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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 be the concentrations of Ag⁺ and Pb²⁺ at equilibrium? (Ksp for AgCl = 1.8 × 10⁻¹⁰, Ksp for PbCl₂ = 1.7 × 10⁻⁵)

1. [Ag⁺] = 1.8 × 10⁻¹⁰ M; [Pb²⁺] = 1.7 × 10⁻¹ M
2. [Ag⁺] = 1.8 × 10⁻⁹ M; [Pb²⁺] = 8.5 × 10⁻⁵ M
3. [Ag⁺] = 1.8 × 10⁻¹¹ M; [Pb²⁺] = 1.7 × 10⁻⁴ M
4. [Ag⁺] = 1.8 × 10⁻¹⁴ M; [Pb²⁺] = 8.5 × 10⁻⁵ M

Correct answer: [Ag⁺] = 1.8 × 10⁻⁹ M; [Pb²⁺] = 8.5 × 10⁻⁵ M

Solution

Ag⁺ will precipitate as AgCl first due to its lower Ksp. At equilibrium, [Ag⁺] = Ksp(AgCl)/[Cl⁻] = (1.8 × 10⁻¹⁰)/(0.10) = 1.8 × 10⁻⁹ M. Pb²⁺ remains mostly in solution, and its concentration is determined by the solubility of PbCl₂: [Pb²⁺] = √(Ksp(PbCl₂)/[Cl⁻]²) = √(1.7 × 10⁻⁵/(0.10)²) = 8.5 × 10⁻⁵ M.

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