# How to determine the equivalents of OH- in a titration curve

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:

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

• 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. – Poutnik Jun 11 at 6:33
• The answer can also be: Because log(1)=0 – Poutnik Jun 22 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 $$\ce{OH-}$$.