How do you derive the Henderson-Hasselbalch equation?

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Answer 1 · 2015-11-17T14:13:07

See 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:

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

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

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

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