Question

A solution contains 0.0330 M Pb2+and 0.0210 M Ag+. Can 99% of Pb2+be precipitated by chromate(CrO42-), without precipitating Ag+? What will the concentration of Pb2+be, when Ag2CrO4begins to precipitate?

Answer 1

Remember that I'm using values for Ksp that I obtained from the web. Be sure to use the values in your text or notes from your class.
Two ions in solution. 0.0330 M Pb2+and 0.0210 M Ag+. Write the reaction with CrO4^2- and solve for (CrO4^2-) at equilibrium; i.e., just when one more drop of CrO4^- will ppt PbCrO4 or Ag2CrO4.
....................Pb^2+ + CrO4^2- ==> PbCrO4
Ksp = 3E-13 = (Pb^2+)(CrO4^2-)
(CrO4^2-) = 3E-13/0.0330 = 9.09E-12 M

Same for Ag2CrO4
Ksp = 1.12E-12 = (Ag^+)^2(CrO4^2-)
(CrO4^2-) = 1.12E-12/(0.0210)^2 = 2.54E-9 M
These two values tell us that if CrO4^2- is dripped slowly into the solution that PbCrO4 will start pptng first. It will continue a bit before Ag2CrO4 starts to ppt. Can 99% of the Pb^2+ be pptd before Ag2CrO4 starts? We can calculate that. Pb^2+ = 0.0330 x 0.01 = 0.000330 Pb^+ remaining if 99% has been ppt. What will the CrO4^2- be at that point?
(CrO4^2-) = 3E-13/0.000330 = 9.09E-10 so PbCrO4 will start pptng first and it will continue pptng until (CrO4^2-) reaches 2,54E-9 M (calculated above) Since 9.09E-10 is smaller than 2.54E-9 then yes, we can ppt 99% of the PbCrO4 before the Ag2CrO4 starts contaminating it.
Next question. What will be (Pb^2+) when Ag2CrO4 first starts to ppt?
First, remember that the (CrO4^2-) = 5.4E-9 M when Ag2CrO4 first starts. Plug this into the Ksp for PbCrO4 and solve for Pb^2+.
(Pb^2+) = Ksp/(CrO4^-) = 3E-13/2.54E-9 = 1.18E-4 M which is rather small. Calculations like this tell us that a mixture of Pb^2+ and Ag^+ can be completely separated with Na2CrO4 and determined independently of each other. Post your work if you get stuck.

Answer 2

Thank you so much for your help.