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SOLUTIONS - CHAPTER 9 Problems
NOTE: Exam 3 will be on Tuesday, November 22 nd, from 8:30pm to 10:30pm. Room
assignments for the exam are the same as for the first two exams. The exam will cover
the following chapters: Chapter 6, sections 6.4 to 6.6; Chapter 7, all; Chapter 8, all;
Chapter 9, all).
1) (Burdge, 9.2) What is the difference between a nonelectrolyte and an electrolyte?
Between a weak electrolyte and a strong electrolyte?
An electrolyte is a substance that produces ions when dissolved in water. A nonelectrolyte is a substance that does not produce ions when dissolved in water (such as
sugar). Note that water itself is considered a nonelectrolyte, although it does produce a
very small (~ 10-7 mol/L) concentration of ions at room temperature due to selfionization.
A strong electrolyte completely ionizes when dissolved in water to form two or
more ions per formula unit of compound. A weak electrolyte ionizes to only a small
extent, producing much less than two ions per formula unit of dissolved compound.
Strong acids and ionic compounds are strong electrolytes, while weak acids and weak
bases are weak electrolytes. Note that “insoluble” ionic compounds are strong
electrolytes, because for whatever amount of the compound dissolves in water two or
more ions are formed.
Figure 9.2 on page 322 gives a useful flow chart for classifying substances as
strong electrolytes, weak electrolytes, or nonelectrolytes.
2) While an aqueous solution of sodium chloride (NaCl) conducts electricity, solid
sodium chloride does not. Explain this observation.
Solid sodium chloride consists of cations and anions arranged in a crystal
structure. Since the ions cannot move, they cannot transfer electricity, and so solid
sodium chloride is an insulator.
When sodium chloride is dissolved in water it forms free cations and anions
(Na+(aq) and Cl-(aq)) that can freely move throughout the solution. The ions can
therefore be used to transport electricity. A solution of NaCl in water is therefore an
electrical conductor.
3) (Burdge, 9.12) Identify each of the following substances as a strong electrolyte, a weak
electrolyte, or a nonelectrolyte.
a) H2O
nonelectrolyte
b) KCl
strong electrolyte (ionic compound)
c) HNO3
strong electrolyte (strong acid)
d) HC2H3O2
weak electrolyte (weak acid)
e) C12H22O11
nonelectrolyte (a disaccharide, a form of sugar)
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4) Identify each of the following compounds as a strong electrolyte, a weak electrolyte, or
a nonelectrolyte.
a) C6H12O6
nonelectrolyte (a monosaccharide, a molecular compound)
b) HNO2
weak electrolyte (a weak acid)
c) KOH
strong electrolyte (a strong soluble base)
d) Fe(OH)3 strong electrolyte (an insoluble strong base. Recall that all ionic
compounds are considered strong electrolytes, because even if only a small amount of the
compound dissolves in water, the amount that does dissolve breaks apart to form Fe +3 and
OH-.)
e) CH3CH2OH nonelectrolyte (an alcohol, and so a molecular compound)
5) (Burdge, 9.14) Describe hydration. What properties of water enable its molecules to
interact with ions in solution?
Hydration is the process where particles (often ions) are surrounded by water
molecules. Because water molecules have a partial positive charge on the hydrogen
atoms and a partial negative charge on the oxygen atom, the molecules can strongly
interact with anions and cations. Water molecules can pull ions out of a crystal structure,
and by surrounding the ions, minimize interactions between cations and anions.
6) (Burdge, 9.20) Characterize the following compounds as soluble or insoluble in water:
a) CaCO3
Insoluble
b) ZnSO4
Soluble (most sulfate compounds are soluble)
c) Hg(NO3)2 Soluble (nitrate compounds are soluble)
d) HgSO4
Soluble (most sulfate compounds are soluble. Note that
Hg2SO4 (mercury (I) sulfate) is insoluble)
e) NH4ClO4 Soluble (both ammonium and perchlorate compounds are soluble)
7) Classify the following compounds as soluble or insoluble in water.
a) AgCl
Insoluble (most silver compounds are insoluble in water)
b) Ba(C2H3O2)2
Soluble (compounds containing acetate ion are soluble)
c) Cu(OH)2
Insoluble (most hydroxide compounds are insoluble)
d) CaF2
Insoluble (most compounds containing F- ion are insoluble)
e) ZnCl2
Soluble (most chloride compounds are soluble)
f) Ni(ClO3)2
Soluble (most chlorate compounds are soluble)
8) (Burdge, 9.22) Write total ionic and net ionic equations for the following reactions:
a) Na2S(aq) + ZnCl2(aq)  2 NaCl(aq) + ZnS(s)
TOTAL
NET
2 Na+(aq) + S2-(aq) + Zn2+(aq) + 2 Cl-(aq)
 2 Na+(aq) + 2 Cl-(aq) + ZnS(s)
Zn2+(aq) + S2-(aq)  ZnS(s)
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b) 2 K3PO4(aq) + 3 Sr(NO3)2  6 KNO3(aq) + Sr3(PO4)2(s)
TOTAL
NET
6 K+(aq) + 2 PO43-(aq) + 3 Sr2+(aq) + 6 NO3-(aq)
 6 K+(aq) + 6 NO3-(aq) + Sr3(PO4)2(s)
2+
33 Sr (aq) + 2 PO4 (aq)  Sr3(PO4)2(s)
c) Mg(NO3)2(aq) + 2 NaOH(aq)  2 NaNO3(aq) + Mg(OH)2(s)
TOTAL
NET
Mg2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 OH-(aq)
 2 Na+(aq) + 2 NO3-(aq) + Mg(OH)2(s)
Mg2+(aq) + 2 OH-(aq)  Mg(OH)2
9) (Burdge, 9.27) Give the Arrhenius and the Bronsted definition of an acid and a base.
Why are the Bronsted definitions more useful in describing acid-base properties?
A Bronsted acid acts as a proton donor (H+ donor) in a chemical reaction, while a
Bronsted base acts as a proton acceptor. For example, in the equation
HCl(aq) + NH3(aq)  NH4+(aq) + Cl-(aq)
HCl is donating a proton, and so is a Bronsted acid, and NH3 is accepting a proton, and so
is a Bronsted base.
The Bronsted definition of acids and bases is usually (although not always) more
useful than the Arrhenius definition because it is less restrictive. It allows classification
of reactions as acid-base reactions (in the Bronsted sense) even for processes not taking
place in water.
10) (Burdge, 9.30) What factors qualify a compound as a salt? Specify which of the
following compounds are salts:
A salt is an ionic compound that can be formed by an acid-base reaction. In
practical terms, all common ionic compounds except hydroxides and oxides are salts.
In the answers below, I have given an acid-base reaction that will form the
compounds classified as salts.
a) CH4
b) NaF
c) NaOH
d) CaO
e) BaSO4
f) HNO3
g) NH3
h) KBr
not a salt (a molecular compound)
a salt ( NaOH + HF  NaF + H2O)
not a salt (although an ionic compound)
not a salt (although an ionic compound)
a salt (Ba(OH)2 + H2SO4  BaSO4 + 2 H2O)
not a salt (a strong acid)
not a salt (a weak base)
a salt (KOH + HBr  KBr + H2O)
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11) (Burdge, 9.31) Identify each of the following as a weak or strong acid or base:
a) NH3
weak base
b) H3PO4
weak (polyprotic) acid
c) LiOH
strong soluble base
d) HCOOH (formic acid)
weak acid
e) H2SO4
strong acid
f) HF
weak acid
g) Ba(OH)2
strong soluble base
12) (Burdge, 9.32) Identify each of the following species as a Bronsted acid, a Bronsted
base, or both.
a) HI
Bronsted acid.
b) C2H3O2-
Bronsted base. To see this, consider the reaction of this ion with
water.
C2H3O2-(aq) + H2O(l)  HC2H3O2(aq) + OH-(aq)
c) H2PO4-
Both (and so amphoteric). Consider the following reactions
H2PO4-(aq) + H2O(l)  HPO42-(aq) + H3O+(aq)
H2PO4-(aq) + HCl(aq)  H3PO4(aq) + Cl-(aq)
In the first reaction the hydrogen phosphate ion donates a proton and so acts as a
Bronsted acid. In the second reaction the ion accepts a proton and so acts as a base.
d) HSO4-
Both (and so amphoteric).
13) (Burdge, 9.34) Balance the following unbalanced equations, and write the molecular
equations, total ionic equations, and net ionic equations (if appropriate):
a) HBr(aq) + NH3(aq) 
MOLECULAR
HBr(aq) + NH3(aq)  NH4Br(aq)
TOTAL IONIC
H+(aq) + Br-(aq) + NH3(aq)  NH4+(aq) + Br-(aq)
NET IONIC
H+(aq) + NH3(aq)  NH4+(aq)
b) Ba(OH)2(aq) + H3PO4(aq) 
MOLECULAR
3 Ba(OH)2(aq) + 2 H3PO4(aq)  Ba3(PO4)2(s) + 6 H2O(l)
TOTAL IONIC
3 Ba2+(aq) + 6 OH-(aq) + 2 H3PO4(aq)
 Ba3(PO4)2(s) + 6 H2O(l)
NET IONIC
the same as the total ionic (no spectator ions)
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c) HClO4(aq) + Mg(OH)2(s) 
MOLECULAR
2 HClO4(aq) + Mg(OH)2(s)  Mg(ClO4)2(aq) + 2 H2O(l)
TOTAL IONIC
2 H+(aq) + 2 ClO4-(aq) + Mg(OH)2(s)
 Mg2+(aq) + 2 ClO4-(aq) + 2 H2O(l)
NET IONIC
2 H+(aq) + Mg(OH)2  Mg2+(aq) + 2 H2O(l)
14) Balance the following unbalanced equations, and write the molecular equations, total
ionic equations, and net ionic equations (if appropriate):
a) AgNO3(aq) + CuCl2(aq) 
MOLECULAR
TOTAL IONIC
NET IONIC
2 AgNO3(aq) + CuCl2(aq)  Cu(NO3)2(aq) + 2 AgCl(s)
2 Ag+(aq) + 2 NO3-(aq) + Cu2+(aq) + 2 Cl-(aq)
 Cu2+(aq) + 2 NO3-(aq) + 2 AgCl(s)
+
Ag (aq) + Cl (aq)  AgCl(s)
b) CuCl2(aq) + NaClO4(aq) 
MOLECULAR
TOTAL IONIC
NET IONIC
CuCl2(aq) + 2 NaClO4(aq)  Cu(ClO4)2(aq) + 2 NaCl(aq)
Cu2+(aq) + 2 Cl-(aq) + 2 Na+(aq) + 2 ClO4-(aq)
 Cu2+(aq) + 2 Cl-(aq) + 2 Na+(aq) + 2 ClO4-(aq)
no reaction
All of the ions are spectator ions, and so no reaction occurs.
15) (Burdge, 9.42) For the complete redox reactions given here, write the half-reactions
and identify the oxidizing and reducing agents:
a) 4 Fe + 3 O2  2 Fe2O3
oxidation
4 Fe  4 Fe3+ + 12 ereduction
3 O2 + 12 e-  6 O2oxidizing agent is O2; reducing agent is Fe
b) Cl2 + 2 NaBr  2 NaCl + Br2
oxidation
2 Br-  Br2 + 2 ereduction
Cl2 + 2 e-  2 Cloxidizing agent is Cl2; reducing agent is Brc) Si + 2 F2  SiF4
oxidation
Si  Si4+ + 4 ereduction
2 F2 + 4 e-  4 Foxidizing agent is F2; reducing agent is Si
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d) H2 + Cl2  2 HCl
oxidation
H2  2 H+ + 2 ereduction
Cl2 + 2 e- 2 Cloxidizing agent is Cl2; reducing agent is H2
16) (Burdge, 9.42) Phosphorus forms many oxoacids (ternary acids containing H, O, and
P). Indicate the oxidation number of phosphorus in each of the following acids:
Note that in all of the compounds below hydrogen has an oxidation number of +1
and oxygen has an oxidation number of -2.
a) HPO3
+5
b) H3PO2
+1
c) H3PO3
+3
d) H3PO4
+5
e) H4P2O7
+5
f) H5P3O10
+5
17) (Burdge, 9.48) Give the oxidation number of the underlined atoms in each of the
following molecules and ions:
a) Mg3N2
Mg is +2, and so N is -3
b) CsO2
Cs is +1, and so O is -1/2 (weird, but fractional oxidation
numbers occasionally occur)
c) CaC2
Ca is +2, and so C is -1
d) CO32O is -2, and so C is +4
e) C2O42O is -2, and so C is +3
2f) ZnO2
Zn is +2, and so O is -2
g) NaBH4
Na is +1, H is -1, and so B is +3 (I got this answer using a
more sophisticated method for finding oxidation numbers,
and so would not expect you to be able to do this one)
2h) WO4
O is -2, and so W is +6
18) Give the oxidation numbers for all atoms in the following molecules or ions:
a) C2H5OH
H is +1, O is -2, and so C is -2
b) CO2
O is -2, and so C is +4
c) ClO4O is -2, and so Cl is +7
d) PbCl2
Cl is -1, and so Pb is +2
e) XeF6
F is -1, and so Xe is +6
f) H3PO4
H is +1, O is -2, and so P is +5
g) IO3
O is -2, and so I is +5
h) H2O2
H is +1, and so O is -1 (this occurs in peroxide compounds)
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19) (Burdge, 9.52) Predict the outcomes of the reactions represented by the following
equations by using the activity series, and balance the equations:
a) Cu(s) + HCl(aq) 
no reaction (copper is below hydrogen in the
activity series)
b) Au(s) + NaBr(aq) 
no reaction (gold is below sodium in the
activity series)
c) Mg(s) + CuSO4(aq) 
MgSO4(aq) + Cu(s)
d) Zn(s) + KBr(aq) 
no reaction
20) (Burdge, 9.54) Classify the following redox reactions as combination, decomposition,
or displacement:
a) P4 + 10 Cl2  4 PCl5
combination
b) 2 NO  N2 + O2
decomposition
c) Cl2 + KI  2 KCl + I2
displacement (Cl displaces I)
d) 3 HNO2  HNO3 + H2O + 2 NO
decomposition
21) (Burdge, 9.66) Calculate the molarity of each of the following solutions:
a) 6.57 g of methanol (CH3OH) in 1.50 x 102 mL of solution.
M(CH3OH) = 32.04 g/mol
moles CH3OH = 6.57 g 1 mol = 0.2051 mol
32.04 g
[CH3OH] = 0.2051 mol = 1.37 mol/L
0.150 L
b) 10.4 g of calcium chloride (CaCl2) in 2.20 x 102 mL of solution
M(CaCl2) = 110.98 g/mol
moles CaCl2 = 10.4 g
1 mol = 0.09371 mol
110.98 g
[CaCl2] = 0.09371 mol = 0.426 mol/L
0.220 L
c) 7.82 g of naphthalene (C10H8) in 85.2 mL of benzene solution
M(C10H8) = 128.2 g/mol
moles C10H8 = 7.82 g 1 mol = 0.0700 mol
128.2 g
[C10H8] = 0.0700 mol = 0.716 mol/L
0.0852 L
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22) (Burdge, 9.70) Water is added to 25.0 mL of a 0.866 M KNO3 solution until the
volume of the solution is exactly 500. mL. What is the concentration of the final
solution?
McLc = MdLd
Md = (Lc/Ld) Mc = (25.0 mL/500. mL) (0.866M) = 0.0433 M
23) (Burdge, 9.98) A sample of 0.6760 g of an unknown compound containing barium
ions (Ba2+) is dissolved in water and treated with excess Na 2SO4. If the mass of the
BaSO4 precipitate formed is 0.4105 g, what is the percent by mass of Ba in the original
unknown compound?
The reaction taking place is
Ba2+(aq) + SO42-(aq)  BaSO4(s)
M(BaSO4) = 233.4 g/mol
M(Ba2+) = 137.3 g/mol
moles precipitate = 0.4105 g
1 mol =1.759 x 10-3 mol
233.4 g
grams Ba2+ ion = 1.759 x 10-3 mol BaSO4 1 mol Ba2+ 137.3 g =0.2415 g Ba2+
1 mol BaSO4 1 mol
% Ba by mass = 0.2415 g 100. % = 35.7 %
0.6760 g
24) For each of the following aqueous solutions find the pH of the solution, and classify
the solution as acidic, basic, or neutral.
a) [H+] = 3.7 x 10-3 M
pH = - log10(3.7 x 10-3) = 2.43
acidic
b) [H+] = 9.2 x 10-12 M
pH = - log10(9.2 x 10-12) = 11.04
basic
c) [OH-] = 5.9 x 10-5 [H+] = (1.00 x 10-14)/(5.9 x 10-5) = 1.7 x 10-10 M
pH = - log10(1.7 x 10-10) = 9.77
8
basic
25) (Burdge, 9.102) Calculate the concentration (in molarity) of an NaOH solution if 25.0
mL of the solution is needed to neutralize 17.4 mL of a 0.312 M HCl solution.
The neutralization reaction is
HCl + NaOH  NaCl + H2O
moles NaOH = 0.0174 mL HCl soln 0.312 mol HCl 1 mol NaOH = 5.429 x 10-3 mol
L soln
1 mol HCl
[NaOH] = 5.429x 10-3 mol = 0.2172 mol/L
0.0250 L
26) (Burdge, 9.114) Sodium carbonate (Na2CO3) is available in very pure form and can
be used to standardize acid solutions. What is the molarity of an HCl solution if 28.3 mL
of the solution is required to react with 0.256 g of Na2CO3?
The reaction taking place is
2 HCl + Na2CO3  2 NaCl + H2CO3
M(Na2CO3) = 106.0 g/mol
moles HCl = 0.256 g Na2CO3 1 mol Na2CO3 2 mol HCl
= 4.830 x 10-3 mol
106.0 g
1 mol Na2CO3
[HCl] = 4.830 x 10-3 mol = 0.1707 mol/L
0.0283 L
27) (Burdge, 9.146) A 0.8870 g sample of a mixture of NaCl and KCl is dissolved in
water, and the solution is then treated with an excess of AgNO 3, to yield 1.913 g of AgCl.
Calculate the percent by mass of each compound in the mixture.
This problem requires a bit of thought. The reaction of interest involves chloride ion
Ag+(aq) + Cl-(aq)  AgCl(s)
Based on the mass of precipitate formed we can determine the grams of Cl-. Since the
only sources of chloride ion are the NaCl and KCl in the mixture, it follows that
total mass Cl- = mass from NaCl + mass from KCl
mass Cl- from NaCl = (mass of NaCl) (fraction Cl in NaCl)
mass Cl- from KCl = (mass KCl) (fraction Cl in KCl)
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M(AgCl) = 143.4 g/mol
M(NaCl) = 58.44 g/mol
M(KCl) = 74.55 g/mol
fraction Cl in NaCl) = 35.45 g Cl/mol NaCl = 0.6066
58.44 g NaCl/mol NaCl
fraction Cl in KCl) =
35.45 g Cl/mol KCl = 0.4755
74.55 g NaCl/mol KCl
total mass Cl = 1.913 g AgCl 1 mol AgCl 1 mol Cl 35.45 g Cl = 0.4729 g Cl
143.4 g 1 mol AgCl 1 mol Cl
Now, let x = mass of NaCl. Then (0.8870 - x) = mass of KCl. Using the equation for the
total mass of Cl, we get
0.4729 g = x (0.6066) + (0.8870 - x) (0.4755) = 0.4218 g - 0.1311 x
x = (0.4729 g - 0.4218 g) = 0.3898 g
0.1311
The percent by mass NaCl in the sample is
% NaCl(by mass) = 0.3898 g NaCl 100. % = 43.9 % NaCl by mass
0.8870 g sample
28) (Burdge, 9.164) The recommended procedure for preparing a very dilute solution is
not to weigh out a very small mass or measure a very small volume of a stock solution.
Instead, it is done by a series of dilutions. A sample of 0.8214 g of KMnO 4 was
dissolved in water and made up to a volume of 500.0 mL in a volumetric flask. A 2.000
mL sample of this solution was transferred to a 1000.0 mL volumetric flask and diluted to
the mark with water. Next, 10.00 mL of the diluted solution was transferred to a 250.0
mL flask and diluted to the mark with water.
a) Calculate the concentration (in molarity) of the final solution.
We may use McLc = Md Ld , which gives Md = (Lc/Ld) Mc , in all of the dilution steps.
For the initial concentration
M(KMnO4) = 158.0 g/mol
moles KMnO4 = 0.8214 g 1 mol = 5.199 x 10-3 mol
158.0 g
Initial concentration =
5.199 x 10-3 mol = 1.0397 x 10-2 mol/L
0.500 L
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1st dilution
Md = (2.000 mL/1000.0 mL)(1.0397 x 10-2 M) = 2.079 x 10-5 mol/L
2nd dilution
Md = (10.00 mL/250.0 mL) (2.079 x 10-5 M) = 8.32 x 10-7 mol/L
This is the molarity of the final solution
b) Calculate the mass of KMnO4 needed to prepare the final solution if it was
prepared directly.
The mass of KMnO4 needed to prepare 250.0 mL of a 8.32 x 10-7 M solution directly is
mass KMnO4 = 0.250 L soln 8.32 x 10-7 mol 158.0 g KMnO4 = 3.29 x 10-5 g
L soln
mol
It would be difficult to precisely measure out such a small mass of this compound
directly, which is why the dilute solution is prepared by serial dilution.
29) (Burdge, 9.76) The maximum level of fluoride ion that the EPA allows in U.S.
drinking water is 4.0 mg/L. Convert this concentration into molarity.
[F-] = 4.0 x 10-3 g
1. L
1 mol = 2.11 x 10-4 M
19.00 g
30) A 20.00 mL sample of a stock solution of nitric acid (HNO 3) is titrated with a 0.2108
M solution of potassium hydroxide (KOH). After the addition of 14.82 mL of the KOH
solution the neutralization reaction is complete. What is the concentration of HNO3 in
the stock solution?
The balanced neutralization reaction is
HNO3(aq) + KOH(aq)  KNO3(aq) + H2O(l)
The moles of KOH present when the neutralization reaction is complete is
moles KOH = 14.82 x 10-3 L KOH 0.2108 mol KOH = 3.124 x 10-3 mol KOH
L soln
moles HNO3 = 3.124 x 10-3 mol KOH 1 mol HNO3 = 3.124 x 10-3 mol HNO3
1 mol KOH
molarity of stock HNO3 soln = 3.124 x 10-3 mol HNO3 = 0.1562 M HNO3 soln
20.00 x 10-3 L
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