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