Calculate the pH of the following solutions?

Question

(a) 0.1 M HONH2 (Kb = 1.1 * 10-8)
(b) A mixture containing 0.1 M HONH2 and 0.1 M HONH3Cl

Please can you explain each step. I would be greatly appreciated

Answer 1

Since you have Kb, it is convenient to work with pOH and convert.
For hydroxylamine, [AOH], dissociation is
AOH -> OH- + A- , and
[OH-][A+]/[AOH] = Kb.
For a Kb of the order of 10-8 and 0.1 M AOH, the dissociation of AOH produces [A+]<<[AOH] at equilibrium. Then if X = [OH-],
X^2/0.1 = 1.1x10-8
X^2 = 11x10-10 and X=3.6x10-5 appx
pOH = - Log X = Log 6 - Log 3.6 = 5.45 appx
pH= 14-pOH=8.55 appx

In part 2, we deal with a "buffer solution". The reaction is the same, but the conc of [A+] is changed since the chloride salt totally dissociates. Then,
0.1* X/0.1 = Kb, so
X= 1.1x10-8 appx,
pOH= 9- Log 1.1 = 8.95 appx
and pH= 5.05 appx