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:

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

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

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

Two assumptions are required:

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

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

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) Template:Unicode 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.

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

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

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

becomes

$K_a=\frac{[{\text{H}}^{+}\text{(aq)}{]}^2}{[\text{HA}\text{(aq)}]}$

The effect of assumption 2 is that

$K_a=\frac{[{\text{H}}^{+}\text{(aq)}{]}^2}{[\text{HA}\text{(aq)}]}$

becomes

$K_a=\frac{[{\text{H}}^{+}\text{(aq)}{]}^2}{[\text{acid}]}$

which can be rearranged to give

$[{\text{H}}^{+}\text{(aq)}{]}^2=K_a\left[\text{acid}\right]$

and therefore

$[{\text{H}}^{+}\text{(aq)}]=\sqrt{K_a\left[\text{acid}\right]}$

By definition,

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

so

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