Q. The solubility of $AgCl_{(s)}$ with solubility product $1.6 \times 10^{-10}$ in 0.1 M NaCl solution would be
A) $1.25 \times 10^{-5}$ M
B) $1.6 \times 10^{-9}$ M
C) $1.6 \times 10^{-11}$ M
D) zero
Explanation
[Solution] Common ion: $s = K_{sp}/[Cl^-] = 1.6 \times 10^{-10}/0.1 = 1.6 \times 10^{-9}$ M. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.