Question

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.

Answer 1

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.

Answer 2

@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
8.17x10-8mol 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?

Answer 3

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 1.8E-5 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.