I just had my biology test paper back from my teacher, and my teacher insisted that when functioning as a buffer in an alkaline solution, the amino acid deprotonates by the ammonium group instead of the carboxyl group, because amino acids in solution are mostly zwitterionic. I thought zwitterions only consist a small portion of the mixture, but checking Wikipedia only further confirms what my teacher said.

In an attempt to find mathematical proof, I tried to derive the equilibrium constant between zwitterionic and completely neutral forms of amino acids, with the assumption that the solution is pure and only constitutes the said acid(and its zwitterionic isomer) and water. This was my working:

Using $K_\mathrm{a}$ for carboxyl group and $K_\mathrm{b}$ for amine group. $$\ce{H2N-CHR-COOH <=> H3N+ -CHR-COO-}$$ \begin{align} [\ce{H2N-CHR-COOH}] &= [\ce{H2N\bond{-}}] = [\ce{-COOH}]\\ [\ce{H3N+ -CHR-COO-}] &= [\ce{H3N+\bond{-}}] = [\ce{-COO^-}] \end{align}

\begin{align} K_c &= \frac{[\ce{H3N+ -CHR-COO-}]}{[\ce{H2N-CHR-COOH}]}\\ &=\sqrt{\frac{[\ce{H3N+-CHR-COO^-}][\ce{H3N+-CHR-COO^-}]}{[\ce{H2N-CHR-COOH}][\ce{H2N-CHR-COOH}]}}\\ &=\sqrt{\frac{[\ce{H3N+\bond{-}}][\ce{-COO^-}]}{[\ce{H2N\bond{-}}] [\ce{-COOH}]}}\\ &=\sqrt{\frac{[\ce{H3N+\bond{-}}]}{[\ce{H2N-}][\ce{H+}]} \times \frac{[\ce{-COO^-}][\ce{H+}]}{[\ce{-COOH}]}}\\ &=\sqrt{\frac{[\ce{H3N+\bond{-}}][\ce{OH-}]}{[\ce{H2N\bond{-}}][\ce{H+}][\ce{OH-}]} \times \frac{[\ce{-COO^-}][\ce{H+}]}{[\ce{-COOH}]}}\\ &\text{(now assuming $[\ce{H2N\bond{-}}] = [\ce{H2O.H2N\bond{-}}]$)}\\ &=\sqrt{\frac{[\ce{H3N+\bond{-}}][\ce{OH-}]}{[\ce{H2O.H2N\bond{-}}][\ce{H+}][\ce{OH-}]} \times \frac{[\ce{-COO^-}][\ce{H+}]}{[\ce{-COOH}]}}\\ &=\sqrt{\frac{K_\mathrm{b}}{K_\mathrm{w}} \times K_\mathrm{a}}\\ &=\sqrt{\frac{K_\mathrm{a} K_\mathrm{b}}{K_\mathrm{w}}} \end{align} With $K_\mathrm{w}$ being the denominator, it is easy to see that $K_c$ would be sufficiently large that nearly all amino acids in aqueous solution exists in the zwitterionic form, that is contrary to my initial belief; bringing in the $K_\mathrm{a}$ and $K_\mathrm{b}$ values further confirms that.

However, I am not content with just that. I'd like to ask:

  1. What are the faults (calculation errors, false/inappropriate assumptions, etc) in my working, if any?
  2. What is the main reaction mechanism that mediates this equilibrium? Is it an intramolecular transition or is it brought about by the solvent, or something else?
  3. Can this be considered as a form of tautomerization? If so, is it prototropy? Hence, is the $K_c$ above the so-called tautomeric ratio?
  • 1
    $\begingroup$ Let’s start by saying that there are two $\mathrm{p}K_\mathrm{a}$ values: one corresponding closely to that of a carboxylic acid, one corresponding close to that of an amino group. Also, since protonation and reprotonation of both groups is completely independent you should have an easier way of arriving at this solution, too. Maybe I’ll write something up later. $\endgroup$
    – Jan
    Commented Oct 1, 2015 at 12:18
  • $\begingroup$ @Jan the pKa above was that of the carboxyl group, and pKb that of the amino group. Or are you saying that I need the amino pKa too, though IMHO that seems very insignificant. $\endgroup$
    – busukxuan
    Commented Oct 1, 2015 at 12:33
  • $\begingroup$ No, that was meant along the lines of ‘each amino acid can be deprotonated (at least) twice. The first deprotonation chemically corresponds nicely to a carboxylic acid, the second to an amino group’. I didn’t read through your argument well enough (yet) to be sure if it’s good or bad ^^' $\endgroup$
    – Jan
    Commented Oct 1, 2015 at 12:38
  • $\begingroup$ @Jan Well that is what I was referring to in the 2nd sentence in my last comment. IIRC the pKa of the amine group was around 9 so I guess I could omit that, but maybe I'm wrong. $\endgroup$
    – busukxuan
    Commented Oct 1, 2015 at 12:46

1 Answer 1


If you assume that the groups react independently, you have two equilibria with equilibrium constants $K_a(\ce{-NH3+})$ and $K_a(\ce{-COOH})$:

$$\ce{^+(H3N)-R <=> (H2N)-R + H+}$$


$$\ce{R-COOH <=> R-COO- + H+}$$

You are interested in the overall reaction

$$\ce{^+(H3N)-CHR-COO- <=> (H2N)-CHR-COOH}.$$

You can take a two step route to achieve this:

$$\ce{^+(H3N)-CHR-COO- <=> (H2N)-CHR-COO- <=> (H2N)-CHR-COOH}$$

The equilibrium constant for the first step is $K_a(\ce{-NH3+})$, and for the second step is $1/K_a(\ce{-COOH})$. Combined, you get a result of

$$ K_\text{combined} = \frac{K_a(\ce{-NH3+})}{K_a(\ce{-COOH})}$$

Because the $K_a$ values of ammonium and carboxylic acid are different by about four orders of magnitude, the uncharged species occurs about ten thousand times less often than the zwitter ionic form in aqueous solution. In general, comparing p$K_a$ values of two groups competing for a proton tells you which group will usually have the proton (the one with the larger p$K_a$) and how unlikely a proton transfer to the other group would be ($10^{\Delta pK_a}$).

What are the faults (calculation errors, false/inappropriate assumptions, etc) in my working, if any?

I don't understand line 2 and 3 of your argument, and how you get from line 5 to 6. The result you get differs from mine in a square root.

I thought zwitterions only consist a small portion of the mixture

This depends on the solvent. In aqueous solution, zwitterions are the common species, whereas in organic solvent, the uncharged species is the common one.


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