For a tribasic acid, , , and . Match the pH (Column II) of the resulting solution (Column I), at . — Ionic Equilibrium Chemistry Question
Question
For a tribasic acid, $\text{H}_3\text{A}$, $K_{a1} = 10^{-4}$, $K_{a2} = 10^{-8}$ and $K_{a3} = 10^{-12}$. Match the pH (Column II) of the resulting solution (Column I), at $25^\circ\text{C}$.
💡 Solution & Explanation
(A) Equimolar $\text{H}_3\text{A}$ and $\text{NaH}_2\text{A}$ form a buffer governed by $K_{a1}$. $\text{pH} = pK_{a1} = 4.0$. Matches Right C. (B) Equimolar $\text{NaH}_2\text{A}$ and $\text{Na}_2\text{HA}$ form a buffer governed by $K_{a2}$. $\text{pH} = pK_{a2} = 8.0$. Matches Right B. (C) Equimolar $\text{Na}_2\text{HA}$ and $\text{Na}_3\text{A}$ form a buffer governed by $K_{a3}$. $\text{pH} = pK_{a3} = 12.0$. Matches Right A. (D) Equimolar $\text{H}_3\text{A}$ and NaOH react completely to form the amphiprotic salt $\text{NaH}_2\text{A}$. $\text{pH} = (pK_{a1} + pK_{a2})/2 = (4.0 + 8.0)/2 = 6.0$. Matches Right D. (E) Equimolar $\text{NaH}_2\text{A}$ and NaOH react completely to form the amphiprotic salt $\text{Na}_2\text{HA}$. $\text{pH} = (pK_{a2} + pK_{a3})/2 = (8.0 + 12.0)/2 = 10.0$. Matches Right E. Therefore, correct answer is 1-C; 2-B; 3-A; 4-D; 5-E.