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
8.17x10-8mol 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?
To determine the number of grams of (NH4)2SO4 that should be present to prevent precipitation of Mg(OH)2, we need to calculate the maximum amount of Mg(OH)2 that can dissolve based on its solubility product constant (Ksp).
Step 1: Write the balanced equation for the dissolution of Mg(OH)2:
Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)
Step 2: Write the expression for the solubility product constant (Ksp) using the concentrations of Mg2+ and OH- ions:
Ksp = [Mg2+][OH-]^2
Step 3: Calculate the concentration of Mg2+ ions in the solution:
Given that the volume of MgCl2 solution is 0.140 L and the concentration is 0.100 M:
Concentration of Mg2+ = 0.100 M
Step 4: Calculate the concentration of OH- ions in the solution:
Since 1 molecule of Mg(OH)2 produces 2 OH- ions, the concentration of OH- ions is:
Concentration of OH- = 2 * concentration of Mg2+ = 2 * 0.100 M = 0.200 M
Step 5: Determine the maximum amount of Mg(OH)2 that can dissolve:
Since Ksp = [Mg2+][OH-]^2, we can solve for [Mg2+]:
[Mg2+] = Ksp / [OH-]^2
[Mg2+] = (1.8 × 10^-11) / (0.200)^2
[Mg2+] = 9.0 × 10^-10 M
Step 6: Calculate the moles of Mg(OH)2 that can dissolve:
Since [Mg2+] = 9.0 × 10^-10 M and the volume of the solution is 0.140 L:
Moles of Mg(OH)2 = [Mg2+] * Volume
Moles of Mg(OH)2 = 9.0 × 10^-10 M * 0.140 L
Moles of Mg(OH)2 = 1.26 × 10^-10 mol
Step 7: Calculate the molecular weight of (NH4)2SO4:
(NH4)2SO4 = 2(N) + 8(H) + (S) + 4(O) = 2*14.01 + 8*1.008 + 32.07 + 4*16.00
Molecular weight of (NH4)2SO4 = 132.14 g/mol
Step 8: Calculate the mass of (NH4)2SO4 required:
Mass of (NH4)2SO4 = Moles of Mg(OH)2 * Molecular weight of (NH4)2SO4
Mass of (NH4)2SO4 = 1.26 × 10^-10 mol * 132.14 g/mol
Mass of (NH4)2SO4 = 1.67 × 10^-8 g
Therefore, approximately 1.67 × 10^-8 grams of (NH4)2SO4 should be present to prevent precipitation of Mg(OH)2.
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.