Consider the ionization of a weak acid HA which has some pK_{a}. It is often
convenient to
be able to relate the pH of a solution of a weak acid to the pK_{a} of the acid
and the extent
of ionization. The reaction would be

HA (reversible arrows) H^{+} + A^{-}

The
acid dissociation constant (K_{a}) for this reaction would be given by the equation

This equation can be rearranged to isolate the hydrogen ion concentration
on the left,
because, remember, we want an equation relating the pH of the solution to the
pK_{a} and
the extent of ionization of the weak acid. The rearranged form of the equation
is

By definition, log 1**/** [*H*^{+}] = *pH* , and
log 1**/***K*_{a} = *pK*_{a} , so that by taking the log of the equation
above, we get the equation

This is the well-known Henderson-Hasselbalch equation that is often used to perform the calculations required in preparation of buffers for use in the laboratory, or other applications. Notice several interesting facts about this equation.

First,
if the pH = pK_{a},
the log of the ratio of dissociate acid and associated acid will be zero, so
the
concentrations of the two species will be the same. In other words, when the
pH equals
the pK_{a}, the acid will be half dissociated.

Second, notice that
as the pH increases or
decreases by one unit relative to the pK_{a}, the ratio of the dissociate
form to the associated
form of the acid changes by factors of 10. That is, if the pH of a solution
is 6 and the pK_{a} is 7, the ratio of
[ A-]**/**[ HA]
will be 0.1, will if the pH were 5, the ratio would be 0.01 and if
the pH were 7, the ratio would be 1.

Also, note that if the pH is below the
pK_{a}, the ratio
will be < 1, while if the pH is above the pK_{a}, the ratio will be >1. In
short, there is a lot of
information in the Henderson-Hasselbalch equation. You would be wise to study
this
equation to understand its various ramifications.