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Offline pgk

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Re: benzoin condensation
« Reply #15 on: July 04, 2015, 03:47:42 PM »
The KOH 0.1M is a quite low concentration, when compared to ROH 16M (1/160). Therefore, autoionization of both ROH and water must also be taken into account.
ROH + KOH → ROK + H2O 
KOH → K(+) + OH(-)
ROK  → RO(-) + K(+)
ROH → RO(-) + H(+)
H2O → HO(-) + H(+)
In addition, positive and negative charges must be equilibrated.
[K(+)] + [H(+)]  =  [OH(-)] + [RO(-)]
So, start re-calculating by tacking all above equilibria constants, into account and by the assumption that the total volume does not change, when mixing ROH with water (if needed) and additionally, that no thermal expasion occurs during exothermic solvolysis.
Good luck.
« Last Edit: July 04, 2015, 04:48:33 PM by pgk »

Offline Borek

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Re: benzoin condensation
« Reply #16 on: July 04, 2015, 05:56:43 PM »
The KOH 0.1M is a quite low concentration, when compared to ROH 16M (1/160). Therefore, autoionization of both ROH and water must also be taken into account.
ROH + KOH → ROK + H2O 
KOH → K(+) + OH(-)
ROK  → RO(-) + K(+)
ROH → RO(-) + H(+)
H2O → HO(-) + H(+)
In addition, positive and negative charges must be equilibrated.
[K(+)] + [H(+)]  =  [OH(-)] + [RO(-)]
So, start re-calculating by tacking all above equilibria constants, into account and by the assumption that the total volume does not change, when mixing ROH with water (if needed) and additionally, that no thermal expasion occurs during exothermic solvolysis.
Good luck.

Burden of proof you are right lies on you. We know the general ideas about how to deal with the equilibrium calculations, so your post adds nothing new to the discussion. Either show results of your calculations, or stop posting at all, as you are just trolling now.
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Offline pgk

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Re: benzoin condensation
« Reply #17 on: July 05, 2015, 04:42:01 AM »
Dear Sir,
Please accept the following comments:
1). Burden of proof me are right lies on me but according the forum policy rules, I have to show only the insights but not the full answer.
Are there different policy rules for the forum stuff than the forum members?
2). These are not general ideas but the basic principles of chemical equilibria.
3). Considering basic principles as nothing new to the educational forum, consists a violation of the first paragraph of the forum’s registration agreement.
Are there different registration terms for the forum stuff than the forum members? 
4). The results of calculations are already shown above:
K = Ka/Kw → K = 10exp(-16)/10exp(-14) → K= 10exp(-2)  →  [RO(-)]/[OH(-)] = 1/100
Please, read my posts more carefully before being the strict lecturer.
5). Please, feel free to restrict my posts that you do not want to judge.
6).”Trolling”? – Please, go back to the comments No1 and No3.
Sincerely yours
pgk
« Last Edit: July 05, 2015, 06:17:23 AM by pgk »

Offline Dan

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Re: benzoin condensation
« Reply #18 on: July 05, 2015, 06:18:54 AM »
K = Ka/Kw → K = 10exp(-16)/10exp(-14) → K= 10exp(-2)  →  [RO(-)]/[OH(-)] = 1/100
Please, read my posts more carefully before being the strict lecturer.

Your equation is incorrect:

K = Ka(ROH)/Ka(water)

Ka(water) ≠ Kw


Ok, if 0.1 M is too dilute, let's try 1 M.

The same calculation as my previous gives x = 0.9, i.e. 90:10 alkoxide:hydroxide

--

I have presented several calculations now that demonstrate that the alkoxide is the major anion present in alcoholic hydroxide solutions, despite the (generally) weaker acidity of alcohols compared to water (i.e. despite the fact that K < 1)

The reason for this is the vast excess of alcohol present. It's a quantitative demonstration of Le Chaterlier's principle.

You appear to dismiss these calculations as invalid on the basis of the assumptions made (i.e. because autodissociation was not considered). While the assumptions I have made in my calculations will affect the accuracy of the result, I think it is safe to say that they have qualitative value in predicting the major anions present. If the result of my calculations ([RO-] >> [HO-] in dilute alcoholic hydroxide solution) is reversed by a more sophisticated mathematical model, then you have to demonstrate it. That's not a forum rule, it's just that unless you can demonstrate it, I'm not going to take your word for it.

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Offline Borek

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Re: benzoin condensation
« Reply #19 on: July 05, 2015, 06:31:42 AM »
Edit: due to circumstances it took me a long time to write this post. Dan was faster, and his post addresses basically the same problems.

1). Burden of proof me are right lies on me but according the forum policy rules, I have to show only the insights but not the full answer.
Are there different policy rules for the forum stuff than the forum members?

These rules are designed for helping those solving homework questions, we are long past this point.

Quote
2). These are not general ideas but the basic principles of chemical equilibria.

Which is why they don't add anything new. At the moment it is not a discussion on the HS level, so we can safely assume we all know basic principles. No need to list them as if they were eye-openers.

Quote
4). The results of calculations are already shown above:
K = Ka/Kw → K = 10exp(-16)/10exp(-14) → K= 10exp(-2)  →  [RO(-)]/[OH(-)] = 1/100

And they are wrong. K is not [RO-]/[OH-]. You have ignored concentrations of ROH and H2O, and their ratio can substantially change the result:

[tex]\frac{[RO^-]}{[OH^-]} = K \frac {[ROH]}{[H_2O]}[/tex]

If you start with anhydrous ethanol, ratio [itex]\frac {[ROH]}{[H_2O]}[/itex] is quite large. Dan calculations took it into account and he have shown to you how to find the result he got. If you think he is wrong, show precisely where Dan is wrong, or show your full calculations. So far your arguments are handwavy and don't look reasonable:

The KOH 0.1M is a quite low concentration, when compared to ROH 16M (1/160). Therefore, autoionization of both ROH and water must also be taken into account.

Autoionization of water and alcohol yields concentrations of products several orders of magnitude lower than the concentrations of RO- and OH- from the equilibria taken into account so far. Experience tells me they will not change anything in the full picture. Feel free to solve the full system and prove me wrong.
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Offline pgk

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Re: benzoin condensation
« Reply #20 on: July 05, 2015, 07:09:10 AM »
Indeed, Kw is the ionization expression of per number of moles, contrary to Ka that is the ionization expression per molarity (even better, normality) of water and therefore, pkw = 14 and pKa = 15.7.
In other words, Kw measures the water ionization degree, contrary to pKa that measures the acidity of water and helps the comparison with other compounds. But when being in mixtures and solutions, the ratio of ionized/noninized molecules of water, remains 10exp(-14) and therefore, Kw is the predominant and valuable constant, for further mathematical applications.
So, let’s do it again:
ROH + OH(-)  →  RO(-)  +  H2O
Kreaction = [RO(-)][H2O]/ [HO(-)][ROH]
If you multiply and divide that function by [H+], you finally get:
K = Ka(ROH diss.)/Kw
ATTENTION: The ratio of concentrations [RO(-)]/[HO(-)] per volume unit does not change, regardless the changes of concentrations of ROH and water that occured during the alkali exchange reaction.
« Last Edit: July 05, 2015, 08:16:16 AM by pgk »

Offline pgk

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Re: benzoin condensation
« Reply #21 on: July 05, 2015, 07:23:01 AM »
The experimental verification of all above seems quite easy because alkoxides have different α- and β- chemical shifts in 1H-NMR and different ipso-C shift in 13C-NMR than the corresponding alcohols.
I will really appreciate any literature and experimental data, regarding the issue.

Offline Borek

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Re: benzoin condensation
« Reply #22 on: July 05, 2015, 08:18:02 AM »
So, let’s do it again:
ROH + OH(-)  → RO(-)  +  H2O
Kreaction = [RO(-)][H2O]/ [HO(-)][ROH]
If you multiply and divide that function by [H+], you finally get:
K = Ka(ROH diss.)/Kw

No matter how many times you will repeat it, K will not become a ratio of concentrations of RO- and OH-. Some things do cancel, [ROH] and [H2O] do not, they are left in the equation. Yes, you have shown how to calculate K for the reaction we are interested in from known Ka and Kw, no, it doesn't show what you claim it shows.

Quote
ATTENTION: the ratio of concentrations [RO(-)]/[HO(-)] per volume unit does not change, regardless the changes of concentrations of ROH and water

This is exactly the problem with your reasoning. Why do you think the above is true? The correct expression describing the ratio of concentrations, derived from the ROH/OH- equilibrium, is

[tex]\frac{[RO^-]}{[OH^-]} = K \frac {[ROH]}{[H_2O]}[/tex]

and the ratio of concentrations of water and alcohol plays an important role. You can't ignore it, unless you have a good reason to do so. So far you have failed to show what the reason is.
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Offline pgk

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Re: benzoin condensation
« Reply #23 on: July 05, 2015, 08:23:21 AM »
Because the total volume does not change (or practically, it does not significantly change).

Offline Dan

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Re: benzoin condensation
« Reply #24 on: July 05, 2015, 08:53:06 AM »
So, let’s do it again:
ROH + OH(-)  →  RO(-)  +  H2O
Kreaction = [RO(-)][H2O]/ [HO(-)][ROH]
If you multiply and divide that function by [H+], you finally get:
K = Ka(ROH diss.)/Kw

I don't think you do.

Kreaction = [RO-][H2O]/ [HO-)][ROH]
              = ([RO-)][H+]/[ROH])*([H2O]/[HO-][H+])
              = Ka(ROH)*Ka(H2O)

The term in red is Ka(H2O), not Kw. Kw = [HO-][H+] in dilute aqueous solution - I don't think it's applicable here. You are incorrectly ignoring [H2O].
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Offline pgk

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Re: benzoin condensation
« Reply #25 on: July 05, 2015, 10:43:04 AM »
Or and in other words, the function Kw = [HO-][H+] = 10exp(-14) is not valuable and applicable for concentrated aqueous solutions, as well as for pure water. Of course, it is!
Please, note that the dissolution medium is not water, hereby and thus, [H2O] is not the basis for calculations of [H+] and [OH-].
Besides and for similar reasons:
KaKb = kw
and:
KaKb ≠ Ka(water)
Once again, the terms Ka(water) and pKa(water) refer to pure water (free of electrolytes and solutes) and they are the expression of water acidity, in comparison with Arrhenius acids.

« Last Edit: July 05, 2015, 10:56:24 AM by pgk »

Offline Dan

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Re: benzoin condensation
« Reply #26 on: July 05, 2015, 12:10:10 PM »
Or and in other words, the function Kw = [HO-][H+] = 10exp(-14) is not valuable and applicable for concentrated aqueous solutions, as well as for pure water. Of course, it is!

Kw is based on the assumption that [H2O] is constant. That is not necessarily true for concentrated solutions.

Kw = [H2O]*Ka(H2O]) = 10-14 if [H2O] = 55.5 M (i.e. pure water)

But that is really beside the point. We are not discussing aqueous solutions, we are discussing alcoholic hydroxide solutions. Kw has no place in these calculations. [H2O] is certainly not constant in this scenario, and it is certainly not 55.5 M.

I think you have misused Kw to incorrectly calculate the equilibrium constant. I calculate K = 0.5, but you get K = 0.01 because you have erroneously divided by the concentration of pure water.

If you can show a valid calculation that supports your assertion that alkoxide formation in alcoholic KOH is practically negligible, please do so. I don't mind being proved wrong, it wouldn't be the first time, but otherwise I think it's time to stop.
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Offline Borek

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Re: benzoin condensation
« Reply #27 on: July 05, 2015, 12:31:54 PM »
Actually Kw is sometimes expressed not as 10-14, but as 1.8×10-16 (10-14/55.5) (and in this case that would be the more correct value).
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Offline Borek

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Re: benzoin condensation
« Reply #28 on: July 05, 2015, 12:42:12 PM »
Because the total volume does not change (or practically, it does not significantly change).

How is it related to problem in question? Are you saying that because the volume doesn't change significantly, concentrations don't change as well? If you start with an anhydrous alcohol containing several ppm of water, and you add 1 mL of water to 1 L of alcohol, the concentration ratio change is hundredfold, while the volume change is for most practical purposes insignificant.
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Offline pgk

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Re: benzoin condensation
« Reply #29 on: July 05, 2015, 02:10:13 PM »
Assuming that the total volume does not significantly change, when mixing ROH with water:
Mixing aqueous KOH 1N and alcoholic KOH 1N, we get an initial mixture of KOH 1N per total volume unit that will immediately be transformed (ionic reaction) to (say) KOH 0.995 N and ROK 0.005 N per total volume and (say) KOH 1.98 N and ROK 0.02 N per ROH volume or a ratio 0.99/0.01
The relative concentrations of KOH and ROK may change by further dilution with either water of ROH or their mixture but the concentration ratio 0.99/0.01 will remain constant, when calculated per ROH partition. 
As a reminder, the latter discussion refers to anhydrous ethanolic KOH.
« Last Edit: July 05, 2015, 02:24:51 PM by pgk »

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