# Finding pKa from equivalence point on titration curve [closed]

Is there a mathematical proof/explanation of why $\mathrm{p}K_\mathrm{a}$ corresponds to the $\mathrm{pH}$ at $\text{Volume of titrant}/2$ at the equivalence point?

A concise proof will suffice.

## closed as off-topic by MaxW, Gaurang Tandon, Mithoron, aventurin, TyberiusJun 11 '18 at 3:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

You've got a weak acid, since you're contemplating a positive pKa, which means when you're halfway to the end point you're in the buffer region and you can use the Henderson-Hasselbalch equation:

pH = pKa + log [A-]/[HA]

You've titrated half your initial HA, so half of it is still around and half got turned into A-, which means [A-] = [HA]. For example, if you started with 0.1 mol/L of HA, you now have 0.05 mol/L HA and 0.05 mol/L of A-.

That means [A-]/[HA] = 1, and the log of 1 is zero. The HH equation then tells you pH = pKa.