How does one go about determining the equivalents of $\ce{OH-}$ needed for the titration of a weak acid. For example here is the titration curve for Aspartate:

The titration curve for Aspartate

Why is the equivalents for $\mathrm{p}K_\mathrm{a1} = 0.5$? Why not $0.75$ or $0.80$? How was the answer 0.5 derived. How were the equivalents for $\mathrm{p}K_\mathrm{a2} = 1.5$? derived? I understand the $\mathrm{pI}$ is 1 because there's a balance of charges but why the 0.5, 1.5, and 2.5

  • 2
    $\begingroup$ It is trivial, as by definition of Ka and the derived pKa, pKa is pH where the ratio of respective concentrations ( exactly rather activities ) is 1:1, therefore the half of the equivalence. $\endgroup$
    – Poutnik
    Jun 11, 2021 at 6:33
  • 1
    $\begingroup$ The answer can also be: Because log(1)=0 $\endgroup$
    – Poutnik
    Jun 22, 2021 at 11:31

1 Answer 1


Let's say you are titrating a weak acid $\ce{H3A}$ with a strong base. $$\ce{H3A + OH- -> H2A- + H2O}$$ $$\ce{H2A- + OH- -> HA^{2-} + H2O}$$ $$\ce{HA^{2-} + OH- -> A^{3-} + H2O}$$ When all $\ce{H3A}$ reacts with $\ce{OH-}$ to form $\ce{H2A-}$, the first equivalence point is reached.

Before the first equivalence point is reached, the $\ce{H3A}$ and $\ce{H2A-}$ buffer exists, whose pH is given by the Henderson-Hasselbalch equation. $$\mathrm{ pH = pK_{a1} + log\frac{\ce[H_2A^-]}{\ce{[H3A]}}}$$ When $\mathrm{pH = pK_{a1}}$, $\ce{[H2A-]}$=$\ce{[H3A]}$ which means the first half-equivalence point is reached, which requires $\pu{0.5 equivalents}$ of $\ce{OH-}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.