# How do you derive the Henderson-Hasselbalch equation?

Nov 17, 2015

See explanation.

#### Explanation:

Consider the general dissociation reaction of an acid in water:

HA(aq) rightleftharpoons H^+(aq)+underbrace(A^(-)(aq))_(color(green)("conjugate base")

The dissociation equilibrium constant ${K}_{a}$ can be written as follows:

${K}_{a} = \frac{\left[{H}^{+}\right] \left[{A}^{-}\right]}{\text{[HA]}}$

if we take the $- \log$ of both sides:

$- \log \left({K}_{a}\right) = - \log \text{([H^+][A^-])/"[HA]}$

$\implies {\underbrace{- \log \left({K}_{a}\right)}}_{\textcolor{b l u e}{p {K}_{a}}} = {\underbrace{- \log \left[{H}^{+}\right]}}_{\textcolor{b l u e}{p H}} - \log \text{ ("[A"^(-)"]")/"[HA]}$

$\implies p {K}_{a} = p H - \log \text{ ("[A"^(-)"]")/"[HA]}$

$\implies \textcolor{p u r p \le}{p H = p {K}_{a} + \log \text{ ("[A"^(-)"]")/"[HA]}}$