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# 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 = pKa + log10[conjugate base/weak acid]

where,

pH: pH value

Ka: 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 H2O [2-6].

HA + H2O ⇄ A + H3O+

The acid dissociation constant is given by,

Ka = [A] [H3O+]/[HA]

or, [H3O+] = Ka[HA]/ [A]

or, log10 [H3O+] = log10 (Ka[HA]/ [A])

or, log10 [H3O+] = log10 Ka + log10 ([HA]/[A])

or, – log10 [H3O+] = – log10 Ka – log10 ([HA]/[A])

or, pH = pKa + log10 ([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 pKa.

• When pH < pKa, [A]/[HA] < 1.
• When pH > pKa, [A]/[HA] > 1.

2. When the pH changes relatively to pKa by one unit, then pH – pKa = 1. Therefore,

1 = log10 ([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 = pKa

The pH value of the solution is equal to the pKa 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 + H2O  ⇄  OH + HB+

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

pOH = pKb + log10 ([HB+]/[B])

where,

pOH: pOH value

Kb: 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 (CH3CO2H) in acetate ion (CH3CO2) is given as follows:

CH3CO2H + H2O  ⇄  CH3CO2 + H3O+

Calculate the pH of a buffer solution made from 0.25 M of CH3CO2H and 0.45 M of CH3CO2 that has an acid dissociation constant for CH3CO2H of 1.76 x 10-5 at 25  ͦC.

Soln. Given,

[CH3CO2H] = 0.25 M

[CH3CO2] = 0.45 M

Ka = 1.76 x 10-5

The Henderson-Hasselbalch equation is given by,

pH = pKa + log10 ([A]/[HA])

Therefore,

pH = -log10 (1.76 x 10-5) + log10 (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. Ka = 1.77 x 10¯4.

Soln. Given,

[HCOONa] = 0.5 ML-1

pH = 3

Ka = 1.77 x 10¯4

The Henderson-Hasselbalch equation is given by,

pH = pKa + log10 ([A]/[HA])

Therefore,

3 = – log10 (-1.77 x 10¯4) + log10 ([A]/0.5)

or, 3 = 3.752 + log10 ([A]/0.5)

or, log10 ([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

### One response to “Henderson-Hasselbalch Equation”

1. Garba lawan Karaye says:

Very good formula in chemistry and biochemistry.