Manipal MET2010ChemistryIonic Equilibrium
The solubility of AgCl in 0.2 M NaCl solution is (K_ sp . of . AgCl =1.20 10⁻¹⁰ )
Options
- A6.0 10⁻¹⁰ M
- B0.2 M
- C1.2 10⁻¹⁰ M
- D0.2 10⁻¹⁰ M
Correct answer
A. 6.0 10⁻¹⁰ M
Step-by-step solution
Given, concentration of NaCl =0.2 M K_ sp ( AgCl )=1.20 10⁻¹⁰ Let the solubility of AgCl in NaCl =x AgCl Ag ⁺+ Cl ⁻ Solubility x 0.2 NaCl x 0.2 Na ⁺ + x 0.2 Cl ⁻ [ Ag ⁺ ]=x and [ Cl ⁻ ]=(x+0.2) K_ sp ( AgCl )= [ Ag ⁺ ] [ Cl ⁻ ] aligned & =x(x+0.2) & =x^2+0.2 x aligned K_ sp =0.2 x (x^2 1 ) or 1.2 10⁻¹⁰=0.2 x x=6 10⁻¹⁰