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I sent a packet of extra problems from another instructor covering
ch 17 <br>
problems (17.4-17.6) on Friday. Here are the solutions from them by
that<br>
instructor. I haven't checked all of them. I've also included a few
comments<br>
below.<br>
<br>
Under the pH & solubility section for #3 (the Cr(OH)3 problem)
it states <br>
a pH = 10.00. In the solution it states a solution buffered at a pH
of 10.00. <br>
Simply stating at a pH of 10 and buffered at a pH of 10 are
different questions. <br>
<br>
You might also want to look at table 17.1 in the 14th ed (I think
it's the same #<br>
in the other editions). It lists some of the more common complexes
you'll run<br>
into (and the Kf values). Note that NH3 and CN- are involved in a
number of<br>
these complexes. The OH- anion is also involved in complex ion
formation,<br>
particularly in amphoteric hydroxides (the book lists Al<sup
class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>3</sup>+,
Cr<sup class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>3</sup>+,
Zn<sup class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>2</sup>+
and Sn<sup class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>2</sup>+
<br>
as examples). You do NOT have to memorize all these complexes and
Kf<br>
values. You will be told in some way in the problem what complex
forms and<br>
the associated Kf.<br>
<span id="OLK_SRC_BODY_SECTION"><span id="OLK_SRC_BODY_SECTION"><br>
There are many questions concerning whether something would be
more soluble<br>
in the presence of different substances.<br>
<br>
a) several questions involve complex-ion formation. We
discussed this in lecture<br>
in chapter 17 and will do so again in ch 23. There
are some common ligands<br>
</span></span><span id="OLK_SRC_BODY_SECTION"><span
id="OLK_SRC_BODY_SECTION"><span id="OLK_SRC_BODY_SECTION"><span
id="OLK_SRC_BODY_SECTION">you </span></span>should know
which form complexes with metal cations. You'll see halides<br>
(F-, Cl-, Br-, I-), CN-, NH3, SCN-. See tables 17.1
and 23.4 in the book.<br>
<br>
b) OH- acts like a ligand (Ch 17 & 23). Think of
amphoteric oxides and hydroxides.<br>
</span></span><span id="OLK_SRC_BODY_SECTION"><span
id="OLK_SRC_BODY_SECTION"><span id="OLK_SRC_BODY_SECTION"><span
id="OLK_SRC_BODY_SECTION">These </span></span>react with
both H+ (acid) and OH- (base) to dissolve. The most<br>
common cations whose oxides and hydroxides are
amphoteric are Al^3+,<br>
Cr^3+, Zn^2+ and Sn^2+ (end of section 17.5).
Al(OH)3 is pretty insoluble<br>
hydroxide. It's soluble in acid because all
hydroxides react with acid. It also<br>
dissolves in a basic solution (as the OH- conc.
inc.) because it forms the complex<br>
ion, Al(OH)4^-, which is soluble.<br>
<br>
c) If the anion of the insoluble salt is the conj. base
of a weak acid it will act<br>
as a weak base and the salt will be more soluble in
acid. For instance,<br>
CaF2 is more soluble in an acidic soln. because the
F- ion reacts with a<br>
strong acid to form HF and the CaF2 becomes more
soluble.<br>
<br>
CaF2(s) <==> Ca^2+(aq) + 2 F-(aq)<br>
<br>
The added H+ reacts with F- to form HF(aq),<br>
<br>
H+ + F- <==> HF<br>
<br>
This removes the F- from the top reaction causing it
to shift to the right and<br>
more CaF2 dissolves.<br>
<br>
All insoluble hydroxides & oxides are more
soluble in acid, as stated above.<br>
<br>
d) Question #3 in the pH & solubility section.<br>
<br>
The question asks for the molar solubility of
Cr(OH)3 at a pH = 10.00. It<br>
doesn't state in a buffered soln with a pH of 10.00
like #1 in this section.<br>
However, the solution given treats it as a buffered
soln and thus the conc.<br>
of the OH- in the ICE table doesn't change. You can
see this in the ICE<br>
table in the solution. <br>
<br>
If this were done with a pH of 10.00 but the
solution wasn't buffered you<br>
would need to put a "-x" in the Change line of the
ICE table under the OH-<br>
and at equil you would have (1.0 x 10^-4 - x). That
would still be easily<br>
solved w/o having to ignore the "x".<br>
<br>
Dr. Zellmer</span></span>
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