I infer you didn't pay any attention to my post. Thank you.
For problem 4: you have a titration with a weak acid and a strong base. You have reached the equivalent point, so:
Kb = [HNO2][OH-] / [NO2-] = [OH-]2
[NO2-] = 0.05 M Kb = 2.2*10-11
pH = 8.02
Now, problem 5. You are doing a titration with a weak base (NH3) and a strong acid (HCl). You haven't reached the equivalent point yet: what you have is a buffer solution.
HCl + NH3 -> NH4+Cl- + NH3 ...because of the NH3 surplus.
After a bit of calculations you get: [NH4+] = 0.04 M and [NH3] = 0.02 M
There is also pKa = 9.26
So: pH = pKa + log [Base] / [Acid] -> pH = 8.95
P.S.: I think this topic should be moved to the high school forum. :coffee1: