The solubility of AgCl will be minimum in

Options

A. `0.001 " M " AgNO_(3)`
B. Pure water
C. `0.01 " M " CaCl_(2)`
D. `0.01 " M " NaCl`

Answer

C

Explanation

`0.01 " M " CaCl_(2)` gives maximum `Cl^(-)` ions to keep `K_(sp)` of AgCl constant , decrease in `[Ag^(+)] ` will be maximum .