# Equivalence point pH derivation of a tribasic acid

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

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 $$\ce{[HA^2-]}$$.

• 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 Apr 8, 2021 at 15:29
• Thanks for pointing out @Poutnik Apr 8, 2021 at 16:33
• Write down ion balance at the partial equivalence and consider conditions leading to approximation where concentrations of H3A and HA^2- are equal. Apr 8, 2021 at 16:46
• See this and this Qs/As Apr 9, 2021 at 7:37