A solution contains 0.0390 M Pb2 (aq) and 0.0390 M Sr2 (aq). If we add SO42–(aq), what will be the concentration of Pb2 (aq) when SrSO4(s) begins to precipitate?

Question

I am just completely lost on this one.

Answer 1

........PbSO4 ==> Pb^2+ + SO4^2-
........SrSO4 ==> Sr^2+ + SO4^2-

You know Ksp for each and you know metal is 0.039. I like to solve for OH^- needed to ppt each. Using my values for Ksp (which won't be the same as yours probably since texts differ),
SO4^2- for Pb = 4.1E-7 M.
SO4^2- for Sr = 8.2E-6 M.

This shows that if you 0.039 M solution of the two metals then add SO4^2- drop wise, the Pb ppts first. It will continue to ppt until the sulfate concn is what? When SO4^2- exceeds 8.2E-6, then Ksp for SrSO4 is exceeded and SrSO4 will start; i.e., (0.039)(8.2E-6) =or> 3.2E-7.
So what will that do to PbSO4.
Ksp PbSO4 = (Pb^2+)(SO4^2-)
1.6E-8 = (Pb^2+)(8.2E-6)
Solve for (Pb^2+). I get something around 0.002 M for Pb^2+. By the way this shows you can't separate (by pptn anyway) two metals with Ksp values so close together, especially if you want to do a CLEAN separation.

Answer 2

Hey Dr. Bob. I'm not sure where you got the 1.6E-8 from?

Answer 3

Also I found the Ksp values from my text. It says Pb = 2.53E–8 and Sr = 3.44E–7. Could you re-write with those values so that I can understand it better please? I really appreciate your help.

Answer 4

I'm stuck.