Q. In qualitative analysis, the metals of group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag⁺ and Pb²⁺ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl⁻ concentration is 0.10 M. What will the concentrations of Ag⁺ and Pb²⁺ be at equilibrium? (Ksp for AgCl = 1.8 × 10⁻¹⁰, Ksp for PbCl₂ = 1.7 × 10⁻⁵)
A) [Ag⁺] = 1.8 × 10⁻⁹ M, [Pb²⁺] = 1.7 × 10⁻³ M
B) [Ag⁺] = 1.8 × 10⁻¹¹ M, [Pb²⁺] = 1.7 × 10⁻⁴ M
C) [Ag⁺] = 1.8 × 10⁻⁷ M, [Pb²⁺] = 1.7 × 10⁻⁶ M
D) [Ag⁺] = 1.8 × 10⁻¹¹ M, [Pb²⁺] = 8.5 × 10⁻⁵ M
Explanation
[Solution from source] At [Cl⁻] = 0.10 M: [Ag⁺] = Ksp/[Cl⁻] = 1.8 × 10⁻¹⁰/0.1 = 1.8 × 10⁻⁹ M and [Pb²⁺] = Ksp/[Cl⁻]² = 1.7 × 10⁻⁵/0.01 = 1.7 × 10⁻³ M. Choices that use [Cl⁻] instead of [Cl⁻]² for PbCl₂ give 1.7 × 10⁻⁴ M, which is wrong. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.