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9: Energy and Our World

9: Energy and Our World

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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







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







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





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



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



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



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)



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



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



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



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