The Henderson Hasselbalch equation plays a pivotal role in teaching acid-base equilibrium and therefore receives considerable attention in general, analytical, and biochemistry courses. Titration curves, buffer problems, and a host of related phenomena can be discussed with relative ease using the equation. In 1908, Lawrence Henderson was the first to derive an equation that can calculate the pH of a buffer solution. It led to the extensive use of Henderson’s equation. Subsequently, in 1917, Karl Hasselbalch changed the formula to get it in logarithmic terms. Which eventually led to the formation of the Henderson Hasselbalch equation. We will now be going through the formula and derivation of the Henderson Hasselbalch equation.

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The image depicts the application of Henderson-Hasselbalch equation.

So, what is the Henderson Hasselbalch equation? The Henderson Hasselbalch equation is an approximate equation that shows the relationship between the pH or pOH of a solution (aqueous solution) and the (acid dissociation constant) and the ratio of the concentrations of the dissociated chemical species. We can only use this equation if we know the acid dissociation constant. The formula for Henderson Hasselbalch is as follows:

\[p^{H}=p^{K_{a}}+log_{10}(\frac{[A^{-}]}{[HA]})\]

Where, \[p^{K_{a}}\] is the acid dissociation constant, \[[A^{-}]\] is the molar concentration of the conjugate base, and \[[HA]\] is the molar concentration of the weak acid. Now let us take a look at Henderson Hasselbalch equation derivation.

We will be going through two Henderson Hasselbalch derivations, the first one will be the derivation of the Henderson Hasselbalch Equation for base, and the second one will be for acid. Let's consider all the assumptions to be made for the derivation before we start with the derivations.

Assumption 1: The acid is monobasic.

Assumption 2: We can ignore the self-ionization of water.

Assumption 3: The salt MA is completely dissolved in the solution

Derivation 1) For the derivation of Henderson Hasselbalch equation for acid, let us take an example of ionization of acid HA:

HA + H2O <-->H+ + A-

The first step is to use the formula for acid dissociation constant, Ka

Ka = \[\frac{[H^{+}][A^{-}]}{[HA]}\]

Taking negative logs on both sides,

-log Ka = - log \[\frac{[H^{+}][A^{-}]}{[HA]}\]

-log Ka = - log [ H+ ] - log \[\frac{[A^{-}]}{[HA]}\]

We know that - log [ H+ ] = \[p^{H}\] and -log Ka = \[p^{K_{a}}\] ,

We get the following equation,

\[p^{K_{a}}\] = \[p^{H}\] - log \[\frac{[A^{-}]}{[HA]}\]

After rearranging the equation, we get,

\[\frac{[A^{-}]}{[HA]}\] = \[p^{K_{a}}\] + log \[\frac{[A^{-}]}{[HA]}\]

Therefore, we have derived the Henderson Hasselbalch equation for acid.

Now, If, [A-] = [HA],

We get log \[\frac{[A^{-}]}{[HA]}\] = 0

Therefore, we get \[p^{H}\] = \[p^{K_{a}}\] which states that both the species are the same, and the acid will be half dissociated.

Derivation 2) For the derivation of Henderson Hasselbalch equation for base, let us take an example of ionization of base:

B + H2O <---> OH+ + HB-

The first step is to use the formula for acid dissociation constant, Ka

\[K_{b} = \frac{[BH^{+}][OH^{-}]}{[B]}\]

Taking negative logs on both sides,

-log \[K_{b} =- log \frac{[BH^{+}][OH^{-}]}{[B]}\]

-log\[K_{b}\] = -log [OH-] - log \[\frac{[BH^{+}]}{[B]}\]

We know that - log [ OH-] = \[p^{OH}\] and -log \[K_{b}\] = \[p^{K_{b}}\]

We get the following equation,

\[p^{K_{b}}\] = \[p^{OH}\] - log \[\frac{[BH^{+}]}{[B]}\]

After rearranging the equation, we get,

\[p^{OH}\] = \[p^{K_{b}}\] + log \[\frac{[BH^{+}]}{[B]}\]

Therefore, we have derived the Henderson Hasselbalch equation for base. Now that we have gone through the Henderson Hasselbalch equation derivation, we will solve some Henderson Hasselbalch equation examples for better understanding.

Question 1) Estimate the pH of a buffer with 0.2M acetic acid (CH3COOH) and 0.5M acetate (CH3COO-), the acid dissociation constant is given as 1.8 * \[10^{-5}\]

Solution 1) Given:

[A-] = [CH3COO-] = 0.5M

[HA] = [CH3COOH] = 0.2M

\[K_{a}\] = 1.8 * \[10^{-5}\]

we need to calculate \[p^{K_{a}}\],

\[p^{K_{a}}\] = - log \[K_{a}\]

-log \[K_{a}\] = - log 1.8 * \[10^{-5}\]

\[p^{K_{a}}\] = 4,7

Now we write the Henderson Hasselbalch equation,

\[p^{H}\] = \[p^{K_{a}}\] + log \[\frac{[A^{-}]}{[HA]}\]

\[p^{H}\] = 4.7 + log \[\frac{0.5}{0.2}\]

\[p^{H}\] = 4.7 + log 2.5

\[p^{H}\] = 4.7 + 0.398

\[p^{H}\] = 5.098

Therefore, by using the equation we get the pH of the solution as 5.098.

Question 2) Estimate the pH of a buffer with 0.1M HC2H3O2 and 0.6M C2H3O2-, the acid dissociation constant is given as 1.8 * \[10^{-5}\].

Solution 2) Given:

[A-] = [C2H3O2-] = 0.6M

[HA] = [HC2H3O2] = 0.1M

Ka = 1.8 * \[10^{-5}\]

Now we substitute the values in the equation,

\[p^{H}\] = \[p^{K_{a}}\] + log \[\frac{[A^{-}]}{[HA]}\]

\[p^{H}\] = 4.7 + log \[\frac{0.6}{0.1}\]

\[p^{H}\] = 4.7 + log 6

\[p^{H}\] = 4.7 + 0.78

\[p^{H}\] = 5.48

Therefore, by using the equation we get the pH of the solution as 5.48.

It gives the formula for pH in terms of acidity in chemical and biological systems.

It is widely used to calculate the isoelectric point of proteins (the position at which a protein can neither accept or yield protons).

FAQ (Frequently Asked Questions)

1. Why is the Henderson Hasselbalch Equation Only Used for Weak Acids or Weak Bases?

The Henderson Hasselbalch equation only works with weak acids or weak bases. The reason why it only works for weak acids and bases is that this is a simplified way of defining the acid dissociation constants. Usually, the amount of H_{2}O does not change significantly, and thus can be thought of as constant and omitted.

Additionally, the word weak means the acid does not react 100% with water. Instead, it sets up an equilibrium with water. A strong acid does not set up an equilibrium with the water, and it reacts 100%.

2. Explain Why Buffer Solution Resists Change in p^{H}?

A buffer solution can resist pH change when we add any acidic or basic component. Buffers contain acidic components, HA, which neutralizes OH^{-} ions, and it also includes basic components, A^{-}, which neutralizes H^{+ }ions. It will always keep the pH of the solution stable as it will always neutralize small amounts of acid or base.

HA + H_{2}O <--> H^{+ }+ A^{-}

H^{+ }+ A^{-} <--> HA