basmoth messaged me his solution (my comments
in blue) :
4a)
Find pOH
Kb = 10^-4.76 = [OH-]^2/0.20
[OH-] = 0.001864 (4 s.f)
pOH = 2.729
pH = 11.27
b) (i)
Kb = [OH-][NH4+]/[NH3]
-lgKb = -lg[OH-] + (-lg[NH4+]/[NH3])
pKb - pOH = (-lg[NH4+]/[NH3])
lg[NH4+]/[NH3] = 0.34
[NH4+]/[NH3] = 10^0.34
[NH4+] = 0.4375 = concentration of NH4Cl added (assuming NH4+ from
NH3 is negligible)
In 1 dm3, mass of NH4Cl added = 23.4g (3 s.f)
b) (ii) I would assume that a large amount of NH3 would act as a
buffer? Change in pH should be almost negligible. I don't know if I
should use Le Chatelier's Principle in saying that as the [NH4+]
increases, equilibrium shifts to the right to reduce it and hence,
the [OH-] drops too and pOH increases a little. Thus, pH should be
slightly lower.
While this is not strictly
required by the question here, (I always advise students
to) go ahead and write down whatever you suspect may be
relevant, as long as you know it's correct (in terms of chemistry)
on its own, you won't be penalized. Often, questions are perceived
(fairly or unfairly) as being ambiguous, and every mark is
precious, so the exam-smart students would go ahead to write down
everything possibly relevant, as long as it's correct.
And also, qualify your answers whenever you are compelled to
give alternative answers. See the 1st post of my thread here for examples on question ambiguity and
alternative answers.
Lastly, I have no idea how to account for the hydrolysis of
NH4+Cl- (NH4+Cl- + H2O -> NH3 + H3O+ + Cl-)
Consider #1
Using Kb and Ka values, compare
the strength of NH3 as a base, versus NH4+ as an acid. (Also notice
that the factor difference in Kb/Ka values outweighs the factor in
molarity difference).
Consider #2
The majority basic species present
is NH3 (rather than OH- from water). For every proton (H+) you
transfer from NH4+ to NH3, you end up back with the same no. of
moles hence molarity of NH3 and NH4+.
Consider #3
A pH of 8.9 (at 25 deg C) is
stated by the question. If the solution is acidic, the Ka
expression is more relevant; if it's alkaline, the Kb
expression is more relevant.
I assumed that NH3 reacted with HCl to give 0.01 mols of
NH4+
New concentration of NH4+ = 0.447/1.05 = 0.426 mol/dm3
New concentration of NH3 = 0.19/1.05 = 0.180 mol/dm3
[OH-] = Kb ([NH3])/[NH4+] = 7.343 x 10^-6 mol/dm3
pOH = 5.134
pH = 8.67 (3 s.f)
Change is -0.23.
You made a careless mistake here?
At 25 deg C, 14 - 5.134 = 8.87 (3 sig fig). My figures
are slightly different from yours, but no difference at 3 sig fig.
To answer the question, you need to comment about "the change in pH
of the solution is small (pH lowered by only
approximately 0.032), due to the buffer action of the weak
base NH3 and its conjuagte acid salt NH4Cl".
c) My graph would start at 0.699 (-lg(0.2)) and end at 11.3
(from Kb expression). There would be a region of rapid change
around pH values 3.5-6.5 as the solution is acidic due to the
hydrolysis of the salt. The end-point should have a pH of 5.5.
As stated by the question, the
x-axis shouuld be volume of HCl(aq) added. Hence, your graph should
start with pH 11.27, maximum buffer capacity at 9.24 for 25cm3
HCl(aq), equivalence point* at 5.12 for 50cm3 HCl(aq) due to
hydrolysis of acidic NH4Cl(aq) salt, and the graph levels off
approaching (but never reaching) pH 0.699 at excess HCl(aq)
added.
(Note that stoichiometric
"equivalence point" is approximated by "end point" via use of
suitable indicator during practical. Some students confuse the two
terms.)
*Sorry if I got everything wrong. We've just started on physical
chemistry and I'm still horrible at it. I'm trying to get better,
though. :p
You got virtually everything
correct. You're currently an undergrad like dkcx and d3sT1nY eh? Enjoy your days in the University
while you can! Every phase in life brings to the table different
joys, pain, issues, priorities, purpose and meaning. Make the most
of life (Carpe Diem) and enjoy each and all of these to the max,
the best you can. Live in The Now. And live it well. 
