Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is widely used in the fields of biochemistry and chemistry in order to estimate the pH value of a given buffer solution when the concentration of the acid and its conjugate base (or the base and the corresponding conjugate acid) are known.

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.

Henderson-Hasselbalch Equation Derivation

The ionization constants of strong acids and strong bases can be easily calculated with the help of the direct methods. However, when it comes to weak bases and weak acids, the same methods cannot be used since the extent of ionization of these acids and bases are very low, i.e. weak acids and bases hardly ionize.

Thus, 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:

\(HA~ +~ H_{2}O~ ⇋ ~H^{+}~ +~ A^{−}\)

Acid dissociation constant, K a can be given as:

K a = \(\frac{[H^+][A^-]}{[HA]}\)

Taking, negative log of RHS and LHS:

\(-log~K_a\) = \(-log~\frac{[H^+][A^-]}{[HA]}\)

\( \Rightarrow -log~K_a\) = \(-log~[H^+] ~ – ~log~\frac{[A^-]}{[HA]}\)

As we know, \(-log~[H^+]\) = \(pH\) and \(-log~K_a\) = \(pKa\),

Above equation can be written as,

\(pK_a\) = \(pH~ -~ log~\frac{[A^-]}{[HA]}\)

Rearranging the equation,

\(\Rightarrow pH\) = \(pK_a ~+ ~log\frac{[A^-]}{[HA]}\)

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 \(pH\) = \(pK_a\)

\(log~\frac{[A^-]}{[HA]}\) = \(0\)

\([A^-]\) = \([HA]\)

That is, when \(pH\) = \(pK_a\), concentration of both the species are same or in other words, acid will be half dissociated.

Similarly, for a weak base B:

\(B~ +~ H_2O~ ⇋ ~ OH^− ~+~ HB^+\)

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

\( K_b\) = \(\frac{[BH^+][OH^-]}{[B]} \)

Taking negative log of RHS and LHS

\(-logK_b\) = \(-log \frac{[BH^+][OH^-]}{[B]}\)

\(\Rightarrow -logK_b\) = \(-log~[OH^-]-log\frac{[BH^+]}{[B]}\)

As we know, \(-log~[OH^-]\) = \(pOH\) and \(-logK_b\) = \(pKb\),

Above equation can be written as,

\(pK_b\) = \(pOH ~- ~log\frac{[BH^+]}{[B]}\)

Rearranging the equation,

\(\Rightarrow pOH\) = \(pK_b + log~\frac{[BH^+]}{[B]}\)

Limitations of Henderson-Hasselbalch equation:

The Henderson-Hasselbalch equation makes correct predictions for weak acids and bases but fails to predict accurate values for the strong acids and strong bases as it assumes that the concentration of the acid and its conjugate base at chemical equilibrium will remain the same as the formal concentration.

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Practise This Question

1. Match the followings – Column IColumn IIA. Ranitidine(Zantac)(i)   Enzyme       B. Inhibitors(ii)   HistamineC. Aspirin(iii)   ReceptorD. Antagonist(iv)   Prostaglandins     Which of the followings is correct?