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Calculate the solubility of silver chromate, Ag2CrO4 (Ksp = 8.8 × 10^-12) in the presence of 0.0052 M K2CrO4.?
2 Answers
- Anonymous5 years ago
Don't u ever do your homework by yourself?
- ?Lv 75 years ago
Let s be the solubility of.
Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq)
Initial concentrations :
[Ag⁺]ₒ = 0 M, [CrO₄²⁻]ₒ = 0.0052 M
At equilibrium :
[Ag⁺] = 2s M, [CrO₄²⁻] = (0.0052 + s) M ≈ 0.0052 M, assuming that 0.0052 ≫ s.
Ksp = [Ag⁺]² [CrO₄²⁻]
8.8 × 10⁻¹² = (2s)² × 0.0052
s = √[(8.8 × 10⁻¹²)/(4 × 0.0052)]
s = 2.1 × 10⁻⁵ (The assumption that 0.0052 ≫ s holds.)
The solubility = 2.1 × 10⁻⁵ M
