# How does pH relate to pKa in a titration?

Aug 7, 2015

The $p H$ at half-equivalence when a weak acid is titrated with a strong base is precisely the $p {K}_{a}$ of the acid.

#### Explanation:

Consider the dissociation of a weak acid, $H A$:
$H A r i g h t \le f t h a r p \infty n s {H}^{+} + {A}^{-}$. As we know,

${K}_{a} = \frac{\left[{H}^{+}\right] \left[{A}^{-}\right]}{\left[H A\right]} ,$

$\mathmr{and} {\log}_{10} \left\{{K}_{a}\right\} = {\log}_{10} \left[{H}^{+}\right] + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$;

Equivalently (multiplying each by $- 1$,
$- {\log}_{10} \left\{{K}_{a}\right\} = - {\log}_{10} \left[{H}^{+}\right] - {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$.

But, by definition, $- {\log}_{10} \left\{{K}_{a}\right\} = p {K}_{a}$, and $- {\log}_{10} \left[{H}^{+}\right] = p H$.

Therefore, $p H$ = $p {K}_{a}$ $+$ ${\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$. This is a form of the buffer equation, with which you are going to get very familiar.

Now at half-equivalence, by definition, $\left[H A\right] = \left[{A}^{-}\right]$, and since ${\log}_{10} 1$ = 0, when plugged back into the equation, $p H = p {K}_{a}$. So, in order to measure $p {K}_{a}$ values of weak acids we plot a titration curve with a $p H$ meter, and note value at half-equivalence.