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Author: Subject: Make Potassium (from versuchschemie.de)
condennnsa
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[*] posted on 11-12-2011 at 00:22


I ran out of KOH , but have a lot of KNO3
what is the best way to make KOH from KNO3?

I figured, I'd add a stoichiometric amount of 35% H2SO4 to KNO3, then boiling the solution to dryness outside should boil off all the nitric acid and leave me with K2SO4 / KHSO4
Then dissolving this back in water and mixing with a slurry of excess Ca(OH)2, stirring it for a couple of hours , will the calcium precipitate as CaSO4? leaving the KOH in solution

I know that from potassium carbonate , CaCO3 precipitates with no problem, but I'm not sure about the sulfate since wikipedia lists CaSO4's solubility at some 0.2g/100ml , much more than CaCO3 ?... thanks
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blogfast25
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[*] posted on 12-12-2011 at 10:22


It's one thing to make fairly weak solutions of KOH, quite another to try and obtain solid KOH which is very hard to do.

Go and buy some more KOH and don't waste your time trying to convert KNO3...




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[*] posted on 23-12-2011 at 05:18


soulless spam ghoul reported.
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[*] posted on 23-12-2011 at 09:45


I guess i could use Magnesium Ribbon
to remplace Magnesium turnings
wonder if the wield is better when using Sodium Hydroxide




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[*] posted on 24-12-2011 at 05:48


@ garry21

Magnesium ribbon has a large surface area, and the fine oxide formed on the surface might impede the reaction, I believe it was determined in previous posts that a source of Magnesium devoid of surface oxidation would react more readily with the potassium hydroxide. Some have even suggested grinding a Magnesium block on a file under mineral oil to ensure less oxidation on the surface of the metallic particles.

I have some magnesium ribbon also and never thought of using it for that reaction... perhaps it could work. Still haven't tried it even though I have gathered all of the necessary reagents and glassware.

Also, it was determined that this particular experiment doesn't appear to work with Sodium, only Potassium.

Robert


[Edited on 24-12-2011 by Arthur Dent]

[Edited on 24-12-2011 by Arthur Dent]




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blogfast25
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[*] posted on 24-12-2011 at 08:36


Quote: Originally posted by Arthur Dent  
@ garry21

Magnesium ribbon has a large surface area, and the fine oxide formed on the surface might impede the reaction, I believe it was determined in previous posts that a source of Magnesium devoid of surface oxidation would react more readily with the potassium hydroxide. Some have even suggested grinding a Magnesium block on a file under mineral oil to ensure less oxidation on the surface of the metallic particles.

I have some magnesium ribbon also and never thought of using it for that reaction... perhaps it could work. Still haven't tried it even though I have gathered all of the necessary reagents and glassware.

Also, it was determined that this particular experiment doesn't appear to work with Sodium, only Potassium.

Robert


[Edited on 24-12-2011 by Arthur Dent][Edited on 24-12-2011 by Arthur Dent]


A couple of corrections.

Just about any old magnesium should work but the greatly reduced surface area for ribbon should make this reaction a slow boat to china. The state of 'passivation' of the Mg may play some part but it's far from clear. All Mg is passivated to some degree anyway, the element being very reactive (electropositive) towards oxygen.

Accoding to the patent, sodium synthesis using this method also works but it takes much more time. That is why some of us are looking into tertiary alcohol catalysts with much longer chain lengths: the sodium alkoxides of these are believed to be more soluble in paraffinic solvents. No definite conclusions have yet been formulated though.



[Edited on 24-12-2011 by blogfast25]




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[*] posted on 24-12-2011 at 08:49


I stand corrected. I'll put away my ribbon. ;)

I'll look into a Mg block then.

Have a great Xmas break, blogfast25! :D

Robert




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blogfast25
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[*] posted on 24-12-2011 at 11:02


U 2! ;)

Some pencil sharpeners are made of Mg (sounds crazy but it's true). Grinding even under water should be good enough for this reaction.




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condennnsa
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[*] posted on 28-12-2011 at 08:25


nurdrage put out his video on potassium : http://www.youtube.com/watch?v=gHgyn-wsxFw&feature=g-u&a...
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blogfast25
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[*] posted on 28-12-2011 at 12:48


Excellent video! As per usual from NR…

A few comments.

Tetrahydronaphthalene (trademark: Tetralin) is a very expensive solvent (Sigma Aldrich $417/20 L) and few, if any, will sell it to hobbyists. I wonder also whether the cost is justified here. Earlier up this thread NR claimed he’d done a reaction with Tetralin as solvent that took only an hour to complete but that doesn’t seem to be the case here. Also the claimed faster coalescence of the formed potassium isn’t really that apparent. I remain happy to use the much more OTC and much cheaper kerosene (lamp oil).

NR remains dedicated to the rather ritualistic addition of the catalyst in stages, as prescribed by the source patent but my own experiences with adding all the catalyst at once and at the very beginning (so-called ‘one pot’) are very positive and indistinguishable from the ‘phased addition’ results.

I’m surprised to see some of his potassium globules go black while the reaction is still ongoing. Could this be the result of prolonged exposure to the solvent, which contains double bonds? I've never experienced that with kerosene or Shellsol D.



[Edited on 28-12-2011 by blogfast25]




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[*] posted on 28-12-2011 at 15:37


Quote: Originally posted by condennnsa  
nurdrage put out his video on potassium : http://www.youtube.com/watch?v=gHgyn-wsxFw&feature=g-u&a...


Incredible! I'm glad he got the video going




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[*] posted on 29-12-2011 at 12:26


Has anyone calculated the enthalpy change of:
2Mg + 2KOH ---> 2K + 2MgO + H<sub>2</sub> ?

I calculated it and came up with +790KJ/mol. That's quite an energy barrier! And yet the synthesis seems to work...

Can anyone double check the enthalpy change of this reaction? I may have made a mistake (the first time I calculated this enthalpy change I made a mistake).




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[*] posted on 29-12-2011 at 14:03


WY:

That's wrong. It's been done higher up: about - 200 kJ/mol of K, off the top of my head. Per mol of K it's basically:

HoF(298, MgO) - HoF(298, KOH) = - 602 - (-425) = - 177 kJ/mol of K (@ 398 K). The escaping hydrogen helps drive the equilibrium further to the right.

There is a kinetic barrier, even at 200 C but it is overcome by the catalyst: that's what catalysts do.

[Edited on 29-12-2011 by blogfast25]




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[*] posted on 30-12-2011 at 07:41


in the old days (1810's) they used to prepare K by passing molten KOH on iron turnings heated over 1000C in the canon of a riffle. one end sealed by a mercury bath. on the other side, pure potassium metal would come running down .
i might try to prepare large quantity of K that way one day...sounds like a fun challenge
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[*] posted on 30-12-2011 at 10:13


Quote: Originally posted by blogfast25  
WY:

That's wrong. It's been done higher up: about - 200 kJ/mol of K, off the top of my head. Per mol of K it's basically:

HoF(298, MgO) - HoF(298, KOH) = - 602 - (-425) = - 177 kJ/mol of K (@ 398 K). The escaping hydrogen helps drive the equilibrium further to the right.


Ahh, I see.
I used Hess's law to find the heat of reaction, I completely forgot about the heat of formation (products)- heat of formation (reactants) equation.




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blogfast25
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[*] posted on 30-12-2011 at 12:46


Hmmm... Hess' Law is basically what I'm using.



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[*] posted on 12-1-2012 at 07:08


Quote: Originally posted by neptunium  
in the old days (1810's) they used to prepare K by passing molten KOH on iron turnings heated over 1000C in the canon of a riffle. one end sealed by a mercury bath. on the other side, pure potassium metal would come running down .
i might try to prepare large quantity of K that way one day...sounds like a fun challenge


Wtf? That sounds incredible and likely not quite so easy. What was the predominant alloy used in rifle barrels in the cusp ofthe seventeeeeeeenth century, it may be pertinent. Using mercury as a condensing fluid is quite clever, i had never thought of it before, all that cooling capacity. How did they stop the potassium vapour from shooting molten koh into the eye of th operator?

Reference or did your grandfather teach you this?




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[*] posted on 12-1-2012 at 08:57


Panache:

Sounds like an old wives tale to me...




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[*] posted on 12-1-2012 at 21:54


I just tried this experiment today on making potassium. My ingredients were:

1. small magnesium chips
2. paraffin oil (bp 215-240 C) Lamplighter brand.
3. 90% Food Grade KOH
4. t-butanol

I added 2 g of magnesium chips into 50ml flask. I added 5g of KOH and then added 25ml of paraffin oil and boiled it under reflux. After it started to boil, I injected 0.5ml of a mixture of (paraffin oil and t-butanol) through the condenser. The addition of which produced vigorous bubbles of hydrogen. Every 10 minutes I added 0.4ml of t-butanol mixture for a total of 5 times.

I allowed it to boil for 1hr. I turned off heating and when the refluxing stopped, I then stirred the mixture so that the magnesium and KOH would be in better contact.

I again started the boil and injected t-butanol mixture several more times. I allow the mixture to boil over 220C for over 4 hrs..

The overall mixture was too cloudy and the magnesium/potassium mixture was completely covered with magnesium oxide or KOH to notice any potassium.

I wasn't sure if the reaction worked, so I took some of the metal mixture and added it to water and noticed, small, very tiny flames of purple. awesome!

However, the mixture did not seem like all potassium but a mixture of potassium and magnesium. I boiled the solvent for a total of about 5 hrs which should have been enough time for the magnesium to be converted to potassium.

I have several questions:

1. do i need to boil it longer to completely react the magnesium and convert it to potassium? isn't 5 hrs enough?
I didn't notice any change, no spheres of potassium or anything.

2. is paraffin oil (msds bp 220-240c) ok for this reaction?

3. can you put too much t-butanol? I injected over 3ml in total over the coarse of 5 hrs.

I definitely produced potassium but how can you clean it up, it seems like a complete mess of magnesium oxide, magnesium, little potassium, and KOH?

What did I do wrong and/or how can I improve it. thanks.



[Edited on 13-1-2012 by jamit]

[Edited on 13-1-2012 by jamit]

[Edited on 13-1-2012 by jamit]

[Edited on 13-1-2012 by jamit]
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[*] posted on 13-1-2012 at 07:01


Jamit:

3 ml t-butanol is <i>far too much</i>: t-butanol reacts with potassium, forming K t-butoxide (it's part of the reaction mechanism!) Your K was simply redissolving!

Try again with the right quantity of t-butanol, occasionally lightly swirling the flask during the run: after about 2 h small globules of molten K will form, longer times (again, occasionally swirling) will make them coalesce into larger ones.




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[*] posted on 19-1-2012 at 11:27


Where did you get t-butanol? Best I've heard for a hobbyist is as an octane booster.



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[*] posted on 19-1-2012 at 12:03


Read the thread.



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[*] posted on 24-1-2012 at 00:16


Forgive me if this has been stated before, it is a long thread and I have not read all of it. Has anyone tried the reaction and been successful with alcohols other than tert-butanol?

And has anyone tried using aluminum instead of magnesium?

It might be interesting to add both NaOH (3 parts) and KOH (7 parts) to the reaction so the resulting product would be a K-Na alloy that is liquid at room temperature...
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[*] posted on 24-1-2012 at 06:53


Quote: Originally posted by AndersHoveland  
Forgive me if this has been stated before, it is a long thread and I have not read all of it. Has anyone tried the reaction and been successful with alcohols other than tert-butanol?

And has anyone tried using aluminum instead of magnesium?

It might be interesting to add both NaOH (3 parts) and KOH (7 parts) to the reaction so the resulting product would be a K-Na alloy that is liquid at room temperature...


Yes. They have tried other alcohols.

Yes. They have tried Aluminum.

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[*] posted on 24-1-2012 at 07:07


The ones we know to work are t-butanol and 2-methyl-2-butanol.

’Shortly’ I’ll be trying dihydromyrcenol, which also has that typical 2-methyl-2-ol structure of that kind of tertiary alocohols but with a longer alkyl chain length.

We are fairly sure only tertiary alcohols work: the alkoxides of primary and secondary alcohols are presumably not stable enough at 200 C.




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