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Question #45013

Difficulty: 8/10 | 🟢 Free

The Question

A $100 \text{ mL}$ aqueous solution of $\ce{NaBH4}$ of unknown molarity $x$ is divided into two equal $50 \text{ mL}$ aliquots, A and B.

Aliquot A is adjusted to $\text{pH} = 12$ . $50 \text{ mL}$ of $0.1 \text{ M}$ $\ce{KIO3}$ is added, and the mixture is heated to ensure complete reaction. The solution is then cooled, acidified, and treated with excess $\ce{KI}$ . The liberated $\ce{I2}$ requires $100 \text{ mL}$ of $0.1 \text{ M}$ $\ce{Na2S2O3}$ for complete titration.

Aliquot B is adjusted to $\text{pH} = 2$ and allowed to stand for 30 minutes in an open flask. Then, $50 \text{ mL}$ of the same $0.1 \text{ M}$ $\ce{KIO3}$ and excess $\ce{KI}$ are added. The liberated $\ce{I2}$ requires $V \text{ mL}$ of $0.1 \text{ M}$ $\ce{Na2S2O3}$ for complete titration.

Assuming no evaporation losses, what are the values of $x$ and $V$ ?

Options

A.
$x = 0.05 \text{ M}, V = 300 \text{ mL}$
B.
$x = 0.10 \text{ M}, V = 300 \text{ mL}$
C.
$x = 0.05 \text{ M}, V = 100 \text{ mL}$
D.
$x = 0.10 \text{ M}, V = 100 \text{ mL}$

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

Equivalent Concept in Redox|Half-Reaction Media Sensitivity|Reduction Selectivity (Hydrides)

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Half-Reaction Media Sensitivity' of hydrides.

Students often memorize n-factors as static properties and fail to account for kinetic competition.

In basic media, $\ce{NaBH4}$ is stable and acts as a powerful 8-electron reducing agent.

However, in acidic media, it undergoes rapid hydrolysis to yield $\ce{H2}$ gas.

Because the flask is open, the $\ce{H2}$ escapes, dropping the effective reducing equivalents in the solution to zero.

A secondary trap is the n-factor of $\ce{NaBH4}$ : students frequently miscalculate it as 4 by assuming the hydrides merely form $\ce{H2}$ gas during the redox reaction with $\ce{IO3-}$ , rather than being fully oxidized to +1.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Redox Reactions - Class 11 Part 2
Concept	Half-Reaction Media Sensitivity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free