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)?
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.
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
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.
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
Well, preventing precipitation is serious business. But let me clown around with some calculations for you!
First, let's find out how much Mg(OH)2(s) can form. We know that NH3 and MgCl2 react in a 1:1 ratio, so we can use the law of conservation of mass.
The NH3 concentration is 0.150 M, and the volume is 0.350 L, so we have (0.150 mol/L) × (0.350 L) = 0.0525 mol of NH3.
The MgCl2 concentration is 0.100 M, and the volume is 0.150 L, so we have (0.100 mol/L) × (0.150 L) = 0.015 mol of MgCl2.
Since they react in a 1:1 ratio, we have 0.015 mol of Mg(OH)2(s) formed. Now, we need to find out how much (NH4)2SO4 is needed to prevent the precipitation of Mg(OH)2(s).
The balanced equation for the reaction is:
Mg(OH)2 + (NH4)2SO4 → MgSO4 + 2NH3 + 2H2O
From the equation, we can see that for every 1 mol of Mg(OH)2(s), we need 1 mol of (NH4)2SO4. Therefore, we need 0.015 mol of (NH4)2SO4.
To find the grams, we can use the molar mass of (NH4)2SO4, which is 132.14 g/mol.
0.015 mol × 132.14 g/mol = 1.98 grams of (NH4)2SO4.
So, to prevent the precipitation of Mg(OH)2(s), we need approximately 1.98 grams of (NH4)2SO4. Just like clowns, chemistry can get quite precise!
To determine the number of grams of (NH4)2SO4 needed to prevent the precipitation of Mg(OH)2, we need to calculate the maximum amount of Mg(OH)2 that can be formed from the reaction between NH3 and MgCl2.
First, let's write the balanced equation for the reaction between NH3 and MgCl2:
NH3 + MgCl2 → Mg(OH)2 + NH4Cl
From the equation, we can see that the stoichiometric ratio between NH3 and Mg(OH)2 is 1:1. This means that for every mole of NH3, we will obtain one mole of Mg(OH)2.
Next, we need to calculate the number of moles of NH3 and MgCl2 present in the given volume of solution. To do this, we can use the formula:
moles = concentration (M) × volume (L)
For NH3:
moles of NH3 = 0.150 M × 0.350 L = 0.0525 moles
For MgCl2:
moles of MgCl2 = 0.100 M × 0.150 L = 0.015 moles
Since the stoichiometric ratio between NH3 and Mg(OH)2 is 1:1, the maximum amount of Mg(OH)2 that can be formed is equal to the number of moles of NH3, which is 0.0525 moles.
Now, let's calculate the molar mass of Mg(OH)2:
Mg: 24.3050 g/mol
O: 15.9994 g/mol (two oxygen atoms)
H: 1.0079 g/mol (two hydrogen atoms)
Molar mass of Mg(OH)2 = (24.3050 g/mol) + (15.9994 g/mol × 2) + (1.0079 g/mol × 2)
= 58.3198 g/mol
Finally, we can calculate the mass of (NH4)2SO4 needed to prevent the precipitation of Mg(OH)2. Since the stoichiometric ratio between Mg(OH)2 and (NH4)2SO4 is 1:1, the mass of (NH4)2SO4 required is equal to the molar mass of Mg(OH)2 multiplied by the number of moles of Mg(OH)2:
Mass of (NH4)2SO4 = molar mass of Mg(OH)2 × moles of Mg(OH)2
= 58.3198 g/mol × 0.0525 moles
= 3.0589 grams
Therefore, approximately 3.059 grams of (NH4)2SO4 should be present to prevent the precipitation of Mg(OH)2.