# How to rationalise the protonation state of amino acids

I am learning about amino acids in various $\ce{pH}$ environments and how their groups are protonated or deprotonated in these environments. I don't need to mention a specific acid for my question, I will just use the generic form of one: $$\ce{H2N-CH(R)-COOH}$$

The $\ce{H2N}$ group has a $\mathrm{p}K_a$ of $8$, I believe, and the acid has a $\mathrm{p}K_a$ of $3.1$.

If I drop these into a solution with $\ce{pH} = 12$, the $\ce{COOH}$ will donate it's hydrogen ion, but the $\ce{H2N}$ will not. It will only add another hydrogen ion if it is basic compared to the solution to become $\ce{H3N+}$. Why is it, that it won't donate its hydrogen ion like $\ce{COOH}$, when it is acidic compared to the solution, to become $\ce{HN-}$?

• You can visit this page to ‎find out how to make your future posts better. – M.A.R. Feb 6 '15 at 12:41
• Oh, nothing wrong. I just wanted to note that \ce{} is used to write chemical formulas and equations. You would be free of lots of formatting with it as it automatically renders numbers and ion charges in favorable behavior. (No, not really :) – M.A.R. Feb 6 '15 at 12:48
• @BjarniJóhannsson Meta contains much more information than the help center. Please have a look here and here. Even if it is tempting, please do not use markup in the title field, see here for details. – Martin - マーチン Feb 13 '15 at 4:25

Let's assume you drop the amino acid in water at $\ce{pH} = 7$. You'll have $$\ce{ H3N^+{}-CH(R)-COO^-}.$$ Then you can go to $\ce{pH}= 12$ and the amino acid loose the proton.
You don't get rid of the second proton because the $\mathrm{p}K_a$ is defined as $$\mathrm{p}k_{a}=-\lg\frac{[\ce{NH2}]\cdot[\ce{H+}]}{[\ce{NH3+}]}.$$ This means $\ce{NH3+ -> NH2}$.
You could define another $\ce{p}K_a$ for $\ce{NH2 -> NH^{-}}$. The $\mathrm{p}K_a$ for this would be much higher than 12 - at least around 35. You may look at this site: How to use a $\mathrm{p}K_a$ table You can compare the amine $\ce{NH3}$ (without the rest of the amino acid) with your $\ce{NH2}$ group of the amino acid.