Polyprotic acids and ionization states

Last time we saw how a buffer system works and how to design one according to the type of conditions we want to mantain during a particualr reaction. All our analyses were done for monoprotic acids/bases. However, in biomolecules we usually have many groups in a single molecule that can give away or take up protons from solution and therefore act as acids or bases. These are called polyprotic acids.

In particular, free amino acids have two groups that will become deprotonated or deprotonated (or charged, or ionized) under different conditions. Proteins will also have two end groups which can become charged, as well as the side chains of several amino acids with functional groups that can also become charged. The main groups we have to deal with in amino acids are the amino (-NH₂) and carboxyl (COOH) groups. For a general case involving a diprotic acids we will have the following reactions:

Unless one of the reactions is almost complete upon dissolution of the diprotic acid, like the first dissociation of H₂SO₄, each reaction will have its own equilibrium constants determined by the concentration of species in solution:

As it is evident from the 'formula' of this acid, the first proton will be easier to pull out than the second one. In other words, K₁ will be larger than K₂. Also, if we have a diprotic acid in solution, the proportions of AH₂, AH^-, and A^2- will depend on the pH.

Now, as we saw last time, it is more convenient to compare the pK_a values to the pH of the solution. The analyses for each ionization of the acid is done in the same way as we did for monoprotic acids. Then, if the pH is below 1 unit below pK₁ (and therefore pK₂), both ionizable groups will be protonated. Remember that one pH unit below the pK_a a weak acid is almost fully protonated, and one pH unit above, almost fully ionized.

As was the case with monoprotic acids, in the (pK₁ - 1) to (pK₁ + 1) range we will have a buffer system that behaves according to the equilibrium equation defined by K₁, which would be the AH₂ / AH^- buffer system.

As we go further up the pH scale, we will go pass the buffer region of the AH₂ / AH^- pair (pH > pK₁ + 1), and the acid corresponding to the K₁ equilibrium will have donated all of its protons. We then move into the buffer region of the second acid/base pair that we can have in the molecule, AH^- / A^2-. This system will also buffer the pH, this time  between the (pK₂ - 1) to (pK₂ + 1) pH range.

Again, the higher  the [OH^-], the more we protons we will pull from the second acid. After (pH > pK₂ + 1), the second buffer system breaks down, and the pH is again controlled exclusivelly by K_w, the ion-product of water.

'Going up' in the pH scale means adding more [OH^-] in the solution, which is basically titrating the acid. A typical titration curve for a diprotic acid is the process we just described, and looks like this (alanine in this case):

Since most of the cases of polyprotic acids we will encounter are amino acids, we can rewrite the above for for alanine, which is the simplest amino acid that has a 'real' side chain (a methyl group):

Some things to pay attention to: First, the amine group, which in alanine has a pK_a of 9.69, is fully protonated (-NH3⁺) at pHs below 8.69. Below 1.34, so is the carboxyl group (-COOH), which has a pK_a of 2.34. So below 1.34, both groups are in their acid, fully protonated, form.

As we move into the buffering range of the carboxyl group in alanine (1.34 ~ 3.34), the carboxyl (-COOH) will gradually become the carboxylate (-COO^-), but, as we have said 10,000 times already, there will be an equilibrium between the two governed by K₁: The pH will be buffered in this region.

As we move beyond 3.34 and towards and before we reach (pK₂ - 1),  the carboxyl will be almost exclusivelly as -COO^-, and the amine as -NH₃⁺. In this region of the pH scale we have two species of opposite charges coexisting in the same molecule. This is called a Zwitterion (which in German means an ion that is zwitter...). The net charge is zero, but the two charged species are there.

Now, once the pH starts getting into the 8.69 to 10.69 range, the -NH₃⁺ will gradually start donating its protons to OH^- (remember that this is what is actually happening all along). Again, there will be an equilibrium between R-NH3+ and R-NH2 which will buffer the pH changes in the ~8.69 to ~10.69 pH range controlled by the values of K₂. The carboxyl end is by now way past ionized, and it will always be as COO-.

Going over all this, we can draw some general conclusions:

1) While pH < (pK₁ + 1) we will have both groups in the acid, protonated form
2) In the (pK₁ - 1) to (pK₁ + 1) pH range, the first acid (-COOH) will be partially protonated, according to pK₁, while the second (-NH3⁺) will be fully protonated.
3) Between the range pK₁ < pH < pK₂, the first acid will be deprotonated (-COO^-), and the second protonated (-NH3⁺). In amino acids, this usually means we have two opposite charges in the same molecule, a Zwitterion.
4) In the (pK₂ - 1) to (pK₂ + 1) pH range, the second acid (-NH3⁺) will be partially protonated, according to pK₂, while the first acid (COOH) will be fully deprotonated.
5) After pH > pK₂ + 1, both acids will be entirelly deprotonated, and the pH is regulated exclusivelly by the ionic product of water, K_w.

As we know, the amine is polar, while the proton is not. In alanine we have two polar groups with opposite 'charges' - If both groups are ionized (-COO^- and -NH3⁺), there will be a favourable electrostatic interaction between them. Trying to protonate the -COO^- or deprotonate the -NH3⁺ will break this interaction. Therefore, it will be easier to deprotonate the carboxylic acid (that is, it will be a stronger acid), and it will be harder to deprotonate the amine (i.e., it will be a stronger base) because both will want to be ionized. Therefore, the pK_a for the carboxylate will drop, and the pK_a for the amine group will rise.

This brings about another important concept/idea. The pK_a of an acid can change depending on its chemical environment. If we have a caroxyl group in a protein, it may be participating in salt bridges. In order to protonate the acid we would have to break this salt bridge, which is pretty stable (~15 Kcal/mol). This means that the carboxylate will rather be ionized (-COO^-) in order to participate in the slat bridge. As we know, the more an acid likes to be ionized means that it will donate its protons more esily, which translates in a smaller pK_a. Therefore, the pK_as of acids and bases in amino acids (or for other groups) will vary from their 'standard' values according to the non-bonded interactions they are involved in.

Finding ionization states of polyprotic acids

In many reactions occuring in proteins, a functional group has to have a certain charge to act either as an acid, or a base, or a nucleophile, in a reaction, or to form a particular salt bridge that will stabilize certain conformational feature of the molecule. As we saw above, the ionization state of groups in amino acids will depend with the pH, so:

a) A protein will function as it is supposed depending on the pH, and therefore
b) We have to be able to calculate the ionization state of amino acids and proteins as a function of pH.

We also have to keep in mind K_w, the ion product of water. Now, for pHs way below pK₁, we are beyond the buffering region of either group: Both groups will be fully protonated (-COOH and -NH3+), so the net charge will be (+1).

As we move into the buffering region of the ⁺H₃N-CH(CH₃)-COOH / ⁺H₃N-CH(CH₃)-COO^- pair, the ratio of [⁺H₃N-CH(CH₃)-COOH] and [⁺H₃N-CH(CH₃)-COO^-] will vary according to pK₁, and therefore the net charge of this group will vary. To calculate the ratios, we can use the Henderson-Hasselbach equation:

If the pH was 3.00, we get:

Do these values make sense? Pretty much. We are above the pK_a value, so the ionized form will predominate. Now, chargewise, the ratio is 0.41 / 0.5, so 82% of the -COOHs will be ionized (-1). In this pH range we will be way below the buffering region of the ⁺H₃N-CH(CH₃)-COO^- / H₂N-CH(CH₃)-COO^- pair, so the amine stays fully protonated (+1). If we add this to the -0.82 charge from the -COOH group, we get a net charge of +0.18 at pH 3.00.

Now, while we are in the 3.34 < pH < 8.69 region, using the Henderson-Hasselbach equation is a lot dicier, because we are not in the buffering region of either acid. However, we saw before that the carboxyl will be fuly deprotonated (-COO-) and the amine fully protonated (-NH3+). This means that we will have a full (-1) charge from one group, and a full (+1) charge from the other. The net charge here is very close to 0. As we will see later, this is the isoelectric point (pI) of the amino acid, and it is reached when pH = (pK₁ + pK₂) / 2. This makes sense, because this point is when we are the further away from either buffer system, but in between both of them.

Now, if we move into the pH region controlled by the ⁺H₃N-CH(CH₃)-COO^- / H₂N-CH(CH₃)-COO^- system, we can calculate the [⁺H₃N-CH(CH₃)-COO^-] / [H₂N-CH(CH₃)-COO^-] ratio with the Henderson-Hasselbach equation, this time using pK₂. For example, for a pH of 9.50 we will have:

Doing the same math we did before:

Now, this means that the -NH₂ group will be 60% ionized (+0.6). Since the carboxyl group is completely ionized at this pH (-COO-), the net charge will be (-0.4). What a joy!

As we had said before, for pH > (pK₂ + 1), both groups will be fully deprotonated, so the total charge ot the system will be equal to (-1).

Next time, we finally start with amino acids...

Prepared by Guillermo Moyna, 1999.