Derive the equation for the calculation of $pH$ of an acidic buffer solution.
(N/A) An acidic buffer solution consists of a weak acid and its salt with a strong base,such as a mixture of acetic acid $(CH_3COOH)$ and sodium acetate $(CH_3COONa)$.
The equilibrium of ionization of a weak acid $HA$ in water is represented as:
$HA + H_2O \rightleftharpoons H_3O^+ + A^-$
The acid dissociation constant $K_a$ is given by:
$K_a = \frac{[H_3O^+][A^-]}{[HA]}$
Rearranging for $[H_3O^+]$:
$[H_3O^+] = K_a \times \frac{[HA]}{[A^-]}$
Taking the negative logarithm on both sides:
$-\log[H_3O^+] = -\log K_a - \log \frac{[HA]}{[A^-]}$
Since $pH = -\log[H_3O^+]$ and $pK_a = -\log K_a$,we get:
$pH = pK_a - \log \frac{[HA]}{[A^-]}$
By inverting the log term:
$pH = pK_a + \log \frac{[A^-]}{[HA]}$
This is the Henderson-Hasselbalch equation,where $[A^-]$ is the concentration of the conjugate base and $[HA]$ is the concentration of the acid.
The equilibrium of ionization of a weak acid $HA$ in water is represented as:
$HA + H_2O \rightleftharpoons H_3O^+ + A^-$
The acid dissociation constant $K_a$ is given by:
$K_a = \frac{[H_3O^+][A^-]}{[HA]}$
Rearranging for $[H_3O^+]$:
$[H_3O^+] = K_a \times \frac{[HA]}{[A^-]}$
Taking the negative logarithm on both sides:
$-\log[H_3O^+] = -\log K_a - \log \frac{[HA]}{[A^-]}$
Since $pH = -\log[H_3O^+]$ and $pK_a = -\log K_a$,we get:
$pH = pK_a - \log \frac{[HA]}{[A^-]}$
By inverting the log term:
$pH = pK_a + \log \frac{[A^-]}{[HA]}$
This is the Henderson-Hasselbalch equation,where $[A^-]$ is the concentration of the conjugate base and $[HA]$ is the concentration of the acid.
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