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

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

Two assumptions are required:

1 The concentrations of H⁺(aq) and A⁻(aq) are equal, or in symbols:

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

The reason this is an approximation is that a very slightly higher concentration of H⁺(aq) exists in reality, due to the autodissociation of water, H₂O(l) ⇌ H⁺(aq) + A⁻(aq). We neglect this effect since water produces a far lower concentration of H⁺(aq) than most weak acids. If you were studying an exceptionally weak acid (you won't at A-level), this assumption might begin to cause big problems.

2 The amount of HA at equilibrium is the same as the amount originally added to the solution.

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

This cannot be quite true, otherwise HA wouldn't be an acid. It is, however, a close numerical approximation to experimental observations of the concentration of HA in most cases.

The effect of assumption 1 is that

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

becomes

${\displaystyle K_{a}={\frac {[{\mbox{H}}^{+}{\mbox{(aq)}}]^{2}}{[{\mbox{HA}}{\mbox{(aq)}}]}}}$

The effect of assumption 2 is that

${\displaystyle K_{a}={\frac {[{\mbox{H}}^{+}{\mbox{(aq)}}]^{2}}{[{\mbox{HA}}{\mbox{(aq)}}]}}}$

becomes

${\displaystyle K_{a}={\frac {[{\mbox{H}}^{+}{\mbox{(aq)}}]^{2}}{[{\mbox{acid}}]}}}$

which can be rearranged to give

${\displaystyle [{\mbox{H}}^{+}{\mbox{(aq)}}]^{2}=K_{a}\left[{\mbox{acid}}\right]}$

and therefore

${\displaystyle [{\mbox{H}}^{+}{\mbox{(aq)}}]={\sqrt {K_{a}\left[{\mbox{acid}}\right]}}}$

By definition,

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

so

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