Ch 17 packet - practice problems and their solutions

[Zellmer, Robert]

I got the following packet of additional practice problems for ch 17 (17.4-17.6)
from another instructor a number of years ago.  Quiz 9 this coming weekend
deals with Ksp problems (sections 17.5-17.6).  This will give you more practice
for these Ksp problems. Normally I send this out before the midterm as a
review.  You can use it now and before MT 3.

I know this stuff is tough and you might need the extra practice.

I've also included the solutions as given to me.  Below I have some comments
about this packet and the solutions.

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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