into a vial.
I use this alloy to dry ether. It's very efficient due to its fine dispersion when vigorously shaken.




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[*] posted on 18-12-2010 at 16:30


Quote: Originally posted by garage chemist  

However, I've read that potassium cannot be stored under aromatics like toluene since it can abstract an alpha hydrogen, forming potassium benzyl which undergoes further reactions, slowly consuming the potassium already at room temperature.


I think if there's no chlorobenzene it's OK. Xylene is often recommended when handling K, like in OS. I wonder why xylene, specifically, instead of only slightly more volatile toluene.

Working with K under any hydrocarbon can really suck, under certain conditions:
http://pubs.acs.org/cen/safety/potassium.html




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[*] posted on 18-12-2010 at 16:30


I have coalesced purified potassium and sodium under toluene and xylene. And although I finally preferred to do it under D70 or paraffin, due to the inflamability of toluene and zylene, there was no visible reaction at up to the boiling points of these liquid (time scale of several days). Over the course of a year there is some reaction in D70, although this maybe be due to some non-parafinic impurities in the latter - it is jot an analytic reagent after all.

I have tried Stefan purifying Na or K under D70 with isopropanol, but could never get reproducible results that way, so I now developped other means.

With regards to the reaction, the key argument seems to be whether t-BuOK dehydrates Mg(OH)2. If it does hydrogen and MgO will be the reaction products, if not there is no gas released by the main reaction at all (contrary to the precise measurement in the patent).

This reaction should also be sensitive to temperature - clearly dehydration will occur spontaneously at about 580C. Just perhaps, at lower temperatures t-BuOK does not dehydrate MgOH2. So that the latter would be the final product. While at higher temperatures it would, and fine MgO sand would form. If this is true it would explain why Pok produced no MgO sand - he certainly could not run his experiment at high reflux due to the nature of his 'condenser'.
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[*] posted on 18-12-2010 at 17:15


I just confirmed i can get decent potassium balls (1-3mm) using parafin oil.

Procedure was:
20mL paraffin oil, 5g KOH, 2g Mg heated to 170C, then 1mL of t-amyl alcohol added and condensors and bubblers were put on. Heating was brought up to 220C and continued until unambigous shiny potassium balls were visible. (~3hr)

Next i'm going to try pharmacy mineral oil.

After that i might try gamsol or varsol. Even though the boiling point is lower i find the temperature isn't too critical, just as long as you're willing to wait long enough for it to go completion.
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[*] posted on 18-12-2010 at 17:38


paraffin oil as in kerosene?
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[*] posted on 18-12-2010 at 17:50


mine was higher density. the bottle reads 0.86g/mL, i think commercial kerosene is lower.
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[*] posted on 18-12-2010 at 18:39
Unusual observation


Is the alcohol required?

During one run i hadn't added the alcohol yet but had to cancel the reaction.

So i started to clean up my fume hood and tossed the reaction mixture of paraffin oil, KOH and Mg into a water bath and to my surprise i saw some tiny but distinct flashes of lilac flames.

I had heated the mixture to 170C for ten minutes but *did not yet add alcohol* before i had to stop and cancel the run.

It might be that my parafin oil, KOH, or Mg was contaminated with some t-alcohol. and produced a minute amount of potassium. But if that wasn't the case, then my observation might mean the alcohol is not critical to the reaction.

I won't have time to try again for another few days so if anyone wants to do a test run without any alcohol it would be interesting to see the results.

[Edited on 19-12-2010 by NurdRage]
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[*] posted on 18-12-2010 at 18:41


The other thing of course is that formation Mg(OH)2 in the overall reaction

Mg+2KOH-> Mg(OH)2 + 2K

consumes half the magnesium than the reaction claimed in the patent

2Mg+2KOH -> 2MgO + H2 + 2K

which would mean the magnesium is present in over *2 excess in the patent.

Just judging by the amount of Mg in the final product - which was little, it appears the first reaction is either excluded, or forms a minor component. Hence I think the t-BuOK dehydrates the Mg(OH)2 - but will now check.

With regard to nurdrage's result, Im not surprised. Obviously the direct reaction as written can also occur. Provided some fresh Mg surface survived the water evolution, some K will form directly by the overall reaction. Its just that there is bad contact between the Mg and KOH, and my guess is you wont get an 80% yield. The t-BuOH provides and indirect route which gets around the kinetic impediment.

But here is an interesting observation. The KOH liquifies at about 150C in the D70 - at the point where initial reaction with Mg is observed - so this is a liquid-solid, not solid-solid reaction, and so is expected to go a lot better. The liquification is due to the crystalization water in the KOH - which however is not evolved at this temperature (in the dry it starts dehydrating at about 250C). So Mg is needed to dehydrate the liquid KOH, whereupon it is locked inside a dry KOH lattice. Any quality Mg I believe will work in the dehydration stage.

[Edited on 19-12-2010 by len1]
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[*] posted on 18-12-2010 at 20:03
More potassium porn


Selected highlights from today's run using paraffin oil

http://www.youtube.com/watch?v=AgfxmCAS4hs

First segment is the reaction mixture timelapsed over an hour. You can see layer separation with the metals rising on top of the oxides/hydroxides.

Second segment is about 3 hours into the run and potassium balls are clearly visible in the mixture.

Last segment is after hydrogen evolution completely ceased. Large potassium spheres (>5mm) are now visible.


I apologize for the poor video quality. My HD camera wasn't available so i had to use my pocket camera.

[Edited on 19-12-2010 by NurdRage]
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[*] posted on 19-12-2010 at 00:13


Well Ive made another run and there are some conclusions.
The gas given out during the reaction was collected in a gasometer and its quantity eliminates Mg(OH)2 as any sizable byproduct of the reaction. It must be MgO.

Using the previous amounts, gas evolution prior to the addition of butanol amounted to 1740ml. Towards the end the evolution slowed down considerably - but never really stopped. The amount is 5% in excess of whats expected from

Mg + H2O -> MgO + H2

with 10% nominal H2O in KOH in the reaction, but thats well within expectations. There is no support for any quantifiable amount of K being evolved prior to the addition of butanol.

Addition of t-Butanol did not immediately change things. However within 10mins the H2 livened up again, and after 1/2hr it was gathering in the gazometer at a sizeable rate. The appearance of the reaction mixture also changed. The D70 - fairly clear to this point became turbid, and fine MgO sand started gathering at the bottom.

This supprots my previous theory. After all water has been reduced, there seems to be a period of initiation, similar to what happens with grignards. This is when the first amounts of t-BuOK form and reach their final concentration. After this it rapidly reduces potassium and oxidizes magnesium. And the reason why my attempt from 3-years ago failed appears to come out of this. With less fresh magnesium at the start, it was all covered with oxide when the t-BuOH was added, and the initiation failed. Holding back some fresh Mg might have helped here.

With the reaction simmiring, stirring, judging by the rate of gas evolution does not seem to have any influence on the reaction rate. And during the initiation phase it seems to hinder it - with no bubbles evolved for about 30s each time the contents were stirred. Its only role I think is to coalesce the K.

The reaction appears complete in 3hrs after commencement of heating - with the last stages only coalescing potassium - after the balls have reached 3-4mm in diameter no new potassium is formed.

The gas evolved during the second stage of the reaction amounted to 2250ml. There were 1.4gm t-BuOH added, contributing 180ml to this. Theoretical is 2.43L by the below reaction

2Mg + 2KOH -> 2MgO + 2K + H2

112->24l

11.3 -> 2.43l

so the yield when I come to weighing the potassium should be 85%. Much as I had before.

Once again I got plenty of MgO sand. The balls are not clean and will need purifying. I would be interested to hear from anyone who succeeded in getting just crust.

And heres another interesting point that seems to flow out of that. Sodium should be more difficult to form by this method. Its butoxide (and hydroxide) are far less soluble in non-polar solvents than the potassium salts.

[Edited on 19-12-2010 by len1]
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[*] posted on 19-12-2010 at 00:53


alright, i guess my potassium observations before the addition of alcohol are probably just a cross-contamination issue.

thanks for checking len1.

The gas production in support of the Mg + H2O -> MgO + H2 theory is important though.


As for "crusts" if you look at my latest video (http://www.youtube.com/watch?v=AgfxmCAS4hs ) at the end of the reaction, instead of fine sand for the white MgO, i get "chunkier" solid MgO. Is this what you're getting? or do you think my stuff qualifies as "crusts"?

[Edited on 19-12-2010 by NurdRage]
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[*] posted on 19-12-2010 at 01:37


Hi Nurdrage, I looked at your video, but its hard to tell due to the resolution. Maybe you can tell me. If its crust, then its rock hard, you cant put a stirring rod through it, if its sand, then you can. Your K balls also float - I guess you are using paraffin - near 80C K actually floats in it, but at higher and lowerr Ts it settles. In D70 it is alsways heavier. The rising and falling might actually be benificial, with the lumps getting rid of the MgO that way. Your balls are certainly much cleaner than mine. Maybe paraffin is better - what kind of yield are you getting?

As for the reaction - my gazometer results are accurate to only a few percent. The KOH could easily have 9% or 11% water. SO it does not preclude a small amount of potassium forming before the t-BuOH is added. But it does preclude large amounts.
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[*] posted on 19-12-2010 at 02:06


Quote: Originally posted by len1  
Your balls are certainly much cleaner than mine.


Lol.

On a more serious note, in all the excitement it looks as if some serial doubleposting has gone un-noticed. How hard is it to actually use the edit button... geez.

But excellent work with the potassium. Its nice to see that this has come to a nice conclusion, unlike many other patented procedures.
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[*] posted on 19-12-2010 at 02:08


I destroyed the mixture with alcohol before i could really test its physical properties so i don't know for certain if it's rock hard. i'll check again in a few days when i have time to run another reaction.

As for the yield, i didn't do quantitative measurements but by eye it seems to be better than 50%. There were several large clean balls in the final product.
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