MCQ
Solubility of $AgCl$ will be minimum in
- A$0.001\,M\,AgN{O_3}$
- BPure 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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