# 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

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