Reminder: Ch 17 packet - practice problems and their solutions
Zellmer, Robert
zellmer.1 at osu.edu
Mon Jul 19 14:11:18 EDT 2021
I sent this out a week ago. This packet has a lot of additional Ksp questions
if you're still having trouble with this. They are kind of separated by sections.
Pay attention to what I've written below as well.
Something else dealing with the complex-ion formation section. For #1, the
first table they've written should technically be done in moles. They've used
molarity. That's okay if you're assuming 1 L of solution. Also, the concentrations
given in the question are the concentrations when the solution was made, not the
equilibrium concentrations. That's just like for acids and bases. When someone
states you have a 1.0 M HF solution that is not the equil. conc. of HF. That's what
the conc. was when the solution was made, i.e. putting enough moles of HF in
the solution to make it 1.0 M in HF to start (e.g. 1.0 moles HF in 1 L soln). You
then use an ICE table to find the equilibrium conc. of HF, H+ and F-.
Dr. Zellmer
From: Zellmer, Robert <zellmer.1 at osu.edu>
Sent: Monday, July 12, 2021 2:30 PM
To: cbc-chem1220 at groups.asc.ohio-state.edu
Cc: cbc-chem1220-ta at groups.asc.ohio-state.edu
Subject: Ch 17 packet - practice problems and their solutions
I got the following packet of additional practice problems for ch 17 (17.4-17.7)
from another instructor a number of years ago. Quiz 5 this coming weekend
deals with Ksp problems (sections 17.5-17.6, as well as Ch 19). This will give
you more practice for these Ksp problems.
I've also included the solutions as given to me. Below I have some comments
about this packet and the solutions.
I know this stuff is tough and you might need the extra practice. Keep in mind
though you can't spend all of your time on this stuff at the expense of everything
else. Midterm 3 will probably have around 26 questions again and it covers
17.3-17.6, Ch 19, 20.1-20.6. That means there won't be 10 questions over Ksp
material. While the material from Ch 19 and 20 is generally easier than the
Ksp material you still need to spend time on the former and not spend all your
time on the Ksp stuff.
I'm not sure when the practice exams will be ready. I've e-mailed the office
because what I see there right now (you can't see it) is very different than last
semester.
1) There are many questions concerning whether something would be more soluble
in the presence of different substances.
a) several questions involve complex-ion formation. The complexing ions
are called ligands (Ch 23). There are some common ligands you
should know which form complexes with metal cations. You'll see halides
(F-, Cl-, Br-, I-), CN-, NH3, SCN-. See tables 17.1 and 23.4 in the book
(just be concerned with those in Ch 17 at this point).
b) OH- acts like a ligand. Think of amphoteric oxides and hydroxides. These
react with both H+ (acid) and OH- (base) to dissolve. The most common
cations whose oxides and hydroxides are amphoteric are Al^3+, Cr^3+,
Zn^2+ and Sn^2+ (end of section 17.5). Al(OH)3 is pretty insoluble hydroxide.
It's soluble in acid because all hydroxides react with acid. It also dissolves in
a basic solution (as the OH- conc. inc.) because it forms the complex ion,
Al(OH)4^-, which is soluble.
ALL hydroxides and oxides are more soluble in acidic soln than in water.
Amphoteric hydroxides & oxides are also more soluble in basic soln compared
to pure water.
c) If the anion of the insoluble salt is the conj. base of a weak acid it will act
as a weak base and the salt will be more soluble in acid. For instance,
CaF2 is more soluble in an acidic soln. because the F- ion reacts with a
strong acid to form HF and the CaF2 becomes more soluble.
CaF2(s) <==> Ca^2+(aq) + 2 F-(aq)
The added H+ reacts with F- to form HF(aq),
H+ + F- <==> HF
This removes the F- from the top reaction causing it to shift to the right and
more CaF2 dissolves.
d) Question #3 in the pH & solubility section.
The question asks for the molar solubility of Cr(OH)3 at a pH = 10.00. It
doesn't state in a buffered soln with a pH of 10.00 like #1 in this section.
However, the solution given treats it as a buffered soln and thus the conc.
of the OH- in the ICE table doesn't change. You can see this in the ICE
table in the solution.
If this were done with a pH of 10.00 but the solution wasn't buffered you
would need to put a "-x" in the Change line of the ICE table under the OH-
and at equil you would have (1.0 x 10^-4 - x). That would still be easily
solved w/o having to ignore the "x".
2) There's a clarification in the section about Precipitation and Separation of Ions.
a) Under Precipitation and Separation of Ions section #2 might not be
written so that it's completely clear what to do, at least based on the
solutions.
One could read it as 7.5 mL of a soln containing 0.025 M NaCl and
0.025 M Na3PO4. One could also read it, as it was intended, as a
solution made by adding 7.5 mL of 0.025 M NaCl and 7.5 mL of 0.025 M
Na3PO4 together and then adding the Ag+ ion to this resulting solution of
15 mL. This means you have to do a dilution problem for the NaCl and
Na3PO4 first and then do the calculations dealing with the [Ag+] which
would ppt AgCl and Ag3PO4.
Dr. Zellmer
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