Consider the step-wise neutralization mechanism of the polyprotic acid . A student mixes of with exa — Ionic Equilibrium Chemistry Question
Question
Consider the step-wise neutralization mechanism of the polyprotic acid $H_3PO_4$. A student mixes $50 \text{ mL}$ of $0.2 \text{ M } H_3PO_4$ with exactly $50 \text{ mL}$ of $0.2 \text{ M } Na_3PO_4$. Assuming the reaction proceeds to form the most stable intermediate conjugate pairs, calculate the exact pH of the final mixture. (Given $pK_{a1} = 2.12, pK_{a2} = 7.2, pK_{a3} = 12.0$).
💡 Solution & Explanation
Initial millimoles: $H_3PO_4 = 10 \text{ mmol}$, $Na_3PO_4 = 10 \text{ mmol}$. The neutralization mechanism strictly favors the comproportionation reaction between the extreme acidic and basic species: $H_3PO_4 + PO_4^{3-} \rightarrow H_2PO_4^- + HPO_4^{2-}$. Since we have $10 \text{ mmol}$ of each, they react completely to form exactly $10 \text{ mmol}$ of $H_2PO_4^-$ and $10 \text{ mmol}$ of $HPO_4^{2-}$. These two products form a perfect equimolar conjugate acid-base buffer pair ($H_2PO_4^- / HPO_4^{2-}$). The relevant equilibrium is the second dissociation step, governed by $pK_{a2}$. Since $[Salt] = [Acid]$, the $\log(1) = 0$, and thus the final pH perfectly equals $pK_{a2} = 7.2$.