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. 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:

Henderson-Hasselbalch Equation

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:

\(HA~ +~ H_{2}O~ ⇋ ~H^{+}~ +~ A^{−}\)

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 \(-log~K_a\) = \(pKa\),

The equation above can also 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,

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

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.

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.

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