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Offline Schmittchen

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approximate degree of dissociation
« on: June 18, 2020, 04:46:18 AM »
Dear experts,
The (calculated acc. to PubChem) pKa of 2-amino acetophenone is 2,31. I want to know to which degree this molecule is dissociated in an aqueous solution of pH 3.
Can someone please help me in getting this information? E. g. does there exist something like a calculator page or something to that effect?
Thank you very much in advance & regards from Germany
Schmittchen

Offline Borek

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Re: approximate degree of dissociation
« Reply #1 on: June 18, 2020, 07:08:23 AM »
This is trivial to calculate, just start with the dissociation constant definition and solve for the ratio of both forms.
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Offline Schmittchen

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Re: approximate degree of dissociation
« Reply #2 on: June 18, 2020, 09:50:52 AM »
Dear Borek,
Thank you for the comment.
Dissociation of the molecule is: R-NH2 + H2O --> R-NH3+ + OH-.
As pKa of 2-aminoacetophenone (subsequently called 2AAP) is said to be be 2,31 I think in plain water and with a low concentration of 2AAP (┬Ám range) we will see roughly 1/200 part of the 2AAP to be dissociated.
Am I correct so far?

However, the problem (that is to say my problem) is what happens in a solution with an pH of 3,0? Where does the pH of the solvent come in in my dissociation formula above?
So, again, I am ashamed to admit that I am unable to do the calculation. Thats why I was asking for help, in the best case by providing the result.

Please excuse my being so hard of grasp...
With kind regards
Schmittchen

Offline Corribus

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Re: approximate degree of dissociation
« Reply #3 on: June 18, 2020, 10:24:25 AM »
Is 2-acetophenone the only thing in solution?
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Schmittchen

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Re: approximate degree of dissociation
« Reply #4 on: June 18, 2020, 01:05:48 PM »
@ Corribus - No, not at all. We are talking about wine, where 2AAP, if present above a certain limit, imparts a wine fault called "atypical ageing" (abbr. ATA, sometimes also UTA).
So because 2AAP coexists with a whole zoo of other constituents things are a bit complex, also the pH varies. Hence a rough estimation of the degree of dissociation would fully suffice.
Thank you Schmittchen

Offline Corribus

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Re: approximate degree of dissociation
« Reply #5 on: June 18, 2020, 02:43:47 PM »
I guess I'm a little confused. You want to know the degree to which the molecule is dissociated, but in the post above you frame the molecule as an association reaction, not a dissociation. 2AAP would appear to be a base, not an acid. So how is the pKa only about 2.3? Where are you getting your pKa value from? Are you sure this isn't a pKb value? The molecule also has two pH-active functional groups (carbonyl, amino group) so there will be two pKa/pKb values, not one. You will need to be more specific about the chemistry.

Based on the structure alone, I would imagine this is almost completely protonated at pH = 3. But maybe I'm missing something here.

It gets even more complicated when you have a lot of other pH-active substances in your solution, especially if they have similar pKa/pKb values. They will all participate in the equilibrium.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Borek

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Re: approximate degree of dissociation
« Reply #6 on: June 18, 2020, 03:05:58 PM »
Is 2-acetophenone the only thing in solution?

I assume he is talking about pH 3 solution, then other things don't matter, as we are not talking about finding pH but just about finding forced dissociation equilibrium (so all that matters is pH and pKa or pKb).

I agree with you on other things, it looks a bit confusing.
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Offline Borek

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Re: approximate degree of dissociation
« Reply #7 on: June 18, 2020, 03:10:39 PM »
Where does the pH of the solvent come in in my dissociation formula above?

Broadly speaking (as Corribus pointed out some things don't add up here) when you have an acid HA, dissociating

HA ::equil:: H+ + A-

dissociation constant is

[tex]K_a = \frac {[H^+][A^-]}{[HA]}[/tex]

If you know pH you know [H+], Ka is also given, so you can directly plug them into the formula, and then solving for the ratio of HA/A- is just a trivial math.

Similar approach applies to base dissociation and hydrolysis.
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Offline Schmittchen

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Re: approximate degree of dissociation
« Reply #8 on: June 19, 2020, 05:34:08 AM »
@ Corribus: Thank you for the comment. A correction is necessary: the source of the (predicted) pKa is ChemicalBook, not Pubchem. Please see: https://www.chemicalbook.com/ChemicalProductProperty_DE_CB5383885.htm. Like you I was wondering why the give a pKa for a base, but I guess they mean pKb.
Anyway I now see that the majority is dissociated , so my question is answered.
Tthank you & kind regards

Offline AWK

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Re: approximate degree of dissociation
« Reply #9 on: June 19, 2020, 06:17:30 AM »
Quote
but I guess they mean pKb.
Nonsense.
pKa and pKb can be defined for each acid and base, and the relationship that connects them is pKa + pKb = pKw
NH3 + H2O = NH4+ + OH-
NH4+ + H2O = NH3 + H3O+
These two equations define Kb and Ka respectively.
Knowing the amine pKa and pH even without using a calculator, it can be estimated that the free amine will be 20-30%.
But it probably doesn't solve your problem in any way, because creating 2AAP is a problem, not its forms in solution.
AWK

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