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Considering $\pu{100 mL} \ \pu{0.1 M} \ \ce{H3A}$ and $\pu{0.1 M} \ \ce{NaOH}$ titration curve:

Titration Curve of a tribasic acid

I understood the half equivalence points, but couldn't understand the reason why $\mathrm{pH} = \frac{1}{2}[\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2}]$ at equivalence points.

So I tried to derive this and for that I did following:

$$\ce{H3A -> H2A- + H+} \quad \mathrm{}K_\mathrm{a1}$$

$$\ce{H2A- -> HA^2- + H+} \quad \mathrm{}K_\mathrm{a2}$$

$$\therefore \ \ce{H3A -> HA^2- + 2H+} \quad K = \mathrm{}K_\mathrm{a1} \cdot \mathrm{}K_\mathrm{a2}$$

So, $\mathrm{}K_\mathrm{a1} \cdot \mathrm{}K_\mathrm{a2} = \frac{[\ce{HA^2-}][\ce{H+}]^2 }{[\ce{H3A}]}$.

This would give $\mathrm{pH} = \frac{1}{2}[\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2}]$ but I don't know how $\ce{[H3A]}$ gets cancelled with $\ce{[HA^2-]}$.

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    $\begingroup$ Try to search the site, it was discussed several times. For eventual writing and formatting of chemical or mathematical formulas or equations, see how to use MathJax $\endgroup$
    – Poutnik
    Apr 8, 2021 at 15:29
  • $\begingroup$ Thanks for pointing out @Poutnik $\endgroup$ Apr 8, 2021 at 16:33
  • $\begingroup$ Write down ion balance at the partial equivalence and consider conditions leading to approximation where concentrations of H3A and HA^2- are equal. $\endgroup$
    – Poutnik
    Apr 8, 2021 at 16:46
  • $\begingroup$ See this and this Qs/As $\endgroup$
    – Poutnik
    Apr 9, 2021 at 7:37

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