# A-level Chemistry/OCR (Salters)/Weak acids

## Calculating the pH of a weak acid solution

The pH of a weak acid solution can be calculated approximately using the following formula:

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

For any equilibrium

$a{\mbox{A}}+b{\mbox{B}}\rightleftharpoons c{\mbox{C}}+d{\mbox{D}}\,\!$ the equilibrium constant, K, is defined as

$K={\frac {[{\mbox{C}}]^{c}[{\mbox{D}}]^{d}}{[{\mbox{A}}]^{a}[{\mbox{B}}]^{b}}}$ Therefore, for the dissociation equilibrium of any acid

${\mbox{HA}}{\mbox{(aq)}}\rightleftharpoons {\mbox{H}}^{+}{\mbox{(aq)}}+{\mbox{A}}^{-}{\mbox{(aq)}}\,\!$ the acid dissociation constant, Ka, is defined as

$K_{a}={\frac {[{\mbox{H}}^{+}{\mbox{(aq)}}][{\mbox{A}}^{-}{\mbox{(aq)}}]}{[{\mbox{HA}}{\mbox{(aq)}}]}}$ Two assumptions are required:

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

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

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

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

$K_{a}={\frac {[{\mbox{H}}^{+}{\mbox{(aq)}}]^{2}}{[{\mbox{HA}}{\mbox{(aq)}}]}}$ The effect of assumption 2 is that

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

$K_{a}={\frac {[{\mbox{H}}^{+}{\mbox{(aq)}}]^{2}}{[{\mbox{acid}}]}}$ which can be rearranged to give

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

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

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

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