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Question #44085

Difficulty: 8/10 | 🟢 Free

The Question

What is the minimum initial concentration of $\ce{NH_3}$ required to completely dissolve $0.05\text{ moles}$ of $\ce{AgCl(s)}$ in $1.0\text{ L}$ of a $0.15\text{ M } \ce{NaCl}$ aqueous solution at $298\text{ K}$ ?

Given: $K_{sp}(\ce{AgCl}) = 1.0 \times 10^{-10}$ Formation constant ( $\beta_2$ ) for $\ce{[Ag(NH_3)_2]^+} = 2.5 \times 10^7$ Assume no change in volume upon the addition of solid $\ce{AgCl}$ and $\ce{NH_3}$ .

Options

A.
$2.1\text{ M}$
B.
$2.0\text{ M}$
C.
$1.1\text{ M}$
D.
$1.0\text{ M}$

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

Common Ion Effect on Solubility|Solubility Product (Ksp) vs. Ionic Product (Qsp)|Ligand Denticity Variations

Why This Question Is Tricky

Common Mistakes

This problem tests the interplay between heterogeneous equilibrium (Ksp) and homogeneous complexation (Kf) under the influence of a common ion.

Students typically fall into two traps: (1) Ignoring the common ion effect from the initial NaCl, which leads to an incorrect [Cl-] of 0.05 M and a free [NH3] of 1.0 M (Option C). (2) Forgetting the mass balance of the ligand, where they calculate the free equilibrium [NH3] as 2.0 M but fail to add the 0.1 M of NH3 consumed in forming the complex (Option B).

The correct approach requires synthesizing Le Chatelier's Principle, common ion stoichiometry, and ligand mass balance.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Equilibrium - Class 11 Part 1
Concept	Common Ion Effect on Solubility
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free