Why does pKa of a acid-base indicator equal to the pH when the equivalence point is reached?
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} &amp;= \mathrm pK_{\mathrm a} + \log 1\\  &amp;=\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 $\mathrm{pH} = \mathrm pK_{\mathrm a}$