Ionic Equilibrium
The solubility product (Ksp) of AgCl at 25°C is 1.8 × 10⁻¹⁰. The maximum concentration of Ag⁺ ions that can exist in a 0.01 M NaCl solution is:
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Solution
1.8 × 10⁻⁸ mol L⁻¹
For AgCl, Ksp = [Ag⁺][Cl⁻] = 1.8 × 10⁻¹⁰. In a 0.01 M NaCl solution, [Cl⁻] ≈ 0.01 M (since NaCl dissociates completely and the contribution from AgCl dissolution is negligible). Therefore, [Ag⁺] = Ksp/[Cl⁻] = 1.8 × 10⁻¹⁰ / 0.01 = 1.8 × 10⁻⁸ mol L⁻¹. This demonstrates the common ion effect: the presence of Cl⁻ from NaCl suppresses the dissolution of AgCl, drastically reducing the Ag⁺ concentration compared to dissolution in pure water where [Ag⁺] would be √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol L⁻¹.
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About This Question
- Subject
- chemistry
- Chapter
- equilibrium
- Topic
- ionic equilibrium
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
1.8 × 10⁻⁸ mol L⁻¹
For AgCl, Ksp = [Ag⁺][Cl⁻] = 1.8 × 10⁻¹⁰. In a 0.01 M NaCl solution, [Cl⁻] ≈ 0.01 M (since NaCl dissociates completely and the contribution from AgCl dissolution is negligible). Therefore, [Ag⁺] = Ksp/[Cl⁻] = 1.8 × 10⁻¹⁰ / 0.01 = 1.8 × 10⁻⁸ mol L⁻¹. This demonstrates the common ion effect: the presence of Cl⁻ from NaCl suppresses the dissolution of AgCl, drastically reducing the Ag⁺ concentration compared to dissolution in pure water where [Ag⁺] would be √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol L⁻¹.
This medium difficulty chemistry question is from the chapter equilibrium, covering the topic of ionic equilibrium. It appeared in the 2025 exam.
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