Look at the following equation:
$$\mathrm{pH} = \mathrm pK_{\mathrm a} + \log\frac{[\ce{A^-}]}{[\ce{HA}]}$$
At the half equivalence point, say we have 10 moles of $\ce{WA},$ and so there will be 5 moles of $\ce{SB}$ as the name suggests ("half equivalence"). So we know that the 10 moles will be neutralized to 5 moles of $\ce{WA}$ by the $\ce{SB}.$ If we do this then the $\ce{CB}$ is also now equal to the $\ce{WA}$ since it will increase by 5 moles.
$$\ce{[WA]~ =~ [CB]}.$$
Thus,
\begin{align}\mathrm{pH} &= \mathrm pK_{\mathrm a} + \log 1\\ &=\mathrm pK_{\mathrm a}\;.\end{align}
So this is why the the $\mathrm{pH} = \mathrm pK_{\mathrm a}$ at the Half-Equivalence Point.
Vocabulary:
- SB : Strong Base. (eg. $\ce{NaOH}$)
- CB : Conjugate Base. (eg. $\ce{Ac^{-}}$)
- WA : Weak Acid. (eg. $\ce{HAc}$)
NOTE: We can never be 100% accurate here, this is just used to derive that $\mathrm{pH} = \mathrm pK_{\mathrm a}$