1.5: Henderson-Hasselbalch Approximation

It is useful to be able to predict the response of the $$HAc$$ system to changes in $$H^+$$ concentration. The Henderson-Hasselbalch equation defines the relationship between pH and the ratio of $$Ac^-$$ and $$HAc$$. It is as follows

$pH = pK_a + \log \left(\dfrac{[Ac^-]}{[HAc]}\right) \label{1.5.1}$

This simple equation defines the relationship between the pH of a solution and the ratio of Ac- and HAc in it. The new term, called the pKa, is defined as

$\text{pKa} = -\log \text{Ka} \label{1.5.2}$

just as

$\text{pH} = -\log [\text{H}^+] \label{1.5.3}$

The Ka is the acid dissociation constant and is a measure of the strength of an acid. For a general acid, HA, which dissociates as

$\text{HA} \leftrightharpoons \text{H}^+ + \text{A}^- \label{1.5.4}$

$\text{Ka} = [\text{H}^+][\text{A}^-] / \text{[HA]} \label{1.5.5}$

Thus, the stronger the acid, the more protons that will dissociate from it and the larger the value its Ka will have. Large values of Ka translate to lower values of pKa. As a result, the lower the pKa value is for a given acid, the stronger the acid is.

Please note that pKais a constant for a given acid. The pKafor acetic acid is 4.76. By comparison, the pKafor formic acid is 3.75. Formic acid is therefore a stronger acid than acetic acid. A stronger acid will have more protons dissociated at a given pH than a weaker acid.

Now, how does this translate into stabilizing pH? The previous figure shows a titration curve. In this curve, the titration begins with the conditions at the lower left (very low pH). At a this pH, the HAc form predominates, but as more and more OH- is added (moving to the right), the pH goes up, the amount of Ac- goes up and (correspondingly), the amount of HAc goes down. Notice that the curve “flattens” near the pKa (4.76).

What this tells us is that the pH is not changing much (not going up as fast) as it did earlier when the same amount of hydroxide was added. The system is resisting a change in pH (not stopping the change, but slowing it) in the region of about one pH unit above and one pH unit below the pKa. Thus, the buffering region of the acetic acid/acetate buffer is from about 3.76 to 5.76. It is maximally strong at a pH of 4.76.

Now it starts to become apparent how the buffer works. HA can donate protons when extras are needed (such as when $$OH^-$$ is added to the solution. Similarly, A- can accept protons when extra $$H^+$$ are added to the solution (adding HCl, for example). The maximum ability to donate or accept protons comes when

$[\text{A}^-] = \text{[HA]} \label{1.5.6}$

To understand how well a buffer protects against changes in pH, consider the effect of adding .01 moles of HCl to 1.0 liter of pure water (no volume change) at pH 7, compared to adding it to 1.0 liter of a 1M acetate buffer at pH 4.76. Since HCl completely dissociates, in 0.01M ($$10^{-2}$$ M) HCl you will have 0.01M $$H^+$$. For the pure water, the pH drops from 7.0 down to 2.0 (pH = -log(0.01M)).

By contrast, the acetate buffer’s pH is 4.74. Thus, the pure water solution sees its pH fall from 7 to 2 (5 pH units), whereas the buffered solution saw its pH drop from 4.76 to 4.74 (0.02 pH units). Clearly, the buffer minimizes the impact of the added protons compared to the pure water.

It is important to note that buffers have capacities limited by their concentration. Let’s imagine that in the previous paragraph, we had added the 0.01 moles HCl to an acetate buffer that had a concentration of 0.01M and equal amounts of Ac- and HAc. When we try to do the math in parallel to the previous calculation, we see that there are 0.01M protons, but only 0.005M A- to absorb them. We could imagine that 0.005M of the protons would be absorbed, but that would still leave 0.005M of protons unbuffered. Thus, the pH of this solution would be approximately

$\text{pH} = -\log 0.005\text{M} = 2.30 \label{1.5.7}$

Exceeding buffering capacity dropped the pH significantly compared to adding the same amount of protons to a 1M acetate buffer. Consequently, when considering buffers, it is important to recognize that their concentration sets their limits. Another limit is the pH range in which one hopes to control proton concentration.

Now, what happens if a molecule has two (or more) ionizable groups? It turns out, not surprisingly, that each group will have its own pKa and, as a consequence, will tend to ionize at different pH values. The figure above right shows the titration curve for a simple amino acid, alanine. Note that instead of a single flattening of the curve, as was seen for acetic acid, alanine displays two such regions. These are individual buffering regions,each centered on the respective pKa values for the carboxyl group and the amino group.

If we think about alanine, it can have three possible charges: +1 (alpha carboxyl group and alpha amino group each has a proton), 0 (alpha carboxyl group missing a proton and alpha amino group has a proton) and -1 (alpha carboxyl group and alpha amino group each lacking a proton).

How does one predict the charge at a given pH for an amino acid? A good rule of thumb for estimating charge is that if the pH is more than one unit below the pKa for a group (carboxyl or amino), the proton is on. If the pH is more than one unit above the pKa for the group, the proton is off. If the pH is NOT more than one or less than one pH unit from the pKa, this simple assumption will not work.

Further, it is important to recognize that these rules of thumb are estimates only. The pI (pH at which the charge of a molecule is zero) is an exact value calculated as the average of the two pKa values on either side of the zero region. It is calculated at the average of the two pKa values around the point where the charge of the molecule is zero.