# 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

$\mbox{pH} = -\log_{10}{ \left ( \sqrt{K_a \left [ \mbox{acid} \right ]} \right ) }$