Question

The solubility of AgCl is 4 × 10^–10 at 298K. The solubility of AgCl in 0.04 m CaCl₂ will be

(a)  2 × 10^–5 m

(b)  5 × 10^–9 m

(c) 10^–4 m

(d)  2.2 × 10^–4 m

Answer 1

Correct option  (b) 5 × 10^–9 m

Explanation:

[Cl^–] due to CaCl₂  = 2 x .04  = .08molL^-1