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

Calculating the pH of a buffer solutionEdit

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

DerivationEdit

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 ) }

Last modified on 17 June 2007, at 10:48