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At the isoelectric point amino-acids are neutral to the outside. It is however not known to my knowledge if the molecules are indeed in a zwitterionic form or just in its "normal" totally uncharged form, i.e. the dissociation constant for this reaction:

a busy cat

is not known. Does anyone know how they actually look like? What form is more probable, how realistic is the zwitterionic form?

Edit: I did a test calculation for cysteine with DFT, using a PBE functional and implicit water and found that the zwitterionic form is about 7 kcal/mol more stable, which would rather strongly indicate that the zwitterionic form is indeed dominating a lot and you would not at all find the structure on the left as the bars from wikipedia indicate (supposing that I found low lying minima in both calculations and not only local ones). Are there some other studies proving this?

Edit2: Ok the zwitterionic form is definitely more stable, that one can somehow reason from chemical knowledge or pKa's. But how much more stable, can one somehow get a dissociation constant for the reaction above?

Edit3: Since there is much confusion about my question. I want a dissociation constant for the reaction (COOH)(NH2) -> (COO-)(NH3+). Until now I think only the dissociation constants for (COOH)(NH3+) -> (COO-)(NH3+) and (COO-)(NH3+) -> (COO-)(NH2) are known if I am right and I cannot tell how these would help to calculate the constant I want...

Edit4: To clear up my question even a bit more, here is a diagram:

a busy cat

The red arrow and question mark is what I want to know about. The pks1 and pks2 are known, that was clear to me, but how to get any information about the equilibrium in the middle?

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    $\begingroup$ What studies? -COOH group in amino acids is just like that in other carboxylic acids, and as such, should be expected to have about the same pKa, so in less acidic media than $\rm pH\approx4$ it should be mostly deprotonated. -NH2 group in amino acids is just like that in primary amines, and as such, should be expected to have about the same pKb, so below $\rm pH\approx9$ it should be mostly protonated. Putting these observations together gives... $\endgroup$
    – Ivan Neretin
    Commented Nov 22, 2016 at 19:08
  • $\begingroup$ Ok thanks for the comment, I have reformulated the question a bit $\endgroup$
    – Guiste
    Commented Nov 22, 2016 at 19:19
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    $\begingroup$ Does this answer your question? Zwitterion (amino acid) equilibrium and the equilibrium constant $\endgroup$
    – Mithoron
    Commented Dec 20, 2022 at 20:43

3 Answers 3

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While the other answers are undoubtedly correct with respect to water, the better answer is that it depends on your solvent.

The zwitterionic form is the one preferred in solid state and in solutions where the solvent can accomodate charges well, i.e. polar solvents. In less polar solvents, the neutral form can predominate to allow dissolution (rise in entropy) in spite of non-dissociation (not as favourable). However, cyclic hydrogen bonds are to be expected.

But in the end, this complicated discussion boils back down to what the other answers have said again. The only difference when changing solvent is that the $K_\mathrm{a}$ equation becomes different. Water is no longer the co-acting principle, it is a different solvent resulting in different $\mathrm{p}K_\mathrm{a}$ values (which are solvent-dependent to a certain degree).

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  • $\begingroup$ That's of course true, it will depend on the solvent also and I did not specificy this in the question, although I was only thinking of water. It does however not provide an answer, how to get such equilibrium constant or estimate it... $\endgroup$
    – Guiste
    Commented Nov 22, 2016 at 23:23
  • $\begingroup$ @Guiste That’s what the other answer say (or maybe: gloss over). $\endgroup$
    – Jan
    Commented Nov 22, 2016 at 23:35
  • $\begingroup$ Thanks for the answer anyways, did not think about the entropy factor for less polar solvents. $\endgroup$
    – Guiste
    Commented Nov 22, 2016 at 23:37
  • $\begingroup$ The other answers however assume that the first pKa of a protonated aminoacid is the same as that of the neutral species we are talking about... $\endgroup$
    – Guiste
    Commented Nov 22, 2016 at 23:39
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By Henderson relation we have

$$\mathrm{pH=pI=p}K_\mathrm{a}+\log\left(\frac{[\ce{A-}]}{[\ce{AH}]}\right)$$

So if $\mathrm{pH - p}K_\mathrm{a} < 0$ then $[\ce{A-}] < [\ce{AH}]$

And if $\mathrm{pH - p}K_\mathrm{a} > 0$ then $[\ce{A-}] > [\ce{AH}]$

If you want then to know the exact proportion you need to know the concentration of your amino-acid.

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  • $\begingroup$ Thanks for the comment, but that was not my question. It is totally clear that at the isoelectric point, we have a neutral species to the outside. But how much of this neutral species is the also partially neutral species and how much is the zwitterion (the two species in the chemical reaction I showed above) $\endgroup$
    – Guiste
    Commented Nov 22, 2016 at 19:18
  • $\begingroup$ @Guiste and if you have the exact proportions you have the equilibrium constant then. $\endgroup$
    – ParaH2
    Commented Nov 22, 2016 at 19:28
  • $\begingroup$ I think you still don't understand the question, please read my Edit 3. Of course knowing the concentrations of both species (COOH)(NH2) and (COO-)(NH3+) would give us the equilibrium constant I want, but is it possible to measure this concentrations? And has someone done this? $\endgroup$
    – Guiste
    Commented Nov 22, 2016 at 19:34
  • $\begingroup$ @Guiste read my answer correctly. I have not said to know the concentration of both, I just said the total concentration of your amino acid. $\endgroup$
    – ParaH2
    Commented Nov 22, 2016 at 19:36
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    $\begingroup$ The pKa value is measured. I don't understand what you are saying... $\endgroup$
    – Zhe
    Commented Nov 22, 2016 at 20:06
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This is totally known based on the pKa values of the carboxylic acid and the ammonium. The zwitterion is the predominant form.

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  • $\begingroup$ Then what would be the pKa of the carboxylic acid for cysteine? I don't think that you can find it, since the ammonium group will be always protonated, so the constant you are mentioning is an artificial equilibrium constant which you might not be able to get... $\endgroup$
    – Guiste
    Commented Nov 22, 2016 at 19:11
  • $\begingroup$ What I am meaning is, if you want to use a thermodynamic cycle to evaluate an equilibrium constant for the reaction I showed above, you would need to start from the left educt... $\endgroup$
    – Guiste
    Commented Nov 22, 2016 at 19:13
  • $\begingroup$ Hexacoordinate-C actually has numbers. Upvote his answer. :) $\endgroup$
    – Zhe
    Commented Nov 22, 2016 at 19:20
  • $\begingroup$ He does not. I want a dissociation constant for the chemical reaction I presented above. To my knowledge this is not known so far! The pKa he presents denote the dissociations (COOH)(NH3)+-> (COO)-(NH3)+, where AH2 is the twicely protonated amino acid. To calculate a dissociation constant for the reaction I presented you would need a constant for the reaction (COOH)(NH2)-> (COO)-(NH2), which is not the same! $\endgroup$
    – Guiste
    Commented Nov 22, 2016 at 19:27
  • $\begingroup$ Since you can change the reaction diagram from the one you want to one that branches to both (NH3+)(COOH) and (NH2)(CO2-) then reconvenes to the final zwitterion, we should be able to use the simple relationships between free energy summed over those reactions (a kJ is a kJ after all) to obtain a valid pKa for the net reaction. I am inclined to believe Zhe is correct in asserting that said math has been done and favors the zwitterion (in fact I think I learned this in AP chem), but even if it not so certain, the method for calculating it is trivial. $\endgroup$
    – sqykly
    Commented Nov 23, 2016 at 4:50

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