Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

# Henderson-Hasselbalch Equation

## What is the Henderson-Hasselbalch Equation?

The Henderson-Hasselbalch equation provides a relationship between the pH of acids (in aqueous solutions) and their pKa (acid dissociation constant). The pH of a buffer solution can be estimated with the help of this equation when the concentration of the acid and its conjugate base, or the base and the corresponding conjugate acid, are known.

## Equation of Henderson-Hasselbalch

The Henderson-Hasselbalch equation can be written as:

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

Where [A] denotes the molar concentration of the conjugate base (of the acid) and [HA] denotes the molar concentration of the weak acid. Therefore, the Henderson-Hasselbalch equation can also be written as:

An equation that could calculate the pH value of a given buffer solution was first derived by the American chemist Lawrence Joseph Henderson. This equation was then re-expressed in logarithmic terms by the Danish chemist Karl Albert Hasselbalch. The resulting equation was named the Henderson-Hasselbalch Equation.

## Derivation of the Henderson-Hasselbalch Equation

The ionization constants of strong acids and strong bases can be easily calculated with the help of direct methods. However, the same methods cannot be used with weak acids and bases since the extent of ionization of these acids and bases are very low (weak acids and bases hardly ionize). Therefore, in order to approximate the pH of these types of solutions, the Henderson-Hasselbalch Equation is used.

Let us take an example of ionization of weak acid HA:

$$\begin{array}{l}HA~ +~ H_{2}O~ ⇋ ~H^{+}~ +~ A^{−}\end{array}$$

Acid dissociation constant, K a can be given as:

K a =

$$\begin{array}{l}\frac{[H^+][A^-]}{[HA]}\end{array}$$

Taking, negative log of RHS and LHS:

$$\begin{array}{l}-log~K_a\end{array}$$
=
$$\begin{array}{l}-log~\frac{[H^+][A^-]}{[HA]}\end{array}$$

$$\begin{array}{l} \Rightarrow -log~K_a\end{array}$$
=
$$\begin{array}{l}-log~[H^+] ~ – ~log~\frac{[A^-]}{[HA]}\end{array}$$

As we know,

$$\begin{array}{l}-log~[H^+]\end{array}$$
=
$$\begin{array}{l}pH\end{array}$$
and
$$\begin{array}{l}-log~K_a\end{array}$$
=
$$\begin{array}{l}pKa\end{array}$$
,

The equation above can also be written as,

$$\begin{array}{l}pK_a\end{array}$$
=
$$\begin{array}{l}pH~ -~ log~\frac{[A^-]}{[HA]}\end{array}$$

Rearranging the equation,

$$\begin{array}{l}\Rightarrow pH\end{array}$$
=
$$\begin{array}{l}pK_a ~+ ~log\frac{[A^-]}{[HA]}\end{array}$$

The above equation is known as Henderson-Hasselbalch equation, popularly known as Henderson equation. It is very useful for estimating the pH of a buffer solution and finding the equilibrium pH in acid-base reactions. From the equation we can infer when

$$\begin{array}{l}pH\end{array}$$
=
$$\begin{array}{l}pK_a\end{array}$$

$$\begin{array}{l}log~\frac{[A^-]}{[HA]}\end{array}$$
=
$$\begin{array}{l}0\end{array}$$

$$\begin{array}{l}[A^-]\end{array}$$
=
$$\begin{array}{l}[HA]\end{array}$$

That is, when

$$\begin{array}{l}pH\end{array}$$
=
$$\begin{array}{l}pK_a\end{array}$$
, concentration of both the species are same or in other words, acid will be half dissociated.

Similarly, for a weak base B:

$$\begin{array}{l}B~ +~ H_2O~ ⇋ ~ OH^− ~+~ HB^+\end{array}$$

Base dissociation constant, Kb, of the base can be given as,

$$\begin{array}{l} K_b\end{array}$$
=
$$\begin{array}{l}\frac{[BH^+][OH^-]}{[B]} \end{array}$$

Taking negative log of RHS and LHS

$$\begin{array}{l}-logK_b\end{array}$$
=
$$\begin{array}{l}-log \frac{[BH^+][OH^-]}{[B]}\end{array}$$

$$\begin{array}{l}\Rightarrow -logK_b\end{array}$$
=
$$\begin{array}{l}-log~[OH^-]-log\frac{[BH^+]}{[B]}\end{array}$$

As we know,

$$\begin{array}{l}-log~[OH^-]\end{array}$$
=
$$\begin{array}{l}pOH\end{array}$$
and
$$\begin{array}{l}-logK_b\end{array}$$
=
$$\begin{array}{l}pKb\end{array}$$
,

Above equation can be written as,

$$\begin{array}{l}pK_b\end{array}$$
=
$$\begin{array}{l}pOH ~- ~log\frac{[BH^+]}{[B]}\end{array}$$

Rearranging the equation,

$$\begin{array}{l}\Rightarrow pOH\end{array}$$
=
$$\begin{array}{l}pK_b + log~\frac{[BH^+]}{[B]}\end{array}$$

## Important Points to Remember

• When exactly half of the acid undergoes dissociation, the value of [A]/[HA] becomes 1, implying that the pKa of the acid is equal to the pH of the solution at this point. (pH = pKa + log10(1) = pKa).
• For every unit change in the pH to pKa ratio, a tenfold change occurs in the ratio of the associated acid to the dissociated acid. For example, when the pKa of the acid is 7 and the pH of the solution is 6, the value of [A]/[HA] is 0.1 but when the pH of the solution becomes 5, the value of [A]/[HA] becomes 0.01.
• The value of [A]/[HA] is dependent on the value of the pH and pKa. When pH < pKa; [A]/[HA] < 1. When pH > pKa; [A]/[HA] > 1.

## Limitations of the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation fails to predict accurate values for the strong acids and strong bases because it assumes that the concentration of the acid and its conjugate base at chemical equilibrium will remain the same as the formal concentration (the binding of protons to the base is neglected).

Since the Henderson-Hasselbalch equation does not consider the self-dissociation undergone by water, it fails to offer accurate pH values for extremely dilute buffer solutions.

## Solved Example

A buffer solution is made from 0.4M CH3COOH and 0.6M CH3COO. If the acid dissociation constant of CH3COOH is 1.8*10-5, what is the pH of the buffer solution?

As per the Henderson-Hasselbalch equation, pH = pKa + log([CH3COO]/[CH3COOH])

Here, Ka = 1.8*10-5 ⇒ pKa= -log(1.8*10-5) = 4.7 (approx.).

Substituting the values, we get:

pH = 4.7 + log(0.6M /0.4M) = 4.7 + log(1.5) = 4.7 + 0.17 = 4.87

Therefore, the pH of the solution is 4.87.

## Frequently Asked Questions – FAQs

### Does Henderson Hasselbalch equation work for bases?

When it includes equilibrium concentrations of an acid and conjugate base, the Henderson-Hasselbalch equation is valid. Equilibrium concentrations can be far from those expected by neutralisation stoichiometry in the case of solutions containing not-so-weak acids (or not-so-weak bases).

### Is NaOH a strong base?

A solid base is one that is fully ionic, such as sodium hydroxide or potassium hydroxide. In solution, you should conceive of the compound as being 100% divided into metal ions and hydroxide ions. Each mole of sodium hydroxide dissolves in order to give the solution a mole of hydroxide ions.

### What is the difference between KA and pKa?

The acid dissociation constant is Ka. PKa is essentially this constant’s-log. Similarly, the base dissociation constant is Kb, while pKb is the constant’s -log. The constants for acid and base dissociation are usually expressed in terms of moles per litre (mol / L).

### Is bleach an acid or base?

Chlorine bleach is extremely simple. Actually, we make it in the following equilibrium by dissolving chlorine gas in a condensed solution of sodium hydroxide, which forms sodium hypochlorite and sodium chloride.

### Is pKa the same as pH?

The pH is a measure of the hydrogen ion concentration of an aqueous solution. It is similar to pKa (acid dissociation constant) and pH, but pKa is more precise in that it lets you determine what a molecule at a given pH would do. The Henderson-Hasselbalch equation explains the relationship between pH and pKa.

To learn more about the Henderson-Hasselbalch equation and the calculations involving buffer solutions, register with BYJU’S and download the mobile application on your smartphone.

Take up a quiz on Henderson-Hasselbalch Equation