What volume of water is necessary to dissolve 0.010 mole of #AgCl#?

#K_(sP) is #1.7*!0^-12#.

1 Answer
Dec 19, 2016

Answer:

Around #77# #m^3#; an impossibly large volume.

Explanation:

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.