MHT CET Chemistry Question (2026) — Solution

Step-by-step solution

The dissociation of AgCl is given by: AgCl(s) Ag^ + (aq) + Cl^ - (aq) Let the solubility of AgCl in 0.1 M NaCl be S mol/L. The concentration of Ag^ + is S and the concentration of Cl^ - is S + 0.1. Since K_ sp is very small, S is negligible compared to 0.1, so [Cl^ - ] 0.1 M. The solubility product expression is: K_ sp = [Ag^ + ][Cl^ - ] 1.8 10^ -10 = S 0.1 S = 1.8 10^ -10 0.1 = 1.8 10^ -9 M Answer: 1.8 10^ -9 M