Before You Begin

Question #44077

Difficulty: 8/10 | 🟢 Free

The Question

Anilinium hydrogen sulfate, \ce{[C6H5NH3+][HSO4-]}, is dissolved in water at $363 \text{ K}$ to form a $0.10 \text{ M}$ solution. At this temperature, the ionic product of water is $K_w = 1.0 \times 10^{-12}$ . The base dissociation constant of aniline is $K_b = 1.0 \times 10^{-10}$ and the acid dissociation constant of the hydrogen sulfate ion is $K_a = 1.0 \times 10^{-2}$ . Assuming ideal behavior and neglecting the autoionization of water, what is the exact hydrogen ion concentration $[\ce{H+}]$ of the solution?

Options

A.
4.0 \times 10^{-2} \text{ M}
B.
1.0 \times 10^{-2} \text{ M}
C.
4.47 \times 10^{-2} \text{ M}
D.
2.7 \times 10^{-2} \text{ M}

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

Ionic Product of Water (Kw)|Salt Hydrolysis Logic|Amphiprotic Ion pH|Aniline Resonance Pitfalls

Why This Question Is Tricky

Common Mistakes

This problem tests three critical exceptions and traps in ionic equilibrium.

First, the 'Amphiprotic Trap': Students blindly apply $[\ce{H+}] = \sqrt{K_{a1}K_{a2}}$ (Option B) for complex salts, failing to recognize that both ions are proton donors here.

Second, the 'Temperature-Dependent Kw Trap': Failing to use $K_w = 10^{-12}$ at $363 \text{ K}$ leads to an incorrect anilinium $K_a$ of $10^{-4}$ (Option D).

Third, the 'Weak Acid Approximation Trap': Using the simplified formula $[\ce{H+}] = \sqrt{K_{a1}C + K_{a2}C}$ (Option C) assumes negligible dissociation ( $\alpha \ll 1$ ).

Here, $\alpha = 40\%$ , necessitating the exact quadratic solution derived from charge and mass balances.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Equilibrium - Class 11 Part 1
Concept	Ionic Product of Water (Kw)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free