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Question #44083

Difficulty: 8/10 | 🟢 Free

The Question

Aniline ( $pK_a$ of $\ce{PhNH3+} = 4.6$ ) reacts with benzaldehyde to form a Schiff base. The reaction is carried out in an aqueous solution of $10^{-8}\text{ M}$ $\ce{HCl}$ at $298\text{ K}$ . Calculate the true pH of the reaction medium and the ratio of unprotonated to protonated aniline ( $\frac{[\ce{PhNH2}]}{[\ce{PhNH3+}]}$ ) present at equilibrium.

(Given: $K_w = 10^{-14}$ , $\sqrt{401} \approx 20.02$ , $\log_{10} 1.05 \approx 0.02$ , $10^{0.38} \approx 2.4$ , $10^{0.4} \approx 2.5$ )

Options

A.
$6.98$ and $2.4 \times 10^2$
B.
$8.00$ and $2.5 \times 10^3$
C.
$6.98$ and $4.2 \times 10^{-3}$
D.
$8.00$ and $4.0 \times 10^{-4}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

pH of Extremely Dilute Solutions|Schiff Base pH Dependence

Why This Question Is Tricky

Common Mistakes

The question tests the 'Hidden Variable' of water's auto-ionization in extremely dilute acidic solutions, which prevents the pH from crossing into the basic regime (pH 8).

Students often blindly apply the $-\log[\ce{H+}]$ formula, yielding an impossible pH of 8 for an acid, which cascades into an incorrect calculation of the nucleophile's protonation state.

Furthermore, it tests the correct application of the Henderson-Hasselbalch equation to determine the active fraction of the amine nucleophile required for Schiff base formation, trapping students who invert the acid/base ratio.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Equilibrium - Class 11 Part 1
Concept	pH of Extremely Dilute Solutions
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free