We have the K_"sp", and clearly we must convert this to a solubility in g*L^-1.
AgCl(s) rightleftharpoons Ag^(+) + Cl^(-)
K_(sp)=[Ag^+][Cl^-]=1.7xx10^-12
If we call S the solubility of AgCl, then S=[Ag^+]=[Cl^-], and S=sqrt(K_(sp))=S=sqrt(1.7xx10^-12)=1.303xx10^-6*mol*L^-1.
And then we convert this molar solubility into a solubility in g*L^-1:
"Solubility" = 1.303xx10^-6*cancel(mol)*L^-1xx143.32*g*cancel(mol^-1)
=1.87xx10^-5*g*L^-1. Again, this is a low value.
You required a solution of a molar quantity of 0.01*mol, i.e. (1.43*g)/(1.87xx10^-5*g*L^-1)=76.4xx10^3*L=76.4*m^3, which is a rather large volume, and the problem is completely impractical.
So what would a chemist do if he or she wanted to get silver ion into solution from silver halide? They would probably use a complexing ligand, for instance, S_2O_3^(2-) or NH_3 or something that would allow the silver ion to dissolve in solution in the presence of the halide that would normally precipitate it out.
