Example Problem

Problem: A volume of 50 mL of 1.8 M NH3 is mixed with an equal volume of a solution containing 0.95 g of MgCl2. What mass of NH4Cl must be added to the resulting solution to prevent the precipitation of Mg(OH)2? Ksp of Mg(OH)2 = 1.5 E-11, Ka of NH4+ = 5.6 E-10

Solution: Let's take a look at what's going on here.

  1. MgCl2 --> Mg2+ + 2Cl¯ (runs to completion)
  2. NH3 + H2O --> NH4+ + OH¯
    Kb = Kw/Ka = 1.8 E-5
  3. Mg2+ + OH¯ --> Mg(OH)2
    Ksp = 1.5 E-11

Okay, so we have three reactions. The first one is the dissociation reaction for MgCl2, which will go to completion, since MgCl2 is a metal salt, and metal salts dissociate completely. The second one tells us how much OH¯ will be in solution after NH3 reacts with water. The third one is the formation reaction for Mg(OH)2.

So what do we do now? First let's convert the mass of MgCl2 to moles and find the total volume of the solution.

0.95 g MgCl2 * (1 mol MgCl2)/(95 g MgCl2) = 0.010 moles MgCl2
V = 50 mL + 50 mL = 100 mL = 0.100 L

Now let's look at the Ksp formula for Mg(OH)2.

Ksp = [Mg][OH¯]2
1.5 E-11 = [0.01/0.100][OH¯]2
[OH¯] = 1.2 M (This is the concentration of OH¯ that will cause Mg(OH)2 to precipitate)

Now that we know what concentration of OH¯ will precipitate Mg(OH)2, we just plug that into the formula for reaction # 2:

Kb = 1.8 E-5 = [NH4+][OH¯]/[NH3]
1.8 E-5 = [NH4+][1.2]/[1.8/2] ([NH3 is halved because we doubled the volume)
[NH4+] = 1.4 E-5

So, as long as you have more than 1.4 E-5 NH4+ in solution, Mg(OH)2 will not precipitate. From here on it's a Mickey Mouse stoich problem.

1.4 E-5 M NH4+ =
1.4 E-5 M NH4Cl * 0.100 L * (53.5 g NH4Cl)/(1 mol NH4Cl) = 7.49 E-5 g NH4Cl.

And there you have it. Now that you've conquered your first Type 1 problem, let's move on to Type 2 problems.

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