[plu]

On a similar note to my question on zinc hydroxide solubility:

How many grams of lead(II) sulfide, PBS, will precipitate from 1.00 L of a saturated solution of lead(II) sulfate, PbSO₄, if the concentration of sulfide ion, S^2-, is adjusted to give a concentration of 1.00 x 10^-17 M?
K_{sp PbSO4} = 1.6 x 10^-8
K_{sp PbS} = 2.5 x 10^-27

I can find the concentration of lead ions in the solution at equilibrium but I don't know how to find the amount of PbS precipitate    *delete me*

[Borek]

I can find the concentration of lead ions in the solution at equilibrium but I don't know how to find the amount of PbS precipitate    *delete me*

Really? Ask your students

[plu]

 Stop mocking me borek and give me a hand here!

[Borek]

Well, you understand that it is against forum rules?

Saturated PbSO₄ is 1.3*10^-4M, after the solution is treated with S^2- you are left with 2.5*10^-27/10^-17=2.5*10^-10.

How much was precipitated?

1L*(1.3*10^-4-2.5*10^-10)=1.3*10^-4 mole.

About 30mg.

[plu]

Thanks again borek.  I'm just a lowly non-science undergrad guy who's no good with this kind of stuff.  My "students" are just a few high school kids I'm helping out.  I'm no prof