Question

What is the [CrO4] in a 2.5 L saturated solution of lead (II) chromate to which 0.05 mol Pb(NO3)2 is added?

Answer 1

PbCrO4 ==> Pb^2+ + CrO4^2-
...x........x.......x
...........Pb(NO3)2 ==> Pb^2+ + 2NO3^-
initial.....0.05 mol.........0........0
change.....-0.05.......0.05......2*0.05
equil........0..........0.05.....0.10
(Pb^2+) = 0.05/2.5L = 0.02M
Ksp = (Pb^2+)(CrO4^-)
Ksp you know.
(Pb^2+) = x + 0.02 [x from the PbCrO4 and 0.02 from Pb(NO3)2]
(CrO4^2-) = x
Solve for x.