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Author: Subject: Make Potassium (from versuchschemie.de)
len1
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[*] posted on 17-12-2010 at 19:07


Removing the potassium from the conical this morning I experienced some difficulty in that the largest ball could not get through the 26/19 neck (see picture and note the characteristic blue tinge on the potassium, and the flattened shape <1cm height of the largest blob, which was about 1.4-2cm diameter), so a 24/29 would be better.

The potassium weighed in at 5.8gm - so with KOH being 10% water as an average figure - thats a 75% yield based on the KOH. If we recall that 1.5gm t-BuOH was added as catalyst - which ended up consuming ~0.8gm K as t-BuOK, thats an 84% yield of theoretically expected. A very good figure. The yield based on Mg (which is only illustrative since it is used in excess by about *2) is 33%.

Pok its not true that I do not trust anyone, if you read what I wrote I believed your reports at first. Its only after you wrote some other stuff which wasnt true that I thought the whole thing was bunkum. I believe that you got the MgO in crust form - so did I, but I also got sand. At the moment there is no explanation to why, and I will try changing a few things.

Anyway I thank you for showing us that the patent can be made to work. I

[Edited on 18-12-2010 by len1]

K6.JPG - 75kB
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[*] posted on 17-12-2010 at 19:16


Perhaps you might have luck heating to 70degC in that dish and forcing the droplets to coalesce while you skim off the 'slag'? Making the potassium is one thing, cleaning it up is another... some people have done good work actually *trying* something. Its not like making K is the holy grail of chemistry, but its still pretty neat that it can be done so readily.
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[*] posted on 17-12-2010 at 19:26


Coalescing and cleaning is no problem - I have done that a lot with electrolysis potassium and this is no different. My biggest worry at the moment is that this might be better than my electrolysis method, there are so few balls that coalescing and purifying should be a matter of 1/2 hour. The only advantage to the electrolysis I see now is that its much faster once youve got the setup, but the setup takes far longer.
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[*] posted on 17-12-2010 at 20:00



Noteworthy results from my latest attempt:


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

The first half of the video is drops of molten potassium floating in hot solvent. The second half is me testing one of the solidified drops.
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[*] posted on 17-12-2010 at 20:30


Quote: Originally posted by NurdRage  


The first half of the video is drops of molten potassium floating in hot solvent. The second half is me testing one of the solidified drops.


Very nice! Those moving droplets are hypnotizing.




The single most important condition for a successful synthesis is good mixing - Nicodem
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[*] posted on 17-12-2010 at 21:02


Even with a large yield from the experiment, the white solid by product from the reaction in D70 (MgO and some unreacted magnesium) contains some initially invisble specs of potassium.

Unlike sodium, potassium pieces of ~0.5gm or more can explode on contact with water and start a fire. Even if there is no reaction on first adding water, there can still be potassium in the mix because water, and D70 protecting the potassium, dont mix. And a fire can start later on.

I quenched the byproduct by adding ethanol dropwise - which reacts with potassium quickly (t-butanol is too slow) and mixes with D70. The ethanol must not be added too quickly - the mixture can burst into flames - as mine just did. Thats a problem because a fire with burning alcohol and potassium can not be extinguished with water. The lab CO2 extinguisher however made short work of the inflamation.
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[*] posted on 17-12-2010 at 21:52


Hi,

Good experiment, I will be trying it soon with different solvents, I don't think I can get Shellsol D70 here anyway. I just need to buy more magnesium, I'm out of it. So it may take until next year.


Oh, by the way, nurdrage, you didn't use d70 did you? If not could you explain a little further your procedure?
What would be candle wax solvent? I don't think I've ever heard about it here.

[Edited on 18-12-2010 by Cuauhtemoc]
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[*] posted on 17-12-2010 at 21:57


Wow, this thread got stickied real fast! I can't wait to try this out in the new year. If it works, I'll aim to ampoule some potassium metal under argon (or nitrogen if I can't get argon)
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[*] posted on 17-12-2010 at 23:20


Quote: Originally posted by Cuauhtemoc  
Hi,

Oh, by the way, nurdrage, you didn't use d70 did you? If not could you explain a little further your procedure?
What would be candle wax solvent? I don't think I've ever heard about it here.

[Edited on 18-12-2010 by Cuauhtemoc]


Candle wax solvent is exactly the title. Candle wax.

Basically you find white unscented candles. These are extremely plentiful this time of year around christmas. The absolutely cheapest brand you can find is what you want as it will be just paraffin wax.

Its melted directly into your reaction vessel and you can heat it to 200 celsius (or more) without boiling.

The major problem with candle wax is that you have to work with it while it's hot, around 80 celsius. Working with it too close to it's melting point, around 60 celsius, isn't good because it solidifies quickly around the neck and sides of your reaction vessel. but at 80 celsius the potassium is molten and its an extreme fire hazard. Even at 80 celsius though, when you pour it, a thin film of wax remains in your container.

As len1 pointed out, tiny bits of potassium are still left in the solvent even after you remove the major balls.

During cleanup this is still essentially "devil's C4" because the bits of solidified wax on the various pieces of glassware you use don't react with water or alcohol unless the wax is cracked or deformed to reveal the potassium bits. This however is extremely unpredictable. heating the apparatus to melt the wax off is also nearly suicidal, i had several tiny fires when i did so.

While the candle wax solvent is viable, i think it's too hazardous because it solidifies onto everything you use.

I'm working on a safer approach but for now i strongly recommend NOT to use candle wax unless you have full fire safety protocols and equipment in place.
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[*] posted on 17-12-2010 at 23:43


I see, I was thinking about using Kerosene, which is also mentioned in the patent. Although I need to be careful because I've read that kerosene may easily ignite at such high temperatures.


I agree that candle wax looks very dangerous, so I won't be trying it.
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[*] posted on 17-12-2010 at 23:55


Well done len1 and Nurdrage.

I have only had success like woelen. For me the key seems to be nice fresh oxide-free Mg. This seems to be the common thread with Pok, len1 and Nurdrage.

Anyone have any ideas how to remove surface oxide from Mg turnings. Or like Pok am I going to have to sit there with an Mg block and file?

Again well done
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[*] posted on 18-12-2010 at 03:29


Ok now how do we go about making best use of this groundbreaking discovery? :D

I suggest we go on and overcome another difficulty for home experimenters, the aquisition of inorganic azides. It should be relatively straight forward to manufacture potassium azide from metallic K. First you need to react it with dry NH3 to form potassium amide. Next, dinitrogen monoxide (laughing gas = whipped cream propellant) is used to convert it into potassium azide. This is a well known procedure with good yield and no special apparatus or exotic chemicals are needed.
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[*] posted on 18-12-2010 at 05:57


The patent says there is vigorous evolution of hydrogen at 130C. There is certainly a vigorous evolution, but did they check what it is? Certainly Mg reacts casually with hot water to give hydrogen, and with steam at 130C the reaction can be assumed to be vigorous, but the reaction is heterogeneous here, as the water can not dissolve in the D70, meaning some water must escape unreacted.

Now thats were an interesting point comes in. If there is relatively little exposed 'fresh' magnesium surface, such as was the case in my shavings experiment, there will be enough water to react with all exposed magnesium surface - and indeed from memory all magnesium was dulled at the end of my experiments three years ago. If the magnesium is in the form of fresh filings - some magnesium will retain its shine - ie it will not be covered in oxide.

This reaction appears to be one of t-BuOK which from the experiment is clearly soluble to an extent in D70, and Mg. It is likely, as in the case of grignard reagents, that the reaction will not occur with dulled magnesium. But unlike the Grignard case, it is not sufficient to have some active Mg surface to get some reaction. You must have sufficient active Mg surface AFTER the reaction with water at 130C, to get any reaction at all. So the surface area appears to be the key.

A new option however also opens up. If the above is correct, and one does not have fine magnesium, one can hold back part of the fresh Mg until 200C has been reached. It appears also advantageous to let any free water escape and not employ a reflux condenser until after the evolution at 130C

[Edited on 18-12-2010 by len1]
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[*] posted on 18-12-2010 at 07:24


len1:

Are we now sure the byproduct is MgO, not Mg(OH)2? Could you verify that: wash with white spirit or such like, dry at low temp., then determine weight less at 500C or so?

We now need to try and understand the mechanism better, in particular why the t-BuOH seems to have a preference to react with the KOH, rather than with the formed K. And the role of water in all this…

I think it’s safe to say that t-butanol is the catalyst and that the reaction depends physically on some short range – boundary layer diffusion of the activated species (T-BuOK).

The yields are fine!

And nurdrage’s experiments seems to clearly indicate the solvent is not that important, within reason…
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[*] posted on 18-12-2010 at 07:43


Quote: Originally posted by len1  
The patent says there is vigorous evolution of hydrogen at 130C. There is certainly a vigorous evolution, but did they check what it is? Certainly Mg reacts casually with hot water to give hydrogen, and with steam at 130C the reaction can be assumed to be vigorous, but the reaction is heterogeneous here, as the water can not dissolve in the D70, meaning some water must escape unreacted.
This same issue has vexed me. It's seems clear that both H2 and H2O are products of various reaction pathways, but I've seen nothing about when and how much these products are produced. I would guess that the original patent authors did originally measure it to understand the science, but then didn't disclose it for the patent because they didn't need to. Patents, after all, are commercial recipes; you have to disclose how to make it go but not why it works. It would seem straightforward enough to put a condenser on a separate collection flask and a syringe to measure non-condensed, evolved gas.
Quote:
Now thats were an interesting point comes in. If there is relatively little exposed 'fresh' magnesium surface, such as was the case in my shavings experiment, there will be enough water to react with all exposed magnesium surface [...] You must have sufficient active Mg surface AFTER the reaction with water at 130C, to get any reaction at all. So the surface area appears to be the key.

A new option however also opens up. If the above is correct, and one does not have fine magnesium, one can hold back part of the fresh Mg until 200C has been reached.
I had a similar thought about a second addition of Mg; see below. The sum of the observations you're making here imply that the solvent-based step of the process is divided into two parts: (1) dehydration of KOH with Mg and (2) KOH / Mg substitution with t-BuO moiety. Previously, these had been done in a single setup, but there's no reason to think of them as two distinct steps. For example, inert gas cover seems far more important for the second step than the first. If the goal is to remove water in the first step, then the right thing to do is to use a different condenser set up in that step. It's mechanically more difficult to change setups will a flask full of hot oil, but it hardly seems insurmountable.

As to the Mg, given that the Mg is reacting in two different ways here, there's no reason to assume that the same mechanical form of the Mg reagent is appropriate to both steps. For the second step, contiguous surface seems necessary, or at least greatly desirable. For the first step, however, total surface area would maximize the rate and completion of dehydration. So it would seem that a protocol might recommend (but not require) very fine mesh Mg powder for the first step and require shavings or turnings for the second.

My previous idea was a variation on this, which is to add fresh Mg to the reaction with each addition of tert-butanol. I didn't mention it because there's no readily available equipment to add large particles a little bit at a time under inert atmosphere; it's not like there's a glass screw feeder that operates in a hermetic atmosphere on most (any?) chemist's shelves. The suggestion to add Mg before the first t-butanol seems to meet my concern and has the advantage of being feasible.
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[*] posted on 18-12-2010 at 07:55


watson:

What about imagining a wholly different way of drying the KOH in situ? Some sacrificial K perhaps?
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[*] posted on 18-12-2010 at 07:58


Quote: Originally posted by blogfast25  
Are we now sure the byproduct is MgO, not Mg(OH)2? Could you verify that: wash with white spirit or such like, dry at low temp., then determine weight less at 500C or so?
I recommend a fire assay on this one. Weigh the crucible and Mg(OH)2 / MgO contents first, after dissolving out any Mg(t-BuO)2 and drying out any residual solvent. Then calcine the crucible and dehydrate Mg(OH)2 --> MgO + H2O. The mass difference between final and initial state yields the amount of hydroxide in the initial charge.
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[*] posted on 18-12-2010 at 08:13


Quote: Originally posted by blogfast25  
What about imagining a wholly different way of drying the KOH in situ? Some sacrificial K perhaps?
That's a good idea. I'm guessing, without having done a total energy computation, that it's more economical to dehydrate with Mg. But it might be feasible to dehydrate with Al as well, although I'm guessing that free Al post-dehydration might poison the reaction. So start with an estimated amount of Al and run to completion. Then add more KOH (with new water) if there's unreacted aluminum still present. Heat that up to dispose of metallic aluminum. Finally add fine Mg to dehydrate. This protocol uses less Mg per unit KOH. I'm not sure it works, though, and I'm not sure it's worth the bother.

It's probably worth doing the experiment to see how much water can be driven off with just heat.

I don't know if molecular sieves would work for this, but they might. I doubt all the equipment required to regenerate the sieves makes their use better than electrolysis, though.
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[*] posted on 18-12-2010 at 08:47


I'm glad all the paranoia and the ridiculous (though very entertaining) flame wars are finally settled away by experimental confirmation.
Congratulations to all who finally got their balls! It was fun reading about these adventures.

Quote: Originally posted by blogfast25  
We now need to try and understand the mechanism better, in particular why the t-BuOH seems to have a preference to react with the KOH, rather than with the formed K. And the role of water in all this…

I don't think it is proper to say that t-BuOH would have a preference to react with KOH rather than K. Contrary to the reaction of K with water, the reaction

KOH(s) + t-BuOH(solution) <=> t-BuOK(solution+KOH phase) + H2O(KOH phase)

is reversible and in such a multiphasic system depends on various effects. For example, due to solubility issues, the t-BuOH can only react with KOH at the phase boundary with the slight solubility of t-BuOK in alkanes and hydration of KOH being the driving force (anyone who ever used solid KOH for deprotonation of weak organic acids of pKa > 15 in aprotic organic solvents would know what effect I talk about - very useful in organic synthesis!).
It appears from the above described experiments that the moisture in KOH somehow reacts with Mg (prior to t-BuOH addition). I can only imagine this happening at the phase contacts between KOH pellets and Mg metal (assuming the evolved gas is H2 at all). The reaction between Mg and t-BuOH should be much slower and I doubt much K(s + Mg phase) forms before all the protic species (t-BuOH and H2O) are consumed or steam distilled away. Thus, my suggestion to the next motivated enough experimenter is to measure the gas phase temperature of the refluxing mixture as this should give an indication of whether K globules start forming before all t-BuOH is converted to metal akoxides or later only.

Quote:
I think it’s safe to say that t-butanol is the catalyst and that the reaction depends physically on some short range – boundary layer diffusion of the activated species (T-BuOK).

It would be more proper to call it a precatalyst and t-BuOK being the catalyst, or more properly phase transfer catalyst. The most likely explanation in my opinion is that there is a

2 t-BuOK(solution) + Mg(s) <=> (t-BuO)2Mg(solution) + 2 K(s)

redox equilibrium in the solution/solid system. If extrapolating redox potentials form aq. chemistry, one would tend to believe the equilibrium is much in favour of the left side, though generalizing among such very different reaction media and species involved is not reliable at all. Experimentally, it might be possible to evaluate this equilibria by measuring the potential of a galvanic cell having Mg and K electrodes immersed in the "electrolyte solution" of either metal akoxide in a alkane solution (the conductivity would be close to zero, but with a voltmeter of very high internal resistance it might actually be feasible to measure something).
Nevertheless, even if the equilibrium constant at 200 °C is only in the range of a couple of negative magnitudes, an efficient removal of (t-BuO)2Mg from the solution should drive the reaction toward the right (forming more K). Among the (un)available possibilities, the most obvious such reaction could occur at the phase boundary of KOH:

(t-BuO)2Mg(solution) + 2 KOH(s) => Mg(OH)2(s) + 2 t-BuOK

This one should be near to irreversible and can be experimentally verified or disproved.

I don't put much faith in such a catalysis mechanism, but if you need a preliminary hypothesis, that's the best I can offer with my limited knowledge of inorganics. Someone from the inorganic field ought to give a better one. Thus the catalytic cycle would be the combination of the above equations:

2 t-BuOK(solution) + Mg(s) <=> (t-BuO)2Mg(solution) + 2 K(s)
(t-BuO)2Mg(solution) + 2 KOH(s) => Mg(OH)2(s) + 2 t-BuOK

Quote:
And nurdrage’s experiments seems to clearly indicate the solvent is not that important, within reason…

I would not call it unimportant, as the solvent apparently needs to fulfil quite some criteria. It is obvious that it needs to be completely inert to KOH, Mg and K at the reaction temperature (+ high bp). So I would rule out most solvents already on this basis. Only fully saturated alkanes seem appropriate, if nothing else, due to potassium. Making assumption based on the above hypothesis it also appears that the solvent must be as non-polar as possible, even though this means lower solubility of the alkoxides. Higher t-alkoxides (t-amyl alcohol and above) might speed up the reaction if the limited solubility of t-butoxides is found to be an issue.




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[*] posted on 18-12-2010 at 08:59


Quote: Originally posted by Nicodem  
Higher t-alkoxides (t-amyl alcohol and above) might speed up the reaction if the limited solubility of t-butoxides is found to be an issue.


On this point i'd like to add my purely qualitative observation that t-amyl alcohol did seem to give me greater rate. the hydrogen bubbles seem to go twice as fast than my trials with t-butanol.

I only tried t-amyl alcohol once though and thus i can't say definitively that it's better. But from what i did see, i think it's worth exploring longer chain alcohols.
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[*] posted on 18-12-2010 at 11:28


Quote: Originally posted by NurdRage  
Quote: Originally posted by Nicodem  
Higher t-alkoxides (t-amyl alcohol and above) might speed up the reaction if the limited solubility of t-butoxides is found to be an issue.


On this point i'd like to add my purely qualitative observation that t-amyl alcohol did seem to give me greater rate. the hydrogen bubbles seem to go twice as fast than my trials with t-butanol.

I only tried t-amyl alcohol once though and thus i can't say definitively that it's better. But from what i did see, i think it's worth exploring longer chain alcohols.


Nurdy, how do you add your t-alcohol? All at once? Slowly? In stages?
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biggrin.gif posted on 18-12-2010 at 11:40
I'VE GOT BALLS!


Right now the ShellSol D70 is happily refluxing for 2 hours already and I now already have potassium balls ranging in size from 1 mm to 3 mm (pictures are made of these, they will be shown in a final write up). I'll let it reflux for another 2 hours, but it seems to work.

I have taken 50 ml of ShellSol D70 and added 7 grams of KOH and 3.5 grams of Mg powder (appr. 100 um size). I put this in a sand bath. When temperature reaches 150 C, then a vigorous boiling starts and white fumes are produced. At this point I opened the apparatus and allowed the white fumes to escape. I left it open till temperature of the sand bath reached 170 C. This must be water vapor, condensing in the cold air.
At 170 C the boiling almost comes to a halt and at that point in time I put on the reflux cooler. Temperature went up further and around 200 C there was slight simmering, at 220 C there was a nice constant simmering. This must be the boiling ShellSol D70.

In the meantime I prepared 0.7 ml of tert-butanol with 7 ml of ShellSol D70. When temperature reached 225 C I added appr. 2 ml of this mix. Five minutes later I added another 2 ml. Ten minutes later I added a third portion of 2 ml and almost 10 minutes after that I added the final amount. So, in total I added 0.7 tert-butanol dissolved in 7 ml of ShellSol D70. Each time when this was added, a vigorous boil started, which quickly subsided, however. I added these small amounts of liquid through the reflux cooler, so that it is not necessary to open it and loosing lots of vapor.

After the addition of the last amount of tert-butanol I left it refluxing for 1.5 hours and in this time the balls of potassium formed.

To be continued....




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[*] posted on 18-12-2010 at 11:50


@blogfast25

I added my alcohol all at once through the top of the condensor without dilution. not really possible to dilute the alcohol since my solvent is candle wax.


@woelen

Excellent work! you should make videos and put them on youtube, you'd do great.
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[*] posted on 18-12-2010 at 13:23


Yep! The K is mine too!

I didn’t get ‘Great balls of K!’ but loads and loads of little ones: about 1 mm or less, which react very vigorously with water.

‘Apparatus’: same as pok, same quantities. Nice and glistening 99 + % Mg:



Syringe with long tube of chemical Nylon tube, allows to inject the t-butanol deep into the refluxer (cooled with iced kitchen roll):



The usual gas evolution started quite early and kept going for 15 – 20 mins (I held at about 170C). Then the ritual of t-butanol/Shellsol mixture over about 15 mins, commenced at 190C – 200C. Almost immediately the solvent started to cloud over and this continued basically all throughout out the experiment with sediment forming gradually and the KOH disappearing. I cranked up the heat till the temperature was outside the range of the thermocouple (200C) and kept it there, all the while tending to the refluxer with iced water. Start time 16.45, end of test 19.15. After cooling this is what it looks like, lots of white powder at the bottom, some flakes (unreacted KOH?) also to be seen:



Feeling pretty sure there was K there, I decided on a partial work up by rinsing the metal with fresh Shellsol a few times: the blueish colour of the metal then became very apparent, although this is a particularly bad shot:



Quite a few were captured on a plastic tea sieve and treated with water: the ensuing reaction produced gas, lilac flames and burnt holes in my sieve! Complete dissolution, no residue.

Tomorrow I’ll attempt to work up some more… and run another test.

But there is no doubt to me: this work but needs improvement…


[Edited on 18-12-2010 by blogfast25]
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[*] posted on 18-12-2010 at 13:31


Thanks Nurd, will keep that in mind for next test.

@woelen: fantastic and neat little idea to mix the t-butanol into a larger volume of Shellsol.

All doubts gone now :cool:. Bizarre way of making potassium and a victory for backyard science!
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