How Do You Solve This NH3 and H2O Chemical Equilibrium Problem?

[TheManw/theGoldenGun]

Hi all, this is the question:
Question data:
Reaction at STP: NH3 (aq) + H2O (l)  NH4+ (aq) + OH- (aq)
[NH3] = 0.0180 M
[OH-] = 5.6E-4 M

The answers I tried are in blue.

a. Write Keq expression for reaction above.
Keq = ([NH4][OH-])/([NH3][H2O])

b. Find pH of 0.0180 M NH3.
Do I use the Henderson Hasselbach equation for this? pH = pKa + log(Base/Acid)? =

c. Find Kb for NH3.
Kb = Kw/Ka right? How do I find Ka then?

d. Find % ionization of NH3 in 0.0180 M NH3.
I guess this is [NH4+]/0.0180*100% right? How do I find NH4?

e. 20.0 mL of 0.0180 M NH3 was titrated beyond equivalence point using 0.0120 M HCl.

a. Find volume of 0.0120 M HCl added to reach equivalence point.
So this happens when moles NH3 = mol HCl. So, 0.0180 M NH3 * .020 L = .00036 mol NH3. We would then need .00036 mol HCl, which could be obtained from doing .00036/.0120 = .030 L or 30.0 mL, yes?

b. Find pH of total solution after 15.0 mL of 0.0120 M HCl was added.
.015 L * .012 M HCl = .00018 mol HCl reacting with .00036 mol NH3 in 0.035 total L solution...do I use Henderson-Hasselbach for this too?

c. Find pH of total solution after 40.0 mL of 0.0120 M HCl was added.
Since you have .04 L * .012 M HCl = .00048 mol HCl reacting with .00036 mol NH3, you have .00012 mol worth of excess H+, divided by total solution of 0.06 L = .002 M, take -log(.002) = 2.70 = pH? Yes?​

Help please!

 

[stamba]

a. true
b. true
c. Kb = [NH4+]*[OH-]/[NH3], [NH4+] = [OH-], [H2O] doesn't goes in equation, because concentracion of water is constant, 55.555...mol/dm3
d. [NH4+] = [OH-]