1.

Solubility of AgCl will be minimum in

Answer»

0.001 M `AgNO_(3)`
Pure water
0.01 M `CaCl_(2)`
0.01 M NaCl

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


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