How do you derive the Henderson-Hasselbalch equation?

1 Answer
Nov 17, 2015

Answer:

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]")#