the solubility product of PbSO4 at 298K is 1.6x10^-8 mol^2dm^-6. Calculate the solubility of PbSO4 in 0.10mol dm^-3 sulphuric acid.Thanks!

Answer 1

PbSO4==> Pb^2+ + SO4^2-
...x.......x......x

Ksp = 1.6E-8 = (Pb^2+)(SO4^2-)

How advanced is this class? This is quite a difficult problem.
The solubility is increased due to the presence of H^+ that reacts with the SO4^2- to produce HSO4^-. To counter that, the SO4^2- from the H2SO4 decreases the solubility because of the common ion effect. Finally, the SO4^2- from the H2SO4 is not as concentrated as one might think. For example, it is closer to 0.00985M in 0.1M H2SO4 and not 0.1M.
My guess is that your prof expects you to use 0.1M for (SO4^2-) and to ignore the effect of the increased H^+. If that is the case, then
Ksp = (Pb^2+)(SO4^2-) = 1.6E-8
For (Pb^2+) we substitute x.
For (SO4^2-) we substitute x+0.1 and solve for x. That's the simple way of doing it but I'm guessing that is what your prof wants.

Answer 2

Im a first year student in tertiary education,started this course for less than 3 months,thanks for helping,it really helps me to better understand the method to use for the common ion effect.