Ag++Cl- = AgCl
Ionic product (Q) of AgCl =[Ag+][Cl-]
Concect of solubility if Ksp value is <Q value then it is precipitated .
If Q value = Ksp then it is just precipitated.
Here in the problem [Cl-] is come from NaCl.=1.0×10-2
[Ag+] =?
Given also Ksp of AgCl =1.6×10-10
According to the given rule
Q= Ksp
>Ksp =[Ag+] [Cl-]
[Ag+] =Ksp/Cl-
= 1.6×10-10/1.0×10-2
=1.6×10-8 M
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