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.
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)?
5 answers
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
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
sorry i posted it under my friends name
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.
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.
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
NH4+ conc came out to be 0.0772M,
Mass of ammonium sulphate needed = (0.0772/2)* 132 = 4.62 gram