# How to determine the charge of amino acid at certain pH? [duplicate]

### Problem

What would charge would you expect on alanine when placed in a solution with a pH of 1.00?

+1. Since alanine is nonpolar, we know that the only parts of the amino acid that can be charged are the N-terminus and the C-terminus.

In an acidic solution, there is an excessive amount of protons available to protonate the amino acid. As a result, the carboxylic acid end and the amine end will both be fully protonated. This will result in an overall charge of +1, due to the nitrogen having three hydrogens attached.

### Question

Let's say I am given a certain pH of 2.00 rather than 1.00 for the "acidic solution" and the pH in this example is of that of a non-polar acid (alanine) changes to +1.

Does the pH dictate how many protons are added to amino or carboxyl groups? What determines the number of protons being added to that amino acid on the carboxylic end? I know a carboxylic acid can take up to four protons being added on the molecule.

• 4 protons to carboxyl? Where did you get it? Commented Jan 11, 2023 at 7:46
• A carboxyl group can accept four protons. bonding abilitys to the oxygens (which are apart of the group carboxyl) Commented Jan 12, 2023 at 20:58
• No, it cannot. You cannot have $\ce{R-COH2^{2+}-OH3^2+}$. Neither you can have $\ce{H4O^2+}$ Commented Jan 12, 2023 at 21:03
• What does "neither you can have h4O2+" is that even english? please elaborate on that last thought. Commented Jan 15, 2023 at 19:56
• Even if it was not proper English, it could not be such a big deal to realize there is no ion $\ce{H4O^2+}$. // You cannot have A. Neither you can have B. IMHO it is correct, but if it is not, I apologize. Commented Jan 15, 2023 at 20:23

Behaviour of amino acids in water solution

Amino acids exist in the solution in the form of zwitterions (molecules that contain an equal number of positively and negatively charged functional groups). Thus, the following equilibrium is established in the alanine solution: $$\ce{H2N-(CH3)CH-COOH <=> ^+H3N-(CH3)CH-COO-}$$ Alanine as an ampholyte

Since actually there are two equilibria in the solution, alanine can be considered as an ampholyte: $$\ce{^+H3N-(CH3)CH-COO- + H2O <=> H2N-(CH3)CH-COO- + H3O+}$$ $$\ce{^+H3N-(CH3)CH-COO- + H2O <=> ^+H3N-(CH3)CH-COOH + OH-}$$ At $$\textrm{pH}\ 2$$ there is an excess of $$\ce{H3O+}$$ and a lack of $$\ce{OH-}$$ so the first equilibrium shifts to the left while the second equilibrium shifts to the right according to Le Chatelier's principle. Therefore, the main form in which alanine exists in water solution at $$\textrm{pH}\ 2$$ is $$\ce{^+H3N-(CH3)CH-COOH}$$. The overall charge of this molecule is $$+1$$. And vice versa, in an alkali media ($$\textrm{pH}>7$$) alanine will have the main form $$\ce{H2N-(CH3)CH-COO-}$$ and the net charge equal to $$-1$$.

It is a possible explanation from the point of view of acid-base theory. Please, check out references for further reading.

References

1. Skoog, Douglas A., West, Donald M., Holler, F. James, Crouch, Stanley R. (2014). Fundamentals of Analytical Chemistry (9th ed.). Singapore: Cengage Learning. p. 371-372.
2. Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). Oxford University Press. p. 167.

The change of $$\mathrm{pH}$$ from $$+1$$ to $$+2$$ does not change much in the main formula of alanine, which is and stays $$\ce{H3\overset{+}{N}−(CH3)CH−COOH}$$, up to $$2.34$$ which is the first $$\mathrm{p}K_\mathrm{a}.$$

• This isn't quite correct: there is only a certain fraction of given protonated species, not unity. Commented Jan 11, 2023 at 21:54
• @andselisk. OK for the fraction. I thought it was obvious. Commented Jan 12, 2023 at 9:20