Name: ___
[ KEY ]________________________________
Date: ________________ Period: ____
REVIEW Unit 4 Test (Chp 4): Aqueous Reactions and Solution Stoichiometry
Multiple Choice polyatomic ions (names, charges, formulas)
6 strong acids (what does it to say a “strong” “acid” ?)
strong bases (group I and Ca, Ba, Sr with hydroxide, OH–) (what is a “strong” “base” ?)
electrolytes (strong, weak, non) (and what particles are actually in solution)
ALWAYS soluble (group I , NH4+ , NO3–)
Typically soluble (Cl– , Br– , I– except with AP: Ag+ & Pb2+)
NET ionic equations (comp – diss – cross – net – bal)
oxidation numbers of elements in compounds (like what is the ox. # of C in oxalate ion C2O42– ?)
redox reactions (LEO says GER)
activity of metals (which metals are least active – or – which metals do not lose electrons as well)
gas forming reactions ( ____ reacted with ____ will form which gas – H2(g) or CO2(g) )
solution stoichiometry ( L  g ) ( g  L ) ( g and L  M ) ( dilutions M1V1 = M2V2 )
Free Response
1) lab question: solution preparation, gravimetric analysis of precipitate to find % by mass (8 pts)
2) redox reactions, oxidation numbers, and oxidation/reduction half reactions (3 pts)
3) lab question: qualitative analysis of unknown solutions by precipitation (9 pts)
4) write 3 NET ionic equations from verbal descriptions, predict products, and balance (15 pts)
Section I Multiple Choice
NO Calculator Allowed
1
C) molecular (not an acid or base)
A) molecular  acid  not strong  must be weak
C) molecular  not acid, BUT IS a weak base
According to ALWAYS soluble rules,
B & C & E are soluble.
A) most S2– are insoluble (except for always list)
D) most CO32– are insoluble, CaCO3 is a very common precipitate
PbCl2 precipitates
KCl + Pb(NO3)2  PbCl2 + KNO3
2
A) Because…
NH4+ is soluble with all the listed anions, and…
and it contains S2– which makes H2S when H+ is added.
C) K+ and NO3- are spectator ions
that do not change charge or
state.
A) Cu(OH)2 precipitates
D) NiCl2 seems like a
possible precipitate, but
most chlorides are soluble
except for Ag, Pb, and Hg.
E) ALL NH4+ compounds are
soluble
3
E) BaSO4 precipitates, but Na+ and I– do not changes charge or state
C) strong acid plus strong base
produces a dissolved salt (Na2SO4)
not in the net ionic equation
(Na+ and SO42– are crossed out as
spectator ions)
Strong Base = completely dissociate in water (all group I and CBS from group II with OH–)
HNO3 is strong
KCl is a salt, NOT
and acid (no H)
A) True b/c NH4+ has an extra H+ it can donate as an acid (though not a strong acid)
and NH3 can accept an H+ acting like a base (though not a strong base)
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C) for example:
CaCO3 + H2SO4  CaSO4 + H2O + CO2
OR
NaHCO3 + HCl  NaCl + H2O + CO2
B) for example:
Mg + 2 HCl  MgCl2 + H2
D) Ag cannot replace H from acids (not active enough) (below hydrogen in activity series)
so it must be below Ni as well on the same series.
D) Cu cannot replace H from acids
(not active enough)
(below hydrogen in activity series)
E) S has a +6 in the sulfate ion in K2SO4
2K + S + 4O = 0
2(+1) + S + 4(-2) = 0
+2 + S + -8 = 0
+2 + S = +8
S = +6
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C) Cl goes from a +5 in chlorate in KClO3 to a –1 as chloride in KCl
(K+ does not change)
(O gets oxidized from –2 in chlorate in KClO3 to 0 in O2)
A) H goes from a +1 in water to a 0 as hydrogen gas
(still a +1 in NaOH)
C) Zn goes from a 0 to a +2
while Ag goes from +1 to 0
All others are double- replacement rxns
D) 0.20 mol BaI2 x 2 mol I– = 0.40 mol I– (assume 1 L)
1 mol BaI2
OR
BaI2(aq)
0.20 mol
I–
Ba2+
I–
 Ba2+(aq) +
 0.20 mol
 Ba2+
2 I–(aq)
0.40 mol
I–
I–
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1.00g BeO x 1 mol BeO = 0.040 mol BeO
25 g BeO
0.040 mol = 0.20 M
0.200 L
You should do the math with simple numbers, then move decimal places in your answers:
100/25 = 4  0.04
(move left 2 b/c 1.00 not 100)
4/2 = 2  0.02  0.2
(move left 2 b/c .04 not 4, right 1 b/c div. by .2 not 2)
M1V1 = M2V2 (key word dilution, though it won’t always be used)
(0.500)(40.0) = M2(200)
20.0 = M2
200
(leave in mL is fine)
M2 = 0.100 M
How many grams of methanol, CH3OH (MW = 32.0 g∙mol–1) are there in 500. mL of a 0.500 M CH3OH
solution?
0.500 L CH3OH x 0.500 mol CH3OH x 32.0 g CH3OH = 8.00 g CH3OH
1 L CH3OH
1 mol CH3OH
You should do the math with simple numbers, then move decimal places in your answers:
5/5 = 25  0.250
(move left 2 b/c 0.5 & 0.5 not 5 & 5)
0.250 x 32 = ¼ of 32 = 8
10.0g HClO4 x 1 mol HClO4 = 0.100 mol HClO4
100 g HClO4
0.100 mol = 0.40 M
0.250 L
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27. Beaker X and beaker Y each contain 1.0 L of solution, as shown above. A student combines the
solutions by pouring them into a larger, previously empty beaker Z and observes the formation
of a white precipitate.
Assuming that volumes are additive, which of the following sets of solutions could be
represented by the diagram above?
Beaker X
Beaker Y
Beaker Z
(A) 2.0 M AgNO3
2.0 M MgCl2
4.0 M Mg(NO3)2 and AgCl(s)
(B) 2.0 M AgNO3
2.0 M MgCl2
2.0 M Mg(NO3)2 and AgCl(s)
(C) 2.0 M AgNO3
1.0 M MgCl2
1.0 M Mg(NO3)2 and AgCl(s)
(D) 2.0 M AgNO3
1.0 M MgCl2
0.50 M Mg(NO3)2 and AgCl(s)
The concentration of Mg2+ in Y is half the concentration of Ag+ in X.
The concentration of Mg2+ in Z is half of what it is in Y because the volume doubled.
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Section II Free Response
Calculator Allowed
CLEARLY SHOW THE METHODS USED AND STEPS INVOLVED IN YOUR ANSWERS. It is to
your advantage to do this, because you may earn partial credit if you do and little or no credit if you do
not. Attention should be paid to significant figures.
1.
A student is assigned the task of preparing 50. mL of 6 M HNO3 to use in an experiment.
(a) The student is provided with a stock solution of 16 M HNO3 , two 100 mL graduated cylinders
that can be read to ±1 mL, a 100 mL beaker that can be read to ±10 mL, safety goggles, rubber
gloves, a glass stirring rod, a dropper, and distilled H2O.
(i) Calculate the volume, in mL, of 16 M HNO3 that the student should use for preparing 50. mL
of 6 M HNO3 . (1)
M1V1 = M2V2
(16 M)(Vi) = (6 M)(50. mL)
Vi= 19 mL or 20 mL (to one significant figure)
(ii) Briefly list the steps of an appropriate and safe procedure for preparing the 50. mL of
6 M HNO3. Only materials selected from those provided to the student (listed above)
may be used. (2)
Wear safety goggles and rubber gloves.
Then measure 20 mL of 16M HNO3 using a 100 mL graduated cylinder.
Measure 30 mL of distilled H2O using a 100 mL graduated cylinder.
Transfer the water to a 100 mL beaker.
Add the acid to the water with stirring.
(AAA : “always add acid” to minimize exothermic reaction and splashing from adding water to acid)
(iii) Explain why it is not necessary to use a volumetric flask (calibrated to 50.00 mL ±0.05 mL)
to perform the dilution. (1)
The graduated cylinders provide sufficient precision in volume measurement to provide
two significant figures, making the use of the volumetric flask unnecessary.
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In a second experiment, a student is given 2.94 g of a mixture containing anhydrous MgCl2 and KNO3.
To determine the percentage by mass of MgCl2 in the mixture, the student uses excess AgNO3(aq) to
precipitate the chloride ion as AgCl(s).
(b) Starting with the 2.94 g sample of the mixture dissolved in water, briefly describe the steps
necessary to quantitatively determine the mass of the AgCl precipitate. (2)
Add excess AgNO3 .
- Separate the AgCl precipitate (by filtration).
- Wash the precipitate and dry the precipitate completely.
- Determine the mass of AgCl by difference.
(c) The AgCl precipitate is filtered, washed, dried, and weighed to constant mass in a filter crucible.
The student determines the mass of the AgCl precipitate to be 5.48 g.
On the basis of this information, calculate each of the following.
(i) The number of moles of MgCl2 in the original mixture (2)
5.48 g AgCl × 1 mol AgCl = 0.0382 mol AgCl
143.32 g AgCl
0.0382 mol AgCl × 1 mol Cl × 1 mol MgCl2 = 0.0191 mol MgCl2
1 mol AgCl
2 mol Cl
(ii) The percent by mass of MgCl2 in the original mixture (1)
0.0191 mol MgCl2 × 95.20 g MgCl2 = 1.82 g MgCl2
1 mol MgCl2
1.82 g MgCl2 × 100% = 61.9% MgCl2 by mass
2.94 g sample
2. An aluminum metal container, Al(s), filled with an aqueous solution of copper(II) sulfate, CuSO4(aq)
eventually developed a leak. It is proposed that a redox reaction occurred that dissolved the
aluminum metal forming soluble aluminum sulfate, Al2(SO4)3(aq) and solid copper, Cu(s).
(a) Write a balanced equation for the overall reaction described above. Indicate whether the
Cu in CuSO4 is oxidized or whether it is reduced in the reaction. Justify your answer. (2)
2 Al + 3 CuSO4  Al2(SO4)3 + 3 Cu
Cu is reduced from +2 as an ion in CuSO4 to 0 as an element by gaining two electrons.
(b) Write the balanced half reaction for the oxidation reaction showing electrons. (1)
OX:
2 Al  2 Al3+ + 6 e–
RED: 3 Cu2+ + 6 e–  3 Cu
(remember to balance it with 2 Al therefore losing 6 e – NOT 3 e–)
(no need to write this RED half reaction, though it may help you balance e–)
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3. The identity of an unknown solid is to be determined. The compound is one of the seven salts in the
following table.
Al(NO3)3•9H2O
BaCl2•2H2O
CaCO3
NaCl
BaSO4
Ni(NO3)2•6H2O
CuSO4•5H2O
Use the results of the following observations or laboratory tests to explain how each compound in
the table may be eliminated or confirmed. The tests are done in sequence from (a) through (e).
(a) The unknown compound is white. In the table below, cross out the two compounds that can be
eliminated using this observation. Be sure to cross out these same two compounds in the tables
in parts (b), (c), and (d). (2)
(blue, seen in lab)
Al(NO3)3•9H2O
BaCl2•2H2O
NaCl
BaSO4
CaCO3
CuSO4•5H2O
–
Ni(NO3)2•6H2O (many transition metals have d orbitals for e
to move in emitting colors (freq’s) of light)
(b) When the unknown compound is added to water, it dissolves readily. In the table below, cross
out the two compounds that can be eliminated using this test. Be sure to cross out these same
two compounds in the tables in parts (c) and (d). (2)
(both CaCO3 & BaSO4 are insoluble precipitates)
Al(NO3)3•9H2O
BaCl2•2H2O
CaCO3
NaCl
BaSO4
Ni(NO3)2•6H2O
CuSO4•5H2O
(c) When AgNO3(aq) is added to an aqueous solution of the unknown compound, a white
precipitate forms. In the table below, cross out each compound that can be eliminated using
this test. Be sure to cross out the same compound(s) in the table in part (d). (1)
(NaCl and BaCl2 will form
AgCl(s) with AgNO3, but
Al(NO3)3 will not)
Al(NO3)3•9H2O
BaCl2•2H2O
CaCO3
NaCl
BaSO4
Ni(NO3)2•6H2O
CuSO4•5H2O
(d) When the unknown compound is carefully heated, it loses mass. In the table below, cross out
each compound that can be eliminated using this test. (1)
(losing mass must be water
loss from dehydrating a
hydrated salt)
Al(NO3)3•9H2O
BaCl2•2H2O
CaCO3
NaCl
BaSO4
Ni(NO3)2•6H2O
CuSO4•5H2O
(e) Describe a test that can be used to confirm the identity of the unknown compound identified in
part (d). Limit your confirmation test to a reaction between an aqueous solution of the unknown
compound and an aqueous solution of one of the other soluble salts listed in the tables above.
Describe the expected results of the test; include the formula(s) of any product(s).
Mix an aqueous solution of BaCl2∙ 2H2O with an aqueous solution of CuSO4∙ 5H2O .
BaSO4 will precipitate.
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4. For each of the following three reactions, in part (i) write a balanced equation for the reaction and in
part (ii) answer the question about the reaction. In part (i), coefficients should be in terms of lowest
whole numbers. Assume that solutions are aqueous unless otherwise indicated. Represent
substances in solution as ions if the substances are extensively ionized. Omit formulas for any ions
or molecules that are unchanged by the reaction. You may use the empty space for scratch work, but
only equations that are written in the answer boxes provided will be graded.
(a) A solution of barium hydroxide is added to a solution of acetic acid (ethanoic acid).
(i) Balanced equation: (4) (weak acids are not dissociated)
–
–
OH + HC2H3O2  H2O + C2H3O2
or
OH– + CH3COOH  H2O + CH3COO–
(ii) Which substance is a weak electrolyte. Explain. (1)
___ HC2H3O2 is a weak electrolyte because it exists mostly as molecules in solution__
rather than as dissociated ions______________________________________________
or
Ba(OH)2 + HC2H3O2  H2O + Ba(C2H3O2)2
Ba(OH)2 + CH3COOH  H2O + Ba(CH3COO)2
(b) A solution of potassium sulfide is added to a solution of iron(III) chloride.
(i) Balanced equation: (4)
3 S2– + 2 Fe3+  Fe2S3
(ii) Classify this reaction in two ways. (1)
___double-replacement and precipitation_____________________________________
______________________________________________________________________
3 K2S + 2 FeCl3  Fe2S3 + 6 KCl
(c) Excess nitric acid is added to solid sodium bicarbonate.
(i) Balanced equation: (4) (solids are not dissociated)
H+ + NaHCO3  Na+ + H2O + CO2
(ii) If this reaction takes place in a rigid sealed container, what change (if any) is observed in the
air pressure inside the container? (1)
___The pressure would increase because a gas is produced.______________________
______________________________________________________________________
HNO3 + NaHCO3  NaNO3 + H2O + CO2
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ANSWER KEY
1. 2011
2008
#2(a)
(dilution)
#2(d)(e)
(gravimetric analysis)
2. 2006B #8(a) (edited)
3. 2008B #5
4.
(a) NH4+ b/c it donates a proton (H+) (Bronsted-Lowry)
(b) precipitation, double replacement
(c) pressure increases due to gas produced in reaction.
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