Derivation

For any equilibrium

${\displaystyle a{\mbox{A}}+b{\mbox{B}}\rightleftharpoons c{\mbox{C}}+d{\mbox{D}}\,\!}$

the equilibrium constant, K, is defined as

${\displaystyle K={\frac {[{\mbox{C}}]^{c}[{\mbox{D}}]^{d}}{[{\mbox{A}}]^{a}[{\mbox{B}}]^{b}}}}$

Therefore, for the dissociation equilibrium of any acid

${\displaystyle {\mbox{HA}}{\mbox{(aq)}}\rightleftharpoons {\mbox{H}}^{+}{\mbox{(aq)}}+{\mbox{A}}^{-}{\mbox{(aq)}}\,\!}$

the acid dissociation constant, K_a, is defined as

${\displaystyle K_{a}={\frac {[{\mbox{H}}^{+}{\mbox{(aq)}}][{\mbox{A}}^{-}{\mbox{(aq)}}]}{[{\mbox{HA}}{\mbox{(aq)}}]}}}$

This equation can be rearranged to make [H⁺(aq)] the subject:

${\displaystyle [{\mbox{H}}^{+}{\mbox{(aq)}}]=K_{a}{\frac {[{\mbox{HA}}{\mbox{(aq)}}]}{[{\mbox{A}}^{-}{\mbox{(aq)}}]}}}$

Two assumptions are required:

1 Every A⁻ ion comes from the salt

Although this is not quite true, it is a close enough that the pH value we get from the final equation is very close to that found experimentally. It allows us to assume that

${\displaystyle \left[{\mbox{HA}}{\mbox{(aq)}}\right]=\left[{\mbox{acid}}\right]}$

2 Every HA molecule remains undissociated

Again, despite being slightly inaccurate, this assumption creates the following useful equation

${\displaystyle \left[{\mbox{A}}^{-}{\mbox{(aq)}}\right]=\left[{\mbox{salt}}\right]}$

The equations in assumptions 1 and 2 allow us to replace [A⁻(aq)] with [salt] and [HA(aq)] with [acid] as follows.

The effect of assumption 1 is that

${\displaystyle [{\mbox{H}}^{+}{\mbox{(aq)}}]=K_{a}{\frac {[{\mbox{HA}}{\mbox{(aq)}}]}{[{\mbox{A}}^{-}{\mbox{(aq)}}]}}}$

becomes

${\displaystyle [{\mbox{H}}^{+}{\mbox{(aq)}}]=K_{a}{\frac {[{\mbox{HA}}{\mbox{(aq)}}]}{[{\mbox{salt}}]}}}$

The effect of assumption 2 is that

${\displaystyle [{\mbox{H}}^{+}{\mbox{(aq)}}]=K_{a}{\frac {[{\mbox{HA}}{\mbox{(aq)}}]}{[{\mbox{salt}}]}}}$

becomes

${\displaystyle [{\mbox{H}}^{+}{\mbox{(aq)}}]=K_{a}{\frac {[{\mbox{acid}}]}{[{\mbox{salt}}]}}}$

By definition,

${\displaystyle {\mbox{pH}}=-\log _{10}{\left(\left[{\mbox{H}}^{+}{\mbox{(aq)}}\right]\right)}}$

so

${\displaystyle {\mbox{pH}}=-\log _{10}{\left(K_{a}{\frac {\left[{\mbox{acid}}\right]}{\left[{\mbox{salt}}\right]}}\right)}}$