posted by john .
To a 100.0 mL solution containing 0.0100 M Ba(NO3)2 and 0.0100 M Pb(NO3)2 was added 101.0 mL of 0.0100 M 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 effects can be ignored
iii) the system immediately achieves equilibrium
and given that, under the above conditions:
i) Ksp (BaSO4) = 1.00 × 10-10 M2
ii) Ksp (PbSO4) = 1.70 × 10-8 M2
Determine the following:
a) The concentrations 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.
(Ba^2+)(SO4^2-) = 1E-10
(Pb^2+)(SO4^2-) = 1.7E-8
How many mmols Ba(NO3)2 do you have. That's 1 mmol.
How much Pb(NO3)2 = 1 mmol.
(Ba^2+) initially = 1 mmol/301 mL = 3.22E-3M
(Pb^2+) = 3.22E-3M
If you start dripping H2SO4 drop wise into the solution the BaSO4 will start pptng first. What is the (SO4^2-) at that point? That's 1E-10/3.33E-3 = 3.1E-8. BaSO4 will continue to ppt as each drop H2SO4is added until Ksp for PbSO4r is reached.What must the (SO4^2-) be when Pb^2+ starts to ppt?
(SO4^2-) = 1.7E-8/3.22E-3 = 5.28E-6 M.
What is the (Ba^2+) at this point? It is (Ba^2+) = 1E-10/5.28E-6 = 1.89E-5 M. Convert that to millimols Ba^2+ in solution, subtract from 1 mmol initially to find BaSO4 pptd and convert to mols BaSO4.
Calculate mmols H2SO4 used to ppt the BaSO4 to that point and subtract from 1.01 mmols H2SO4 initially to see how much H2SO4 is left. Then write
Pb^2+ + SO4^2- ==> PbSO4
Knowing you have 1 mmols Pb and 0.016 mmols SO4, I would assume all of the SO4 is used to form PbSO4. Convert that to mols PbSO4 and calculate mole fraction PbSO4 and XBaSO4 from that.
To fine Pb^2+ at the end, 1 mmols Pb - ? mmols SO4 = mmols Pb^2+ left.
All of that gives you Ba^2+, Pb^2+, SO4^2- and XBaSO4 and XPbSO4. Post your work if you get stuck.