Posted by Fred on Sunday, October 14, 2012 at 6:07pm.
......PbCl2(s) ==> Pb^2+ + 2Cl^-
I......solid.......0.1.......0
C......solid.........x.......2x
E......solid.....0.1+x.......2x
Ksp = (Pb^2+)(Cl^-)^2 = 1.7E-5
.........AgCl(s)==> Ag^+ + Cl^-
I........solid......0.1.....0
C........solid......+x.......x
E.........solid...0.1+x......x
Ksp = (Ag^+)(Cl^-) =1.8E-10
These are done this way.
When NaCl is added, AgCl will ppt first because it has the smaller Ksp. It will continue pptng until the Ksp for PbCl2 is exceeded. What is the (Cl^-) when PbCl2 first ppts? That is
1.7E-5 = (Pb^2+)(Cl^-)^2
Substitute and solve for Cl^-. Remember Pb^2+ is 0.1M. I get something like 0.013 M but you need to confirm that.
Then plug this Cl^- into Ksp for AgCl to find Ag^+ when PbCl2 just reaches that point.
Ksp AgCl = (Ag^+)(Cl^-)
1.8E-10 = (Ag^+)(0.013)
Ag^+ = 1.8E-10/0.013 = 1.38E-8 M which rounds to 1.4E-8M
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