The Henderson-Hasselbalch equation is a mathematical expression used to find out the pH of buffer solutions. This equation relates pH, dissociation of a weak acid, and the ratio of the concentrations of its salt of the conjugate base and the concentration of acid.

The equation is as follows:

where,

- pH is the negative log of H
^{+}ions concentration - Pk
_{a}is the negative log of the acid dissociation constant

In 1908, Lawrence Joseph Henderson derived this equation. Later, this equation became famous as the Henderson-Hasselbalch equation. It shows how two factors affect the pH of the buffer solution. The first factor is the pK_{a} of the acid and base used, while the second factor is the ratio of the concentration of salt and acid or a base.

This equation is used to estimate the pH of buffer solutions which are important in many biological as well as industrial applications.

Outline

## Derivation of Henderson-Hasselbalch equation for acidic buffer

Consider a weak acid HA and its salt NaA with a strong base. The dissociation reaction of HA are as follows:

HA ⇌ H^{+} + A^{_}

The dissociation constant of a weak acid HA is as follows:

K_{a} =[ H^{+}][A^{–}] ⁄ [HA]

On rearranging the equation;

[H^{+}] = K_{a} [HA] / [A^{_}]

By taking a log of the above equation,

log [H^{+}] = log K_{a} [HA] / [A^{_}]

Or, log [H^{+}] = log K_{a} + log [HA] / [A^{_}]

Multiply the negative sign on both sides,

– log [H^{+}] = – log K_{a} – log [HA] / [A^{_}]

Since

– log [H^{+}] = pH

– log K_{a}= pK_{a}

So,

= pK_{a} – log [HA] / [A^{_}]

Where,

- [A
^{–}] = concentration of salt - [HA] = concentration of the weak acid

pH=pK_{a} – log [acid] / [salt]

Further, on reversing the numerator and denominator the sign of the log changes at the above equation becomes;

pH = pK_{a} + log [salt] / [acid]

### How is the buffer solution prepared using the Henderson-Hasselbalch equation?

The Henderson-Hasselbalch equation tells us how the pH of the buffer solution changes with the pK_{a} of a weak acid. The best buffer solution is prepared when we take an equal amount of salt and acid. The pH is controlled by the pK_{a} of the acid.

For example, for the buffer of acetic acid and sodium acetate, the pH will be:

[CH_{3}COOH] = [CH_{3}COONa]

pH = pK_{a} + log [CH_{3}COONa] / [CH_{3}COOH]

pH = pK_{a} + log (1)

pH = pK_{a} + 0

pH = pK_{a}

pK_{a} of the acid is 4.74. So the pH of this buffer is equal to the pK_{a} of the acid. Hence, the pH of the buffer is 4.74.

### Solved Example

**If [CH _{3}COOH] is equal to the 1.0 mole dm^{-3} and [CH_{3}COONa] is 0.1 mole dm^{-3} then find the pH of the buffer solution.**

**Solution:**

pH = 4.74 + log 0.1/1

= 4.74 + log 1/10

= 4.74 + log 10^{-1}

‘= 4.74 – 1

pH = 3.74

## Henderson-Hasselbalch equation for basic buffers

The basic buffer has also the Henderson equation. However, this equation is used to find the pH of the basic buffer solution. Let us take the weak base and its salt of strong acid. The dissociation of a weak base is as follows:

NH_{3} + H_{2}O ⇌ NH_{4}^{+} + OH^{–}

K_{b} = [ NH_{4}^{+}][OH^{–}] ⁄ [NH_{3}]

Similarly, by taking the log, multiplying it with a negative sign, and rearranging we get,

pOH = pK_{b} + log [salt] / [base]

where,

- K
_{b}is the base dissociation constant

**Limitations**

The Henderson equation fails to explain the accurate values of the strong acids and strong bases. In addition, it can explain the pH values for very dilute solutions.

## Applications

The following applications of the Henderson-Hasselbalch equation:

- It helps in the calculation of pH if the ratio of salt to acid is known.
- To find the change in pH, when the strong base is added to the solution of a weak acid.
- It is used in the determination of the pK
_{a}of a weak acid. - It is extensively applied in the pharmaceutical industry and drug synthesis.

**Related resources**

## Concepts Berg

**What is the Henderson-Hasselback equation? Give example.**

Henderson hasselbalch equation is the mathematical derivation that relates pH with the acid dissociation constants and the concentrations of the conjugate base. For example, if a buffer solution of pH 1 is needed, by using this equation we can determine the amount of required acid and conjugate base.

**What is the importance of the Henderson-Hasselbalch equation?**

It is of great importance in the pharmaceuticals and chemical industry for the preparation of buffer solutions.

**What is meant by buffer solution?**

A buffer solution is a complex solution based on the principle of the common ion effect. When this solution is added to the medium, it helps to retain the pH upon the addition of a small amount of acid or base.

**What is a basic buffer solution?**

A basic buffer solution has a pH from 7 to 14 in the basic region. For example, ammonia buffer has a pH of 10. It is added to maintain the specific pH to some extent.

**What is the Henderson Hasselbalch constant?**

In this equation pKa is the -log of the acid dissociation constant.

**What do you mean by buffer capacity?**

The buffer capacity is the tendency of the buffer solution to which it can resist the pH.

**What is the relationship between pKa and pH? **

Both depend upon the ratio of concentrations of conjugate base and acid.

**Why is the Henderson Hasselbalch equation only used for weak acids?**

It cannot be applied to strong acids because the acid dissociation constant ka for strong acids is very high.

**How to find the ratio using the Henderson Hasselbalch equation?**

The ratio can found out by inserting the values of the acid dissociation constant and required pH. Then rearrange the equation to make a ratio subject. Finally, the anti-log is taken on both sides to calculate the ratio of salt and acid to be used.

**References**

- Essential Medical Physiology by Leonard R. Johnson
- Experiment 9 (ulm.edu)