JEE Main 2021 · 20 July · evening · Single correct · +4 / −1
A solution is 0.1 M in Cl⁻ and 0.001 M in CrO₄²⁻. Solid AgNO₃ is gradually added to it. Assuming that the addition does not change in volume and Kₛₚ(AgCl) = 1.7 × 10⁻¹⁰ M² and Kₛₚ(Ag₂CrO₄) = 1.9 × 10⁻¹² M³.
Select correct statement from the following :
- (a) AgCl precipitates first because its Kₛₚ is high.
- (b) Ag₂CrO₄ precipitates first as its Kₛₚ is low.
- (c) Ag₂CrO₄ precipitates first because the amount of Ag⁺ needed is low.
- (d) AgCl will precipitate first as the amount of Ag⁺ needed to precipitate is low.
Answer and solution
Answer (d) AgCl will precipitate first as the amount of Ag⁺ needed to precipitate is low.
Solution
Chemistry ma’am · 4 steps
Do not compare the Kₛₚ values; find how much Ag⁺ each salt needs to start precipitating.
- When precipitation starts. A salt starts to precipitate once its ionic product reaches Kₛₚ. As AgNO₃ is added, [Ag⁺] rises slowly, so the salt that needs the smaller [Ag⁺] comes out first.
- Silver ion for AgCl. [Ag⁺][Cl⁻] = Kₛₚ, so [Ag⁺] = (1.7 × 10⁻¹⁰)/0.1 = 1.7 × 10⁻⁹ M.
- Silver ion for Ag2CrO4. Here [Ag⁺]²[CrO₄²⁻] = Kₛₚ, so the silver ion concentration comes from a square root.
[Ag⁺] = √((1.9 × 10⁻¹²)/0.001) = √(1.9 × 10⁻⁹) = 4.36 × 10⁻⁵ M
- So. AgCl needs only 1.7 × 10⁻⁹ M of Ag⁺, far less than the 4.36 × 10⁻⁵ M that Ag₂CrO₄ needs. AgCl precipitates first because it needs less Ag⁺, which is option (d).
The idea to keep. In selective precipitation, work out the [Ag⁺] (or other common ion) each salt needs from its own Kₛₚ and concentration. The smallest requirement precipitates first.
Answer from the NTA final answer key. Solution by PrepWiser.