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 '21 at 6:33
  • 1
    $\begingroup$ The answer can also be: Because log(1)=0 $\endgroup$
    – Poutnik
    Jun 22 '21 at 11:31

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 $\ce{OH-}$.


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