To 0.350L of 0.150M NH3 is added 0.150L of 0.100M MgCl2. How many grams of (NH4)2So4 should be present to prevent precipitation of Mg(0H)2(s)?

Answer 1

Mg(OH)2 ==> Mg^+2 + 2OH^-
Ksp = (Mg^+2)(OH^-)^2
Use the Mg concn added to calculate the (OH^-). Then plug that in to the OH^- from
NH3 + HOH ==> NH4^+ + OH^-
Kb = (NH4^+)(OH^-)/(NH3) and solve for (NH4^+).
Use the (NH4^+) to determine mols NH4^+ needed and from there to grams (NH4)2SO4. Post your work if you get stuck.

Answer 2

the answer is suppose to be 2.7g im not getting that. This is what i did
1. Used KSP expression for Mg(OH)2
1.8*10^-11= (0.1)(OH-)^2
for [OH-]= 0.000013416

then for Kb expression
1.8*10^-5= [NH4](0.000013416)/(0.150)
[NH4]= 0.20125223614

then
0.20125223614(mol/L) * 0.5L (total volume 0.350+0.150)
= 0.100626118mol NH4

then i used mole ratio
0.10062118mol * 1mol (NH4)2SO4/1mol NH4
= 0.050313059 mol (NH4)2SO4

0.050313059mol (NH)2SO4 * 126g (molar mass)
= 6.339445434g

Answer 3

sorry i posted it under my friends name

Answer 4

You have made at least two errors I see and I stopped there.
First, you must recognize that each of the solutions dilute each other; therefore, the concn of the MgCl2 is not 0.1 but 0.1 x (0.150/0.500) = 0.03 M. The total volume is (0.350 L + 0.150 L = 0.500 L) and the concn of the NH3 is similar at 0.150 x (0.350/0.500) = 0.105 M. Those two will change both the (OH^-) you calculated as well as the (NH4^+) you calculated. [I get something like 0.0772 M or so here for concn (NH4^+).]
mols (NH4)2SO4 x molar mass NH4)2SO4 = ??
I get about 5.1 g here; then we divide by 2 since there are two moles NH4 per mol (NH4)2SO4.
It appears to me that you did everything ok after the calculation of the NH4^+. I used 132.14 for the molar mass of (NH4)2SO4. Check my work, especially for typos.

Answer 5

The answer is 4.62 gram.
NH4+ conc came out to be 0.0772M,
Mass of ammonium sulphate needed = (0.0772/2)* 132 = 4.62 gram