......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
A solution is 0.10M Pb(NO3)2 and 0.10M AgNO3. If solid NaCl is added to the solution what is [Ag+] when PbCl2 begins to precipitate? (Ksp PbCl2 = 1.7 x 10^-5; AgCl = 1.8 x 10^-10)
The answers is 1.4 x 10^-8
How do I find the answer
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