Study Guide Key Exam 3
CHEM 109
Remember you will need to show your work for full credit. On the real exam always work the problems you know best
first. If you get hung up on a problem, you should move on and come back to it at the end. If you have time, check over
your work. To test your speed, work this study guide as if it was the exam. How long does it take to finish?
1. Briefly describe a human genetic disease based on oxidation-reduction chemistry of hemoglobin.
A full credit answer should include one or more equations for chemical reactions.
There are inherited blood disorders, methemoglobinemia, that exist because a patient has problems with
maintaining the required charge on the Fe ion in hemoglobin. One type occurs due to a mutation in the beta
chain of Hb that makes the iron heme more easily oxidized: Hb(Fe2+)  Hb(Fe3+) + e-. Hb(Fe3+) doesn’t
bind O2, so the hemoglobin can’t carry oxygen through the body.
Everyone’s Hb(Fe) is occasionally oxidized to the Fe3+ when it releases oxygen. So another type of
methemoglobinemia results from a lack of an enzyme (cytochrome b5 reductase). This enzyme is necessary to
convert Fe3+ in methemoglobin back to Fe2+, so that it can bind oxygen.
2. What is the concentration in molarity units of a solution that has 7.00 g of NaCl dissolved in water to give 520.6 mL of solution?
M = mol/liter, First convert 7.00 g of NaCl into moles of NaCl using the molecular weight of NaCl.
Na 22.98977 g
Cl 35.4527 g
58.44247 g in 1 mol of NaCl
520.6 mL x
0.1198 mol
0.5206 L
1L
1000 mL
7.00 g x (
1 mol NaCl
58.44247 g
) = 0.1197759 mol
= 0.5206 L
= 0.2301 mol/L
or 0.230 M
3. If 83.00 mL of a 3.075 M solution of NaOH is diluted to a final volume of 12.88 L, what will its concentration be?
M1V1 = M2V2 .
Rearrange to get
M1V1
V2
Convert 83.00 mL to liters. 83.00 mL x
= M2. The volume amounts must be expressed in the same units.
1L
1000 mL
= 0.08300 L; M2 =
3.075 𝑀 𝑥 0.08300 𝐿
12.88 𝐿
= 0.01982 M
4. Based on the following reaction, HCl + NaOH  H2O + NaCl,
determine the concentration of hydrochloric acid, HCl, if 10.00 mL of acid required 15.51 mL of 0.2500 M NaOH to reach the
endpoint of the titration.
Follow the four steps for titration. Step 1. The equation is already balanced.
Step 2. Find the moles of NaOH:
(15.51 mL)(
1L
1000 mL
)(
0.2500 mol NaOH
L
) = 0.0038775 mol NaOH
Step 3. Find the moles of HCl: (0.0038775 mol NaOH)(
Step 4. Find the concentration (M) of HCl: (10.00 mL)(
0.0038775 mol HCl
1 mol HCl
1 mol NaOH
1L
1000 mL
) = 0.0038775 mol HCl
) = 0.01000 L
= 0.38775 mol/L  0.3878 M HCl
0.01000 L
5. What volume of a 15% (w/v) solution of LiOH would be produced if you used 8750 g of LiOH?
% (w/v) =
x=
8750
15
g solute
mL solution
x 100
Rearrange to get mL of solution by itself. (mL solution) =
g solute
% (w/v)
x 100
x 100 = 58333 mL or 58,000 mL
6. Which of the following compounds would you predict to have the highest, intermediate, and lowest solubility in water? Explain
your answer, using structures where appropriate.
H
H
H
H
a.
Solubility: a > b> c
C
C
C
C
O
b.
H
H
H
H
H
H
H
H
C
O
H
c.
H
C
H
H
H
C
C
H
H
H
C
H
H
Molecule c. is a non-polar. It would not be soluble in polar water.
Both a. and b. have a non-polar region and a polar region. The non-polar regions have similar surface areas for both
molecules. Next, compare the polar regions.
a. can form hydrogen bonds with water. Each of its propanol molecules has two H-bonding acceptor sites
and 1 donor site.
b. can also form hydrogen bonds w/ water. It has two acceptor sites and no donor sites. This has fewer hydrogen
bonding sites than molecule a.
a. is most soluble. b. has intermediate solubility. c. is least soluble.
Drawings follow.
Indicates hydrophobic surface area
H
H
H
H
H
C
C
C
C
H
H
H
H
H
c. Lowest solubility. All of
it's surface is non-polar.
H
O
Indicates hydrophobic surface
H
H
H
C
C
H
H
a. Highest solubility. It has about
the same size non-polar surface as
c, but it has more sites for
hydrogen-bonding with water.
H
H
O
H
O
O
H
H
H
H
O
H
H
H
C
H
H
b. Intermediate solubility. It has fewer
hydrogen bonding sites than a.
C
H
O
C
H
H
H
O
7. For salt X, determine how much will remain undissolved if 500 g
From the table, it is apparent that at 65 oC, 120 grams of
salt will dissolve in 1 liter of water. If 500 g of this salt
was added to a liter than 500-120 or 380 g would remain
undissolved
Solubility of Salt X in Water at Different Temp.
Grams of salt Dissolved per Liter of H2O
is
mixed into a liter of pure water at 65 °C.
160
140
120
100
80
60
40
20
0
15
25
35
45
Temp oC
55
65
8. a) Determine whether the following would be more soluble in water or CCl4.
First note that water is polar and CCl4 is non-polar. (You can determine this from the Lewis Dot structures.)
Then look at the structure of each solute to determine if it is polar. Ionic compounds form ions in water, and
polar molecules like water form attractive forces with charged particles.
i) CH3CH3 , iv) I2 are non-polar, so CCl4
ii) HCl, iii) NaCl and v) Na3PO4 are polar, so H2O
b) Which of the following are strong electrolytes, weak electrolytes or non-electrolytes?
Remember, strong acids, strong bases and ionic compounds are strong electrolytes.
Strong electrolytes: ii) HCl
iv) NaCl
Weak electrolytes:
i) CH3COOH (acetic acid)
Non-electrolytes :
iii) CH3CH2OH (ethanol)
9. Given these data for the reaction of Fe3+ with I- :
2 Fe3+ (aq) + 2 I- (aq)  2 Fe2+ (aq) + I2 (aq)
a) Write a general rate law for the reaction.
a) Rate = k [Fe3+]m[I-]n
b) Use the experimental data, which looks at initial rates to produce I2, to calculate a rate for the reaction.
Initial Rate Data
Time (s)
10
20
30
40
50
Conc. of I2 (M)
3.23 x 10-3
6.42 x 10-3
9.59 x 10-3
1.31 x 10-2
1.66 x 10-2
I am going to choose to use the 20 s and 50 s time points for this calculation.
Rate =
1.66 x 10-2 - 3.23 x 10-3 M
50-10 s
=
1.337 x 10−2 M
40 s
= 3.34 x 10-4 or 3.3 x 10-4 M/s
(assuming 2 sf for sec)
10. a) In a reaction between NO and O2 to form NO2, the rate law was found to be Rate = 1.45 x104[NO]2[O2]. Determine the initial rate of the
rxn. when 0.035 M NO reacts with 0.027 M O2.
Rate = 1.45 x104[NO]2[O2] = 1.45x104(0.035)2(0.027) = 0.47958 M/s = 0.48 M/s
b) If for this same reaction, the rate was found to be 0.26 M/s when O2 was 0.018 M, what was the concentration of NO?
Rearrange to solve for [NO]:
Rate = 1.45 x104[NO]2[O2]
75
[NO]2 = Rate / (1.45x104 x [O2]) = 0.26 / (1.45 x 104 x 0.018) = 9.96168 x 10-4
[NO] = √9.96168 𝑥 10−4 = 0.03156 = 0.032 M
c) In a kinetics experiment, if I told you that the value of k that you obtained is relatively large, what height hill should be drawn for this reaction
on a reaction progress diagram (vertical axis = G°).
If k is relatively large, than the rate is relatively fast and the activation energy hill would be relatively small.
11. Explain how/why changes in temperature result in a change in rate constant, k. Include an appropriate graph.
Boltzmann Distributions
frequency
Molecules move with a distribution of speeds. When
colder temperature
temperaturechanges, the distribution changes. For higher
temperatures, the average speed and the top speed of a
warmer temperature
population of molecules would increase. The highest
speeds are particularly imprtant because only they have
speeds that could
suffiennt energy to reach the transition state in a collision
overcome Ea barrier.
(in other words reach the top of the energy barrier, the top
of Ea hill). Because a larger fraction of molecules is
speed
moving at a high enough speed to have their collisions
reach the transition state, a larger fraction of collisions will be productive, and the rate of the reaction will
increase. For lower temperatures, there is a decrease in the fraction of molecules that can reach the transition
state.
12. Write equilibrium constant expressions for the following reactions:
a) 2 H2 (g) + 2 NO (g)  2 H2O (g) + N2 (g)
b) Fe (s) + CO2 (g)  FeO (s) + CO (g)
a) Keq =
[H 2 O]2 [N 2 ]
[H 2 ]2 [NO]2
b Keq =
[CO]
[CO 2 ]
13. Shown below is a balanced equation for the decomposition of H2S to form H2 and S2.
2 H2S (g)  2 H2 (g) + S2 (g)
a) Write an equilibrium constant expression for the reaction
.
Keq =
[H 2 ] 2 [S 2 ]
[H 2 S] 2
b) Given the equilibrium concentrations: [H2S] = 0.1007 M, [H2] = 0.0219 M, and [S2] = 3.30 × 10-3 M, calculate the numerical value of Keq.
Keq =
[0.0219]2 [3.30 x 10 -3 ]
= 1.56 x 10-4
[0.1007]2
c) Assume the equilibrium is perturbed. When equilibrium is reestablished, the following concentrations are observed: [H 2] =
0.00287 M and [S2] = 0.171 M. Calculate [H2S] under these new conditions.
Rearrange the equation in (a)
[H 2 ] 2 [S 2 ]
[0.00287]2 [0.171]
[H2S] =
=
= 9.0289 x 10-3 M2
K eq
1.56 x 10 -4
2
[H2S] =
9.0289 x10 3 = 0.0950 M
d) What can you say about the forward and reverse reaction rates when the system is at equilibrium?
At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction.
e) ) How is Keq defined?
kf
kr
= Keq
14. a) Qualitatively, for the reaction shown in the previous problem, would the ∆G° value be positive or negative? Explain your
logic.
∆G would be positive, because Keq < 1.
b) Assuming I told you the reaction in 004, was relatively fast. Draw a reaction coordinate diagram that describes that system. Then draw another
showing how the system would be changed in the presence of a catalyst. Be sure to clearly label all important quantities.
(The Keq determined
in the previous
problem was less than
1, so the reaction is
reactant favored.
transition state (top of hill)
transition state (top of hill)
Eact uncatalyzed
2H2 + S2
2H2 + S2
Eact catalyzed
Go
Change in
G is positive
A relatively fast
reaction means a
relatively small hill.
(Actually write
“relatively small hill”
next to drawing.)
Go
2 H2S
Change in
G is positive
2 H2S
catalyzed reaction (smaller hill)
uncatalyzed reaction (small hill)
Reaction Progress
Reaction Progress
15. What is the definition of a Brønsted-Lowry acid?
A Brønsted-Lowry acid is a substance that can donate an H+.
16. Identify the acid and base on the reactant side of the equation shown below. Predict the productsof the reaction and indicate the
conjugate of each reactant.
NH3 + HBr 
NH3
base
+
HBr
acid

NH4+
+
conjugate
acid
Br conjugate
base
17.(a) Write a balanced chemical reaction for the dissociation of acetic acid (CH 3COOH) in water, and then
(b) write the Ka expression for that rxn. Acetic acid is a relatively weak acid. It is a weaker acid than formic acid.
(c) Would formic acid dissociate to a greater or lesser extent in water compared to acetic acid? (d) Which acid would have the
larger Ka?
a)
CH3COOH + H2O  CH3OO
-
[CH 3 COO - ] [H 3 O  ]
b) Ka =
CH 3 COOH
+
+ H3O
c) Formic acid would dissociate to a greater extent
d) Formic would have the larger Ka
18.(a) Calculate the [H3O+] of a solution that contains an [OH-] = 7.00 x 10-3 M solution. Is this solution acidic, neutral or basic?
14
a)
+
-
Kw = [H3O ] [OH ]
1x10
Kw
-12
Rearrange to get [H3O ] =
=
M
3 = 1.43 x 10
7.00 x10
[OH-]
+
b) Basic. (You can tell because [H3O+] < 1 x 10-7 M and [OH-] > 1 x 10-7 M.)