Ch.4
Solutions to similar example problems
4.19
Refer to Table 4.2 of the text to solve this problem.
(a)
(b)
(c)
(d)
4.20
Ca3(PO4)2 is insoluble.
Mn(OH)2 is insoluble.
AgClO3 is soluble.
K2S is soluble.
Strategy: Although it is not necessary to memorize the solubilities of compounds, you should keep in
mind the following useful rules: all ionic compounds containing alkali metal cations, the ammonium ion,
and the nitrate, bicarbonate, and chlorate ions are soluble. For other compounds, refer to Table 4.2 of the
text.
Solution:
(a) CaCO3 is insoluble. Most carbonate compounds are insoluble.
(b) ZnSO4 is soluble. Most sulfate compounds are soluble.
(c) Hg(NO3)2 is soluble. All nitrate compounds are soluble.
(d) HgSO4 is insoluble. Most sulfate compounds are soluble, but those containing Ag, Ca2, Ba2, Hg2,
and Pb2 are insoluble.
(e) NH4ClO4 is soluble. All ammonium compounds are soluble.
4.21
(a)
Ionic: 2Ag(aq)  2NO3 (aq)  2Na(aq)  SO24 (aq) 
 Ag2SO4(s)  2Na(aq)  2NO3 (aq)
Net ionic: 2Ag(aq)  SO24 (aq) 
 Ag2SO4(s)
(b)
Ionic: Ba2(aq)  2Cl(aq)  Zn2(aq)  SO24 (aq) 
 BaSO4(s)  Zn2(aq)  2Cl(aq)
Net ionic: Ba2(aq)  SO24 (aq) 
 BaSO4(s)
(c)
Ionic: 2NH 4 (aq)  CO32 (aq)  Ca2(aq)  2Cl(aq) 
 CaCO3(s)  2NH 4 (aq)  2Cl(aq)
Net ionic: Ca2(aq)  CO32 (aq) 
 CaCO3(s)
4.22
(a) Na2S(aq)  ZnCl2(aq) 
 ZnS(s)  2NaCl(aq)
The ionic and net ionic equations are:
Ionic: 2Na(aq)  S2(aq)  Zn2(aq)  2Cl(aq) 
 ZnS(s)  2Na(aq)  2Cl(aq)
Net ionic: Zn2(aq)  S2(aq) 
 ZnS(s)
(b)
2K3PO4(aq)  3Sr(NO3)2(aq) 
 Sr3(PO4)2(s)  6KNO3(aq)
The ionic and net ionic equations are:
Ionic: 6K(aq)  2PO34 (aq)  3Sr2(aq)  6NO3 (aq) 
 Sr3(PO4)2(s)  6K(aq)  6NO3 (aq)
Net ionic: 3Sr2(aq)  2PO34 (aq) 
 Sr3(PO4)2(s)
(c)
Mg(NO3)2(aq)  2NaOH(aq) 
 Mg(OH)2(s)  2NaNO3(aq)
The ionic and net ionic equations are:
Ionic: Mg2(aq)  2NO3 (aq)  2Na(aq)  2OH(aq) 
 Mg(OH)2(s)  2Na(aq)  2NO3 (aq)
Net ionic: Mg2(aq)  2OH(aq) 
 Mg(OH)2(s)
4.31
(a)
HI dissolves in water to produce H and I, so HI is a Brønsted acid.
(b)
CH3COO can accept a proton to become acetic acid CH3COOH, so it is a Brønsted base.
(c)
H2 PO4 can either accept a proton, H, to become H3PO4 and thus behaves as a Brønsted base, or
can donate a proton in water to yield H and HPO24 , thus behaving as a Brønsted acid.
(d)
HSO 4 can either accept a proton, H, to become H2SO4 and thus behaves as a Brønsted base, or can
donate a proton in water to yield H and SO 24 , thus behaving as a Brønsted acid.
4.32
What are the characteristics of a Brønsted acid? Does it contain at least an H atom? With the
exception of ammonia, most Brønsted bases that you will encounter at this stage are anions.
Solution:
(a)
PO34 in water can accept a proton to become HPO24 , and is thus a Brønsted base.
(b)
ClO2 in water can accept a proton to become HClO2, and is thus a Brønsted base.
(c)
NH 4 dissolved in water can donate a proton H, thus behaving as a Brønsted acid.
(d)
HCO3 can either accept a proton to become H2CO3, thus behaving as a Brønsted base. Or, HCO3
can donate a proton to yield H and CO32 , thus behaving as a Brønsted acid.
Comment: The HCO3 species is said to be amphoteric because it possesses both acidic and basic
properties.
4.33
(a)
Ionic: H(aq)  Br(aq)  NH3(aq) 
 NH4 (aq)  Br(aq)
Net ionic: H(aq)  NH3(aq) 
 NH4 (aq)
(b)
Ionic: 3Ba2(aq)  6OH(aq)  2H3PO4(aq) 
 Ba3(PO4)2(s)  6H2O(l)
Net ionic: 3Ba2(aq)  6OH(aq)  2H3PO4(aq) 
 Ba3(PO4)2(s)  6H2O(l)
(c)
Ionic: 2H(aq)  2ClO4 (aq)  Mg(OH)2(s) 
 Mg2(aq)  2ClO4 (aq)  2H2O(l)
Net ionic: 2H(aq)  Mg(OH)2(s) 
 Mg2+(aq) + 2H2O(l)
4.47
See Section 4.4 of the text.
(a)
ClF: F 1 (rule 5), Cl 1 (rule 6)
(b)
IF7: F 1 (rule 5), I 7 (rules 5 and 6)
(c)
CH4: H 1 (rule 4), C 4 (rule 6)
(d)
C2H2: H 1 (rule 4), C 1 (rule 6)
(e)
C2H4: H 1 (rule 4), C 2 (rule 6),
(f)
K2CrO4: K 1 (rule 2), O 2 (rule 3), Cr 6 (rule
(g)
K2Cr2O7: K 1 (rule 2), O 2 (rule 3), Cr 6 (rule 6)
(h)
KMnO4: K 1 (rule 2), O 2 (rule 3), Mn 7 (rule 6)
(i)
NaHCO3: Na 1 (rule 2), H 1 (rule 4), O 2 (rule 3), C 4 (rule 6)
(j)
Li2: Li 0 (rule 1)
(k)
(l)
KO2: K 1 (rule 2), O 1/2 (rule 6)
(m) PF6 : F 1 (rule 5), P 5 (rule 6)
(n)
KAuCl4: K 1 (rule 2), Cl 1 (rule 5), Au 3 (rule 6)
6)
NaIO3: Na 1 (rule 2), O 2 (rule 3), I 5 (rule 6)
4.48
All are free elements, so all have an oxidation number of zero.
4.49
(a) Cs2O, 1
4
(f)
3
4.55
(k)
(a)
(c)
4.56
(a)
(c)
MoO24 , 6
SbF6 , 5
(b) CaI2, 1
(g)
PtCl24 ,
(c) Al2O3, 3
2
(h)
PtCl62 ,
4
(d) H3AsO3, 3
(e) TiO2,
(i) SnF2, 2
(j) ClF3,
Disproportionation reaction
Decomposition reaction
(b)
(d)
Displacement reaction
Combination reaction
Combination reaction
Displacement reaction
(b)
(d)
Decomposition reaction
Disproportionation reaction
4.65 Molar mass of C2H5OH  46.068 g/mol; molar mass of C12H22O11  342.3 g/mol; molar mass of
NaCl  58.44 g/mol.
(a)
? mol C2 H5OH  29.0 g C2 H5OH 
Molarity 
(b)
1 mol C12 H 22 O11
 0.04499 mol C12 H 22O11
342.3 g C12 H 22 O11
0.04499 mol C12 H 22 O11
mol solute

 0.608 M
L of soln
74.0  103 L soln
? mol NaCl  9.00 g NaCl 
Molarity 
4.68
0.6295 mol C2 H5OH
mol solute

 1.16 M
L of soln
0.545 L soln
? mol C12 H22 O11  15.4 g C12 H 22 O11 
Molarity 
(c)
1 mol C2 H5OH
 0.6295 mol C2 H5OH
46.068 g C2 H5OH
1 mol NaCl
 0.154 mol NaCl
58.44 g NaCl
mol solute
0.154 mol NaCl

 1.78 M
L of soln
86.4  103 L soln
A 250 mL sample of 0.100 M solution contains 0.0250 mol of solute (mol  M  L). The computation in
each case is the same:
259.8 g CsI
 6.50 g CsI
1 mol CsI
(a)
0.0250 mol CsI 
(b)
0.0250 mol H2SO4 
(c)
0.0250 mol Na 2CO3 
(d)
0.0250 mol K 2 Cr2 O7 
98.086 g H2SO4
 2.45 g H 2SO4
1 mol H 2SO4
105.99 g Na 2CO3
 2.65 g Na2CO3
1 mol Na 2CO3
294.2 g K 2 Cr2 O7
 7.36 g K 2Cr2O7
1 mol K 2 Cr2 O7
158.04 g KMnO4
 3.95 g KMnO4
1 mol KMnO4
First, calculate the moles of each solute. Then, you can calculate the volume (in L) from the molarity and
the number of moles of solute.
(e) 0.0250 mol KMnO4 
4.67
(a)
? mol NaCl  2.14 g NaCl 
L soln 
(b)
1 mol C2 H5OH
 0.09334 mol C2H5OH
46.068 g C2 H5OH
0.09334 mol C2 H5OH
mol solute

 0.0622 L  62.2 mL soln
Molarity
1.50 mol/L
? mol CH3COOH  0.85 g CH3COOH 
L soln 
4.75
mol solute
0.03662 mol NaCl

 0.136 L  136 mL soln
Molarity
0.270 mol/L
? mol C2 H5OH  4.30 g C2 H5OH 
L soln 
(c)
1 mol NaCl
 0.03662 mol NaCl
58.44 g NaCl
1 mol CH3COOH
 0.0142 mol CH3COOH
60.052 g CH3COOH
0.0142 mol CH3COOH
mol solute

 0.047 L  47 mL soln
Molarity
0.30 mol/L
MinitialVinitial  MfinalVfinal
Vinitial 
M final  Vfinal
0.200 M  60.00 mL

 3.00 mL
M initial
4.00 M
To prepare the 0.200 M solution, you would dilute 3.00 mL of the 4.00 M HNO3 solution to a final volume
of 60.0 mL.
4.76
You need to calculate the final volume of the dilute solution. Then, you can subtract 505 mL from this
volume to calculate the amount of water that should be added.
Vfinal 
 0.125 M  505 mL 
M initialVinitial

 631 mL
M final
 0.100 M 
(631  505) mL  126 mL of water
4.82
We want to calculate the mass % of Ba in the original compound. Let's start with the
definition of mass %.
want to calculate
mass % Ba 
need to find
mass Ba
 100%
mass of sample
given
The mass of the sample is given in the problem (0.6760 g). Therefore we need to find the mass of Ba in the
original sample. We assume the precipitation is quantitative, that is, that all of the barium in the sample has
been precipitated as barium sulfate. From the mass of BaSO 4 produced, we can calculate the mass of Ba.
There is 1 mole of Ba in 1 mole of BaSO4.
Solution: First, we calculate the mass of Ba in 0.4105 g of the BaSO4 precipitate. The molar mass of
BaSO4 is 233.4 g/mol.
? mass of Ba  0.4105 g BaSO4 
1 mol BaSO4
1 mol Ba
137.3 g Ba


233.37 g BaSO4 1 mol BaSO4
1 mol Ba
 0.24151 g Ba
Next, we calculate the mass percent of Ba in the unknown compound.
%Ba by mass 
4.89
0.24151 g
 100%  35.73%
0.6760 g
The reaction between KHP (KHC8H4O4) and KOH is:
KHC8H4O4(aq)  KOH(aq)  H2O(l)  K2C8H4O4(aq)
We know the volume of the KOH solution, and we want to calculate the molarity of the KOH solution.
want to calculate
M of KOH 
need to find
mol KOH
L of KOH soln
given
If we can determine the moles of KOH in the solution, we can then calculate the molarity of the solution.
From the mass of KHP and its molar mass, we can calculate moles of KHP. Then, using the mole ratio
from the balanced equation, we can calculate moles of KOH.
1 mol KHP
1 mol KOH

 2.0654  103 mol KOH
204.22 g KHP 1 mol KHP
? mol KOH  0.4218 g KHP 
From the moles and volume of KOH, we calculate the molarity of the KOH solution.
M of KOH 
4.90
mol KOH
2.0654  103 mol KOH

 0.1106 M
L of KOH soln
18.68  103 L soln
The reaction between HCl and NaOH is:
HCl(aq)  NaOH(aq)  H2O(l)  NaCl(aq)
We know the volume of the NaOH solution, and we want to calculate the molarity of the NaOH solution.
need to find
want to calculate
M of NaOH 
mol NaOH
L of NaOH soln
given
If we can determine the moles of NaOH in the solution, we can then calculate the molarity of the solution.
From the volume and molarity of HCl, we can calculate moles of HCl. Then, using the mole ratio from the
balanced equation, we can calculate moles of NaOH.
? mol NaOH  17.4 mL HCl 
0.312 mol HCl 1 mol NaOH

 5.429  103 mol NaOH
1000 mL soln
1 mol HCl
From the moles and volume of NaOH, we calculate the molarity of the NaOH solution.
mol NaOH
5.429  103 mol NaOH

 0.217 M
L of NaOH soln
25.0  103 L soln
In order to have the correct mole ratio to solve the problem, you must start with a balanced chemical
equation.
M of NaOH 
4.91
(a)
HCl(aq)  NaOH(aq) 
 NaCl(aq)  H2O(l)
From the molarity and volume of the HCl solution, you can calculate moles of HCl. Then, using the
mole ratio from the balanced equation above, you can calculate moles of NaOH.
? mol NaOH  25.00 mL 
2.430 mol HCl 1 mol NaOH

 6.075  102 mol NaOH
1000 mL soln
1 mol HCl
Solving for the volume of NaOH:
liters of solution 
volume of NaOH 
moles of solute
M
6.075  102 mol NaOH
 4.278  102 L  42.78 mL
1.420 mol/L
(b)
This problem is similar to part (a). The difference is that the mole ratio between base and acid is 2:1.
H2SO4(aq)  2NaOH(aq) 
 Na2SO4(aq)  H2O(l)
? mol NaOH  25.00 mL 
volume of NaOH 
(c)
4.500 mol H2SO4 2 mol NaOH

 0.2250 mol NaOH
1000 mL soln
1 mol H 2SO4
0.2250 mol NaOH
 0.1585 L  158.5 mL
1.420 mol/L
This problem is similar to parts (a) and (b). The difference is that the mole ratio between base and
acid is 3:1.
H3PO4(aq)  3NaOH(aq) 
 Na3PO4(aq)  3H2O(l)
? mol NaOH  25.00 mL 
volume of NaOH 
4.95
1.500 mol H3PO4
3 mol NaOH

 0.1125 mol NaOH
1000 mL soln
1 mol H3PO4
0.1125 mol NaOH
 0.07923 L  79.23 mL
1.420 mol/L
The balanced equation is given in the problem. The mole ratio between Fe 2 and Cr2O72 is 6:1.
First, calculate the moles of Fe2 that react with Cr2O72 .
26.00 mL soln 
0.0250 mol Cr2O72
6 mol Fe2

 3.90  103 mol Fe2
2

1000 mL soln
1 mol Cr2O7
The molar concentration of Fe2 is:
M 
4.96
3.90  103 mol Fe2
25.0  103 L soln
 0.156 M
We want to calculate the grams of SO2 in the sample of air. From the molarity and volume of
KMnO4, we can calculate moles of KMnO4. Then, using the mole ratio from the balanced equation, we
can calculate moles of SO2. How do we convert from moles of SO2 to grams of SO2?
Solution: The balanced equation is given in the problem.
5SO2  2MnO4  2H2O 
 5SO24  2Mn2  4H
The moles of KMnO4 required for the titration are:
0.00800 mol KMnO4
 7.37 mL  5.896  105 mol KMnO 4
1000 mL soln
We use the mole ratio from the balanced equation and the molar mass of SO 2 as conversion factors to
convert to grams of SO2.
(5.896  105 mol KMnO4 ) 
5 mol SO2
64.07 g SO2

 9.44  103 g SO2
2 mol KMnO4
1 mol SO2
ANSWERS TO REVIEW OF CONCEPTS
Section 4.1 (p. 121)
Section 4.2 (p. 125)
Section 4.3 (p. 129)
Section 4.4 (p. 144)
Section 4.5 (p. 149)
Section 4.6 (p. 151)
Section 4.7 (p. 154)
Strongest electrolyte: AC2 (b). Weakest electrolyte: AD2 (c).
(a)
Weak acid: (b). Very weak acid: (c). Strong acid: (a).
(c)
0.9 M
9.47 g.
(b) H3PO4
(c) HCl
(d) H2SO4