An Introduction to 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 PH 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 pH, 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 pH 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 pH 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 pH 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
pH = pKa + log10 ( [A- ] / [HA])
pH = acidity of a buffer solution
pKa = 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.
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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 pH = 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 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.
FAQs on Henderson Hasselbalch Equation
1. What is the role of the Henderson Hasselbalch equation in acid-base balance?
The major role played by this equation in terms of acid-base balance is, it gives a qualitative description of acid-base physiology. These particular criteria help in the separation of the metabolic components from the respiratory components. It plays major criteria while demonstrating the relationship between blood pH with the component of the H2CO3 buffering system.
2. What is the application of the Henderson Hasselbalch equation in terms of pharmaceutical analysis?
This equation plays an important role in calculating the pH levels in a buffer solution. In terms of pharmaceutical analysis, this helps majorly to calculate the PKA value of a molecule with the help of a chemical moiety. It also acts very usefully while calculating the ratio of its ionised or unionised forms. Though, this includes some other processes involved in it.
3. What do you mean by buffer solution and what are its types?
A buffer solution is simply described as the solution which comprises a weak acid and its conjugate base. On the other hand, it can also be a solution consisting of a strong acid and its conjugate base or vice versa.
There are broadly two types of buffer solution. Those are acidic buffer solution and alkaline buffer solution.
Acidic buffer solutions have a pH of less than 7. Alkaline buffer solutions have a pH of more than 7.
4. What do you mean by the pH of a solution?
The measure of the total number of H+ ions in a particular solution is the pH of the solution. This particular term further helps in the determination of the acidity of the solution.
5. 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 H2O 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%.
6. Explain Why Buffer Solution Resists Change in pH?
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 + H2O <--> H+ + A-
H+ + A- <--> HA
