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_________
Page 1/9
Chemistry 125/126, Exam 2
Tuesday, April 20, 2010
Welcome to the first hourly exam for Chemistry 125/126. This exam consists of 7 questions worth a
total of 75 points plus a bonus question worth 3 points for a possible total of 78 points. It is 10 pages
long; 8 pages of questions including a bonus question and periodic tables (page 10).
To receive credit, your answers must be placed in the indicated spaces or boxes. If calculations are
required to obtain an answer, show your calculations; you will not receive credit for numerical
answers alone. Do not write in any box marked "For use by grader".
The exam should take about l hour, however, you may use up to 1.5 hours to complete the exam. All
exams must be turned in at 7:45 pm. Save any questions you have during the exam for Prof. Kerner.
The GSIs have been told not to answer questions during the exam.
Graded exams will be available starting tomorrow. If, after checking the exam key, you feel there
was a grading mistake, turn your exam in with a statement describing the mistake to Prof. Kerner’s
mailbox in 1500o chem. All re-grades must be turned in by noon, Monday, April 26, 2010.
Course Information
Section
GSI
127
129
131
135
137
139
141
143
T AM
T AM
T AM
T PM1
T PM1
T PM1
T PM1
T PM2
Kevin Hartman
Rahul Rattan
Lauren Soblosky
Zhenxin Lin
Rahul Rattan
Rachel Barnard
Russell Bornschein
Joshua Skodack
Section
GSI
191 T PM2
Zhenxin Lin
237 Th PM2 Blake Erickson
195
199
233
225
227
231
235
Shuwen Sun
Heidi Pedini
Lauren Soblosky
Shuwen Sun
Blake Erickson
Rachel Barnard
Heidi Pedini
239
241
251
253
119
123
T PM2
T PM2
ThPM1
Th PM1
Th PM1
Th PM1
Th PM2
Section
GSI
Th PM2
Th PM2
F PM1
F PM1
F PM2
F PM2
Russell Bornschein
Joshua Skodack
Akiko Kochi
Kevin Hartman
Akiko Kochi
Alaina DeToma
For use by grader
Page
Points
Score
Name:
GSI:
Section:
e-mail:
2
15
3
10
4
09
5
07
6
09
7
10
8
12
9
03 (+03)
TOTAL
75 (+03)
Name:__Grading key
_________
Page 2/9
Question 1 (15 points) requires you to use your knowledge of structure and properties and
periodicity to determine the identity of chemicals. Periodic tables are provided at the back of the
exam for your reference.
A. A white compound dissolves in water (pH 7) and the resulting solution tests acidic.
Circle the white compound testing acidic in water:
Al(NO3)3
Na2CO3
Zn(OH)2
3 points or zero
B. You need to distinguish between solid white compounds. Tests that you can perform are:
1.
Determine if the compounds dissolve in 1M NH3.
2.
3.
Determine if the compounds are soluble in water.
Determine the pH values of aqueous solutions of the compounds.
A desirable test produces notably different results for the two compounds. Select the
number (1, 2, or 3) for the one best test for distinguishing between the two compounds.
White Compounds
Test (1, 2, or 3)
C.
BaCO3
vs
ZnCO3
1
Ca(NO3)2
vs
Pb(NO3)2
3
AgNO3
vs
AgCl
2
2 points or zero each
Predict the comparative base strengths of RbOH, Sr(OH)2, and Ga(OH)3.
Strongest base
RbOH
Weakest base
>
Sr(OH)2
>
Ga(OH)3
3 points or zero
D.
Predict the comparative acid strengths of HBrO3, HIO3, and HClO3.
Strongest acid
HClO3
3 points or zero
Weakest acid
>
HBrO3
>
HIO3
Name:__Grading key
_________
Page 3/9
Question 2 (16 points) deals with a study of acids and bases and their measured pH values:
Sample
pH
Sample
pH
0.10 M pentanoic acid
2.89
0.10 M nitric acid (HNO3)
1.0
(C4H9COOH or HC5H9O2)
0.10 M sulfurous acid (H2SO3)
1.94
0.10 M ammonia (NH3)
11.0
0.10 M boric acid (H3BO3)
4.10
0.10 M potassium hydroxide (KOH)
13.0
A. Based on the pH data compare the acid strengths of pentanoic, sulfurous, and boric acids.
strongest acid
sufurous
weakest acid
>
pentaonoic
>
boric
3 points or zero
B.
The electronegativity of phosphorous (P) is greater than the electronegativity of boron (B).
Predict the comparative pH of 0.10 M H3PO3 and 0.10 M H3BO3.
Indicate (circle) if the pH of 0.10 M H3PO3 is
Less than
equal to
greater than
pH 4.10 (pH of 0.10 M H3BO3)
3 points or zero
C.
Based on the pH data choose the reaction below that best describes the reaction of pentanoic
acid with water. Place an X by the best reaction. (2 points). For the reaction that you choose,
label each species as an “acid” or a “base”, according to how it reacts and draw lines
connecting each conjugate acid-base pair. (2 points)
Best Reaction
HC5H9O2 + H2O
OH-
+ H2C5H9O2+
_________________ HC5H9O2 + H2O
OH-
+ H2C5H9O2+
_________________
_________________
_____ X (2 pts)____
+ H2O
H3O+
+ C5H9O2___________________
___________________________________________
HC5H9O2
Acid
BASE
ACID
Base(2 pts)
HC5H9O2 + H2O
H3O+
+ C5H9O2__________________
___________________________________________
Name:__Grading key
D.
_________
You titrate a sample of pentanoic acid (C4H9COOH or HC5H9O2) with 0.10 M NaOH. What is
the formula for the salt product? Is the salt product acidic, basic, or neutral?
Formula for the salt product:
Indicate (circle) if the salt product is:
basic
acidic
C4H9COONa or NaC5H9O2
2 points or zero
E.
Page 4/9
neutral
2 points or zero
Based on the above pH data, will 0.10 M KHSO3 or 0.10 M NH4HSO3 have a lower pH?
Indicate (circle) the 0.10 M salt solution with the lower pH:
NH4HSO3
KHSO3
2 points or zero
Question 3 (3 points) requires you to use your knowledge of Lewis acids and bases to explain the
acidity metal ion iron (II).
You add Fe(Cl)2(s) to water to form a clear 0.10 M solution. The picture below represents the
Lewis acid iron(II) ion reacting with a sample of the Lewis base water:
2+
Fe
..
..O
H
+
H
..
.. O
H
? products
H
Circle the correctly completed and balanced equation showing the products in the clear,
colorless, acidic solution.
1.
[Fe(H2O)]2+ + H2O → [Fe(OH)]2+ + H3O+
2.
[Fe(H2O)]2+ + H2O → [Fe(H3O)]2+ + OH-
3.
[Fe(H2O)]2+ + H2O
4.
[Fe(H2O)]2+ + H2O → [FeO]
[Fe(H2O)]2+ + H2O → [Fe(OH)]+
5.
3 points or zero
→ [Fe(OH)]+
+ H 3O +
+ 2H+
+
H+
+ H2O
+
OH-
Name:__Grading key
_________
Page 5/9
Question 4 (7 points) deals with your titration of 0.202 g of an amino acid with the molecular
formula C5H9O4N. The end point occurs when you have added 27.50 mL of 0.100 M NaOH. You
determine that the equivalent weight of the amino acid is 73.54.
A.
How many ionizable protons are there per molecule of the amino acid C5H9O4N? For credit
be sure to show your calculations.
Ionizable protons:
= Molecular Wt. = 5 (12) + 9 (1) + 4 (16) + 14 = 147 (1 pt for correct MW;
Equivalent Wt.
73.5
1 pt for dividing by EW)
Ionizable protons = 2 (1 pt. for correct calculation)
# Ionizable protons/molecule = = 2
3 points
B.
Your titration results are correct. Your teammate conducts an identical titration (using 0.202g of
the amino acid and 0.100 M NaOH) but her results are NOT correct.
Error:
1.
Your titration was conducted using the indicator phenolphthalein (pH end point = pH
8.2-10.0). Your teammate used methyl red (pH end point = pH 4.8-6.0).
Determine the impact of the indicator error choice on the volume of NaOH used by your
teammate to reach the end point.
Indicate (circle) the volume of NaOH used by your teammate at the end point.
more than
2 points or zero
equal to
less than
27.50 mL
2. Determine the impact of the indicator error choice on your teammates calculated equivalent
weight value.
Indicate (circle) the effect of the error on the calculated equivalent weight of your
teammate.
greater than
equal to
less than
73.54 (eq.wt)
2 points ONLY awarded for “greater than” choice if “less than” circled in box above. Any
other combos constitute a guess and lack of knowledge/understanding so no credit
Name:__Grading key
_________
Page 6/9
Question 5 (13 points) deals with the reaction of 0.10 M CoCl2 with 0.10 M Na2CO3 and 1.0 M
NH3 where the following observations are recorded:
Step 1.
CoCl2(aq)
Clear red
Step 2. Step 1 mixture
+
+
Na2CO3 (aq) →
colorless
NH3(aq)
colorless
red precipitate (ppt.)
→ red ppt. dissolves; clear yellow solution forms
Information: Co2+(aq) = [Co(H2O)6]2+.
A. When 0.10 M Na2CO3 is added to 0.10 M CoCl2 a Lewis acid-base reaction occurs resulting in
the formation of a red precipitate.
The Lewis base species reacting to form the red precipitate =
CO32-
2 points or zero
The Lewis formula for the red precipitate formed when Na2CO3(aq) reacts with CoCl2(aq)
=
[Co(H2O)5CO3]
3 points or zero
B. In step 2 (above), the addition of NH3(aq) causes the red precipitate formed in step 1 to
dissolve. Record the formula for the species (Lewis acid or base) in the step 1 equilibrium
system reacting with the NH3(aq).
The species reacting upon addition of NH3(aq) =
[Co(H2O)6]2+
2 points or zero
C.
What will you observe if you add drops of 1 M HNO3 to the step 2 product mixture (yellow
solution) and stir till an excess amount of 1 M HNO3 has been added?
Indicate the correct sequence of observations you record (circle a single choice next to each of steps 3, 4,
and 5) as you add drops of 1 M HNO3 to the step 2 product mixture till excess HNO3 is added.
Step 3. The solution remains yellow
A red ppt. appears
Step 4. The solution remains yellow
The red ppt dissolves
Step 5. The solution remains yellow The
2 points or zero
The yellow solution turns clear red
The solution remains clear red
solution turns clear red
The solution remains clear red
Name:__Grading key
D.
_________
Page 7/9
Indicate in five words or less what you will visually observe if you reverse the order of addition of
reagents to the of 0.10 M NiCl2? Note that only the first five words of each recorded observation
will be read and graded.
Visual observation?
Step 6. CoCl2(aq)
Clear red
+
NH3(aq)
colorless
→
Step 7. Step 6 mixture + Na2CO3(aq) →
colorless
Red ppt dissolves to yellow soln#
(1 point or 0)
or (“red solution turns yellow”)#
The solution remains yellow( 3 points)*
( or “ no change” if step 6 correct = 3 pts)
( or “ no reaction” if step 6 correct = 1 pts)
If student indicates “no reaction” it is true but that is not an observation and thus = only 1 pt
# observations depend on if cobalt chloride is added to ammonia or ammonia is added to
cobalt chloride
*Only the first five words of each recorded observation are to be read and graded
4 points
Question 6 (6 points) deals with your analysis of a reaction where you add the metal Pd to a yellowgreen solution of 0.10 M FeCl2(aq) and observe no apparent reaction.
Circle every observation (1and/or 2 and/or 3) that can help confirm that NO reaction took place.
Observation
1.
Add hexane to the reaction mixture and the hexane remains clear and
colorless.
2.
Add the metal Fe to 0.10 M PdCl2 and observe a reaction.
3.
Add 0.10 M FeCl3(aq) to 0.10 M PdCl2 and observe no reaction.
For use of grader only
6 points for circling 1 and 2 and NOT circling 3
Each correct observations (1, 2, and 3) = 2 points for each
Name:__Grading key
_________
Page 8/9
Question 7 (15 points) deals with the analysis of the reaction between CuSO4(aq) and SnCl2(aq):
CuSO4 (aq) +
blue
→
loss of blue color and formation of a white precipitate.
colorless
Reference Tests
1. CuSO4 (aq)
2.
blue
CuSO4 (aq)
3.
blue
Na2SO4 (aq)
colorless
A.
SnCl2 (aq)
+
NaCl (aq)
→
no reaction.
+
colorless
Sn(NO3)2(aq)
→
loss of blue color. A clear, colorless solution forms.
+
colorless
SnCl2 (aq)
→
no reaction.
colorless
What does test 2 , by itself, tell you about the reaction of CuSO4 (aq) and SnCl2 (aq)?
What do you know about the CuSO4 (aq) and SnCl2 (aq) reaction from test 2?
Cl- is critical to the formation of the white precipitate
No credit if the ion Cl- not recorded regardless of the interpretation; 1 pt if ion Clrecorded but answer does not include fact that it is critical to formation of the white
precipitate—e.g. if answer only indicates it is a critical species.
3 points
B.
What does test 3 , by itself, tell you about the reaction of CuSO4 (aq) and SnCl2 (aq)?
What do you know about the CuSO4 (aq) and SnCl2 (aq) reaction from test 3?
Cu2+ is critical to the reaction (or loss of blue color and formation of white ppt)
No credit if the ion Cu2+ not recorded
3 points
C.
What is the identity of the white precipitate formed when CuSO4 (aq) and SnCl2 (aq) react?
The formula of the white precipitate product =
CuCl or (CuCl or SnCl4) (3 points)
SnCl4 only answer = 1 point
D.
What is the identity of the reacting species causing the loss of blue color? Note: For credit do
not record any spectator species!
The reacting species are =
Cu2+
and
Sn2+
3 points or zero (both species must be recorded and correct)
Name:__Grading key
_________
Page 9/9
Based on the reaction, what are the relative strengths of Cu2+ vs. Sn4+ as oxidizing agents?
Strongest
Weakest Oxidizing Agent
Cu2+
Sn4+
>
3 points or zero
BONUS QUESTION (3 points)
Baking powder contains sodium bicarbonate, NaHCO3, and the double sulfate salt,
NaAl(SO4)2. When added to moist dough, small bubbles of CO2 give the finished
baked product a light and porous texture:
____?____
+
H3O+
→
CO2(g)
+ 2 H2O
1. Fill in the missing species in the equation for the reaction of baking powder:
HCO3-
+
H3O+
→
CO2(g)
+ 2 H2O
2. Identify the species in the baking powder that is critical to the production of the hydronium
ions, when the baking powder is added to moist dough:
Al3+(aq) or [Al(H2O)6]3+
3 points or zero (both answers must be correct for credit)