-1
$\begingroup$

How to find the half equivalence point knowing the pH, molarity, titrant added at equivalence point?

This a fairly straightforward and simple question, however I have found many different answers to this question. I originally thought that the half equivalence point was obtained by taking half the pH at the equivalence point. However, I have encountered some sources saying that it is obtained by halving the volume of the titrant added at equivalence point.

$\endgroup$
  • $\begingroup$ Yeah it's not half the pH at equivalence point your other sources are correct $\endgroup$ – Avnish Kabaj Mar 7 '18 at 23:43
0
$\begingroup$

Half equivalence point is exactly what it sounds like. It is the point where the volume added is half of what it will be at the equivalence point.

There are 3 cases

  1. Strong Acid vs Strong Base: Here one can simply apply law of equivalence and find amount of $\ce{H+}$ in the solution. Similar method for Strong base vs Strong Acid.
  2. Weak Acid vs Strong Base: Here one can apply Henderson-Hasselbalch equation as a general result one gets pH = pKa.
  3. Weak Base vs Strong Acid: This case is similar to WA vs SB and one can easily use Henderson-Hasselbalch for bases again one gets pOH = pKb
$\endgroup$

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .