Derive the Henderson-Hasselbalch equation.

(N/A) The Henderson-Hasselbalch equation relates the $pH$ of a buffer solution to the $pK_{a}$ of the weak acid and the ratio of the concentrations of the conjugate base and the acid.
Consider the ionization of a weak acid $HA$ in water:
$HA + H_{2}O ⇌ H_{3}O^{+} + A^{-}$
The acid dissociation constant $K_{a}$ is given by:
$K_{a} = \frac{[H_{3}O^{+}][A^{-}]}{[HA]}$
Rearranging for $[H_{3}O^{+}]$:
$[H_{3}O^{+}] = K_{a} \times \frac{[HA]}{[A^{-}]}$
Taking the negative logarithm $(-\log)$ on both sides:
$-\log [H_{3}O^{+}] = -\log K_{a} - \log \frac{[HA]}{[A^{-}]}$
Since $pH = -\log [H_{3}O^{+}]$ and $pK_{a} = -\log K_{a}$,we get:
$pH = pK_{a} - \log \frac{[HA]}{[A^{-}]}$
By inverting the fraction inside the logarithm,the sign changes:
$pH = pK_{a} + \log \frac{[A^{-}]}{[HA]}$
This is the Henderson-Hasselbalch equation,where $[A^{-}]$ is the concentration of the conjugate base and $[HA]$ is the concentration of the acid.