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Question #44111

Difficulty: 8/10 | 🟢 Free

The Question

Ammonolysis of 2-bromoethanoic acid with excess $\ce{NH3(aq)}$ at 298 K yields an $\alpha$ -amino acid, $\textbf{A}$ . In aqueous solution, $\textbf{A}$ exhibits macroscopic acid dissociation constants $pK_{a1} = 2.30$ and $pK_{a2} = 9.70$ . To probe the internal acid-base micro-environment, the $pK_a$ of the ethyl ester hydrochloride of $\textbf{A}$ is measured to be $7.30$ . Assuming esterification does not significantly alter the intrinsic basicity of the adjacent amino group, calculate the equilibrium constant ( $K_Z$ ) for the internal proton transfer (zwitterion formation): $\ce{NH2-CH2-COOH(aq)} \rightleftharpoons \ce{^{+}NH3-CH2-COO^{-}(aq)}$ at 298 K.

Options

A.
$10^5$
B.
$10^{7.4}$
C.
$10^{2.4}$
D.
$10^{-5}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Polyprotic Acid Dissociation|Acidity and Basicity (pKa/pKb)|Ammonolysis Complexity

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Chemical Exception' of microscopic equilibria in polyprotic zwitterionic systems.

Students typically rely on simple Henderson-Hasselbalch approximations, treating pKa1 and pKa2 as strictly successive, independent dissociations.

They overlook that the macroscopic K1 is actually a composite of two microscopic pathways (forming either the zwitterion or the uncharged neutral species).

By providing the pKa of the esterified analog, the problem introduces a 'Hidden Variable' that isolates the microscopic dissociation constant of the amine (KB).

Failing to recognize the relationship KZ = KA / KB leads students to erroneously subtract the macroscopic pKa values (yielding 10^7.4) or misapply the ester's pKa.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Equilibrium - Class 11 Part 1
Concept	Polyprotic Acid Dissociation
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free