Really tough buffer question?

Question

Describe the preparation of 2 liters of 0.05 M ammonium choloride buffer, pH 9.4, starting with 1 M NH4OH and 2.13 M HCL.

Answer 1

The HCl reacts with NH4OH to form NH4Cl. So we anticipate a mix of the the hydroxide and the salt. Then:
Kb = [NH4+][OH-]/[NH3]
Kb= 1.8x10-5.
We can convert pH to pOH, which is 4.6, and from that determine [OH-], which is 10-4.6= 2.5x10-5.
(all numbers are approximate here)
Then 1.8/2.5 = [NH4+]/[NH3].
So if the buffer is [0.05M] in [NH4+], this relation tell us that [NH3] is about 0.067M at equilibrium. Since all this comes initially from NH4OH, and we want TWO liters, we need 0.117x2 = 0,234 moles of NH4OH to start with. So we would use 234 mL of the NH4OH solution. To produce the 0.1 moles of {NH4+] ion, we require V ml of HCl *2.13 M = 100 or about 46 mL of the HCl. Probably, you should add 2000-234-46 mL of water to this, and then mix the NH4OH stock solution in.

So we would

Answer 2

Where does the .117 that you multiply the 2L come from? I can't seem to figure it out.