MHT CET202616 April 2026Morning ShiftChemistryIonic EquilibriumActual
The solubility of AgCl in 0.1 M NaCl is S mol/L. If the solubility product of AgCl is 1.8 10⁻¹⁰ , then S is approximately:
Options
- A1.8 10⁻⁹ M
- B1.8 10⁻¹⁰ M
- C1.8 10⁻¹¹ M
- D1.8 10⁻¹² M
Correct answer
A. 1.8 10⁻⁹ M
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