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Question #44099

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The Question

Excess $\ce{AgCl(s)}$ is added to $1.0 \text{ L}$ of an aqueous solution at $298 \text{ K}$ . The solution contains a total analytical concentration of $0.20 \text{ M}$ glycine ( $\ce{H2N-CH2-COOH}$ ) and is externally buffered to maintain a constant $\text{pH} = 9.6$ .

Given:

For glycine, $pK_{a1} = 2.3$ and $pK_{a2} = 9.6$ .
$K_{sp}(\ce{AgCl}) = 1.0 \times 10^{-10}$ .
The overall formation constant for the complex $[\ce{Ag(Gly)2}]^-$ is $\beta_2 = 1.0 \times 10^8$ .

Assuming the external buffer has sufficient capacity to maintain the pH exactly at $9.6$ , what is the exact molar solubility of $\ce{AgCl}$ in this solution?

Options

A.
$\frac{1}{100} \text{ M}$
B.
$\frac{1}{110} \text{ M}$
C.
$\frac{1}{120} \text{ M}$
D.
$\frac{1}{60} \text{ M}$

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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

Solubility Product (Ksp) vs. Ionic Product (Qsp)|Amino Acid Zwitterion pH Sensitivity

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of solubility equilibria, complexation, and buffer capacity.

The 'Hidden Variable' is the dynamic re-equilibration of the amino acid buffer.

Students typically fall into three traps: 1) Ignoring the pH and assuming all glycine is deprotonated (Option D). 2) Assuming the glycinate concentration remains constant at $0.10 \text{ M}$ despite complexation (Option A). 3) Subtracting the consumed ligand directly from the initial glycinate concentration ( $0.10 - 2S$ ) without accounting for the buffer's Le Chatelier response, where the zwitterion deprotonates to maintain the $1:1$ ratio dictated by the pH (Option C).

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Equilibrium - Class 11 Part 1
Concept	Solubility Product (Ksp) vs. Ionic Product (Qsp)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free