The question is missing the initial information required to answer but that information is provided below:
"A solution contains three anions with the following concentrations, 0.20 M CrO₄²⁻, 0.10 M CO₃²⁻ and 0.01M Cl⁻."
Ag₂CrO₄ Ksp = 1.2 x 10⁻¹²
Ag₂CO₃ Ksp = 8.1 x 10⁻¹²
AgCl Ksp = 1.8 x 10⁻¹⁰
With the following information we can calculate the minimum amount of Ag⁺ needed to precipitate each solid.
AgCl → Ag⁺ + Cl⁻
Ksp = [Ag⁺][Cl⁻] = 1.8 x 10⁻¹⁰
[Ag⁺] = 1.8 x 10⁻¹⁰ / 0.01 M
[Ag⁺] = 1.8 x 10⁻⁸ M
Ag₂CrO₄ → 2Ag⁺ + CrO₄²⁻
Ksp = [Ag⁺]²[CrO₄²⁻] = 1.2 x 10⁻¹²
[Ag⁺]² = 1.2 x 10⁻¹² / 0.20 M
[Ag⁺] = √(1.2 x 10⁻¹² / 0.20 M)
[Ag⁺] = 2.4 x 10⁻⁶ M
Ag₂CO₃ → 2Ag⁺ + CO₃²⁻
Ksp = [Ag⁺]²[CO₃²⁻] = 8.1 x 10⁻¹²
[Ag⁺]² = 8.1 x 10⁻¹² / 0.10 M
[Ag⁺] = √(8.1 x 10⁻¹² / 0.q0 M)
[Ag⁺] = 9.0 x 10⁻⁶ M
AgCl requires the minimum amount of [Ag⁺] to precipitate with a value of [Ag⁺] = 1.8 x 10⁻⁸ M
Therefore, AgCl will precipitate first.
