To 0.360 L of 0.140 M NH3 is added 0.150 L of 0.100 M MgOH2. How many grams of (NH4)2SO4 should be present to prevent precipitation of Mg(OH)2 (s)?
THANK YOU!!!
Just to clarify the question, I wonder if that 0.150L of 0.100 M MgOH2 was supposed to be 0.150 L of 0.100 M MgCl2?
@DrBob222 - No, it is suppose to be Mg(OH)2
The Ksp to be used for this problem is 1.8*10^-11
Oh wait, nevermind. Sorry! It is suppose to be 0.150L of 0.100M MgCl2 :)
To determine the number of grams of (NH4)2SO4 needed, we need to first calculate the amount of Mg2+ ions present in the solution by using the molarity and volume of Mg(OH)2 solution.
Step 1: Find the moles of Mg(OH)2.
Moles of Mg(OH)2 = Molarity * Volume
Moles of Mg(OH)2 = 0.100 M * 0.150 L
Moles of Mg(OH)2 = 0.015 moles
Step 2: Determine the moles of Mg2+ ions.
Moles of Mg2+ = Moles of Mg(OH)2
Moles of Mg2+ = 0.015 moles
Step 3: Find the required moles of (NH4)2SO4 to react with all the Mg2+ ions.
From the balanced equation: Mg2+(aq) + (NH4)2SO4(aq) → MgSO4(aq) + 2NH4+(aq)
The ratio between Mg2+ and (NH4)2SO4 is 1:1.
Therefore, the required moles of (NH4)2SO4 = Moles of Mg2+ = 0.015 moles
Step 4: Calculate the mass of (NH4)2SO4.
Mass of (NH4)2SO4 = Moles of (NH4)2SO4 * Molar mass of (NH4)2SO4
Mass of (NH4)2SO4 = 0.015 moles * (2*(14.01 g/mol) + 4*(1.01 g/mol) + 32.06 g/mol + 16.00 g/mol)
Mass of (NH4)2SO4 ≈ 1.06 g
Therefore, approximately 1.06 grams of (NH4)2SO4 should be present to prevent precipitation of Mg(OH)2 (s) in the solution.