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:
Select the correct option:
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.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
More ionic equilibrium Practice Questions
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.
Looking for more practice? Explore all chemistry questions or browse equilibrium questions on RankGuru.