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A17
Solutions to Self-Check Exercises
Self-Check Exercise 11.2
Electron
Configuration
Element
Al
1s22s22p63s23p1
[Ne]3s23p1
Si
[Ne]3s23p2
Orbital Diagram
1s
cT
2s
cT
2p
cT cT cT
3s
cT
3p
c
cT
cT
cT cT cT
cT
2
3
cT
cT
cT cT cT
cT
c c c
S
2
4
[Ne]3s 3p
cT
cT
cT cT cT
cT
cT c c
Cl
[Ne]3s23p5
cT
cT
cT cT cT
cT
cT cT c
Ar
[Ne]3s23p6
cT
cT
cT cT cT
cT
cT cT cT
P
[Ne]3s 3p
Self-Check Exercise 12.4
Self-Check Exercise 11.3
F:
Si:
Cs:
Pb:
2
2
5
2
c c
5
1s 2s 2p or [He]2s 2p
1s22s22p63s23p2 or [Ne]3s23p2
1s22s22p63s23p64s23d 104p65s24d 105p66s1 or [Xe]6s1
1s22s22p63s23p64s23d 104p65s24d 105p66s24f 145d106p2 or
[Xe]6s24f 145d 106p2
I: 1s22s22p63s23p64s23d 104p65s24d 105p5 or [Kr]5s24d 105p5
Silicon (Si): In Group 4 and Period 3, it is the second of the “3p
elements.” The configuration is 1s22s22p63s23p2, or [Ne]3s23p2.
Cesium (Cs): In Group 1 and Period 6, it is the first of the
“6s elements.” The configuration is
1s22s22p63s23p64s23d 104p65s24d 105p66s1, or [Xe]6s1.
Lead (Pb): In Group 4 and Period 6, it is the second of the “6p elements.” The configuration is [Xe]6s24f 145d 106p2.
Iodine (I): In Group 7 and Period 5, it is the fifth of the “5p elements.” The configuration is [Kr]5s24d 105p5.
See table on top of page A10.
Self-Check Exercise 12.5
a. NH4ϩ
ϩ
H
The Lewis structure is H
N
H
H
(See Self-Check Exercise 12.4.) There are four pairs of electrons around the nitrogen. This requires a tetrahedral
arrangement of electron pairs. The NH4ϩ ion has a tetrahedral molecular structure (row 3 in Table 12.4), because all
electron pairs are shared.
b. SO42Ϫ
The Lewis structure is
2Ϫ
O
O
O
S
O
Chapter 12
Self-Check Exercise 12.1
Using the electronegativity values given in Figure 12.3, we choose
the bond in which the atoms exhibit the largest difference in electronegativity. (Electronegativity values are shown in parentheses.)
a. HOC
Ͼ
HOP
c.
SOO
Ͼ NOO
(2.1)(2.5)
(2.1)(2.1)
(2.5)(3.5)
(3.0)(3.5)
b. OOI
Ͼ
OOF
d.
NOH
Ͼ SiOH
(3.5)(2.5)
(3.5)(4.0)
(3.0)(2.1)
(1.8)(2.1)
Self-Check Exercise 12.2
H has one electron, and Cl has seven valence electrons. This gives
a total of eight valence electrons. We first draw in the bonding
pair:
H¬Cl, which could be drawn as HϺCl
We have six electrons yet to place. The H already has two electrons,
so we place three lone pairs around the chlorine to satisfy the octet
rule.
H
Cl or H Cl
Self-Check Exercise 12.3
Step 1 O3: 3(6) ϭ 18 valence electrons
Step 2 OOOOO
Step 3 O O O and O O O
This molecule shows resonance (it has two valid Lewis structures).
(See Self-Check Exercise 12.4.) The four electron pairs around
the sulfur require a tetrahedral arrangement. The SO42Ϫ has a
tetrahedral molecular structure (row 3 in Table 12.4).
c. NF3
The Lewis structure is F
N
F
F
(See Self-Check Exercise 12.4.) The four pairs of electrons on
the nitrogen require a tetrahedral arrangement. In this case,
only three of the pairs are shared with the fluorine atoms,
leaving one lone pair. Thus the molecular structure is a trigonal pyramid (row 4 in Table 12.4).
d. H2S
The Lewis structure is H S H
(See Self-Check Exercise 12.4.) The four pairs of electrons
around the sulfur require a tetrahedral arrangement. In this
case, two pairs are shared with hydrogen atoms, leaving two
lone pairs. Thus the molecular structure is bent or V-shaped
(row 5 in Table 12.4).
Ϫ
e. ClO3Ϫ
The Lewis structure is O Cl O
O
(See Self-Check Exercise 12.4.) The four pairs of electrons require a tetrahedral arrangement. In this case, three pairs are
shared with oxygen atoms, leaving one lone pair. Thus the
molecular structure is a trigonal pyramid (row 4 in Table 12.4).
A18 Solutions to Self-Check Exercises
Molecule
or Ion
Total Valence
Electrons
Draw Single
Bonds
a. NF3
5 ϩ 3(7) ϭ 26
Calculate Number
of Electrons
Remaining
Use Remaining
Electrons to
Achieve Noble
Gas Configurations
Check
Atoms
Electrons
N
F
8
8
F
26 Ϫ 6 ϭ 20
F N
F N
F
b. O2
2(6) ϭ 12
c. CO
4 ϩ 6 ϭ 10
d. PH3
5 ϩ 3(1) ϭ 8
O
C
H
F
O
12 Ϫ 2 ϭ 10
O
O
O
8
O
10 Ϫ 2 ϭ 8
C
O
C
O
8
8
H
P
H
8
2
H
S
H
8
2
S
O
8
8
N
H
8
2
Cl
O
8
8
S
O
8
8
H
8Ϫ6ϭ2
P
H
H
e. H2S
2(1) ϩ 6 ϭ 8
H
S
g. NH4ϩ
O
5 ϩ 4(1) Ϫ 1 ϭ 8
S
P
H
H
8Ϫ4ϭ4
H
O
f. SO42Ϫ 6 ϩ 4(6) ϩ 2 ϭ 32
F
S
O
32 Ϫ 8 ϭ 24
O
S
O
O
H
H
H N
2Ϫ
O
H
8Ϫ8ϭ0
H
H
O
ϩ
N
H
H
O
h. ClO3Ϫ 7 ϩ 3(6) ϩ 1 ϭ 26
O
i. SO2
6 ϩ 2(6) ϭ 18
Ϫ
26 Ϫ 6 ϭ 20
Cl
O
O
O
S
O
Cl
O
O
18 Ϫ 4 ϭ 14
O
O
S O
and
S
O
Answer to Self-Check Exercise 12.4.
f. BeF2
The Lewis structure is F
Be
F
The two electron pairs on beryllium require a linear arrangement. Because both pairs are shared by fluorine atoms, the
molecular structure is also linear (row 1 in Table 12.4).
Chapter 13
Solving Boyle’s law (P1V1 ϭ P2V2) for V2 gives
P1
V2 ϭ V1 ϫ
P2
635 torr
ϭ 1.51 L ϫ
ϭ 1.22 L
785 torr
Note that the volume decreased, as the increase in pressure led us
to expect.
Self-Check Exercise 13.1
Self-Check Exercise 13.3
We know that 1.000 atm ϭ 760.0 mm Hg. So
Because the temperature of the gas inside the bubble decreases (at
constant pressure), the bubble gets smaller. The conditions are
525 mm Hg ϫ
1.000 atm
ϭ 0.691 atm
760.0 mm Hg
Self-Check Exercise 13.2
Initial Conditions
Final Conditions
P1 ϭ 635 torr
V1 ϭ 1.51 L
P2 ϭ 785 torr
V2 ϭ ?
Initial Conditions
T1 ϭ 28 °C ϭ 28 ϩ 273 ϭ 301 K
V1 ϭ 23 cm3
Final Conditions
T2 ϭ 18 °C ϭ 18 ϩ 273 ϭ 291 K
V2 ϭ ?
Solutions to Self-Check Exercises
Self-Check Exercise 13.7
Solving Charles’s law,
V1
V2
ϭ
T1
T2
for V2 gives
V2 ϭ V1 ϫ
A19
T2
291 K
ϭ 23 cm3 ϫ
ϭ 22 cm3
T1
301 K
Self-Check Exercise 13.4
Because the temperature and pressure of the two samples are the
same, we can use Avogadro’s law in the form
V1
V2
ϭ
n1
n2
The following information is given:
To solve this problem, we take the ideal gas law and separate those
quantities that change from those that remain constant (on opposite sides of the equation). In this case, volume and temperature
change, and number of moles and pressure (and, of course, R) remain constant. So PV ϭ nRT becomes V/T ϭ nR/P, which leads to
V1
nR
ϭ
T1
P
Combining these gives
V1
nR
ϭ
T1
P
and
ϭ
V2
nR
ϭ
T2
P
V2
V1
V2
or
ϭ
T2
T1
T2
to give n2 ϭ n1 ϫ V2/V1.
We are given
Initial Conditions
T1 ϭ 5 °C ϭ 5 ϩ 273 ϭ 278 K
V1 ϭ 3.8 L
Final Conditions
T2 ϭ 86 °C ϭ 86 ϩ 273 ϭ 359 K
V2 ϭ ?
Thus
(359 K)(3.8 L)
T2V1
V2 ϭ
ϭ
ϭ 4.9 L
T1
278 K
Check: Is the answer sensible? In this case, the temperature was increased (at constant pressure), so the volume should increase. The
answer makes sense.
Note that this problem could be described as a “Charles’s law
problem.” The real advantage of using the ideal gas law is that you
need to remember only one equation to do virtually any problem
involving gases.
Self-Check Exercise 13.5
Self-Check Exercise 13.8
We are given the following information:
P ϭ 1.00 atm
V ϭ 2.70 ϫ 106 L
n ϭ 1.10 ϫ 105 mol
We solve for T by dividing both sides of the ideal gas law by nR:
We are given the following information:
Initial Conditions
P1 ϭ 0.747 atm
T1 ϭ 13 °C ϭ 13 ϩ 273 ϭ 286 K
V1 ϭ 11.0 L
Final Conditions
P2 ϭ 1.18 atm
T2 ϭ 56 °C ϭ 56 ϩ 273 ϭ 329 K
V2 ϭ ?
In this case, the number of moles remains constant. Thus we can
say
P1V1
P2V2
ϭ nR and
ϭ nR
T1
T2
or
Sample 1
V1 ϭ 36.7 L
n1 ϭ 1.5 mol
Sample 2
V2 ϭ 16.5 L
n2 ϭ ?
We can now solve Avogadro’s law for the value of n2 (the moles of
N2 in sample 2):
V2
16.5 L
n2 ϭ n1 ϫ
ϭ 1.5 mol ϫ
ϭ 0.67 mol
V1
36.7 L
Here n2 is smaller than n1, which makes sense in view of the fact
that V2 is smaller than V1.
Note: We isolate n2 from Avogadro’s law as given above by multiplying both sides of the equation by n2 and then by n1/V1,
an2 ϫ
n1 V1
n1 V2
b
ϭ an2 ϫ
b
V1 n1
V1 n2
nRT
PV
ϭ
nR
nR
to give
Tϭ
PV
ϭ
nR
(1.00 atm)(2.70 ϫ 106 L)
(1.10 ϫ 105 mol) a0.08206
L atm
b
K mol
ϭ 299 K
The temperature of the helium is 299 K, or 299 Ϫ 273 ϭ 26 ЊC.
Self-Check Exercise 13.6
We are given the following information about the radon sample:
n ϭ 1.5 mol
V ϭ 21.0 L
T ϭ 33 °C ϭ 33 ϩ 273 ϭ 306 K
Pϭ?
We solve the ideal gas law (PV ϭ nRT ) for P by dividing both sides
of the equation by V:
L atm
(1.5 mol) a0.08206
b (306 K)
K mol
nRT
ϭ
Pϭ
V
21.0 L
ϭ 1.8 atm
P1V1
P2V2
ϭ
T1
T2
Solving for V2 gives
T2
P1
329 K
0.747 atm
V2 ϭ V1 ϫ
ϫ
ϭ (11.0 L) a
ba
b
T1
P2
286 K
1.18 atm
ϭ 8.01 L
Self-Check Exercise 13.9
As usual when dealing with gases, we can use the ideal gas equation PV ϭ nRT. First consider the information given:
P ϭ 0.91 atm ϭ Ptotal
V ϭ 2.0 L
T ϭ 25 °C ϭ 25 ϩ 273 ϭ 298 K
A20 Solutions to Self-Check Exercises
Given this information, we can calculate the number of moles of
gas in the mixture: ntotal ϭ nN2 ϩ nO2. Solving for n in the ideal gas
equation gives
(0.91 atm) (2.0 L)
PtotalV
ϭ
ϭ 0.074 mol
n total ϭ
RT
L atm
a0.08206
b (298 K)
K mol
We also know that 0.050 mole of N2 is present. Because
n total ϭ nN2 ϩ nO2 ϭ 0.074 mol
c
(0.050 mol)
we can calculate the moles of O2 present.
0.050 mol ϩ nO2 ϭ 0.074 mol
nO2 ϭ 0.074 mol Ϫ 0.050 mol ϭ 0.024 mol
Now that we know the moles of oxygen present, we can calculate
the partial pressure of oxygen from the ideal gas equation.
PO2 ϭ
nO2RT
V
ϭ
(0.024 mol) a0.08206
L atm
b (298 K)
K mol
2.0 L
ϭ 0.29 atm
Although it is not requested, note that the partial pressure of the
N2 must be 0.62 atm, because
0.62 atm ϩ 0.29 atm ϭ 0.91 atm
PO2
PN2
Ptotal
Self-Check Exercise 13.10
The volume is 0.500 L, the temperature is 25 ЊC (or 25 ϩ 273 ϭ
298 K), and the total pressure is given as 0.950 atm. Of this total
pressure, 24 torr is due to the water vapor. We can calculate the
partial pressure of the H2 because we know that
Ptotal ϭ PH2 ϩ PH2O ϭ 0.950 atm
c
24 torr
Before we carry out the calculation, however, we must convert the
pressures to the same units. Converting PH2O to atmospheres gives
24 torr ϫ
1.000 atm
ϭ 0.032 atm
760.0 torr
Step 1 Using the atomic mass of zinc (65.38), we calculate the
moles of zinc in 26.5 g.
1 mol Zn
ϭ 0.405 mol Zn
65.38 g Zn
Step 2 Using the balanced equation, we next calculate the moles
of H2 produced.
1 mol H2
ϭ 0.405 mol H2
0.405 mol Zn ϫ
1 mol Zn
Step 3 Now that we know the moles of H2, we can compute the
volume of H2 by using the ideal gas law, where
P ϭ 1.50 atm
Vϭ?
n ϭ 0.405 mol
R ϭ 0.08206 L atm>K mol
T ϭ 19 °C ϭ 19 ϩ 273 ϭ 292 K
L atm
(0.405 mol) a0.08206
b (292 K)
K mol
nRT
Vϭ
ϭ
P
1.50 atm
26.5 g Zn ϫ
ϭ 6.47 L of H2
Self-Check Exercise 13.12
Although there are several possible ways to do this problem, the
most convenient method involves using the molar volume at STP.
First we use the ideal gas equation to calculate the moles of NH3
present:
PV
nϭ
RT
where P ϭ 15.0 atm, V ϭ 5.00 L, and T ϭ 25 ЊC ϩ 273 ϭ 298 K.
(15.0 atm) (5.00 L)
nϭ
ϭ 3.07 mol
L atm
a0.08206
b (298 K)
K mol
We know that at STP each mole of gas occupies 22.4 L. Therefore,
3.07 mol has the volume
22.4 L
ϭ 68.8 L
1 mol
The volume of the ammonia at STP is 68.8 L.
3.07 mol ϫ
Chapter 14
Self-Check Exercise 14.1
Thus
Ptotal ϭ PH2 ϩ PH2O ϭ 0.950 atm ϭ PH2 ϩ 0.032 atm
and
PH2 ϭ 0.950 atm Ϫ 0.032 atm ϭ 0.918 atm
PH2V
ϭ
(0.918 atm) (0.500 L)
a0.08206
L atm
b (298 K)
K mol
ϭ 0.0188 mol ϭ 1.88 ϫ 10 Ϫ2 mol
The sample of gas contains 1.88 ϫ 10Ϫ2 mole of H2, which exerts a
partial pressure of 0.918 atm.
RT
1 mol H2O
ϭ 0.83 mol H2O
18 g H2O
kJ
ϭ 5.0 kJ
0.83 mol H2O ϫ 6.02
mol H2O
Energy to heat the water from 0 °C to 100 °C:
J
4.18
ϫ 15 g ϫ 100 °C ϭ 6300 J
g °C
1 kJ
6300 J ϫ
ϭ 6.3 kJ
1000 J
Energy to vaporize the water at 100 °C:
kJ
0.83 mol H2O ϫ 40.6
ϭ 34 kJ
mol H2O
Total energy required:
15 g H2O ϫ
Now that we know the partial pressure of the hydrogen gas, we can
use the ideal gas equation to calculate the moles of H2.
nH2 ϭ
Energy to melt the ice:
Self-Check Exercise 13.11
We will solve this problem by taking the following steps:
5.0 kJ ϩ 6.3 kJ ϩ 34 kJ ϭ 45 kJ
Self-Check Exercise 14.2
Grams
of
zinc
Moles
of
zinc
Moles
of
H2
Volume
of
H2
a. Contains SO3 molecules—a molecular solid.
b. Contains Ba2ϩ and O2Ϫ ions—an ionic solid.
c. Contains Au atoms—an atomic solid.
Solutions to Self-Check Exercises
Chapter 15
Self-Check Exercise 15.4
Self-Check Exercise 15.1
mass of solute
ϫ 100%
mass of solution
For this sample, the mass of solution is 135 g and the mass of the
solute is 4.73 g, so
4.73 g solute
Mass percent ϭ
ϫ 100%
135 g solution
ϭ 3.50%
Mass percent ϭ
When Na2CO3 and Al2(SO4)3 dissolve in water, they produce ions
as follows:
H2o(l)
Na2CO3(s) ¬¡ 2Na ϩ (aq) ϩ CO2Ϫ
3 (aq)
H2o(l)
Al2(SO4 )3(s) ¬¡ 2Al3ϩ (aq) ϩ 3SO2Ϫ
4 (aq)
Therefore, in a 0.10 M Na2CO3 solution, the concentration of Naϩ
ions is 2 ϫ 0.10 M ϭ 0.20 M and the concentration of CO32Ϫ ions is
0.10 M. In a 0.010 M Al2(SO4)3 solution, the concentration of Al3ϩ
ions is 2 ϫ 0.010 M ϭ 0.020 M and the concentration of SO42Ϫ ions is
3 ϫ 0.010 M ϭ 0.030 M.
Self-Check Exercise 15.2
Self-Check Exercise 15.5
Using the definition of mass percent, we have
When solid AlCl3 dissolves, it produces ions as follows:
H2o(l)
AlCl3(s) ¬¡ Al3ϩ (aq) ϩ 3ClϪ(aq)
Mass of solute
ϭ
Mass of solution
grams of solute
ϫ 100% ϭ 40.0%
grams of solute ϩ grams of solvent
There are 425 grams of solute (formaldehyde). Substituting, we have
425 g
ϫ 100% ϭ 40.0%
425 g ϩ grams of solvent
We must now solve for grams of solvent (water). This will take
some patience, but we can do it if we proceed step by step. First we
divide both sides by 100%.
425 g
100%
40.0%
ϫ
ϭ
ϭ 0.400
425 g ϩ grams of solvent
100%
100%
Now we have
425 g
ϭ 0.400
425 g ϩ grams of solvent
Next we multiply both sides by (425 g ϩ grams of solvent).
425 g
(425 g ϩ grams of solvent) ϫ
425 g ϩ grams of solvent
ϭ 0.400 ϫ (425 g ϩ grams of solvent)
This gives
425 g ϭ 0.400 ϫ (425 g ϩ grams of solvent)
Carrying out the multiplication gives
425 g ϭ 170. g ϩ 0.400 (grams of solvent)
Now we subtract 170. g from both sides,
425 g Ϫ 170. g ϭ 170. g Ϫ 170. g ϩ 0.400 (grams of solvent)
255 g ϭ 0.400 (grams of solvent)
and divide both sides by 0.400.
255 g
0.400
ϭ
(grams of solvent)
0.400
0.400
We finally have the answer:
255 g
ϭ 638 g ϭ grams of solvent
0.400
ϭ mass of water needed
Self-Check Exercise 15.3
The moles of ethanol can be obtained from its molar mass (46.1).
1 mol C2H5OH
ϭ 2.17 ϫ 10 Ϫ2 mol C2H5OH
46.1 g C2H5OH
1L
Volume in liters ϭ 101 mL ϫ
ϭ 0.101 L
1000 mL
moles of C2H5OH
Molarity of C2H5OH ϭ
liters of solution
2.17 ϫ 10 Ϫ2 mol
ϭ
0.101 L
ϭ 0.215 M
1.00 g C2H5OH ϫ
A21
so a 1.0 ϫ 10Ϫ3 M AlCl3 solution contains 1.0 ϫ 10Ϫ3 M Al3ϩ ions
and 3.0 ϫ 10Ϫ3 M ClϪ ions.
To calculate the moles of ClϪ ions in 1.75 L of the 1.0 ϫ 10Ϫ3 M
AlCl3 solution, we must multiply the volume by the molarity.
1.75 L solution ϫ 3.0 ϫ 10 Ϫ3 M Cl Ϫ
3.0 ϫ 10 Ϫ3 mol Cl Ϫ
ϭ 1.75 L solution ϫ
L solution
ϭ 5.25 ϫ 10 Ϫ3 mol Cl Ϫ ϭ 5.3 ϫ 10 Ϫ3 mol Cl Ϫ
Self-Check Exercise 15.6
We must first determine the number of moles of formaldehyde in
2.5 L of 12.3 M formalin. Remember that volume of solution (in
liters) times molarity gives moles of solute. In this case, the volume
of solution is 2.5 L and the molarity is 12.3 moles of HCHO per liter
of solution.
12.3 mol HCHO
ϭ 31 mol HCHO
L solution
Next, using the molar mass of HCHO (30.0 g), we convert 31 moles
of HCHO to grams.
30.0 g HCHO
31 mol HCHO ϫ
ϭ 9.3 ϫ 102 g HCHO
1 mol HCHO
Therefore, 2.5 L of 12.3 M formalin contains 9.3 ϫ 102 g of formaldehyde. We must weigh out 930 g of formaldehyde and dissolve it in
enough water to make 2.5 L of solution.
2.5 L solution ϫ
Self-Check Exercise 15.7
We are given the following information:
mol
mol
M1 ϭ 12
M2 ϭ 0.25
L
L
V1 ϭ ? (what we need to find)
V2 ϭ 0.75 L
Using the fact that the moles of solute do not change upon dilution, we know that
M1 ϫ V1 ϭ M2 ϫ V2
Solving for V1 by dividing both sides by M1 gives
mol
0.25
ϫ 0.75 L
M2 ϫ V2
L
V1 ϭ
ϭ
M1
mol
12
L
and
V1 ϭ 0.016 L ϭ 16 mL
Self-Check Exercise 15.8
Step 1 When the aqueous solutions of Na2SO4 (containing Naϩ
and SO42Ϫ ions) and Pb(NO3)2 (containing Pb2ϩ and NO3Ϫ ions) are
mixed, solid PbSO4 is formed.
Pb2ϩ(aq) ϩ SO42Ϫ(aq) n PbSO4(s)
A22 Solutions to Self-Check Exercises
Step 2 We must first determine whether Pb2ϩ or SO42Ϫ is the limiting reactant by calculating the moles of Pb2ϩ and SO42Ϫ ions
present. Because 0.0500 M Pb(NO3)2 contains 0.0500 M Pb2ϩ ions,
we can calculate the moles of Pb2ϩ ions in 1.25 L of this solution
as follows:
0.0500 mol Pb2ϩ
1.25 L ϫ
ϭ 0.0625 mol Pb2ϩ
L
The 0.0250 M Na2SO4 solution contains 0.0250 M SO42Ϫ ions, and
the number of moles of SO42Ϫ ions in 2.00 L of this solution is
0.0250 mol SO42Ϫ
2.00 L ϫ
ϭ 0.0500 mol SO42Ϫ
L
Step 3 Pb2ϩ and SO42Ϫ react in a 1:1 ratio, so the amount of SO42Ϫ
ions is limiting because SO42Ϫ is present in the smaller number of
moles.
Step 4 The Pb2ϩ ions are present in excess, and only 0.0500 mole
of solid PbSO4 will be formed.
Step 5 We calculate the mass of PbSO4 by using the molar mass
of PbSO4 (303.3 g).
303.3 g PbSO4
0.0500 mol PbSO4 ϫ
ϭ 15.2 g PbSO4
1 mol PbSO4
Self-Check Exercise 15.9
Step 1 Because nitric acid is a strong acid, the nitric acid solution
contains Hϩ and NO3Ϫ ions. The KOH solution contains Kϩ and
OHϪ ions. When these solutions are mixed, the Hϩ and OHϪ react
to form water.
H ϩ (aq) ϩ OHϪ(aq)SH2O(l)
Step 2 The number of moles of OHϪ present in 125 mL of 0.050 M
KOH is
1L
0.050 mol OH Ϫ
125 mL ϫ
ϫ
ϭ 6.3 ϫ 10 Ϫ3 mol OH Ϫ
1000 mL
L
Step 3 Hϩ and OHϪ react in a 1:1 ratio, so we need 6.3 ϫ 10Ϫ3 mole
of Hϩ from the 0.100 M HNO3.
Step 4 6.3 ϫ 10Ϫ3 mole of OHϪ requires 6.3 ϫ 10Ϫ3 mole of Hϩ to
form 6.3 ϫ 10Ϫ3 mole of H2O.
Therefore,
0.100 mol H ϩ
Vϫ
ϭ 6.3 ϫ 10 Ϫ3 mol H ϩ
L
where V represents the volume in liters of 0.100 M HNO3 required.
Solving for V, we have
6.3 ϫ 10 Ϫ3 mol H ϩ
ϭ 6.3 ϫ 10 Ϫ2 L
Vϭ
0.100 mol H ϩ
L
1000 mL
Ϫ2
ϭ 6.3 ϫ 10 L ϫ
ϭ 63 mL
L
Self-Check Exercise 15.10
From the definition of normality, N ϭ equiv/L, we need to calculate (1) the equivalents of KOH and (2) the volume of the solution
in liters. To find the number of equivalents, we use the equivalent
weight of KOH, which is 56.1 g (see Table 15.2).
1 equiv KOH
23.6 g KOH ϫ
ϭ 0.421 equiv KOH
56.1 g KOH
Next we convert the volume to liters.
1L
755 mL ϫ
ϭ 0.755 L
1000 mL
Finally, we substitute these values into the equation that defines
normality.
equiv
0.421 equiv
Normality ϭ
ϭ
ϭ 0.558 N
L
0.755 L
Self-Check Exercise 15.11
To solve this problem, we use the relationship
Nacid ϫ Vacid ϭ Nbase ϫ Vbase
where
equiv
Nacid ϭ 0.50
L
Vacid ϭ ?
equiv
Nbase ϭ 0.80
L
Vbase ϭ 0.250 L
We solve the equation
Nacid ϫ Vacid ϭ Nbase ϫ Vbase
for Vacid by dividing both sides by Nacid.
Nacid ϫ Vacid
Nbase ϫ Vbase
ϭ
Nacid
Nacid
equiv
(0.80
) ϫ (0.250 L)
Nbase ϫ Vbase
L
Vacid ϭ
ϭ
equiv
Nacid
0.50
L
Vacid ϭ 0.40 L
Therefore, 0.40 L of 0.50 N H2SO4 is required to neutralize 0.250 L
of 0.80 N KOH.
Chapter 16
Self-Check Exercise 16.1
The conjugate acid–base pairs are
H2O,
H3O ϩ
Base
Conjugate acid
HC2H3O2,
C2H3O2Ϫ
and
Acid
Conjugate base
The members of both pairs differ by one Hϩ.
Self-Check Exercise 16.2
Because [Hϩ][OHϪ] ϭ 1.0 ϫ 10Ϫ14, we can solve for [Hϩ].
1.0 ϫ 10 Ϫ14
1.0 ϫ 10 Ϫ14
3HϪ 4 ϭ
ϭ
ϭ 5.0 ϫ 10 Ϫ13 M
Ϫ
3OH 4
2.0 ϫ 10 Ϫ2
This solution is basic: [OHϪ] ϭ 2.0 ϫ 10Ϫ2 M is greater than [Hϩ] ϭ
5.0 ϫ 10Ϫ13 M.
Self-Check Exercise 16.3
a. Because [Hϩ] ϭ 1.0 ϫ 10Ϫ3 M, we get pH ϭ 3.00 because pH ϭ
Ϫlog[Hϩ] ϭ Ϫlog[1.0 ϫ 10Ϫ3] ϭ 3.00.
b. Because [OHϪ] ϭ 5.0 ϫ 10Ϫ5 M, we can find [Hϩ] from the Kw
expression.
Kw
1.0 ϫ 10 Ϫ14
ϭ 2.0 ϫ 10 Ϫ10 M
3H ϩ 4 ϭ
Ϫ ϭ
3OH 4
5.0 ϫ 10 Ϫ5
pH ϭ Ϫlog[Hϩ] ϭ Ϫlog[2.0 ϫ 10Ϫ10] ϭ 9.70
Self-Check Exercise 16.4
pOH ϩ pH ϭ 14.00
pOH ϭ 14.00 Ϫ pH ϭ 14.00 Ϫ 3.5
pOH ϭ 10.5
Self-Check Exercise 16.5
Step 1 pH ϭ 3.50
Step 2 ϪpH ϭ Ϫ3.50