pH at the second equivalence point is — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

At the second equivalence point, all the $\text{H}_3\text{PO}_4$ has been converted into the amphiprotic salt $\text{Na}_2\text{HPO}_4$. The pH of this salt solution is determined primarily by the corresponding $pK_a$ values for the adjacent dissociation steps, which are $pK_2$ and $pK_3$. Since $K_2 = 10^{-8}$ and $K_3 = 10^{-13}$, $pK_2 = 8$ and $pK_3 = 13$. The pH is given by $\frac{pK_2 + pK_3}{2} = \frac{8 + 13}{2} = 10.5$. Therefore, correct answer is D.