20.3 Biological Acids and the Henderson–Hasselbalch Equation

Biological Acids and the Henderson–Hasselbalch Equation

The Henderson–Hasselbalch equation lets you predict what form a weak acid takes at any given pH. For carboxylic acids in biological systems, this matters because whether the acid is protonated ($HA$) or deprotonated ($A^-$) affects its solubility, reactivity, and ability to cross cell membranes.

Ratio Calculation with the Henderson–Hasselbalch Equation

The core equation relates pH, $pK_a$, and the ratio of deprotonated to protonated forms:

$pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)$

Where:

• $pH$ = the pH of the solution
• $pK_a$ = the negative log of the acid dissociation constant ($K_a$)
• $[A^-]$ = concentration of the deprotonated (conjugate base) form
• $[HA]$ = concentration of the protonated (acid) form

To find the ratio of deprotonated to protonated species at a given pH:

1. Rearrange by subtracting $pK_a$ from both sides: $\log\left(\frac{[A^-]}{[HA]}\right) = pH - pK_a$

2. Take the antilog (raise 10 to both sides): $\frac{[A^-]}{[HA]} = 10^{pH - pK_a}$

Example: Acetic acid ($pK_a = 4.76$) at pH 5.5

1. $\log\left(\frac{[A^-]}{[HA]}\right) = 5.5 - 4.76 = 0.74$

2. $\frac{[A^-]}{[HA]} = 10^{0.74} \approx 5.5$

This means about 5.5 molecules are deprotonated for every 1 that remains protonated. The pH is above the $pK_a$, so the deprotonated form dominates.

A quick rule of thumb: if $pH > pK_a$, the deprotonated form wins. If $pH < pK_a$, the protonated form wins. At $pH = pK_a$, you get a 1:1 ratio.

Carboxylic Acids as Anions at Physiological pH

Physiological pH is approximately 7.4. Most carboxylic acids have $pK_a$ values in the range of 4–5. That's a gap of roughly 2.4–3.4 pH units, which on a logarithmic scale translates to an enormous excess of the deprotonated form.

The math confirms this directly. If $pH > pK_a$, then $pH - pK_a > 0$, so $10^{pH - pK_a} > 1$, meaning $[A^-] > [HA]$.

Example: Lactic acid ($pK_a = 3.86$) at physiological pH (7.4)

1. $\log\left(\frac{[A^-]}{[HA]}\right) = 7.4 - 3.86 = 3.54$

2. $\frac{[A^-]}{[HA]} = 10^{3.54} \approx 3467$

For every single protonated lactic acid molecule, roughly 3,467 exist as the lactate anion. That's why biological carboxylic acids are almost always drawn and named as their carboxylate forms (lactate, pyruvate, citrate) rather than as the free acids.

This has real consequences: the anionic form is charged and water-soluble, which keeps these molecules dissolved in the aqueous environment of cells and blood. It also means they can't freely diffuse across nonpolar cell membranes without a transporter.

Percentages of Acid Species at a Specific pH

Sometimes you need the actual percentage of each form rather than just the ratio. Here's how to convert:

1. Calculate the ratio: $\frac{[A^-]}{[HA]} = 10^{pH - pK_a}$

2. The total acid concentration is: $[A]_T = [HA] + [A^-]$

3. Percentage of protonated form: $\%HA = \frac{1}{\frac{[A^-]}{[HA]} + 1} \times 100\%$

4. Percentage of deprotonated form: $\%A^- = \frac{\frac{[A^-]}{[HA]}}{\frac{[A^-]}{[HA]} + 1} \times 100\%$

These two percentages will always add up to 100%.

Example: Benzoic acid ($pK_a = 4.20$) at pH 4.0

1. $\frac{[A^-]}{[HA]} = 10^{4.0 - 4.20} = 10^{-0.20} \approx 0.63$

2. $\%HA = \frac{1}{0.63 + 1} \times 100\% \approx 61.3\%$

3. $\%A^- = \frac{0.63}{0.63 + 1} \times 100\% \approx 38.7\%$

The pH is slightly below the $pK_a$ here, so the protonated form is the majority species, but not by a huge margin. Compare this to the lactic acid example at pH 7.4, where the deprotonated form is essentially 100%. The closer the pH is to the $pK_a$, the more significant both species become.

Buffer Solutions and Equilibrium

A buffer solution contains a weak acid and its conjugate base in appreciable amounts. It resists pH changes when small quantities of strong acid or base are added.

Buffers work most effectively when the pH is within about 1 unit of the $pK_a$. In that range, both $HA$ and $A^-$ are present in significant concentrations. If you add a strong acid, the conjugate base ($A^-$) reacts with the added $H^+$ to form $HA$, absorbing the acid. If you add a strong base, $HA$ donates a proton to neutralize the added $OH^-$. This is Le Chatelier's principle in action: the equilibrium shifts to counteract the disturbance.

On a titration curve, the buffering region appears as the relatively flat portion centered around the $pK_a$. At the half-equivalence point, $[A^-] = [HA]$, so $pH = pK_a$ exactly. This is where the buffer has its maximum capacity.