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Solubility of AgCl will be minimum inA. 0.001 M `AgNO_(3)`B. Pure waterC. 0.01 M `CaCl_(2)`D. 0.01 M NaCl

Answer» Correct Answer - C
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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