Henderson-Hasselbalch Equation

Core Concepts – Henderson-Hasselbalch Equation

In this tutorial, you will learn what the Henderson-Hasselbalch equation is, how to use the equation to determine the pH of a buffer, how to derive the equation, and how to solve some example problems

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What is the Henderson Hasselbalch Equation?

The Henderson-Hasselbalch equation is used to determine the pH of a buffer system.

$\begin{gather*} {pH = pK_{a} + log_{10}\frac{[\text{Conjugate Base}]}{[\text{Weak Acid}]}} \end{gather*}$

The $K_{a}$ is the acid dissociation constant, and pK_a is the negative log of $K_{a}$ . [Conjugate Base] is also commonly written as [A-]. And [Weak Acid] may be seen as [HA]. These values refer to the equilibrium concentration.

If you have the same concentration of acid and conjugate base, then the pH will be the same as the pK_a because the $log(1) = 0$ .

If [HA] = [A^–], then

$\begin{gather*} {pH = pK_{a} + log_{10}\frac{1}{1}=pK_{a}} \end{gather*}$

buffer that can be analyzed using the henderson-hasselbalch equation — Phosphoric acid buffer solution that can be analyzed using the Henderson-Hasselbalch Equation

Derivation of the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation can be derived from the $K_{a}$ of a weak acid. Here we will assume we have a weak acid, HA, and its conjugate base, A^–. In water the following dissociation occurs:

$\begin{gather*} {HA + H_{2}O \rightleftarrows A^{-} + H_{3}O^{+}} \end{gather*}$

Therefore, the acid dissociation constant is defined as

$\begin{gather*} {K_{a} =\frac{[A^{-}][H_{3}O^{+}]}{[HA]}} \end{gather*}$

Which can be rearranged to

$\begin{gather*} {[H_{3}O^{+}] = K_{a}\frac{[HA]}{[A^{-}]}} \end{gather*}$

Then take the log of both sides

$\begin{gather*} {log[H_{3}O^{+}] = log\left( K_{a}\frac{[HA]}{[A^{-}]} \right)} \end{gather*}$

Multiply by -1 so that in the next step we can substitute in for pH.

$\begin{gather*} {-log[H_{3}O^{+}] = -log\left( K_{a}\frac{[HA]}{[A^{-}]} \right)} \end{gather*}$

Based on the definition of $pH = -log [H_{3}O^{+}]$ we can substitute into our equation

$\begin{gather*} {pH = -log\left( K_{a}\frac{[HA]}{[A^{-}]} \right)} \end{gather*}$

And then use logarithm rules to simplify the right side

$\begin{gather*} {pH = -log\left( K_{a} \right) + log \left( \frac{[HA]}{[A^{-}]} \right)} \end{gather*}$

Knowing that $pK_{a} = -log(K_{a})$ , we can substitute that into our previous equation to arrive at the Henderson-Hasselbalch equation

$\begin{gather*} {pH = pK_{a} + log_{10}\frac{[\text{Conjugate Base}]}{[\text{Weak Acid}]}} \end{gather*}$

Example Problems of Henderson Hasselbalch Equation

Determining Buffer Ratio

Suppose we are making a buffer solution of ammonium (NH₄⁺) and ammonia (NH₃). In this example, ammonium is our weak acid and ammonia is the conjugate base. The reaction is as follows.

$\begin{gather*} {NH_{4}^{+} \leftrightarrows H^{+} + NH_{3}} \end{gather*}$

If the pK_a of ammonium is 9.3 and we want a pH of 10. What ratio of ammonia to ammonium should you add to the solution?

The problem asks for a ratio, so we are looking for the value of $\frac{[NH_{3}]}{[NH_{4}^{+}]}$ , or $\frac{[A^{-}]}{[HA]}$ for this specific buffer solution. To solve the problem, start by plugging in the known values– pH and pK_a.

$\begin{gather*} 10 = 9.3 + log\frac{[A^{-}]}{[HA]} \end{gather*}$

Subtract 9.3 from each side to simplify

$\begin{gather*} 0.7 = log\frac{[A^{-}]}{[HA]} \end{gather*}$

The ratio we are looking for is within the logarithmic term. To get rid of the log we can use logarithmic rules to simplify.

$\begin{gather*} 10^{0.7} = \frac{[A^{-}]}{[HA]} = \frac{5.01}{1} \end{gather*}$

Therefore, the ratio is about 5 times the amount of ammonia as ammonium to get a buffer solution with a pH of 10.

Determining pH

Use the Henderson-Hasselbalch equation to determine the pH of a buffer solution made up of $0.15 \text{M}$ CH₃COO^– and $0.20 \text{M}$ CH₃COOH. The pKa of acetic acid (CH₃COOH) is 4.75.

To solve this example problem, we want to start with the Henderson-Hasselbalch equation since we are looking at a buffer solution of a weak acid and conjugate base.

$\begin{gather*} {pH = pK_{a} + log_{10}\frac{[\text{Conjugate Base}]}{[\text{Weak Acid}]}} \end{gather*}$

Then plug the known values into our equation

$\begin{gather*} {pH = 4.75 + log_{10}\frac{[0.15]}{[0.20]}} \end{gather*}$

Then simplify and solve

$\begin{gather*} {pH = 4.75 + (-0.12) = 4.63} \end{gather*}$

Our buffer solution of acetic acid has a pH of 4.63.

Limitations of the Henderson-Hasselbalch Equation

When using the Henderson-Hasselbalch to make a buffer solution the ratio of $\frac{[A^{-}]}{[HA]}$ should be close to one. This makes the most stable solution. When the ratio moves far away from one, the capacity to buffer the solution decreases as a small disturbance changes the pH dramatically.
This pH equation does not account for the self-dissociation of water. At very low concentrations of buffer, water can contribute to the pH
If an acid can easily dissociate more than one hydronium ion (H⁺) the equation does not account for that and may not work well.
Practically when making specific pH buffers, it is important to have the pK_a be close to the target pH to have the biggest range of buffering capability.

Fun Facts:

The equation takes its name after Lawrence Joseph Henderson and Karl Albert Hasselbalch. Henderson made the original equation in 1908. Then in 1917 Hasselbalch put the formula into logarithmic terms and the way it is most seen today.
The Henderson-Hasselbalch equation has important biological relevance. Many biological samples require a specific pH. So, when biologists prepare samples, the Henderson-Hasselbalch equation ensures a pH that helps proteins and other pH sensitive molecules maintain their structure.
A similar equation can calculate the pOH of a buffer system.

$\begin{gather*} {pOH = pK_{b} + log_{10}\frac{[\text{Conjugate Acid}]}{[\text{Weak Base}]}} \end{gather*}$

The [Conjugate Acid] is also referred to as [HB⁺]. And the [Weak Base] is also referred to as [B]. Try checking your understanding by going through the derivation for the pOH version of the equation.