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

Calculating the pH of a buffer solution

$\text{pH}=-{\log }_{10}\left(K_a\frac{\left[\text{acid}\right]}{\left[\text{salt}\right]}\right)$

Derivation

For any equilibrium

$a\text{A}+b\text{B}\rightleftharpoons c\text{C}+d\text{D}$

the equilibrium constant, K, is defined as

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

Therefore, for the dissociation equilibrium of any acid

$\text{HA}\text{(aq)}\rightleftharpoons {\text{H}}^{+}\text{(aq)}+{\text{A}}^{-}\text{(aq)}$

the acid dissociation constant, K_a, is defined as

$K_a=\frac{[{\text{H}}^{+}\text{(aq)}][{\text{A}}^{-}\text{(aq)}]}{[\text{HA}\text{(aq)}]}$

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

$[{\text{H}}^{+}\text{(aq)}]=K_a\frac{[\text{HA}\text{(aq)}]}{[{\text{A}}^{-}\text{(aq)}]}$

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

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

2 Every HA molecule remains undissociated

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

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

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

$[{\text{H}}^{+}\text{(aq)}]=K_a\frac{[\text{HA}\text{(aq)}]}{[{\text{A}}^{-}\text{(aq)}]}$

becomes

$[{\text{H}}^{+}\text{(aq)}]=K_a\frac{[\text{HA}\text{(aq)}]}{[\text{salt}]}$

The effect of assumption 2 is that

$[{\text{H}}^{+}\text{(aq)}]=K_a\frac{[\text{HA}\text{(aq)}]}{[\text{salt}]}$

becomes

$[{\text{H}}^{+}\text{(aq)}]=K_a\frac{[\text{acid}]}{[\text{salt}]}$

By definition,

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

so

$\text{pH}=-{\log }_{10}\left(K_a\frac{\left[\text{acid}\right]}{\left[\text{salt}\right]}\right)$