A-level Chemistry/OCR (Salters)/Weak acids

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Calculating the pH of a weak acid solution[edit]

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[edit]

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 ) }