Henderson Hasselbalch Equation

In 1908, in order to calculate the Hydrogen concentration in a buffer solution, one of the leading scientists of the era, Lawrence Joseph Henderson, derived an equation. The particular derivation found by Lawrence was widely used till today in order to calculate the P^H of a particular buffer solution.

Theory for the Henderson Hasselbalch Equation:

A simple buffer solution consists of a combination of an acid and a salt of the conjugate base. The most defining property of a buffer solution is its ability of it to resist the changes occurring in a p^H, even if a small amount of acid or base is added to it.

Let us consider some assumptions 

Assumptions 1:

Let us consider an acid that is monobasic and dissociates itself into H⁺ and A^-. 

H⁺ stands for hydrated hydronium ion. This equation comes into existence only if the difference between the polybasic acid and consecutive p^H value should be at least 3.

Example: phosphoric acid. 

Assumption 2:

Let us consider that the self ionisation of water can be ignored. \But this particular assumption is not valid for pH values which are near pk( water)/2. The pk of water is often 14 at 25°C.

So, the p^H will be found by solving the two mass balance equations that occur simultaneously for the two unknown values for [H⁺] and [A^-].

Assumption 3:

Let us suppose that MA is a salt. This particular salt is completely dissociated from the solution. 

Assumption 4:

The quotient of the activity coefficients is a constant which are defined under the experimental conditions which are often covered by the calculation

Applications of the Henderson Hasselbalch Equation

The major applications of the following equation are:

It helps in the estimation of the p^H value of a solution that contains acid and a base related to it, the solutions are generally called a buffer solution. 

The general formula for finding the above equation is

           p^H = p^Ka + log10 ( [A-  ] / [HA]) 

p^H = acidity of a buffer solution

p^Ka = negative logarithm of Ka

Ka = acid dissociation constant

[A-  ] = Concentration of the conjugate base.

[HA] = concentration of an acid

More About the Henderson Hasselbalch Equation 

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. 

(Image will be added soon)

The image depicts the application of Henderson-Hasselbalch equation.

The Formula for 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:

\[p^H=p^{K_a}+log10(\frac{[A^-]}{[HA]}\]

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.

Derivation of the Henderson Hasselbalch Equation

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 = pKa 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.

Solved Problems

Question 1) Estimate the p^H 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 p^H 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 p^H of the solution as 5.48.