CHEMISTRY HELP PLEASE!!
posted by Marissa .
0.350L of 0.160 M NH3 is added 0.140 L of 0.100 M MgCl2 .
How many grams of (NH4)2SO4 should be present to prevent precipitation of Mg(OH)2 (aq) ? The Ksp of Mg(OH)2 is 1.8 x 10–11.

Mg(OH)2 ==> Mg^2+ + 2OH^
Ksp = (Mg^2+)(OH^)^2
(Mg^2+) = 0.1 x (140/490) = ?
Solve for (OH^).
Then NH3 + H2O ==> NH4^+ + OH^
Kb = (NH4^+)(OH^)/(NH3)
Substitute for Kb and (OH^)[from above] and (NH3) where (NH3) = 0.160 x (350/490) = ? and solve for (NH4^+)
Finally, convert mols NH4^+ to mols (NH4)2SO4, then to grams. grams (NH4)2SO4 = mols (NH4)2SO4 x molar mass. 
@DrBob222
so, for the first part I do,
Ksp = (Mg^2+)(OH^)^2 , which is
1.8x 10^11 = (.0286)(OH)^2
for OH, i get 2.51 x 10^5
then, for the second part, I do,
Kb = (NH4^+)(OH^)/(NH3), which is
1.8x 10^11 = (NH4)(2.51 x10^5)/(.114)
for NH4, i got 8.17x10^8 mol
then I converted this to (NH4)2SO4 by dividing
8.17x108mol NH4/ 2 , for then i get
4.08 x 10^8 mol (NH4)2SO4
I then multiply that by the molar mass,
4.08x10^8 mol X 132.065 g (NH4)2SO4,
then i get 5.4x 10^6.
i put this answer in, but it says that its wrong.
did i do anything wrong? 
yes.
Your Mg from the first part is ok.
At the beginning of the second part you used Ksp for Kb for NH3. I think Kb for NH3 is about 1.8E5 but use whatever is in your text/notes. I know it isn't the same as Ksp for Mg(OH)2. I didn't check past that point.