Question

For a tribasic acid, $H_3A$, $K_{a1}=10^{-4}$, $K_{a2}=10^{-8}$ and $K_{a3}=10^{-12}$. Match the pH (Column II) of the resulting solution (Column I) at 250C.

• A. Equimolar mixture of $H_3A$ and $NaH_2A$.
• B. Equimolar mixture of $NaH_2A$ and $Na_2HA$.
• C. Equimolar mixture of $Na_2HA$ and $Na_3A$.
• D. Equimolar mixture of $H_3A$ and NaOH.
• E. 12.0
• F. 8.0
• G. 4.0
• H. 6.0
• I. 10.0

Answer 1

A → R; B → Q; C → P; D → S

Answer 2

The problem requires matching the pH of solutions formed from a tribasic acid $H_3A$ and its salts, given $K_{a1}=10^{-4}$, $K_{a2}=10^{-8}$, and $K_{a3}=10^{-12}$. This implies $pK_{a1} = 4.0$, $pK_{a2} = 8.0$, and $pK_{a3} = 12.0$.

• (A) Equimolar mixture of $H_3A$ and $NaH_2A$: This is a buffer solution of $H_3A/H_2A^-$. Using the Henderson-Hasselbalch equation, $pH = pK_{a1} + \log \frac{[H_2A^-]}{[H_3A]}$. Since the concentrations are equimolar, $pH = pK_{a1} = 4.0$. Thus, A → R.

• (B) Equimolar mixture of $NaH_2A$ and $Na_2HA$: This is a buffer solution of $H_2A^-/HA^{2-}$. The Henderson-Hasselbalch equation gives $pH = pK_{a2} + \log \frac{[HA^{2-}]}{[H_2A^-]}$. With equimolar concentrations, $pH = pK_{a2} = 8.0$. Thus, B → Q.

• (C) Equimolar mixture of $Na_2HA$ and $Na_3A$: This is a buffer solution of $HA^{2-}/A^{3-}$. The Henderson-Hasselbalch equation gives $pH = pK_{a3} + \log \frac{[A^{3-}]}{[HA^{2-}]}$. With equimolar concentrations, $pH = pK_{a3} = 12.0$. Thus, C → P.

• (D) Equimolar mixture of $H_3A$ and $NaOH$: The reaction $H_3A + NaOH \rightarrow NaH_2A + H_2O$ goes to completion, forming $NaH_2A$. The $H_2A^-$ species is amphoteric. The pH is approximately the average of $pK_{a1}$ and $pK_{a2}$: $pH \approx \frac{pK_{a1} + pK_{a2}}{2} = \frac{4.0 + 8.0}{2} = 6.0$. Thus, D → S.

Therefore, the correct matching is A → R; B → Q; C → P; D → S.