Question

The solubility (in molL−1) of AgCl(Ksp=1.0×10−10) in a 0.1MKCl solution will be:

• A. (A) 1.0×10−9
• B. (B) 1.0×10−10
• C. (C) 1.0×10−5
• D. (D) 1.0×10−11

Answer 1

Answer: (A)

(A) 1.0×10−9

Answer 2

Explanation:
Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium. The solubility product constant describes the equilibrium between a solid and its constituent ions in a solution,Let solubility of AgCl=x mole/LAgCl⇌Ag++Cl−i.e., Ksp(AgCl)=x2KCl→K++Cl− [Cl−] from KCl=0.1 mTotal [Cl−] in solution =x+0.1Ksp(AgCl)=[Ag+][Cl−]=x(x+0.1)1.0×10−10=x(x+0.1)1.0×10−10=0.1x( as x2<<1)x=1.0×10−9 mol/LHence, the correct option is (A).