MCQ
Solubility of $AgCl$ will be minimum in
  • A
    $0.001\,M\,AgN{O_3}$
  • B
    Pure water
  • $0.01 \,M\; CaCl _2$
  • D
    $0.01\,M\,NaCl$

Answer

Correct option: C.
$0.01 \,M\; CaCl _2$
(c) $0.01 \,M $ $CaC{l_2}$ gives maximum $C{l^ - }$ ions to keep ${K_{sp}}$ of $AgCl$ constant, decrease in $[A{g^ + }]$ will be maximum.

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