Download CHAPTER 5 INTRODUCTION TO REACTIONS IN AQUEOUS

CHAPTER 5
INTRODUCTION TO REACTIONS
IN AQUEOUS SOLUTIONS
PRACTICE EXAMPLES
1A
In determining total Cl  , we recall the definition of molarity: moles of solute per liter of solution.
0.438 mol NaCl 1 mol Cl 
from NaCl, Cl =

= 0.438 M Cl 
1 L soln
1 mol NaCl
0.0512 mol MgCl2
2 mol Cl
from MgCl2 , Cl  =
= 0.102 M Cl

1 L soln
1 mol MgCl2

Cl  total = Cl  from NaCl + Cl  from MgCl2 = 0.438 M + 0.102 M = 0.540 M Cl
1B
2A
(a)
1.5 mg F1 g F
1 mol F


= 7.9  10-5 M F

L
1000 mg F
18.998 g F
(b)
1.00  106 L 
7.9 105 mol F1 mol CaF2
78.075 g CaF2
1 kg



= 3.1 kg CaF2
1 mol CaF2
1000 g
1L
2 mol F
In each case, we use the solubility rules to determine whether either product is insoluble.
The ions in each product compound are determined by simply “switching the partners” of the
reactant compounds. The designation “(aq)” on each reactant indicates that it is soluble.
(a)
Possible products are potassium chloride, KCl, which is soluble, and aluminum hydroxide,
Al  OH 3 , which is not. Net ionic equation: Al3+  aq  + 3 OH   aq   Al  OH 3  s 
(b) Possible products are iron(III) sulfate, Fe 2  SO 4 3 , and potassium bromide, KBr, both
of which are soluble. No reaction occurs.
2B
(c)
Possible products are calcium nitrate, Ca(NO3)2, which is soluble, and lead(II) iodide,
PbI 2 , which is insoluble. The net ionic equation is: Pb 2+  aq  + 2 I   aq   PbI 2  s 
(a)
Possible products are sodium chloride, NaCl, which is soluble, and aluminum phosphate,
3
AlPO 4 , which is insoluble. Net ionic equation: Al3+  aq  + PO 4  aq   AlPO 4  s 
(b)
Possible products are aluminum chloride, AlCl 3 , which is soluble, and barium sulfate,
BaSO 4 , which is insoluble. Net ionic equation: Ba 2+  aq  + SO 4
97
2
 aq  
BaSO 4  s 
Chapter 5: Introduction to Reactions in Aqueous Solutions
(c)
Possible products are ammonium nitrate, NH 4 NO 3 , which is soluble, and lead (II) carbonate,
PbCO 3 , which is insoluble. Net ionic equation: Pb 2+  aq  + CO3
2
 aq  
PbCO3  s 
3A
Propionic acid is a weak acid, not dissociated completely in aqueous solution. Ammonia similarly
is a weak base. The acid and base react to form a salt solution of ammonium propionate.


NH 3  aq  + HC3 H 5 O 2  aq   NH 4  aq  + C3 H 5 O 2  aq 
3B
Since acetic acid is a weak acid, it is not dissociated completely in aqueous solution (except
at infinite dilution); it is misleading to write it in ionic form. The products of this reaction
are the gas carbon dioxide, the covalent compound water, and the ionic solute calcium
acetate. Only the latter exists as ions in aqueous solution.

CaCO3  s  + 2 HC 2 H 3 O 2  aq   CO 2  g  + H 2 O  l  + Ca 2+  aq  + 2 C 2 H 3 O 2  aq 
4A
(a) This is a metathesis or double displacement reaction. Elements do not change oxidation
states during this reaction. It is not an oxidation–reduction reaction.
(b) The presence of O2(g) as a product indicates that this is an oxidation–reduction reaction.
Oxygen is oxidized from O.S. = -2 in NO3- to O.S. = 0 in O2(g). Nitrogen is reduced
from O.S. = +5 in NO3- to O.S. = +4 in NO2.
4B
Vanadium is oxidized from O.S. = +4 in VO2+ to an O.S. = +5 in VO2+ while manganese is
reduced from O.S. = +7 in MnO4- to O.S. = +2 in Mn2+.
5A
Aluminum is oxidized (from an O.S. of 0 to an O.S. of +3 ), while hydrogen is reduced
(from an O.S. of +1 to an O.S. of 0).
Oxidation : Al  s   Al3+  aq  + 3 e    2
Reduction:
2 H  aq  + 2 e
+

 H 2  g   3
Net equation : 2 Al  s  + 6 H +  aq   2 Al3+  aq  + 3 H 2  g 
5B
Bromide is oxidized (from 1 to 0), while chlorine is reduced (from 0 to 1 ).
Oxidation : 2 Br   aq   Br2  l  + 2 e 
Reduction: Cl 2  g  + 2 e   2 Cl  aq 
Net equation : 2 Br   aq  + Cl2  g   Br2  l  + 2 Cl  aq 
6A
Step 1: Write the two skeleton half reactions.

MnO 4  aq   Mn 2+  aq  and Fe 2+  aq   Fe3+  aq 
Step 2: Balance each skeleton half reaction for O (with H 2 O ) and for H atoms (with H + ).
MnO 4  aq   8 H +  aq   Mn 2+  aq   4 H 2 O(l)

98
and
Fe 2+  aq   Fe3+  aq 
Chapter 5: Introduction to Reactions in Aqueous Solutions
Step 3: Balance electric charge by adding electrons.
MnO 4

 aq   8 H +  aq   5 e   Mn 2+  aq   4 H 2 O(l)
and
Fe 2+  aq   Fe3+  aq   e 
Step 4: Combine the two half reactions
Fe2+  aq   Fe3+  aq  + e   5
MnO 4  aq  + 8 H +  aq  + 5 e   Mn 2+  aq  + 4 H 2O(l)

MnO 4  aq  + 8 H +  aq  + 5 Fe 2+  aq   Mn 2+  aq  + 4 H 2 O(l) + 5 Fe3+  aq 

6B
Step 1: Uranium is oxidized and chromium is reduced in this reaction. The “skeleton”
2+
2
UO 2+  aq   UO 2  aq  and Cr2 O7 (aq)  Cr 3 (aq)
half-equations are:
Step 2: First, balance the chromium skeleton half-equation for chromium atoms:
2
Cr2 O7  aq   2 Cr 3+  aq 
Next, balance oxygen atoms with water molecules in each half-equation:
2
2
UO 2+  aq  + H 2O(l)  UO 2  aq  and Cr2 O7 (aq)  2Cr 3 (aq) 7H 2 O(l)
Then, balance hydrogen atoms with hydrogen ions in each half-equation:
2
UO 2+  aq  + H 2 O(l)  UO 2  aq  + 2 H +  aq 
2
Cr2 O7 (aq)  14H  (aq)  2Cr 3 (aq) 7H 2 O(l)
Step 3: Balance the charge of each half-equation with electrons.
2
UO 2+  aq  + H 2 O(l)  UO 2  aq  + 2 H +  aq  + 2 e
Cr2 O7
2
 aq  +14 H +  aq  + 6 e  2 Cr 3+  aq  + 7 H 2O(l)
Step 4: Multiply the uranium half-equation by 3 and add the chromium half-equation to it.
2
UO 2+  aq  + H 2 O(l)  UO 2  aq  + 2 H +  aq  + 2 e   3

Cr2 O7

2
 aq  +14 H +  aq  + 6 e  2 Cr 3+  aq  + 7 H 2O(l)
3 UO 2+ (aq)+Cr2 O 7 2- (aq)+14 H + (aq)+3 H 2 O(l)  3 UO 2 2+ (aq)+2 Cr 3+ (aq)+7 H 2 O(l)+6 H + (aq)
Step 5:
Simplify. Subtract 3 H 2 O (l) and 6 H+ (aq) from each side of the equation.
3 UO 2+  aq  + Cr2 O7
7A
2
 aq  + 8 H +  aq   3 UO2 2  aq  + 2 Cr 3+  aq  + 4 H 2O(l)
Step 1: Write the two skeleton half-equations.
2
S(s)  SO 3 (aq ) and OCl  (aq )  Cl  (aq )
Step 2: Balance each skeleton half-equation for O (with H 2 O ) and for H atoms (with H + ).
3 H 2O(l) + S  s   SO3
2
 aq  + 6 H +
OCl (aq)  2H   Cl (aq)  H 2 O(l)
Step 3: Balance electric charge by adding electrons.
3 H 2 O(l) + S  s   SO3
2
 aq  + 6 H + (aq) + 4 e 
OCl  (aq)  2H  (aq)  2e   Cl  (aq)  H 2 O(l)
99
Chapter 5: Introduction to Reactions in Aqueous Solutions
Step 4: Change from an acidic medium to a basic one by adding OH  to eliminate H + .
2
3H 2 O(l) + S  s  + 6 OH  (aq)  SO3  aq  + 6 H + (aq) + 6 OH  (aq) + 4 e
OCl  aq  + 2 H + (aq) + 2 OH  (aq) + 2 e   Cl  aq  + H 2 O(l) + 2 OH  (aq)
Step 5: Simplify by removing the items present on both sides of each half-equation, and
combine the half-equations to obtain the net redox equation.
2
{S  s  + 6 OH  (aq)  SO3  aq  + 3 H 2 O(l) + 4 e  } 1
{OCl  aq  + H 2 O(l) + 2 e   Cl  aq  + 2 OH  (aq)}  2
S  s  + 6 OH  (aq) + 2 OCl   aq   2H 2 O(l)  SO 3
2
 aq  + 3 H 2O(l) + 2 Cl  aq  + 4OH -
Simplify by removing the species present on both sides.
2
Net ionic equation: S  s  + 2 OH   aq  + 2 OCl  aq   SO3  aq  + H 2 O(l) + 2 Cl  aq 
7B
Step 1: Write the two skeleton half-equations.
MnO 4   aq   MnO 2  s  and SO32 (aq)  SO 4 2 (aq)
Step 2: Balance each skeleton half-equation for O (with H 2 O ) and for H atoms (with H + ).
MnO 4  aq  + 4 H +  aq   MnO 2  s  + 2 H 2 O(l)

2
2
SO3 (aq)  H 2 O(l)  SO 4 (aq)  2H  (aq)
Step 3: Balance electric charge by adding electrons.

MnO 4  aq  + 4 H +  aq  + 3 e   MnO 2  s  + 2 H 2 O(l)
SO3
2
 aq  +
H 2 O(l)  SO 4
2
 aq  + 2 H +  aq  + 2 e
Step 4: Change from an acidic medium to a basic one by adding OH  to eliminate H + .
MnO 4
SO3
2

 aq  + 4
H +  aq  + 4 OH   aq  + 3 e   MnO 2  s  + H 2 O(l) + 4 OH   aq 
 aq  + H 2 O(l) + 2 OH   aq   SO4 2  aq  + 2 H +  aq  +
2 OH   aq  + 2 e 
Step 5: Simplify by removing species present on both sides of each half-equation, and
combine the half-equations to obtain the net redox equation.

{MnO 4  aq  + 2 H 2 O(l) + 3 e   MnO 2  s  + 4 OH   aq }  2
 aq  + 2 OH   aq   SO4 2  aq  + H 2O(l) + 2 e } 3

2
2 MnO 4  aq  + 3SO3  aq  + 6 OH - (aq) + 4 H 2 O(l) 
2
2 MnO 2  s  + 3SO 4  aq  + 3H 2 O(l) + 8 OH   aq 
{SO3
2
Simplify by removing species present on both sides.
Net ionic equation:

2
2
2 MnO 4  aq  + 3SO3  aq  + H 2 O(l)  2 MnO 2  s  + 3SO 4  aq  + 2 OH   aq 
100
Chapter 5: Introduction to Reactions in Aqueous Solutions
8A
8B
Since the oxidation state of H is 0 in H2 (g) and is +1 in both NH3(g) and H2O(g), hydrogen
is oxidized. A substance that is oxidized is called a reducing agent. In addition, the
oxidation state of N in NO2 (g) is +4 , while it is 3 in NH 3 ; the oxidation state of the
element N decreases during this reaction, meaning that NO2 (g) is reduced. The substance
that is reduced is called the oxidizing agent.
In  Au  CN 2 

 aq  , gold has an oxidation state of +1; Au has been oxidized and, thus,
Au(s) (oxidization state = 0), is the reducing agent. In OH- (aq), oxygen has an oxidation
state of -2; O has been reduced and thus, O2(g) (oxidation state = 0) is the oxidizing agent.
9A
We first determine the amount of NaOH that reacts with 0.500 g KHP.
1 mol KHP
1 mol OH  1 mol NaOH


n NaOH = 0.5000 g KHP 
= 0.002448 mol NaOH
204.22 g KHP 1 mol KHP 1 mol OH 
0.002448 mol NaOH 1000 mL
[NaOH] =

= 0.1019 M
24.03 mL soln
1 L
9B
The net ionic equation when solid hydroxides react with a strong acid is OH- + H+  H2O.
There are two sources of OH-: NaOH and Ca(OH)2. We compute the amount of OH- from
each source and add the results.
moles of OH  from NaOH:
= 0.235 g sample 
92.5 g NaOH
1 mol NaOH
1 mol OH 


= 0.00543 mol OH 
100.0 g sample 39.997 g NaOH 1 mol NaOH
moles of OH  from Ca  OH 2 :
= 0.235 g sample 
7.5 g Ca  OH 2
100.0 g sample

1 mol Ca  OH 2
74.093 g Ba  OH 2

2 mol OH 
1 mol Ca  OH 2
= 0.00048 mol OH 
total amount OH - = 0.00543 mol from NaOH + 0.00048 mol from Ca  OH 2 = 0.00591 mol OH 
[HCl] =
0.00591 mol OH  1 mol H + 1 mol HCl 1000 mL soln



= 0.130 M
45.6 mL HCl soln 1 mol OH  1 mol H +
1 L soln
10A First, determine the mass of iron that has reacted as Fe 2+ with the titrant. The balanced
chemical equation provides the essential conversion factor to answer this question.
Namely: 5 Fe 2+ aq   MnO 4  aq   8 H  aq  
 5 Fe 3+ aq   Mn 2+ aq   4 H 2 Ol
mass Fe = 0.04125 L titrant 
0.02140 mol MnO 4
1 L titrant
Then determine the % Fe in the ore.
% Fe =


5 mol Fe 2+
1 mol MnO 4


55.847 g Fe
1 mol Fe 2+
= 0.246 g Fe
0.246 g Fe
 100% = 65.4% Fe
0.376 g ore
10B The balanced equation provides us with the stoichiometric coefficients needed for the solution.
10 CO 2  g   2 Mn 2+  aq   8 H 2 O l 
Namely: 5 C2 O 4 2-  aq   2 MnO 4   aq   16 H   aq  
101
Chapter 5: Introduction to Reactions in Aqueous Solutions
1 mol Na 2 C 2 O 4

amount MnO 4 = 0.2482 g Na 2 C 2 O 4 
134.00 g Na 2 C 2 O 4
= 0.0007409 mol MnO 4

1 mol C 2 O 4
2
1 mol Na 2 C 2 O 4

2 mol MnO 4
5 mol C 2 O 4

2


0.0007409 mol MnO 4 1000 mL 1 mol KMnO 4
[KMnO 4 ] =
= 0.03129 M KMnO4



23.68 mL soln
1 L
1 mol MnO 4
INTEGRATIVE EXAMPLE
A.
First, balance the equation. Break down the reaction of chlorate and ferrous ion as follows:
ClO3 +6H  +6e   Cl +3H 2 O
6  Fe 2  Fe3  e  
Net reaction: ClO 3  6Fe 2  6H   Cl  6Fe3  3H 2 O
The reaction between Fe2+ and Ce4+ is already balanced. To calculate the moles of Fe2+ that
remains after the reaction with ClO3-, determine the moles of Ce4+ that react with Fe2+:
mol Ce4+ = 0.01259 L × 0.08362 M = 1.0527×10-3 mol = mol of excess Fe2+
total mol of Fe2+ = 0.0500 L × 0.09101 = 4.551×10-3 mol
Therefore, the moles of Fe2+ reacted = 4.551×10-3 - 1.0527×10-3 = 3.498×10-3 mol. To
determine the mass of KClO3, use the mole ratios in the balanced equation in conjunction
with the molar mass of KClO3.
1 mol ClO3
1 mol KClO3
122.54 g KClO3
3
2


3.498 10 mol Fe 

2
6 mol Fe
1 mol ClO3
1 mol KClO3
= 0.07144 g KClO3
%KClO3 =
B.
0.07144 g
100% = 49.89%
0.1432 g
First, balance the equation. Break down the reaction of arsenous acid and permanganate as
follows:
5  H 3 AsO3 + H 2 O  H3 AsO 4 + 2e- + 2H + 
2  MnO 4 + 8H + 5e-  Mn 2+ + 4H 2 O 
Net reaction: 5H 3AsO3 + 2MnO 4  6H +  5H 3AsO 4 + 2Mn 2+ + 3H 2O
moles of MnO4- = 0.02377 L × 0.02144 M = 5.0963×10-4 mol
102
Chapter 5: Introduction to Reactions in Aqueous Solutions
To calculate the mass of As, use the mole ratios in the balanced equation in conjunction with
the molar mass of As:
5 mol H 3 AsO3
1 mol As
74.922 g As


5.0963 104 mol MnO 4 

2 mol MnO 4
1 mol H 3AsO3
1 mol As
= 0.095456 g As
0.095456 g
mass% As =
100% = 1.32%
7.25 g
EXERCISES
Strong Electrolytes, Weak Electrolytes, and Nonelectrolytes
1.
(a)
Because its formula begins with hydrogen, HC6 H 5O is an acid. It is not listed in
Table 5-1, so it is a weak acid. A weak acid is a weak electrolyte.
(b)
Li 2SO 4 is an ionic compound, that is, a salt. A salt is a strong electrolyte.
(c)
MgI 2 also is a salt, a strong electrolyte.
(d)
 CH3 CH 2 2 O is a covalent compound whose formula does not begin with H.
Thus, it is neither an acid nor a salt. It also is not built around nitrogen,
and thus it does not behave as a weak base. This is a nonelectrolyte.
(e)
Sr  OH 2 is a strong electrolyte, one of the strong bases listed in Table 5-2.
3.
HCl is practically 100% dissociated into ions. The apparatus should light up brightly. A
solution of both HCl and HC2H3O2 will yield similar results. In strongly acidic solutions,
the weak acid HC2H3O2 is molecular and does not contribute to the conductivity of the
solution. However, the strong acid HCl is practically dissociated into ions and is unaffected
by the presence of the weak acid HC2H3O2. The apparatus should light up brightly.
5.
(a)
(c)
Barium bromide: strong electrolyte
Ammonia: weak electrolyte
(b)
Propionic acid: weak electrolyte
Ion Concentrations
7.
0.238 mol KNO 3
1 mol K +

= 0.238 M K +
1 L soln
1 mol KNO 3
(a)
K+ =
(b)
 NO3  =


0.167 mol Ca  NO3 2
1 L soln

2 mol NO3

= 0.334 M NO3

1 mol Ca  NO3  2
103
Chapter 5: Introduction to Reactions in Aqueous Solutions
(c)
(d)
 Al3+  =
Na
+
0.083 mol Al2  SO 4 3
1 L soln

2 mol Al3+
= 0.166 M Al3+
1 mol Al2  SO 4 3
0.209 mol Na 3 PO 4
3 mol Na +
=

= 0.627 M Na +
1 L soln
1 mol Na 3 PO 4
9.
Conversion pathway approach:
 OH   =
0.132 g Ba  OH 2  8H 2 O 1000 mL 1 mol Ba  OH 2  8H 2 O
2 mol OH 



275 mL soln
1 L
315.5 g Ba  OH 2  8H 2 O 1 mol Ba  OH 2  8 H 2 O
= 3.04  103 M OH 
Stepwise approach:
0.132 g Ba  OH  2  8H 2 O 1000 mL
= 0.480 g/L

275 mL soln
1 L
0.00152 mol Ba  OH 2 ×8H 2O
0.480 g 1 mol Ba  OH 2  8H 2 O
=

315.5 g Ba  OH  2  8H 2 O
L
L
0.00152 mol Ba  OH 2 ×8H 2 O
L
11.
(a)
[Ca 2+ ] =
(c)
13

1 L solution
32.8 mg K +
+
(b) [K ] =
14.2 mg Ca 2+
100 mL solution
[Zn 2+ ] =
225  g Zn

2+
1 mL solution


2 mol OH 
1 mol Ba  OH 2  8 H 2 O
1 g Ca 2+
1000 mg Ca 2+
1 g K+
1000 mg K
1 g Zn
+
2+
1 mol Ca 2+
40.078 g Ca 2+
 3.54  104 M Ca 2+
1000 mL solution
1 L solution
2+
1  10  g Zn
6


= 3.04 10-3 M OH -


1000 mL solution
1 L solution
1 mol K +
39.0983 g K

 8.39  103 M K +
+
1 mol Zn 2+
65.39 g Zn
2+
 3.44  103 M Zn 2+
In order to determine the solution with the largest concentration of K+, we begin by
converting each concentration to a common concentration unit, namely, molarity of K+.
0.0850 M K 2 SO 4
2 mol K +

 0.17 M K +
1 L solution
1 mol K 2 SO 4
1000 mL solution
1.25 g KBr
1 mol KBr
1 mol K +



 0.105 M K +
100 mL solution
1 L solution
119.0023 g KBr 1 mol KBr
1000 mL solution
8.1 mg K +
1 g K+
1 mol K +



 0.207 M K +
1 mL solution
1 L solution
1000 mg K + 39.0983 g K +
Clearly, the solution containing 8.1 mg K+ per mL gives the largest K+ of the three solutions.
104
Chapter 5: Introduction to Reactions in Aqueous Solutions
15.
Determine the amount of I  in the solution as it now exists, and the amount of I  in the
solution of the desired concentration. The difference in these two amounts is the amount of
I  that must be added. Convert this amount to a mass of MgI 2 in grams.
1 L

moles of I in final solution = 250.0 mL 

0.1000 mol I 
= 0.02500 mol I 
1000 mL
1 L soln
1 L
0.0876 mol KI 1 mol I 


moles of I  in KI solution = 250.0 mL 
= 0.0219 mol I 
1000 mL
1 L soln
1 mol KI
1 mol MgI 2 278.11 g MgI 2 1000 mg


mass MgI 2 required =  0.02500  0.0219  mol I  
2 mol I 
1 mol MgI 2
1 g
= 4.3  102 mg MgI 2
17.
moles of chloride ion

0.625 mol KCl 1 mol Cl   
0.385 mol MgCl 2
2 mol Cl  



=  0.225 L 
+
0.615
L

 
1 L soln
1 mol KCl  
1 L soln
1 mol MgCl 2 

0.615 mol Cl
= 0.732 M
= 0.141 mol Cl  + 0.474 mol Cl  = 0.615 mol Cl-  Cl  =
0.225 L + 0.615 L
Predicting Precipitation Reactions
19.
In each case, each available cation is paired with the available anions, one at a time, to
determine if a compound is produced that is insoluble, based on the solubility rules of
Chapter 5. Then a net ionic equation is written to summarize this information.
(a) Pb 2+  aq  + 2 Br   aq   PbBr2  s 
(b) No reaction occurs (all are spectator ions).
(c) Fe3+  aq  + 3 OH   aq   Fe  OH 3  s 
21.
(a)
23.
Mixture
Result (Net Ionic Equation)
HI  a  + Zn  NO3 2 (aq):
No reaction occurs.
(b)
CuSO 4  aq  + Na 2 CO3  aq  :
(c)
Cu  NO3 2  aq  +
(a)
Cu 2+  aq  + CO3
 aq   CuCO3  s 
3
Na 3 PO 4  aq  : 3Cu 2+  aq  + 2 PO 4  aq   Cu 3  PO 4 2  s 
2
Add K 2 SO 4  aq  ; BaSO 4  s  will form and MgSO4 will not precipitate.
BaCl2  s  + K 2 SO 4  aq   BaSO 4  s  + 2 KCl  aq 
105
Chapter 5: Introduction to Reactions in Aqueous Solutions
(b)
Add H 2 O  l  ; Na 2 CO3  s  dissolves, but MgCO3 (s) will not dissolve (appreciably).
water
Na 2 CO3  s  
 2 Na +  aq  + CO3
(c)
2
 aq 
Add KCl(aq); AgCl(s) will form, while Cu(NO3)2 (s) will dissolve.
AgNO3  s  + KCl  aq   AgCl  s  + KNO3  aq 
25.
(a)
Mixture
Sr  NO3 2  aq  + K 2SO 4  aq  :
Net Ionic Equation
2
Sr 2+  aq  + SO 4  aq   SrSO 4  s 
(b)
Mg  NO3 2  aq  + NaOH  aq  : Mg 2+  aq  + 2 OH   aq   Mg  OH 2  s 
(c)
BaCl2  aq  + K 2SO 4  aq  :
2
Ba 2+ (aq)  SO 4 (aq)  BaSO4 (s)
(upon filtering, KCl (aq) is obtained)
Acid–Base Reactions
27.
The type of reaction is given first, followed by the net ionic equation.
(a) Neutralization: OH   aq  + HC2 H3 O2  aq   H 2 O  l  + C2 H3 O 2

 aq 
(b) No reaction occurs. This is the physical mixing of two acids.
(c) Gas evolution: FeS  s  + 2 H +  aq   H 2 S  g  + Fe 2+  aq 
 aq  + H +  aq   "H 2 CO3  aq  " 
H +  aq   Mg 2+  aq  + H 2  g 
(d) Gas evolution: HCO3
(e) Redox: Mg  s  + 2
29.
31.

As a salt:
NaHSO4  aq   Na +  aq  + HSO 4
As an acid:
HSO 4

 aq  +
OH   aq  
H 2 O  l  + CO 2  g 
 aq 
2
H 2 O  l  + SO 4  aq 

Use (b) NH3(aq): NH3 affords the OH- ions necessary to form Mg(OH)2(s).
Applicable reactions: {NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)} 2
MgCl2(aq)  Mg2+(aq) + 2 Cl-(aq)
Mg2+(aq) + 2 OH-(aq)  Mg(OH)2(s)
106
Chapter 5: Introduction to Reactions in Aqueous Solutions
Oxidation–Reduction (Redox) Equations
33.
35.
(a)
The O.S. of H is +1, that of O is 2 , that of C is +4 , and that of Mg is +2 on each
side of this equation. This is not a redox equation.
(b)
The O.S. of Cl is 0 on the left and 1 on the right side of this equation. The O.S. of
Br is 1 on the left and 0 on the right side of this equation. This is a redox reaction.
(c)
The O.S. of Ag is 0 on the left and +1 on the right side of this equation. The O.S. of
N is +5 on the left and +4 on the right side of this equation. This is a redox reaction.
(d)
On both sides of the equation the O.S. of O is 2 , that of Ag is +1 , and that of Cr is
+6 . Thus, this is not a redox equation.
(a)
Reduction:
 aq  + 6 H +  aq  + 4 e  S2O32  aq  + 3 H 2O(l)

2 NO3  aq  +10 H +  aq  + 8 e   N 2 O  g  + 5 H 2 O  l 

Al  s  + 4 OH   aq   Al  OH 4  aq  + 3 e 
2SO3
(b) Reduction:
37.
2
(c)
Oxidation:
(a)
Oxidation: { 2 I   aq   I 2  s  + 2 e 
Reduction: { MnO 4

 aq  + 8
}5
H +  aq  + 5 e   Mn 2+  aq  + 4 H 2 O  l 
}2
Net: 10 I   aq  + 2 MnO 4  aq  +16 H +  aq   5 I 2  s  + 2 Mn 2+  aq  + 8 H 2 O  l 

(b) Oxidation: { N 2 H 4  l   N 2  g  + 4 H +  aq  + 4 e 
} 3


+
Reduction: { BrO3  aq  + 6 H  aq  + 6 e  Br  aq  + 3 H 2 O  l 

Net:
(c)
3 N 2 H 4  l  + 2 BrO3  aq   3 N 2  g  + 2 Br   aq  + 6 H 2 O  l 

Oxidation: Fe 2+  aq   Fe3+  aq  + e
Reduction: VO 4
Net:
3
 aq  + 6
Fe 2+  aq  + VO 4
3
H +  aq  + e  VO 2+  aq  + 3 H 2 O  l 
 aq  + 6
H +  aq   Fe3+  aq  + VO2+  aq  + 3 H 2 O  l 
(d) Oxidation: { UO 2+  aq  + H 2 O  l   UO2
Reduction: { NO3  aq  + 4 H +  aq  + 3 e

Net:
39.
(a)
} 2
3 UO 2+  aq  + 2 NO 3

 aq  + 2
 aq  + 2 H +  aq  + 2 e
 NO  g  + 2 H 2 O  l 
2
H +  aq   3 UO 2
2
 aq  + 2
}3
}2
NO  g  + H 2 O  l 
Oxidation: { MnO 2  s  + 4 OH   aq   MnO 4  aq  + 2 H 2 O(l) + 3 e  }  2

Reduction: ClO3  aq  + 3 H 2 O(l) + 6 e   Cl   aq  + 6 OH   aq 

Net:
2 MnO 2  s  + ClO3  aq  + 2 OH   aq   2MnO 4  aq  + Cl  aq  + H 2 O(l)


107
Chapter 5: Introduction to Reactions in Aqueous Solutions
(b) Oxidation: { Fe  OH 3  s  + 5 OH   aq   FeO 4
2
 aq  + 4
H 2 O(l) + 3 e 
Reduction: { OCl   aq  + H 2 O(l) + 2 e   Cl  aq  + 2OH -  aq 
Net: 2 Fe  OH 3  s  + 3 OCl   aq  + 4 OH   aq   2FeO 4
(c)
2
 aq  + 3
}2
}3
Cl   aq  + 5 H 2 O(l)
Oxidation: { ClO 2 (aq) + 2 OH   aq   ClO3  aq  + H 2 O  l  + e  }  5

Reduction: ClO 2 (aq) + 2 H 2 O  l  + 5 e  Cl (aq) + 4 OH   aq 
6 ClO 2 (aq) + 6 OH   aq   5ClO3  aq  + Cl  aq  + 3 H 2 O (l)

Net:
(d)
41.
(a)
Oxidation: (Ag (s)  Ag+ (aq) + 1 e¯ )  3
Reduction: 4 H2O(l) + CrO42- + 3 e¯  Cr(OH)3(s) + 5 OH Net: 3 Ag(s) + CrO42- + 4 H2O(l)  3 Ag+(aq) + Cr(OH)3(s) + 5 OHOxidation: Cl2  g  +12 OH   aq   2 ClO3  aq  + 6 H 2 O(l) +10 e

Reduction: { Cl 2  g  + 2 e   2 Cl   aq 
(b)
43.
(a)
}5
Net:
6 Cl2  g  +12 OH   aq   10 Cl  aq  + 2 ClO3  aq  + 6 H 2 O(l)
Or:
3 Cl2  g  + 6 OH   aq   5 Cl  aq  + ClO3  aq  + 3 H 2 O(l)


 aq  + 2 H 2O(l)  2 HSO3  aq  + 2 H +  aq  + 2 e
2
2
Reduction: S2 O 4  aq  + 2 H +  aq  + 2 e   S2 O3  aq  + H 2 O (l)
2

2
Net:
2 S2 O 4  aq  + H 2 O(l)  2 HSO3  aq  + S2 O3  aq 
Oxidation: S2 O 4
2
 aq  + H 2 O  l   NO3  aq  + 2 H +  aq  + 2 e

+
2+

Reduction: { MnO 4  aq  + 8 H  aq  + 5 e  Mn  aq  + 4 H 2 O  l 
Oxidation: { NO 2

}5
}2
Net: 5 NO 2  aq  + 2 MnO 4  aq  + 6 H +  aq   5 NO3  aq  + 2 Mn 2+  aq  + 3 H 2 O  l 

(b)
(c)


Oxidation: { Mn2+ (aq) + 4 OH- (aq)  MnO2 (s) + 2 H2O (l) + 2 e- }  3
Reduction: { MnO4- (aq) + 2 H2O (l) + 3 e-  MnO2 (s) + 4 OH- (aq) }  2
Net: 3 Mn2+ (aq) + 2 MnO4- (aq) + 4 OH- (aq)  5 MnO2 (s) + 2 H2O (l)
Oxidation: { C2 H 5 OH  CH 3 CHO + 2 H +  aq  + 2 e 
} 3
 aq  +14 H  aq  + 6 e  2 Cr  aq  + 7 H 2 O  l 
Net: Cr2 O 7 2  aq  + 8 H +  aq  + 3 C2 H 5 OH  2 Cr 3+  aq  + 7 H 2 O  l  + 3 CH 3 CHO
Reduction: Cr2 O 7
2

+
108
3+
Chapter 5: Introduction to Reactions in Aqueous Solutions
45.
For the purpose of balancing its redox equation, each of the reactions is treated as if it takes
place in acidic aqueous solution.
(a)
2 H2O(g) + CH4(g)  CO2(g) + 8 H+(g) + 8 e{2 e- + 2 H+(g) + NO(g)  ½ N2(g) + H2O(g) }4
CH4(g) + 4 NO(g)  2 N2(g) + CO2(g) + 2 H2O(g)
(b)
{H2S(g)  1/8 S8(s)+ 2 H+(g) + 2 e}2
4 e- + 4 H+(g) + SO2(g)  1/8 S8(s) + 2 H2O(g)
2 H2S(g) + SO2(g)  3/8 S8(s) + 2 H2O(g) or
16 H2S(g) + 8 SO2(g)  3 S8(s) + 16 H2O(g)
(c)
{Cl2O(g) + 2 NH4+(aq) + 2 H+(aq) + 4 e-  2 NH4Cl(s) + H2O(l)
{2 NH3(g)  N2(g) + 6 e- + 6 H+(aq)
6 NH3(g) + 6 H+(aq)  6 NH4+(aq)
10 NH3(g) +3 Cl2O(g)  6 NH4Cl(s) + 2 N2(g) + 3 H2O(l)
}3
} 2
Oxidizing and Reducing Agents
47.
The oxidizing agents experience a decrease in the oxidation state of one of their elements, while
the reducing agents experience an increase in the oxidation state of one of their elements.
 aq  is the reducing agent; the O.S. of S = +4 in SO32  and = +6 in SO 4 2  .


MnO 4  aq  is the oxidizing agent; the O.S. of Mn = +7 in MnO 4 and + 2 in Mn 2+ .
2
(a)
SO3
(b)
H 2  g  is the reducing agent; the O.S. of H = 0 in H 2  g  and = +1 in H 2 O  g  .
NO 2  g  is the oxidizing agent; the O.S. of N = +4 in NO 2  g  and  3 in NH 3 (g).
(c)  Fe  CN 6 
4
 aq  is the reducing agent; the O.S. of
Fe = +2 in  Fe  CN 6 
3
4
and = +3 in  Fe  CN 6  . H 2O 2  aq  is the oxidizing agent; the O.S. of O = 1
in H 2 O 2 and = 2 in H 2 O .
Neutralization and Acid–Base Titrations
49.
The problem is most easily solved with amounts in millimoles.
VNaOH = 10.00 mL HCl  aq  
0.128 mmol HCl 1 mmol H + 1 mmol OH 


1 mL HCl  aq  1 mmol HCl 1 mmol H +
109
Chapter 5: Introduction to Reactions in Aqueous Solutions

51.
1 mL NaOH  aq 
1 mmol NaOH
= 13.3 mL NaOH  aq  soln


0.0962 mmol NaOH
1 mmol OH
The net reaction is OH   aq  + HC3 H5O 2  aq   H 2 O(l) + C3 H 5O2  aq  .

Conversion pathway approach:
Vbase = 25.00 mL acid 
0.3057 mmol HC3 H 5 O 2
1 mL acid

1 mmol KOH
1 mmol HC3 H 5 O 2

1 mL base
2.155 mmol KOH
= 3.546 mL KOH solution
Stepwise approach:
25.00 mL acid 
0.3057 mmol HC3 H 5 O 2
= 7.643 mmol HC3 H 5 O 2
1 mL acid
1 mmol KOH
7.643 mmol HC3H 5O 2 
= 7.643 mmol KOH
1 mmol HC3 H 5 O 2
7.643 mmol KOH 
53.
1 mL base
2.155 mmol KOH
NaOH  aq  + HCl  aq   NaCl  aq  + H 2 O(l) is the titration reaction.
0.02834 L 
[NaOH] =
55.
= 3.546 mL KOH solution
0.1085 mol HCl 1mol NaOH

1L soln
1 mol HCl
= 0.1230 M NaOH
0.02500 L sample
The mass of acetylsalicylic acid is converted to the amount of NaOH, in millimoles, that
will react with it.
0.32 g HC9 H 7 O 4
1 mol HC9 H 7 O 4
1 mol NaOH
1000 mmol NaOH



 NaOH  =
23 mL NaOH  aq  180.2 g HC9 H 7 O 4 1 mol HC9 H 7 O 4
1 mol NaOH
= 0.077 M NaOH
57.
The equation for the reaction is HNO3  aq  + KOH  aq   KNO3  aq  + H 2 O 1 .
This equation shows that equal numbers of moles are needed for a complete reaction.
We compute the amount of each reactant.
mmol HNO 3 = 25.00 mL acid 
0.132 mmol HNO 3
= 3.30 mmol HNO 3
1 mL acid
0.318 mmol KOH
= 3.18 mmol KOH
1 mL base
There is more acid present than base. Thus, the resulting solution is acidic.
mmol KOH = 10.00 mL acid 
110
Chapter 5: Introduction to Reactions in Aqueous Solutions
59.
Vbase = 5.00 mL vinegar 

61.
1.01 g vinegar 4.0 g HC2 H 3O 2 1 mol HC2 H 3O 2


1 mL
100.0 g vinegar 60.0 g HC2 H3O2
1 mol NaOH
1 L base
1000 mL


= 34 mL base
1 mol HC 2 H 3O 2 0.1000 mol NaOH
1 L
Answer is (d): 120 % of necessary titrant added in titration of NH3
required for
equivalence
point
5 NH3
+
5 HCl
+
1 HCl
5 NH4+ + 6 Cl- + H3O+
(depicted in question's drawing )
20 % excess
Stoichiometry of Oxidation–Reduction Reactions
63.
Conversion pathway approach:
0.1078 g As 2 O 3 

[ MnO 4 ]=
1mol As 2 O 3
197.84 g As 2 O 3
22.15 mL 

4 mol MnO 4

5 mol As 2 O3
1L

1mol KMnO 4
1mol MnO 4
1000 mL
Stepwise approach:
mol KMnO 4
L solution
1mol As 2 O3
0.1078 g As 2 O3 
= 5.449  10-4 mol As 2 O3
197.84 g As 2 O3
[ KMnO 4 ]=
5.449 10-4 mol As 2 O3 
4.359  10-4 mol MnO 4  
22.15 mL 
1L
1000 mL
[ KMnO 4 ]=
4 mol MnO 4 
= 4.359 10-4 mol MnO 4 
5 mol As 2 O3
1mol KMnO 4
1mol MnO 4

= 4.359  10-4 mol KMnO 4
 0.02215 L solution
mol KMnO 4
L solution
=
4.359  10-4 mol KMnO 4
0.02215 L solution
111
= 1.968  10-2 M

= 0.01968 M KMnO 4
Chapter 5: Introduction to Reactions in Aqueous Solutions
65.
First, we will determine the mass of Fe, then the percentage of iron in the ore.
2
0.05051 mol Cr2 O7
1 L
6 mol Fe 2+
55.85 g Fe
mass Fe = 28.72 mL 



2
1000 mL
1 L soln
1 mol Fe 2+
1 mol Cr2 O7
0.4861g Fe
mass Fe = 0.4861 g Fe % Fe =
 100%  53.23% Fe
0.9132 g ore
67.
First, balance the titration equation:
}5
 aq   2 CO2  g  + 2 e

Reduction: { MnO 4  aq  + 8 H +  aq  + 5 e   Mn 2+  aq  + 4 H 2 O  l  }  2
2

Net: 5 C2 O 4  aq  + 2 MnO 4  aq  +16 H +  aq   10 CO 2  g  + 2 Mn 2+  aq  + 8 H 2 O  l 
Oxidation: { C2 O 4
2
mass Na 2C2O4 =1.00 L satd soln Na 2 C2 O 4 

1000 mL 25.8 mL satd soln KMnO 4 0.02140 mol KMnO 4


1L
5.00 mL satd soln Na 2 C2 O 4
1000 mL KMnO4
1 mol MnO 4  5 mol C2 O4 2 1 mol Na 2 C2 O4 134.0 g Na 2 C2 O 4



1 mol KMnO4 2 mol MnO4 
1 mol Na 2 C2 O 4
1 mol C2 O 4 2
mass Na 2C2O4 = 37.0 g Na 2 C2 O 4
Integrative and Advanced Exercises
71. A possible product, based on solubility rules, is Ca 3 (PO 4 ) 2 . We determine the % Ca in this
compound.
 3  40.078 g Ca  2  30.974 g P  8  15.999 g O
 120.23 g Ca  61.948 g P  127.99 g O  310.17 g
120.23 g Ca
% Ca 
 100%  38.763%
310.17 g Ca 3 (PO 4 ) 2
molar mass
Thus, Ca 3 (PO 4 ) 2 is the predicted product. The net ionic equation follows.
2
3 Ca 2  (aq)  2 HPO 4 (aq) 
 Ca 3 (PO 4 ) 2 (s)  2 H  (aq)
74. Let us first determine the mass of Mg in the sample analyzed.
Conversion pathway approach:
1 mol Mg 2 P 2 O7
2 mol Mg
24.305 g


 0.0120 g Mg
222.55 g Mg 2 P 2 O7 1 mol Mg 2 P2 O7 1 mol Mg
0.0120 g Mg
ppm Mg  106 g sample 
 108 ppm Mg
110.520 g sample
mass Mg  0.0549 g Mg 2 P2 O7 
112
Chapter 5: Introduction to Reactions in Aqueous Solutions
Stepwise approach:
0.0549 g Mg 2 P2 O 7 
1 mol Mg 2 P 2 O7
= 2.47 10-4 mol Mg 2 P 2 O7
222.55 g Mg 2 P 2 O 7
2.47 10-4 mol Mg 2 P 2 O7 
4.93 10-4 mol Mg 
2 mol Mg
 4.93  10-4 mol Mg
1 mol Mg 2 P2 O 7
24.305 g
 0.0120 g Mg
1 mol Mg
ppm Mg  106 g sample 
0.0120 g Mg
 108 ppm Mg
110.520 g sample
75. Let V represent the volume of added 0.248 M CaCl2 that must be added.
We know that [Cl ] = 0.250 M, but also,
0.248 mol CaCl 2
0.186 mol KCl 1 mol Cl 
2 mol Cl 

V 

0.335 L
1 L soln
1 mol KCl
1 L soln
1 mol CaCl 2

[Cl ] 
0.335 L  V
0.250 (0.335  V )  0.0838  0.250 V  0.0623  0.496 V
V 
0.0838  0.0623
 0.0874 L
0.496  0.250
80. (a) [FeS2 + 8 H2O → Fe3+ + 2 SO42− + 16 H+ + 15 e−] × 4
[O2 + 4 H+ + 4 e− → 2 H2O] × 15
overall: 4 FeS2(s) + 15 O2(g) + 2 H2O(l) → 4 Fe3+(aq) + 8 SO42−(aq) + 4 H+(aq)
(b) One kilogram of tailings contains 0.03 kg (30 g) of S. We have
moles of FeS2 = 30 g S 
1 mol S 1 mol FeS2

 0.468 mol FeS2
32.07 g S
2 mol S
moles of H+ = 0.468 mol FeS2 
4 mol H +
 0.467 mol H +
4 mol FeS2
moles of CaCO3 = 0.467 mol H + 
1 mol CaCO3
 0.234 mol CaCO3
2 mol H +
mass of CaCO3 = 0.234 mol CaCO3 
100.09 g CaCO3
 23.4 g CaCO3
1 mol CaCO3
113
Chapter 5: Introduction to Reactions in Aqueous Solutions
83. Oxidation : {2 Cl  (aq) 
 Cl 2 (g)  2 e  }
3
2
Reduction : Cr2 O 7 (aq)  14 H  (aq)  6 e  
 2 Cr 3 (aq)  7 H 2 O
2
Net : 6 Cl  (aq)  Cr2 O 7 (aq)  14 H  (aq) 
 2 Cr 3 (aq)  7 H 2 O  3 Cl 2 (g)
We need to determine the amount of Cl2(g) produced from each of the reactants. The limiting
reactant is the one that produces the lesser amount of Cl2..
1.15 g 30.1 g HCl 1 mol HCl 1 mol Cl  3 mol Cl 2




1 mL 100. g soln 36.46 g HCl 1 mol HCl 6 mol Cl 
 1.54 mol Cl 2
amount Cl 2  325 mL 
2
98.5 g K 2 Cr2 O 7
1 mol K 2 Cr2 O 7
1 mol Cr2 O 7
3 mol Cl 2
amount Cl 2  62.6 g 



100. g sample
294.2 g K 2 Cr2 O 7 1 mol K 2 Cr2 O 7 1 mol Cr2 O 7 2
 0.629 mol Cl 2 , the amount produced from the limiting reactant
Then we determine the mass of Cl2(g) produced. = 0.629 mol Cl2 ×
85. Cl 2 (g)  NaClO 2 (aq) 
 NaCl(aq)  ClO 2 (g)
70.91 g Cl2
= 44.6 g Cl2
1 mol Cl2
(not balanced)
Cl 2 (g)  2 NaClO 2 (aq) 
 2 NaCl(aq)  2 ClO 2 (g)
amount ClO 2  1 gal 

88.
2 mol ClO2
67.45 g ClO 2
3.785 L 2.0 mol NaClO 2



1 gal
1 L soln
2 mol NaClO 2 1 mol ClO2
97 g ClO 2 produced
 5.0  102 g ClO 2 (g)
100 g ClO 2 calculated
(a) First, balance the redox equations needed for the calculation.
Oxidation: {HSO3- (aq)+ H2O(l)  SO42- (aq) + 3 H+ (aq) + 2 e- }
×
3
Reduction: {IO3- (aq) + 6 H+ (aq) + 6 e-  I-(aq) + 3 H2O(l) }
×
1
Net: 3 HSO3- (aq) + IO3- (aq)  3 SO42- (aq) + 3 H+ (aq) + I- (aq)
The solution volume of 5.00 L contains 29.0 g NaIO3. This represents
29.0 g/197.9g/mol NaIO3 = 0.147 mol NaIO3.
(b) From the above equation, we need 3 times that molar amount of NaHSO3, which is
3(0.147 mol) = 0.441 mol NaHSO3; the molar mass of NaHSO3 is 104.06 g/mol.
The required mass then is 0.441(104.06) = 45.9 g.
114
Chapter 5: Introduction to Reactions in Aqueous Solutions
For the second process:
Oxidation: {2 I-(aq)  I2(aq) + 2 e- }
× 5
Reduction: {2 IO3- (aq) + 12 H+ (aq) + 10 e-  I2(aq) + 6 H2O(l) }
× 1
Net:
5 I-(aq) + IO3- (aq) + 6 H+ (aq)  3 I2(aq) + 3 H2O(l)
In Step 1, we produced 1 mol of I- for every mole of IO3- reactant; therefore we had
0.147 mol I-.
In step 2, we require 1/5 mol IO3- for every mol of I-.
We require only 1.00 L of the solution in the question instead of the 5.00 L in the first
step.
89.
Mg(OH)2(aq) + 2 HCl(aq) → MgCl2(aq) + 2 H2O(l)
(1)
Al(OH)3(aq) + 3 HCl(aq) → AlCl3(aq) + 3 H2O(l)
(2)
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
(3)
initial moles of HCl =
0.0500 L 
0.500 mol
= 0.0250 mol
1L
moles of HCl that reacted with NaOH =
moles of HCl left over from reaction with active ingredients =
0.0165 L 
0.377 mol NaOH
1 mol HCl

= 6.22 10-3 mol
1L
1 mol NaOH
moles of HCl that react with active ingredients =
0.0250 mol - 6.22 10-3 mol = 0.0188 mol
 # moles HCl that   # moles HCl that 
 react with Mg(OH)  +  react with Al(OH)  = total moles of HCl reacted/used
3

2

# moles HCl that react with Mg(OH)2 =

1 mol Mg(OH) 2
2 mol HCl 

 X grams Mg(OH) 2 

58.32 g Mg(OH) 2
1 mol Mg(OH) 2 

# moles HCl that react with Al(OH)3 =

1 mol Al(OH)3
3 mol HCl 

0.500-X grams Al(OH)3 

78.00 g Al(OH)3
1 mol Al(OH)3 

115
Chapter 5: Introduction to Reactions in Aqueous Solutions
2X
3(0.500  X)

 0.0188
58.32
78.00
X = 0.108, therefore the mass of Mg(OH)2 in the sample is 0.108 grams.
% Mg(OH)2 = (0.108/0.500) × 100 = 21.6
%Al = 100 − %Mg(OH)2 = 78.4
91.
0.1386 g AgI 
1 mol AgI 1 mol CHI3 1 mol C19 H16 O 4 308.33 g C19 H16O 4



234.77 g AgI 3 mol AgI
1 mol CHI3
1 mol C19 H16 O 4
 0.06068 g C19 H16 O 4
% C19 H16 O 4 =
0.06068 g
100  0.4346 %
13.96 g
93. (a) CaO(s) + H2O(l) → Ca2+(aq) + 2 OH−(aq)
H2PO4−(aq) + 2 OH−(aq) → PO43−(aq) + 2 H2O(l)
HPO4−(aq) + OH−(aq) → PO43−(aq) + H2O(l)
5 Ca2+(aq) + 3 PO43−(aq) + OH−(aq) → Ca5(PO4)3OH(s)
10.0  103 g P 1 mol P 1 mol PO34 5 mol Ca 2



1.00  10 L 
L
30.97 g P
1 mol P
3 mol PO34
4
(b)

1 mol CaO 56.08 g CaO

= 301.80 g CaO = 302 g = 0.302 kg
1 mol Ca 2 1 mol CaO
FEATURE PROBLEMS
94.
From the volume of titrant, we can calculate both the amount in moles of NaC5 H 5 and
(through its molar mass of 88.08 g/mol) the mass of NaC5 H 5 in a sample. The remaining mass
in a sample is that of C 4 H 8O (72.11 g/mol), whose amount in moles we calculate. The ratio of
the molar amount of C 4 H 8O in the sample to the molar amount of NaC5 H 5 is the value of x.
116
Chapter 5: Introduction to Reactions in Aqueous Solutions
Conversion pathway approach:
0.1001 mol HCl 1 mol NaOH 1 mol NaC5 H 5


1 L soln
1 mol HCl
1 mol NaOH
= 0.001493 mol NaC5 H 5
moles of NaC5 H 5 = 0.01492 L 

88.08 g NaC5 H 5 
mass of C4 H8O = 0.242 g sample   0.001493 mol NaC5 H 5 

1 mol NaC5 H 5 

= 0.111 g C4 H8O
1mol C4 H8O
72.11g C4 H8O
= 1.03
0.001493 mol NaC5 H 5
0.110 g C4 H8O 
x=
Stepwise approach:
0.1001 mol HCl
0.01492 L 
= 1.493  10-3 mol HCl
1 L soln
1 mol NaOH
1.493 10-3 mol HCl 
= 1.493 10-3 mol NaOH
1 mol HCl
1 mol NaC5 H 5
1.493 10-3 mol NaOH 
 1.493 10-3 mol NaC5 H 5
1 mol NaOH
88.08 g NaC5 H 5
1.493 10-3 mol NaC5 H5 
 0.1315 g NaC5 H 5
1 mol NaC5 H 5
mass of C4 H8O = 0.242 g sample  0.1315 g NaC5 H5 = 0.111 g C4 H8O
0.111g C4 H8O 
1mol C4 H8O
= 1.54  10-3 mol C4 H8O
72.11g C4 H8O
1.54  10-3 mol C4 H8O
= 1.03
0.001493 mol NaC5 H 5
For the second sample, parallel calculations give 0.001200 mol NaC5 H 5 , 0.093 g C 4 H 8 ,
x = 1.1. There is rounding error in this second calculation because it is limited to two
significant figures. The best answer is from the first run x ~1.03 or 1. The formula is
NaC5H5(THF)1.
95.
First, we balance the two equations.
Oxidation: H 2 C2 O 4  aq   2 CO 2  g  + 2 H +  aq  + 2 e 
Reduction: MnO 2  s  + 4 H +  aq  + 2 e   Mn 2+  aq  + 2 H 2 O  l 
Net:
H 2 C 2 O 4  aq  + MnO 2  s  + 2 H +  aq   2 CO 2  g  + Mn 2+  aq  + 2 H 2O  l 
117
Chapter 5: Introduction to Reactions in Aqueous Solutions
Oxidation: { H 2 C2 O 4  aq   2 CO 2  g  + 2 H +  aq  + 2 e 
}5
Reduction: { MnO 4  aq  + 8 H +  aq  + 5 e   Mn 2+  aq  + 4 H 2 O  l 
}2

Net: 5 H 2 C2 O 4  aq  + 2 MnO 4  aq  + 6 H +  aq   10 CO 2  g  + 2 Mn 2+  aq  + 8 H 2 O  l 

Next, we determine the mass of the excess oxalic acid.

0.1000 mol KMnO 4 1mol MnO 4
5 mol H 2 C2 O 4
mass H 2C 2 O 4  2H 2 O  0.03006 L 




1L
1mol KMnO 4
2 mol MnO 4

1 mol H 2 C 2 O 4  2H 2 O 126.07 g H 2 C 2 O 4  2H 2 O
= 0.9474 g H 2C2O4  2H 2O

1 mol H 2 C 2 O 4
1 mol H 2 C 2 O 4  2H 2 O
The mass of H 2 C 2 O 4  2 H 2 O that reacted with MnO2 = 1.651 g  0.9474 g = 0.704 g H 2C2O4  2H 2O
mass MnO 2 = 0.704 g H 2 C 2 O 4  2H 2 O 
1 mol H 2 C 2 O 4

1 mol MnO 2
126.07 g H 2 C 2 O 4  2H 2 O 1 mol H 2 C 2 O 4

86.9 g MnO 2
1 mol MnO 2
= 0.485 g MnO 2
% MnO 2 
97.
0.485g MnO 2
 100%  91.0% MnO 2
0.533g sample
The molecular formula for CH3CH2OH is C2H6O and for CH3COOH is C2H4O2.
The first step is to balance the oxidation–reduction reaction.
Oxidation: [C2H6O + H2O → C2H4O2 + 4 H+ + 4 e−] × 3
Reduction: [Cr2O72− + 14 H+ + 6e− → 2 Cr3+ + 7 H2O] × 2
Overall: 3 C2H6O + 2 Cr2O72− + 16 H+ → 3 C2H4O2 + 4 Cr3+ + 11 H2O
Before the breath test:
0.75 mg K 2 Cr2 O7
1g
1 mol
1000 mL



= 8.498  10-4 M
3 mL
1000 mg 294.19 g
1L
= 8×10-4 M (to 1 sig fig)
For the breath sample:
0.05 g C2 H 6 O
2.38×107 g C2 H 6 O
1 mL blood

=
BrAC =
100 mL blood 2100 mL breath
mL breath
mass C2H6O =
2.38×107 g C2 H 6 O
× 500. mL breath = 1.19 × 10−4 g C2H6O
mL breath
118
Chapter 5: Introduction to Reactions in Aqueous Solutions
Calculate the amount of K2Cr2O7 that reacts:
–
1 mol C2 H 6 O
2 mol Cr2 O72 1 mol K 2 Cr2 O7
1.19  10 g C2 H 6 O 


–
46.068 g C2 H 6 O 3 mol C2 H 6 O 1 mol Cr2 O72
4
= 1.72  106 mol K 2 Cr2 O7
# mol K2Cr2O7 remaining = moles K2Cr2O7 before – moles K2Cr2O7 that reacts
moles K2Cr2O7 before = 0.75 mg K 2 Cr2 O7 
1g
1 mol
= 2.5  10-6 mol

1000 mg 294.19 g
# mol K2Cr2O7 remaining = 2.5 × 10−6 mol − 1.72 × 10−6 mol = 0.78 × 10−6 mol
concentration of K2Cr2O7 after the
breath test = 0.78 × 10−6 mol/0.003 L = 2.6 × 10−4 mol/L = 3 × 10−4 mol/L (to 1 sig fig)
102. The answer is (b).
Conversion pathway approach:
0.0050 mol Ba(OH) 2
2 mol OH 
0.300 L 
= 0.0030 mol

1L
1mol Ba(OH) 2
Stepwise approach:
0.0050 mol Ba(OH) 2
0.300 L 
= 1.5 10-3 mol Ba(OH) 2
1L
2 mol OH 
1.5  10-3 mol Ba(OH) 2 
= 0.0030 mol
1mol Ba(OH) 2
103. The answer is (d), because H2SO4 is a strong diprotic acid and theoretically yields
0.20 mol of H+ for every 0.10 mol of H2SO4.
104. The answer is (c). Based on the solubility guidelines in Table 5-1, carbonates (CO32-) are
insoluble.
105. The answer is (a). Reaction with ZnO gives ZnCl2 (soluble) and H2O. There is no reaction
with NaBr and Na2SO4, since all species are aqueous. By the process of elimination, (a) is
the answer.
106.
Balanced equation: 2 KI + Pb(NO 3 ) 2  2KNO3 + PbI 2
Net ionic equation: 2I- + Pb 2  PbI 2 (s)
119
Chapter 5: Introduction to Reactions in Aqueous Solutions
107.
Balanced equation: Na 2 CO3 + 2HCl  2NaCl + H 2 O + CO 2
Net ionic equation: CO 23  + 2H   H 2O (l) + CO 2 (g)
108.
(a)
(b)
(c)
Balanced equation: 2 Na 3PO 4 + 3 Zn(NO 3 ) 2  6NaNO3 + Zn 3 (PO 4 ) 2
Net ionic equation: 3 Zn 2+ + PO34  Zn 3 (PO 4 ) 2 (s)
Balanced equation: 2 NaOH + Cu(NO 3 ) 2  Cu(OH) 2 + 2 NaNO3
Net ionic equation: Cu 2+ + 2 OH   Cu(OH) 2 (s)
Balanced equation: NiCl2 + Na 2 CO 3  NiCO3 + 2 NaCl
Net ionic equation: Ni 2+ + CO32  NiCO3 (s)
109. (a) Species oxidized: N in NO
(b) Species reduced: O2
(c) Oxidizing agent: O2
(d) Reducing agent: NO
(e) Gains electrons: O2
(f) Loses electrons: NO
110. The answer is (b). The charges need to be balanced on both sides. Using a coefficient of 4,
the charges on both sides of the reaction becomes +12.
111. The answer is (d), 5 ClO- to 1 I2. The work to balance the half-reactions is shown below:
Reduction: 5ClO  + 2H + + 2e   Cl  + H 2 O
Oxidation: I 2 + 6H 2 O  2IO3 + 10e  + 12H 
To combine the above reactions, the oxidation reaction should be multiplied by 5. The
combined equation is:
Combined: 5ClO  + I 2 + H 2O  5Cl  + 2IO3 + 2H 
112. The answer is (a). The balanced half-reaction is as follows:
NpO 2 + 4 H + + e-  Np 4 + 2H 2 O
113. (a)
False. Based on solubility rules, BaCl2 dissolves well in water. Therefore, it is a
strong electrolyte.
(b) True. Since H- is a base, H2O is by necessity an acid. It also reduces H- (-1) to H2 (0).
(c) False. The product of such a reaction would be NaCl and H2CO3, neither of which
precipitates out.
(d) False. HF is among the strongest of weak acids. It is not a strong acid, because it
doesn’t completely dissociate.
120
Chapter 5: Introduction to Reactions in Aqueous Solutions
(e) True. For every mole of Mg(NO3)2, there are 3 moles of ions, in contrast to 2 moles of
ions for NaNO3.
114. (a)
No. Oxidation states of C, H or O do not change throughout the reaction.
(b) Yes. Li is oxidized to Li+ and H in H2O is reduced from +1 to 0 in H2.
(c) Yes. Ag is oxidized and Pt is reduced.
(d) No. Oxidation states of Cl, Ca, H, and O remain unchanged.
121