Before You Begin

Question #44538

Difficulty: 8/10 | 🟢 Free

The Question

An aqueous solution contains $0.10\text{ M}$ of a weak monoacidic base $B$ ( $pK_b = 4.52$ ). The pH of this solution is adjusted by adding a negligible volume of concentrated \ce{H2SO4}. At what pH will this solution become isotonic with a $0.1375\text{ M}$ aqueous solution of glucose at $298\text{ K}$ ?

(Assume \ce{H2SO4} dissociates completely into \ce{2H+} and \ce{SO4^{2-}}, and the solutions behave ideally. Given: $\log 2 = 0.30$ , $\log 3 = 0.48$ )

Options

A.
9.00
B.
9.70
C.
9.96
D.
5.00

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

Isotonic, Hypertonic, Hypotonic|Basicity-Solvation Competition

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of Colligative Properties and Ionic Equilibrium.

Students often blindly apply the van't Hoff factor formula $i = 1 + \alpha$ for weak electrolytes.

However, when pH is adjusted using an external acid like \ce{H2SO4}, the counterion (\ce{SO4^{2-}}) contributes to the total particle count, while the \ce{H+} is consumed by the base.

The 'Hidden Variable' is the stoichiometry of the added acid: \ce{H2SO4} adds only 1 \ce{SO4^{2-}} for every 2 \ce{BH+} formed, making the total particle count $0.10 + y$ rather than $0.10(1+\alpha)$ .

Failing to account for the specific counterion contribution leads to incorrect protonation ratios and erroneous pH values (e.g., pH = 9.70).

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Solutions - Class 12
Concept	Osmotic Pressure vs Diffusion
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free