NEETChemistryIonic Equilibrium
The solubility product ( K_ sp ) of AgCl is 1.6 10⁻¹⁰ . What is the molar solubility of AgCl in a 0.1 M aqueous solution of NaCl ?
Options
- A1.6 10⁻⁹ mol/L
- B1.26 10⁻⁵ mol/L
- C1.6 10⁻¹¹ mol/L
- D1.6 10⁻¹⁰ mol/L
Correct answer
A. 1.6 10⁻⁹ mol/L
Step-by-step solution
In a 0.1 M NaCl solution, the concentration of chloride ions is [ Cl ^-] = 0.1 M (due to complete dissociation of NaCl ). Let the molar solubility of AgCl in this solution be s . The dissociation of AgCl is: AgCl (s) Ag ^+(aq) + Cl ^-(aq) At equilibrium: [ Ag ^+] = s [ Cl ^-] = 0.1 + s 0.1 M (since s is very small compared to 0.1 ) The solubility product expression is: K_ sp = [ Ag ^+][ Cl ^-] 1.6 10⁻¹⁰ = (s)(0.1) s = 1.6 10⁻¹⁰ 0.1 = 1.6 10⁻⁹ mol/L Thus, the molar solubility is 1.6 10⁻⁹ mol/L . Answer: 1.6 10⁻⁹ mol