JEE Main2012ChemistryIonic EquilibriumActual
The solubility (in mol L ⁻¹ ) of AgCl (K_ sp =1.0 10⁻¹⁰ ) in a 0.1 M KCl solution will be
Options
- A1.0 10⁻⁹
- B1.0 10⁻¹⁰
- C1.0 10⁻⁵
- D1.0 10⁻¹¹
Correct answer
A. 1.0 10⁻⁹
Step-by-step solution
Let solubility of AgCl =x ~mole / L aligned & AgCl Ag ⁺+ Cl ⁻ & i.e., K_ sp ( AgCl ) =x x & KCl K ⁺+ Cl ⁻ & [ Cl ⁻ ] from KCl =0.1 ~m & Total [ Cl ⁻ ] in solution =x+0.1 & K_ s p ( AgCl )= [ Ag ⁺ ] [ Cl ⁻ ]=x(x+0.1) & 1.0 10⁻¹⁰=x(x+0.1) & 1.0 10⁻¹⁰=x^2+0.1 x & 1.0 10⁻¹⁰=0.1 x ( as x^2< < 1 ) & x=1.0 10⁻⁹ ~mol / L aligned