In the continuous titration of of with , distinct intermediate buffer regions and equivalence points — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

The titration of triprotic $H_3PO_4$ proceeds in three steps. Step 1 buffer is $H_3PO_4 / H_2PO_4^-$. At Equivalence 1, $NaH_2PO_4$ forms, and its amphiprotic anion $H_2PO_4^-$ determines pH via $\frac{pK_{a1}+pK_{a2}}{2}$. During the second step, the buffer is $H_2PO_4^- / HPO_4^{2-}$, so at half-neutralization of the second proton, $pH = pK_{a2}$, not $pK_{a3}$. Step 3 creates an $HPO_4^{2-} / PO_4^{3-}$ buffer.