[qwerty193]

To a 100.0 mL solution containing 0.01 M Ba(NO3)2 and 0.01 M Pb(NO3)2 was added 101 mL of 0.01M H2SO4 to provide a slight excess of SO42- relative to Ba2+ in the resulting solution. Assuming that:
i) H+ does not bind to SO42-
ii) ionic strength effect can be ignored
iii) the system immediately achieves equilibrium
and given Ksp (Baso4) = 10^-10 M^2, Ksp (PbSO4) = 1.7 x 10^-8 M^2
Determine the following:
a) The concentration of Pb2+, Ba2+ and SO42- in the final solution
b) The composition of the precipitate, reported as the mole fraction of BaSO4 and the mole fraction of PbSO4.
My attempt of solving the question:
New [Ba2+] in 0.201 L = New [Pb2+] = 10^-3 mol/0.201 L= 4.98x 10^-3 M
Figured out [SO42-]max before BaSO4 start to ppt:
[SO42-] = Ksp/[Ba2+] = 10^-10/ 4.98 x 10^-3 M = 2.01 x 10^-8 M
[SO42-] max before PbSO4 start to ppt:
[SO42-] = Ksp/[Pb2+] = 1.7 x 10^-7 / [4.98 x 10^-3 M) = 3.43 x 10^-6 M
[Ba2+] remaining = Ksp / [SO42-] = 10^-10 / 3.43 x 10^-6 M = 2.93 x 10^-5 M.
But I really stuck at figuring out how much Pb2+ is present in the final solution and how to figure out SO42- conc. Any help/suggestion would be greatly appreciated 

[Borek]

I would start assuming precipitation of Ba²⁺ went to completion. Use this assumption to calculate concentration of the excess SO₄^2- and see what it tells you about precipitation of Pb²⁺.