MHT CET2007ChemistryIonic Equilibrium
The solubility of AgCl is 1 10⁻⁵ ~mol / L . Its solubility in 0.1 molar sodium chloride solution is
Options
- A1 10⁻¹⁰
- B1 10⁻⁵
- C1 10⁻⁹
- D1 10⁻⁴
Correct answer
C. 1 10⁻⁹
Step-by-step solution
K_ s p of AgCl =( solubility of AgCl )² = (1 10⁻⁵ )²=1 10⁻¹⁰ Suppose its solubility in 0.1 M NaCl is x ~mol / L array c AgCl Ag ⁺+ Cl ⁻ NaCl Na ⁺+ 0.1 M Cl ⁻ _ 0.1 M [ Cl ⁻ ]=(x+0.1) M K_ s p of AgCl = [ Ag ⁺ ] [ Cl ⁻ ] =x (x+0.1) 1 10⁻¹⁰=x²+0.1 x array Higher power of x are neglected aligned 1 10⁻¹⁰ &=0.1 x x &=1 10⁻⁹ M aligned