Why does the concentration of the acid equals that of the base at the mid-point of titration? Consider this graph of the titration of a weak acid and a strong base:



The buffering region is strongest at the midpoint of titration because there is equal concentrations of acetic acid and its conjugate base.


  My question is why at the midpoint of titration does the concentration of the acid equal the concentration of its conjugate base? Also, why does the $\mathrm{p}K_{a}$ of the acid = the $\mathrm{pH}$ at this point?
At the midpoint of titration, the concentration of acid is just approximately equal to its conjugated base, as there must be considered the acid dissociation and ( in less extend ) salt hydrolysis.

If we take the definition(simplified using concentrations):

$$K_\mathrm{a}=\frac {\ce{[H+][A-]}}{\ce{[HA]}}$$

we can see that by definition:

$$\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log { \frac {\ce{[A-]}}{\ce{[HA]}}}$$

$\mathrm{p}K_\mathrm{a}$ is $\mathrm{pH}$ when both items of the conjugate pair have the same concentration.

$\mathrm{pH}$ near titration midpoint is approximately equal

$$\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log { \frac {n_\mathrm{base}}{
n_\mathrm{acid, init} - n_\mathrm{base}
}}$$

This function has minimal slope = maximal buffer capacity at $$n_\mathrm{acid, init} = 2 \cdot n_\mathrm{base}$$