Errors in Ch 16 - sample and practice exercises, Go Figure, Self-Assessment, etc.

[Zellmer, Robert]

There are some problems with several of the practice exercises that follow the
Sample Exercises in Ch 16.  The following also has corrections for sample
exercises, Go Figure questions and Self-Assessment exercise.

Section 16.1:

Practice Exercise 16.1:

It has the wrong answer.  It asks for the reaction between O2^- and H2O when
Li2O is put in water.

LI2O (aq)  +  H2O(l )  ->  2 Li^+ (aq)  +  2 OH- (aq)

The net ionic eqn is:

O^2- (aq)  +  H2O(l )  ->  2 OH- (aq)


Section 16.2:

Practice Ex 16.2:

Has no "Check Answer" button.

Asks for the conjugate acids of HSO3^-, F^-, CH3NH2, PO4^3-.

This means treat these as bases and add an H+ to get the conj. acid.

H2SO3, HF, CH3NH3^+, HPO4^2-

Practice Ex. 16.3:

This question is asking whether the acid/base reactions lie to the right
(mostly product, K>1) or lie to the left (mostly reactant, K<1).

When you click on the "Check Your Answer" button it gives the answer
to Practice Ex. 16.2.

a) lies to the left (OH-, the stronger base, and stronger acid, H2PO4^-, are
on the product side)

            HPO4^2-  +  H2O(l )  ->    H2PO4^-  +  OH-

            OH- is a stronger base than HPO4^2-
            H2PO4^- is a stronger acid than H2O

            Actually, once you know which side the stronger acid is located
            then the stronger base must be on the same side, and vice versa.

            You can answer this one without Figure 16.4 since OH- is the
            strongest base you can have in water.

b)  lies to the right (OH-, the stronger base, and stronger acid, NH4^+, are
on the reactant side)

            NH4^+  +  OH-  ->  NH3  +  H2O

            OH- is a stronger base than NH3
            NH4^+  is a stronger acid than H2O


Section 16.4:

Practice Ex. 16.5:

It has the answers for PE 16.3.

a)  In a sample of lemon juice, [H+] = 3.8 x 10^-4 M.  What is the pH?

            pH = - log(3.8 x 10^-4) = 3.42

b) [OH-] = 1.9 x 10^-6 M  What is the pH?

            Different ways to do this.  You could use Kw = [H+] [OH-] = 1.0 x 10^-14
            and find [H+] and the pH.  You could use pKw = pH + pOH = 14.00 and
            solve for pOH and then get pH from this latter eqn.

            pOH = - log [OH-] = - log(1.9 x 10^-6) = 5.72

            pH = 14.00 - 5.72 = 8.28  (this makes sense since [OH-] > 10^-7 and
                                                            pH > 7.00 at 25 C means basic).

Go Figure 16.8:

It states the solution is pink when phenolphthalein is added and asks what it
tells you about the pH.  It then states pH > 10.  Actually, it would mean pH >8.2
since it turns a light pink around a pH = 8.2.

Practice Ex. 16.6:

A solution formed by dissolving an antacid table has a pOH of 4.82. What is
the [H+] in this solution?

It gives the answer for PE 16.5.  The correct answer is:

            pH = 14.00 - 4.82 = 9.18

            [H+] = 10^(-pH) = 10^-(9.18) = 6.6 x 10^-10


Section 16.5:

Practice Ex. 16.7:

An aqueous solution of HNO3 had a pH of 2.34.  What is the conc. of the acid?

This has the answer to PE 16.6.

HNO3 is a strong acid so it completely ionizes,

            HNO3 -> H+  +  NO3^-

There's essentially no HNO3 at equilibrium.

            [H+]_eq = [HNO3]_original

            [H+] = 10^-(pH) = 10^-(2.34) = 4.57 x 10^-3 M = 4.6 x 10^-3 M

            Thus, [HNO3]_original = 4.6 x 10^-3 M

Sample Ex. 16.8:

Part (b) Method 1:  the [OH-] in the denominator should be 0.0022 M.  The
answer given for [H+] is correct.


Section 16.6:

Practice Ex. 16.9:

Just incorrectly rounded. They have the answer as 1.5 x 10^-5.  I get
1.552 x 10^-5 M when carrying extra digits and this rounds 1.6 x 10^-5 M.

Go Figure 16.11:

The explanation given may be hard to understand.  I went over this in lecture.
Can be explained by thinking about diluting the solution.  When the volume of
solution is increased by adding H2O all the conc. of reactants and products
decrease (i.e. total molarities in solution decreases).  When that happens, the
reaction shifts to the right, toward more moles in solution to raise the total
moles (molarities), just like for gases when P (and molarity) is decreased by
increasing the volume, the rxn shifts to increase the moles of gas in solution.
In solution, increasing the volume of solution by adding solvent the rxn shifts
to inc the moles of solutes in solution.

Self-Assessment 16.16:

Rounded incorrectly.  I get Ka = 2.579 x 10-3 which rounds to 2.6 x 10-3


Section 16.7:

Sample Exercise 16.13:

Uses rounded value of OH- to determine pOH and pH.  You should not
round in intermediate steps.  You should carry at least one extra digit than
the number of sig. fig. into subsequent steps.  If you don't round at all
you will get pH = 11.22.


Section 16.9:

Practice Ex. 16.17:

The answer of acidic is correct.  The reason is the Ka for HC6H5O7^2-
acting as an acid is bigger than the Kb for it acting as a base.



Dr. Zellmer
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