How do I calculate the ensemble-average net charge of an amino acid at given pH?

Question

I am given an amino acid with an ionizable side chain at a certain pH. How do I determine the net charge of that amino acid when there are mixed protonation states of one or more of the groups at that pH (pKa of side chain, for example, is really close to the pH)?

Amino acids have terminal carboxyl and amino groups; some amino acids have ionizable side chains. When determining the charge of an amino acid, you have to take into consideration the pH and the pKa's of each of these groups. When the pKa of one group (or more) is close enough to the pH, a fraction of the amino acids will be deprotonated at that group and the other fraction of amino acids will be protonated at that group in solution. Thus, when determining the average net charge across the ensemble (or a time-averaged charge of a single particle), you have to take this into account.

I am asking for the expected value of the net charge (which would not be an integer); this number is relevant, for example, for the migration speed of the amino acid (or a protein) in gel electrophoresis or the strength of interaction with ion exchange chromatography media.

For example, a carboxylic acid/carboxylate group at a pH equal to its pKa would have an average charge of minus one half because half of the functional groups would be protonated (charge of zero) and half would be deprotonated (charge of minus one).

Answer 1

The Henderson-Hasselbalch relationship describing each ionizable group is:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{\ce{[A-]}}{\ce{[AH]}}$$

We can solve for the ratio:

$$10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a})}= \frac{\ce{[A-]}}{\ce{[AH]}}$$

However, we really want the fraction of protonated among the total (not the ratio of deprotonated to protonated).

$$10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a})} = \frac{[\mathrm{total}] - \ce{[AH]}}{\ce{[AH]}} = \frac{[\mathrm{total}]}{\ce{[AH]}} - 1$$

Add one to both sides: $$10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a})} + 1 = \frac{[\mathrm{total}]}{\ce{[AH]}}$$

Take the reciprocal: $$\frac{\ce{[AH]}}{[\mathrm{total}]} = \frac{1}{10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a})} + 1}\tag{1}$$

This is still general for any acid/base group. For example, we could use it to calculate the charge of ammonia/ammonium ($\ce{NH3(aq) + H+(aq) <=> NH4+(aq)}$). At very basic pH, the charge would be zero, at very acidic pH, +1. To get the average charge at any pH, we take the charge at very basic pH and add the result of equation [1] using the $\mathrm{p}K_\mathrm{a}$ value of ammonium.

For any amino acid (or any other molecule with ionizable groups with $i$ different $\mathrm{p}K_\mathrm{a}$ values), you take the charge of the species at very basic pH (all groups deprotonated), plus the following:

$$\sum_i \frac{1}{10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a,i})} + 1}\tag{2}$$

This is only an approximation because there might be some cross talk between ionizable groups (i.e. if one group becomes negatively charged, it becomes more "difficult" for the neighboring group to become negatively charged). It also becomes more complicated for polyprotic groups, but all the groups in amino acids are monoprotic with water as a solvent.