# How does pKa relate to acidity?

Jun 30, 2016

Consider the general reaction of an acid in water:

$H A + {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + {A}^{-}$

#### Explanation:

Now it is well known that for a given temperature, the extent of and acid base reaction is a constant:

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \left[{A}^{-}\right]}{\left[H A\right]}$

Clearly, for strong acids, ${K}_{a}$ is large, and ${K}_{a}$ is smaller for weaker acids.

Now this is a mathematical expression, which we are free to manipulate, provided that we do it to BOTH sides of the equation. Now we can take ${\log}_{10}$ of both sides of the equation:

${\log}_{10} {K}_{a}$ $=$ ${\log}_{10} \left(\left[{H}_{3} {O}^{+}\right]\right) + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$

On rearrangement:

$- {\log}_{10} \left\{\left[{H}_{3} {O}^{+}\right]\right\} = - {\log}_{10} {K}_{a} + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$

But by definition $- {\log}_{10} \left\{\left[{H}_{3} {O}^{+}\right]\right\} = p H$ and $- {\log}_{10} {K}_{a} = p {K}_{a}$.

And so $p H = p {K}_{a} + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$

And thus, when we tritrate a weak acid with a base, and plot $p H$ versus volume of titrant, at the point of half equivalence,

$p H = p {K}_{a}$ because ${\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\} = {\log}_{10} \left\{1\right\} = 0 ,$ i.e at half equivalence, by definition $\left[H A\right] = \left[{A}^{-}\right]$.