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Question #44059

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The Question

A $1.0\text{ L}$ aqueous solution at $298\text{ K}$ contains $0.10\text{ mol}$ of a salt $BHCl$ , where $B$ is a weak monoacidic base with $K_b = 10^{-9}$ . To this solution, $0.01\text{ mol}$ of $\ce{KMnO4}$ and $0.025\text{ mol}$ of $\ce{H2O2}$ are added. The ensuing redox reaction proceeds to completion, reducing $\ce{MnO4-}$ to $\ce{Mn^{2+}}$ and oxidizing $\ce{H2O2}$ to $\ce{O2(g)}$ . Assuming no change in volume and ignoring the hydrolysis of $\ce{Mn^{2+}}$ , what is the final pH of the solution? (Given: $\log 3 = 0.48$ , $\log 7 = 0.85$ , $K_w = 10^{-14}$ )

Options

A.
4.63
B.
5.60
C.
3.00
D.
9.37

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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

Reaction Quotient vs. Equilibrium Constant|Salt Hydrolysis Logic|KMnO₄ Redox pH Dependency

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Hidden Variable' of dynamic equilibrium coupling.

Students often memorize the salt hydrolysis formula ( $pH = \frac{1}{2}(pK_w - pK_b - \log c)$ ) and treat it as a static state, yielding a pH of 3.00.

When a redox reaction is introduced, they may erroneously assume it is limited by the free $[H^+]$ .

Furthermore, a major stoichiometric trap exists: students frequently use only the permanganate half-reaction (consuming $8H^+$ per $MnO_4^-$ ) and forget that the oxidation of $H_2O_2$ releases $H^+$ , partially offsetting the consumption.

Falling for this trap leads to $0.08$ mol of $H^+$ consumed, resulting in the distractor pH of 5.60.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Equilibrium - Class 11 Part 1
Concept	Reaction Quotient vs. Equilibrium Constant
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free