I am not sure what you are saying.
I looked up the wikipedia article you coppied from and the whole section was:
At higher temperatures, solid KOH crystallizes in the NaCl crystal structure. The OH group is either rapidly or randomly disordered so that the OH− group is effectively a spherical anion of radius 1.53 Å (between Cl− and F− in size).
At room temperature, the OH− groups are ordered and the environment about the K+ centers is distorted, with K+ —OH− distances ranging from 2.69 to 3.15 Å, depending on the orientation of the OH group. KOH forms a series of crystalline hydrates, namely the monohydrate KOH·H2O, the dihydrate KOH·2 H2O, and the tetrahydrate KOH·4 H2O"
Can you be a bit more specific for us non chemistry people why you feel this is relevant to the discussion.
Are you trying to say there are other impurities in the bag of KOH that are not listed on the MSDS?
Any MSDS I have read, all that is listed is Potassium hydroxide and water which total 100%.
If you think there are other unlisted impurities please give a percentage for these unlisted impurities.
Interesting. You seem to be saying that even if methoxide is not produced, water will be produced by dissolving KOH in methanol?
Please provide a link to your source for this information.
Really the first part is clarifying for Dgs's sake that (and how) in fact H2O or water IS incorporated in the structure of solid KOH, particularly when it is solidified at low temperature.
It's an absolute certainly that H+ and OH- ions are released when KOH is dissolved in CH3OH.
KOH- + CH3OH > CH3OK + H2O where the potassium ion is substituted for the proton in the MeOH molecule, and meanwhile the displaced proton (H+) is free to associate with the hydroxyl (OH-) ion surplus from the lye. So we say "water forms." At a molecular level it is probably more complicated than that.
Thank you for your quick reply.
I am super busy today so I will have a closer look this evening when I get back home
So I have done some googling and think I understand what is happening.
Just to make things a little clearer for everyone who is not chemistry literate like I am not;
KOH= Potassium Hydroxide
MEOH and CH3OH= Methanol
CH3OK= Potassium Methoxide
I agree with you that if you mix Potassium Hydroxide with methanol (KOH- + CH3OH) you get Potassium Methoxide and Water (CH3OK + H2O).
HOWEVER the question is how much Potassium methoxide (CH3OK) and water (H2O) is actually produced when you mix Potassium Hydroxide (KOH) with Methanol (CH3OH)?
The author of this procedure and the chemist neutral both thought that there was actually very little Potassium methoxide (CH3OK) produced and therefore there was very little water (H2O) produced.
32gms of water are produced for every 100 gms of KOH dissolved in methanol.
I suspect that number is based on the assumption that all the KOH is converted into Potassium Methoxid and water which apparently does not happen.
The Sandy Brae tests that I did on the methoxide did indeed confirm the water content was in the range the32gms/100gms suggests.
50gms of KOH in 400mls of methanol, S/B was 3.12% water. Methanol was 2600ppm water.
What did the S/B read on the methanol before adding the KOH?
Have you taken into consideration the 10% water in the KOH?
As my post states Tilly, the water in the methanol was 2600ppm.
The sandy brae doesn't take anything into consideration. It just reads the water content.
I will try to get my chemist friend to add to this thread and try to clear up this anomoly re water in the KOH.
To be meaningful you would have performed an S/B before adding KOH and then an S/B after adding KOH so you would know what the actual water increase was.
If you did not do an S/B test of the Methanol before adding the KOH then you did not do a meaningful test
You have to take the 10% water impurity in the KOH into consideration.
The KOH we use is typically about 10% water which will show up in the S/B test.
This water is an impurity and has nothing to do with the chemical reaction that results by mixing KOH with Methanol.
That's good information Tilly. I did not know that so much water was in my Potassium Hydroxide solid pellets. My KOH is 85%. I never noticed that the listing of impurities does not add up to 100%, not even close.
I can't understand why you keep asking me that?
THE METHANOL WAS 2600PPM WATER
AFTER ADDING THE KOH AND MIXING THE WATER WAS 3.12%
I might be able to assist in this research inquiry. I know calcium oxide reacts with water to form calcium hydroxide, that's calcium bonded to two hydroxide groups. So for each molecule of one calcium bonded to one oxygen, upon the absorption/ reaction with water one molecule of water is removed from solution in methanol. But there is an equilibrium that exists. I can not directly prove it, except, if you add adequate Calcium oxide to wet ethanol, the purest dry alcohol you can distill out of the still pot is 99.5%. Calcium oxide will not capture all the water from the alcohol solution.
so your methoxide went from .26% to 3.12% water after mixing in the KOH. some of the water comes from the reaction some comes from the KOH, why do we care where it came from? How do we get rid of it or how much can we reasonable remove?
" I don't know what I don't know until I know"
1994 GMC 6.5 Tubo 2005 Dodge ram 3500, 3 VW's 2000, 2002, 2005.
Thats what we are trying to find out Tom.
So let me see if I understand what you are saying,
You took some fresh Methanol and did a S/B test and found it to be 2600PPM.
You then immediately added 50g KOH into 400ml of this methanol and after mixing the S/B test showed 3.12% water
Let's see how those numbers work out.
400ml methanol weighs about 320g
3.12% of 320g is about 10g water detected by the S/B test which includes the small amount of water already in the methanol and the water bound in the KOH.
About 10% of the original 50g KOH was water which is 5g.
To find the amount of water produced by the reaction this 5g of water needs to be deducted from the 10g total water detected by the S/B
That leave sbout 5g water actually produced by the reaction.
So that means about 10g Water would be produced if you dissolved 100g KOH in methanol
I read that the SB vessel is anodized aluminium (aluminum). The anodizing is aluminium oxide. Apparently methanol attacks aluminium oxide, creating water. Seeing as the SB test detected a quarter of 1 percent water, I think we should exclude it for this discussion, as it is minor.
So we are talking about adding 50gm dry KOH to 400ml methanol, which having a density of 0.792 at (I presume) 20 deg C means it has a mass of 316.8gm.
We know as Tilly says that 10% of the KOH is H2O, meaning in fact only 45gm KOH is dissolved.
Of this 45gm, K has a mass of 31.358gm and OH a mass of 13.642gm.
However in the production of the methoxide, the methanol consumed gives up an H (or proton) which, with the OH, goes to make HOH or H2O (water).
The ratio of atomic weights of H2O to OH is 1.059 (H2O: 18.01528, OH: 17.01).
So the OH in picking up an H, or proton, becomes 1.059
times more massive - so the water created now has a mass of 14.449gm
Added to the initial 5gm impurity (excluding carbonates) the water is 19.449gm, or 0.388972434 (say 39%) of the initial 50gm.
That's a significant amount.
Now, if we add the mass of the MeOH 316.8gm and the 50gm of impure KOH the total mass of the solution is 366.8gm, of which 19.449gm is water - 5.3%.
This is the minimum percentage.
This page https://en.wikipedia.org/wiki/Methoxide states "The methoxide ion has the formula of CH3O− and is the conjugate base of methanol. It is a strong organic base, even stronger than the inorganic hydroxide ion. As such, methoxide solutions must be kept free of water; otherwise, the methoxide will remove a proton from a water molecule, yielding methanol and hydroxide."
For every K ion introduced, a potential H2O molecule exists, to which we must add the 10% impurity. So is "methoxide" created? Apparently it is, because the transesterification reaction proceeds. These ions all exist in solution. I just think it's a poor man's methoxide.
I have been driving all day thinking about this.
I also realized that the "50g KOH" was in fact 45g KOH and 5g H2O. I wondered if anyone else would spot that so well done for spotting it.
I must admit that a lot of your last post does not make sense to me at the moment, however a few thoughts came to mind today while dodging wombats and thylacines.
Apparently Dgs has been told that theoretically 100g KOH mixed into excess methanol has the ability to produce 32g of water.
His testing shows that 100g KOH mixed into excess methanol actually only created about 10g Water.
If water production is the result of the formation of Potassium methoxide and about 1/3 the water was produced that is the theoretical maximum, that would suggest to me that the reaction producing Potassium methoxide has only gone to around 33% completion.
So "drying" or removing the water as it is formed should move the reaction further to the right- completion.
Unfortunately, the original author of this procedure did very little actual testing and it is only a guess as to how much farther the methoxide reaction will go with this drying.
There was a fellow on another forum who did do a lot of testing but for now I will not introduce his results.
That's about in accordance with my calcs - just under 29gm for 100gm KOH. And that excludes the 10% impurity.
Still doesn't solve the problem of getting rid of the water.
it seems to be about 10g water for 90g KOH if you go by the S/B numbers Dgs posted if you remove the 10% water impurities.
Curious how you arrived at your number.
It would be good if you would run through how you arrived at 29g water per 100g KOH.
So if we calculate this by your weight which takes into consideration the weight of the KOH:
...the mass of the MeOH 316.8gm and the 50gm of impure KOH the total mass of the solution is 366.8gm,"
and calculate the 3.12% total water from Dgs S/B test which includes the water in the methanol and the KOH as well as the water produced in the reaction, that works out to about 11.4g total water of which 5g is from the KOH.
Remove the 5g water from the KOH gives about 6.4g water produced by 45g KOH which becomes 12.8g per 90g KOH.
Increase that by 10% to get the water per 100g KOH brings it to about 14g water produced which is a long ways off the 29g water that you have calculated
I think I see what you have done, you have calculated the maximum theoretical amount of water that would be produced if all the KOH was converted into potassium methoxide and water.This message has been edited. Last edited by: Tilly,
|Powered by Social Strata||Page 1 2 3 4 5 6 7|