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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">I got the following packet of additional practice problems for ch 17 (17.4-17.7)
<br>
from another instructor a number of years ago. Quiz 5 this coming weekend<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">deals with Ksp problems (sections 17.5-17.6, as well as Ch 19). This will give<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">you more practice for these Ksp problems.
<br>
<br>
I've also included the solutions as given to me. Below I have some comments<br>
about this packet and the solutions.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">I know this stuff is tough and you might need the extra practice. Keep in mind<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">though you can’t spend all of your time on this stuff at the expense of everything<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">else. Midterm 3 will probably have around 26 questions again and it covers<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">17.3-17.6, Ch 19, 20.1-20.6. That means there won’t be 10 questions over Ksp<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">material. While the material from Ch 19 and 20 is generally easier than the<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">Ksp material you still need to spend time on the former and not spend all your<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">time on the Ksp stuff.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">I’m not sure when the practice exams will be ready. I’ve e-mailed the office<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">because what I see there right now (you can’t see it) is very different than last<o:p></o:p></span></p>
<p class="MsoNormal" style="margin-bottom:12.0pt"><span style="font-family:"Arial",sans-serif">semester.<br>
<br>
1) 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. The complexing ions<br>
are called ligands (Ch 23). There are some common ligands you<br>
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>
(just be concerned with those in Ch 17 at this point).<br>
<br>
b) OH- acts like a ligand. Think of amphoteric oxides and hydroxides. These<br>
react with both H+ (acid) and OH- (base) to dissolve. The most common<br>
cations whose oxides and hydroxides are amphoteric are Al^3+, Cr^3+,<br>
Zn^2+ and Sn^2+ (end of section 17.5). Al(OH)3 is pretty insoluble hydroxide.<br>
It's soluble in acid because all hydroxides react with acid. It also dissolves in<br>
a basic solution (as the OH- conc. inc.) because it forms the complex ion,<br>
Al(OH)4^-, which is soluble.<br>
<br>
ALL hydroxides and oxides are more soluble in acidic soln than in water.<br>
<br>
Amphoteric hydroxides & oxides are also more soluble in basic soln compared<br>
to pure water.<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>
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".<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">2) There’s a clarification in the section about Precipitation and Separation of Ions.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal" style="text-indent:.5in"><span style="font-family:"Arial",sans-serif">a) Under Precipitation and Separation of Ions section #2 might not be<o:p></o:p></span></p>
<p class="MsoNormal" style="margin-left:.5in"><span style="font-family:"Arial",sans-serif"> written so that it's completely clear what to do, at least based on the<o:p></o:p></span></p>
<p class="MsoNormal" style="margin-left:.5in"><span style="font-family:"Arial",sans-serif"> solutions.
<br>
<br>
One could read it as 7.5 mL of a soln containing 0.025 M NaCl and<o:p></o:p></span></p>
<p class="MsoNormal" style="margin-left:.5in"><span style="font-family:"Arial",sans-serif"> 0.025 M Na3PO4. One could also read it, as it was intended, as a<o:p></o:p></span></p>
<p class="MsoNormal" style="margin-left:.5in"><span style="font-family:"Arial",sans-serif"> solution made by adding 7.5 mL of 0.025 M NaCl and 7.5 mL of 0.025 M<o:p></o:p></span></p>
<p class="MsoNormal" style="margin-left:.5in"><span style="font-family:"Arial",sans-serif"> Na3PO4 together and then adding the Ag+ ion to this resulting solution of<o:p></o:p></span></p>
<p class="MsoNormal" style="margin-left:.5in"><span style="font-family:"Arial",sans-serif"> 15 mL. This means you have to do a dilution problem for the NaCl and<o:p></o:p></span></p>
<p class="MsoNormal" style="margin-left:.5in"><span style="font-family:"Arial",sans-serif"> Na3PO4 first and then do the calculations dealing with the [Ag+] which<o:p></o:p></span></p>
<p class="MsoNormal" style="margin-left:.5in"><span style="font-family:"Arial",sans-serif"> would ppt AgCl and Ag3PO4.
<o:p></o:p></span></p>
<p class="MsoNormal" style="margin-left:.5in"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">Dr. Zellmer <o:p>
</o:p></span></p>
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