# A-level Chemistry/OCR (Salters)/Buffer solutions

## Calculating the pH of a buffer solution

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

$K_{a}={\frac {[{\mbox{H}}^{+}{\mbox{(aq)}}][{\mbox{A}}^{-}{\mbox{(aq)}}]}{[{\mbox{HA}}{\mbox{(aq)}}]}}$ This equation can be rearranged to make [H+(aq)] the subject:

$[{\mbox{H}}^{+}{\mbox{(aq)}}]=K_{a}{\frac {[{\mbox{HA}}{\mbox{(aq)}}]}{[{\mbox{A}}^{-}{\mbox{(aq)}}]}}$ 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
$\left[{\mbox{HA}}{\mbox{(aq)}}\right]=\left[{\mbox{acid}}\right]$ 2 Every HA molecule remains undissociated

Again, despite being slightly inaccurate, this assumption creates the following useful equation
$\left[{\mbox{A}}^{-}{\mbox{(aq)}}\right]=\left[{\mbox{salt}}\right]$ 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

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

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

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

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

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

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