Please help me through this. Summer class going very quickly! You are presented

Question

Please help me through this. Summer class going very quickly! You are presented a solution that is both 1.0 times 10^-4 M in Pb^+2 and 1.0 times 10^-4 M in Ca^+2. You have the following chemicals to use: Na_2CO_3;NaF;Na_2SO_4 What chemical will best precipitate one ion and not the other? Why? What concentration of your anion will just cause each of the ions to precipitate? Which ppt forms first? At what concentration will you stop the precipitation? How much of the precipitating ion remains in solution? What percentage of the precipitating ion is left? Was precipitated out?

Explanation / Answer

Pb2+ and Ca2+ separation

The Ksp for PbCO3 and CaCO3 are maximum apart and thus Na2CO3 would be the best chemical for precipitation of one ion and not the other.

Concentration of CO3^2- to precipitate Ca2+ = Ksp/[Ca2+] = 3.69 x 10^-9/1 x 10^-4 = 3.69 x 10^-5 M

Concentration of CO3^2- to precipitate Pb2+ = Ksp/[Pb2+] = 7.40 x 10^-14/1 x 10^-4 = 7.4 x 10^-10 M

Ksp for PbCO3 is lower than CaCO3 and thus PbCO3 would precipitate first from the solution

Concentration of CO3^2- needed to stop precipitation of PbCO3 = 7.4 x 10^-10 M

Amount of [Pb2+] remained in solution when CaCO3 begins to precipitate = (7.4 x 10^-14/3.69 x 10^-5) = 2 x 10^-9 M

Percentage of Pb2+ precipitating ion left in solution = 2 x 10^-9 x 100/1 x 10^-4 = 0.002%

Percentage of Pb2+ precipitated out = 100 - 0.002 = 99.998%

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