Question

How do you derive the Henderson-Hasselbalch equation?

Answer 1

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=([H^+][A^-])/"[HA]"`

if we take the `-log` of both sides:

`-log(K_a)=-log""([H^+][A^-])/"[HA]"`

`=> underbrace(-log(K_a))_(color(blue)(pK_a))=underbrace(-log[H^+])_(color(blue)(pH))-log"" ("[A"^(-)"]")/"[HA]"`

`=> pK_a = pH -log"" ("[A"^(-)"]")/"[HA]"`

`=> color(purple)(pH = pK_a +log"" ("[A"^(-)"]")/"[HA]")`