Henderson Hasselbalch Equation

The Henderson-Hasselbalch equation is used in situations where both undissociated acids and their conjugate bases are present in a solution. This situation can occur by adding the conjugate base to a solution of a weak acid or by partially neutralizing the weak acid with a strong base. Mixtures of an undissociated weak acid and its conjugate base, or an undissociated weak base and its conjugate acid, are known as buffers.

Derivation Of Henderson–Hasselbalch Equation

Let us examine the equilibrium calculations of the known amount of HA (Ca) and the known amount of its conjugate base as a salt (NaA) (Cb) in a solution. Again, five species are present in the solution.

The Assumptions

  1. Equilibrium:
    $$ (1)\;\;\;K_a=\frac{[H^+][A^-]}{[HA]} $$
    $$ (2)\;\;\;K_w=[H^+][OH^-]=10^{-14} $$
  2. Material balance: The total amount of the weak acid and its conjugate base at equilibrium should be the amount of the weak acid and its conjugate base initially present as:
    $$ (3)\;\;\;C_a+C_b=[HA]+[A^-] $$
    From the complete dissociation of NaA, one can obtain:
    $$ (4)\;\;\;C_b=[Na^+] $$
  3. Electroneutrality: There should be no net charge among the four ionic species as follows:
    $$ (5)\;\;\;[Na^+]+[H^+]=[OH^-]+[A^-] $$

The Henderson–Hasselbalch Equations for a Weak Acid

Substituting Equation (4) into Equation (5) and solving the resulting equation for [A−] followed by substituting it into Equation (3) yields:

$$ (6)\;\;\;[A^-]=C_b+[H^+]-[OH^-] $$

$$ (7)\;\;\;[HA]=C_a-([H^+]-[OH^-]) $$

Substituting Equation (6) and Equation (7) into Equation (1) yields:

$$ (8)\;\;\;K_a=\frac{[H^+](C_b+[H^+]-[OH^-])}{C_a-([H^+]-[OH^-])} $$

Equation (8) is cubic and so is not easy to solve. Let us simplify it under certain assumptions.

If we assume [H+] is much greater than [OH−], Equation (8) becomes:

$$ (9)\;\;\;K_a=\frac{[H^+](C_b+[H^+])}{C_a-[H^+]} $$

Assuming Ca>>[H+] and Cb>>[H+] Equation (9) further reduces to:
$$ (10)\;\;\;[H^+]=K_a\frac{C_a}{C_b} $$

Equation (10) is identical to Equation (1) if one considers that Ca and Cb are the equilibrium concentrations of HA and A−, respectively. Taking the logarithm of Equation (10) for the mixture of a weak acid and its conjugate base gives:
$$ (11)\;\;\;pH=pK_a+log(\frac{[salt]}{[acid]}) $$

For the ionization of a weak acid by a strong base, Equation (11) can be rewritten as:

$$ (12)\;\;\;pH=pK_a+log(\frac{α}{1-α}) $$

where α is the degree of ionization. Equation (10), Equation (11), and Equation (12) are the Henderson-Hasselbalch equations for a weak acid.

The Henderson–Hasselbalch Equations for a Weak Base

For a mixture of a weak base and its conjugate acid, the equivalent equation to Equation (8) is given by:

$$ (13)\;\;\;K_b=\frac{[OH^-](C_a+[OH^-]-[H^+])}{C_b-([OH^-]-[H^+])} $$

Equation (13) can be simplified as follows:

  1. Assumption (1. [OH−]>>[H+]):
    $$ K_b=\frac{[OH^-](C_a+[OH^-])}{C_b-[OH^-]} $$
  2. Assumption (2. Ca>>[OH–] and Cb>>[OH–]):
    $$ [OH^-]=K_b\frac{C_b}{C_a} $$

The Henderson–Hasselbalch equation for a weak base can be written as:

$$ (14)\;\;\;pH=pK_a+log(\frac{[base]}{[salt]}) $$

or.

$$ (15)\;\;\;pH=pK_a+log(\frac{1-α}{α}) $$

Reference:

  • Kim, C. (2004). Advanced pharmaceutics : physicochemical principles. London: CRC Press LLC.

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