# Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is an equation used in chemistry and biochemistry to determine the pH value of a buffer solution during the dissociation of a weak acid. A weak acid does not dissociate completely. A property of the buffer solution is that the mixture can resist changes in pH when a small amount of acid is added ^{[1-5]}.

## How to Use Henderson Hasselbalch Equation

The Henderson-Hasselbalch equation is used to determine the amount of acid and conjugate base needed to make a buffer solution of a certain pH. Therefore, it is often used to perform the calculations necessary to prepare buffers for use in the laboratory or other applications.

**General Formula of the Henderson-Hasselbalch Equation**

A simple buffer solution consists of a weak acid solution and a salt of the acid’s conjugate base. The equation is given by,

pH = pK_{a} + log_{10}[conjugate base/weak acid]

where,

pH: pH value

K_{a}: acid dissociation constant

[conjugate base]: Concentration of the conjugate base in moles per liter

[weak acid]: Concentration of the acid in moles per liter

American chemist Lawrence Joseph Henderson first derived the equation in 1908 and, later, by Danish chemist Karl Albert Hasselbalch in 1917.

## Derivation of Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation can be derived from the equilibrium constant expression for the dissociation of a weak acid. Consider the dissociation of the acid HA in water H_{2}O ^{[2-6]}.

HA + H_{2}O ⇄ A^{–} + H_{3}O^{+}

The acid dissociation constant is given by,

K_{a} = [A^{–}] [H_{3}O^{+}]/[HA]

or, [H_{3}O^{+}] = K_{a}[HA]/ [A^{–}]

or, log_{10} [H_{3}O^{+}] = log_{10} (K_{a}[HA]/ [A^{–}])

or, log_{10} [H_{3}O^{+}] = log_{10} K_{a} + log_{10 }([HA]/[A^{–}])

or, – log_{10} [H_{3}O^{+}] = – log_{10} K_{a} – log_{10 }([HA]/[A^{–}])

or, pH = pK_{a} + log_{10 }([A^{–}]/[HA])

From the above equation, the following points can be noted.

**1.** The value of [A^{–}]/[HA] is dependent on the values of the pH and pK_{a}.

- When pH < pK
_{a}, [A^{–}]/[HA] < 1. - When pH > pK
_{a}, [A^{–}]/[HA] > 1.

**2.** When the pH changes relatively to pK_{a }by one unit, then pH – pK_{a} = 1. Therefore,

1 = log_{10 }([A^{–}]/[HA])

or, [A^{–}]/[HA] = 10

Thus, the ratio of the dissociated form to the associated form changes by a factor of 10.

**3.** When the acid is 50% dissociated, the acid and the conjugate base concentrations are the same.

[HA] = [A^{–}]

or, [HA]/[A^{–}] = 1

Therefore,

pH = pK_{a}

The pH value of the solution is equal to the pK_{a} value of the acid.

## Henderson-Hasselbalch Equation for Base

The Henderson-Hasselbalch equation also applies to a weak base. A weak base dissociates as follows,

B + H_{2}O ⇄ OH^{–} + HB^{+}

The Henderson-Hasselbalch equation can be derived using the same principles as that for acid and gives,

pOH = pK_{b} + log_{10 }([HB^{+}]/[B])

where,

pOH: pOH value

K_{b}: base dissociation constant

[HB^{+}]: Concentration of the conjugate acid in moles per liter

[B]: Concentration of the base in moles per liter

## Limitations of Henderson-Hasselbalch Equation

- Fails to predict the pH values of strong acids because it assumes that the acid and its conjugate base concentrations remain constant at equilibrium.
- Can only be applied to a polybasic acid if its consecutive pK values differ by at least 3.
- Fails to give an accurate pH value for extremely buffer solutions.
- Does not consider the self-dissociation by water.

## Examples and Problems

**P.1.** The dissociation of acetic acid (CH_{3}CO_{2}H) in acetate ion (CH_{3}CO_{2}^{−}) is given as follows:

CH_{3}CO_{2}H + H_{2}O ⇄ CH_{3}CO_{2}^{−} + H_{3}O^{+}

Calculate the pH of a buffer solution made from 0.25 M of CH_{3}CO_{2}H and 0.45 M of CH_{3}CO_{2}^{−} that has an acid dissociation constant for CH_{3}CO_{2}H of 1.76 x 10^{-5} at 25 ͦC.

**Soln.** Given,

[CH_{3}CO_{2}H] = 0.25 M

[CH_{3}CO_{2}^{−}] = 0.45 M

K_{a} = 1.76 x 10^{-5}

The Henderson-Hasselbalch equation is given by,

pH = pK_{a} + log_{10 }([A^{–}]/[HA])

Therefore,

pH = -log_{10} (1.76 x 10^{-5}) + log_{10 }(0.45 M /0.25 M) = 5

**P.2.** How much sodium formate (HCOONa), 68.0069 g/mol) do you need to add to 500 mL of 0.50 M formic acid (HCOONa) for a pH 3 buffer. K_{a} = 1.77 x 10¯^{4}.

**Soln.** Given,

[HCOONa] = 0.5 ML^{-1}

pH = 3

K_{a} = 1.77 x 10¯^{4}

The Henderson-Hasselbalch equation is given by,

pH = pK_{a} + log_{10 }([A^{–}]/[HA])

Therefore,

3 = – log_{10} (-1.77 x 10¯^{4}) + log_{10 }([A^{–}]/0.5)

or, 3 = 3.752 + log_{10 }([A^{–}]/0.5)

or, log_{10 }([A^{–}]/0.5) = -0.752

or, [A^{–}]/0.5 = 0.177

or, [A^{–}] = 0.177 x 0.5 = 0.0885 ML^{-1}

The relationship between the concentration and the molecular weight is,

Concentration x volume = weight/molecular weight

0.0885 ML^{-1 }x 0.5 L = w/ 68.0069 gM^{-1}

Or, w = 0.0885 ML^{-1 }x 0.5 L x 68.0069 gM^{-1} = 3 g

Very good formula in chemistry and biochemistry.