Precipitation from saturated solution

[sparkle123]

Homework Statement

How many grams of PbS will precipitate from 1.00 L of a saturated solution of PbSo4 if the concentration of S2- is adjusted to give a concentration of 1.00E-17?

Homework Equations

PbSO4 Ksp=1.6E-8
PbS Ksp=2.5E-27

The Attempt at a Solution

I assumed that there was no excess PbSO4 in the solution. (can i do this?)
I calculated [Pb2+] from Ksp for PbSO4, so i got [Pb2+]=sqrt(1.6E-8)=1.26E-4
Then I assumed all of the Pb2+ precipitated, so n(PbS)=n(Pb2+)=1.26E-4
finally m=0.0303g

Is this a complete solution? Is this even right? How do I mathematically verify my assumption that all the Pb2+ will precipitate (I know that Ksp is very small, so Kf is very large)? Can I assume that there was no excess PbSO4 in the solution?

 

[Borek]

There were no excess PbSO₄, that's right - solution was saturated, so it contained as much as it could, but not more.

Then it becomes tricky, as wording is lousy to me. There is no S^2- in lead sulfate solution. Is 10^-17] final concentration of S^2-, after precipitation, or is it "initial", before precipitation starts? I guess it the final, after precipitation - if so, you have to use it to calculate concentration of lead left in the solution. You can't assume all lead precipitated, nothing supports this approach.