chapter 8 acid–base equilibrium -...

Copyright © 2003 Nelson Acid–Base Equilibrium 287

CHAPTER 8 ACID–BASE EQUILIBRIUM

Reflect on Your Learning

(Page 526)

1. The concentration of the hydrogen ion in pure water at SATP is equal to 1.0 � 10–7 mol/L. pH � –log [H�

(aq) ]

� –log [1.0 � 10–7]

pH � 7

2. The pH of hydrochloric acid is lower because HCl(aq) is a stronger acid than acetic acid. 3. The products for the neutralization of HCl(aq) and NaOH(aq) are NaCl(aq) and water. Neither the sodium ion nor chlo-

ride hydrolyze to change the pH of water. However, the neutralization of acetic acid with sodium hydroxide produceswater and sodium acetate. The acetate ion is a stronger base than chloride and hydrolyzes to release hydroxide ionsin solution. The production of hydroxide accounts for why the resulting solution is basic.

4. The blood contains a variety of buffering agents, which resist changes in pH.

Try This Activity: Antacid Equilibrium

(Page 527)

(a) Mg(OH)2(s) e Mg2�(aq) � 2 OH–

(aq)

(b) The equilibrium shifts to the right when hydrochloric acid is added. The evidence is the observation that the solutionbecomes clear, indicating that all the magnesium hydroxide had dissolved.

8.1 THE NATURE OF ACID–BASE EQUILIBRIA

PRACTICE

(Page 532)

Understanding Concepts1. (a) HCO3

–(aq) / CO3

2–(aq) HS–

(aq) / S2–(aq)

(b) H2CO3(aq) / HCO3–(aq) H2O(l) / OH–

(aq)

(c) HSO4–(aq) / SO4

2–(aq) H2PO4

–(aq) / HPO4

2–(aq)

(d) H2O(l) / OH–(aq) H3O�

(aq) / H2O(l)

2. Amphoteric substances in question 1 are H2O(l), HCO3–(aq).

3. H2CO3(aq) / HCO3–(aq) HCO3

–(aq) / CO3

2–(aq)

PRACTICE

(Page 537)

Understanding Concepts4. (a) 0 mol/L

(b) [OH–] � �1.0

0�

.3100–14

� � 3.3 � 10–14 mol/L

5. nHCl � �36.

04.637

g/gmol

�

nHCl � 0.010 mol

[H�] � �25

00.0

�

101m0–

o3lL

�

[H�] � 0.040 mol/L

[OH–] � �[H

K�

w

]�

� �1.0

0�

.04100–14

�

[OH–] � 2.5 � 10–13 mol/L

The hydroxide ion concentration is 2.5 � 10–13 mol/L.

6. [H�] � 4.40 � 10–3 mol/L

[OH–] � �[H

K�

w

]�

��4.40

1.�

0 �

101–30–

m

14

ol/L�

[OH–] � 2.3 � 10–12 mol/L

The hydroxide ion concentration in the effluent is 2.3 � 10–12 mol/L.

Applying Inquiry Skills7. Prediction

An increase in the hydrogen ion concentration should shift the equilibrium to the right.

Experimental Design1. Half-fill one test tube with 0.1 mol/L potassium chromate, K2CrO4.2. Add 5 drops of 1.0 mol/L hydrochloric acid, HCl, to the test tube. Stopper the test tube and carefully mix its

contents.3. Continue adding acid, if necessary, until a colour change is observed.

PRACTICE

(Page 540)

Understanding Concepts8. Ca(OH)2(aq) e Ca2�

(aq) � 2 OH–(aq)

[Ca(OH)2(aq)] � 6.9 � 10–3 mol/L

[OH–] � 2[Ca(OH)2(aq)]

[OH–] � 1.38 � 10–2 mol/L

[H�] � �[O

K

Hw

–]�

��1.38

1.�

0 �

101–20–

m

14

ol/L�

[H�] � 7.2 � 10–13 mol/L

9. [OH–] � 0.299 � 10–3 mol/L

[H�] � �[O

K

Hw

–]�

�

[H�] � 3.34 � 10–11 mol/LThe hydrogen ion concentration in the cleaning solution is 3.34 � 10–11 mol/L.

1.00 � 10–14��0.299 � 10–3 mol/L

288 Chapter 8 Copyright © 2003 Nelson

PRACTICE

(Page 546)

Understanding Concepts12. (a) pH � – log [H�]

� –log [0.006]

pH � 2.2

pOH � 14.0 – pH

� 14.0 – 2.2

pOH � 11.8

[OH–] � 10–pOH

� 10–11.8

[OH–] � 2 � 10–12 mol/L

(b) pH � –log [H�]

� –log [0.025]

pH � 1.60

pOH � 14.0 – pH

� 14.00 – 1.60

pOH � 12.40

[OH–] � 10–pOH

� 10–12.40

[OH–] � 4.0 � 10–13 mol/L

(c) pH � –log [H�]

� – log [0.010]

pH � 2.00

pOH � 14.0 – pH

� 14.00 – 2.00

pOH � 12.00

[OH–] � 10–pOH

� 10–12.00

[OH–] � 1.0 � 10–12 mol/L

13. nNaOH � �40.0

206gg/mol�

nNaOH � 0.65 mol

[NaOH] � �00..61550

mLol

�

[NaOH] � 4.33333333 mol/L

pOH � – log [OH–]


� – log [4.333333]

pOH � –0.64

pH � 14.00 – pOH

� 14.00 – (–0.64)

pH � 14.64

14. pOH � 14.0 – 11.5

pOH � 2.5

[OH–] � 3 � 10–3 mol/L

nKOH � [3 � 10–3 mol/L][0.5 L]

nKOH � 1.5 � 10–3 mol

mKOH � [1.5 � 10–3 mol][56.11 g/mol]

mKOH � 0.09 g

Making Connections15. (a) (Solution provided for oranges. Calculations for other fruits are similar.)

Oranges:

[OH–] � �15..05

�

�

1100

–

–

1

3

4�

[OH–] � 1.8 � 10–12 mol/L

pOH � –log [OH–]

� –log [1.8 �10–12]

pOH � 11.74

pH � 14.00 – pOH

� 14.00 – 11.74

pH � 2.26

(b) Oranges and olives would taste the most sour because they have the lowest pH.(c) Blackberries may relieve heartburn caused by excess stomach acid because they are basic.(d) Diet suggestions: The speed of movement of sperm decreases with increasing acidity. Vaginal secretions are

slightly acidic while uterine secretions are slightly alkaline – a more favourable environment to sperm. Foods thatresult in an increase in the acidity of vaginal secretions should be avoided.

16. (a) The hydrogen ion concentration increases after the change in pH.(b) The stomach is made up of layers of tissue, each with a specific function. The outermost layer, called the serosa,

is tough and acid resistant. Specialized cells in an inner layer of the stomach called the epithelium secrete mucus,which protects the epithelium and underlying tissues from attack by stomach acid and digestive proteins.


Food [H+(aq)] [OH–

(aq)] pH pOH

Oranges 5.5 � 10–3 1.8 � 10–12 2.26 11.74

Asparagus 4 � 10–9 3 � 10–6 8.4 5.6

Olives 5.0 � 10–4 2.0 � 10–11 3.30 10.70

Blackberries 4.0 � 10–4 2.5 � 10–11 3.40 10.60

PRACTICE

(49)

Understanding Concepts 17. [OH–

(aq)] � 0.15 mol/L

pOH � –log 0.15

pOH � 0.8239 (extra digits carried)

pH � 14 – pOH

� 14 – 0.8239

pH � 13.18

The pH of the sodium hydroxide solution is 13.18.

18. [OH–(aq)] � 2 � [Ba(OH)2(aq)]

� 2 � 0.032 mol/L

[OH–(aq)] � 0.064 mol/L

pOH � –log 0.064


pH � 14 – pOH

� 14 – 1.194

pH � 12.81

The pH of the barium hydroxide solution is 12.81.

19. nCa(OH)2(aq)� �

74.01.080

g/gmol

�

nCa(OH)2(aq)� 0.0108 mol (extra digits carried)

[Ca(OH)2(aq)] � �0.

00.110080

mL

ol�

[Ca(OH)2(aq)] � 0.108 mol/L

[OH–] � 2 � [Ca(OH)2(aq)]

� 2 � 0.108

[OH–] � 0.216 mol/L

pOH � –log 0.216

pOH � 0.666

pH � 14 – 0.666

pH � 13.33

The pH of the barium hydroxide solution is 13.33.


1. Test all solutions with pH test strips. The strong acid has the lowest pH, the weak acid has the next lowest pH,and the remaining solutions have pH 7.

2. Test the pH 7 solutions with a conductivity tester. The neutral ionic solution will test positive while the neutralmolecular solution will test negative.

Making Connections9. (a) Gastroesophageal reflux disease (GERD) is a condition that results in the movement of stomach or duodenal

contents into the esophagus.(b) GERD can affect all age groups.(c) A thin spaghetti-like tube containing a tiny pH probe is passed down the throat to the esophagus where it meas-

ures acidity over a 24-h period. Data from the probe is recorded on a recording device. Hospitalization is usuallyrequired while the test is being done.

(d) Treatments currently available include Monitoring diet: the following foods can aggravate acid reflux: deep-fried foods, whole milk, chocolate, creamyfoods. The following foods can aggravate an already-inflamed lower esophagus: coffee and other caffeinatedbeverages, carbonated soft drinks, citric juices like orange and grapefruit juice.Surgery: the lower portion of the esophagus is surgically tightened, the end result being a one-way valve, whichallows food to enter the stomach while preventing stomach contents from flowing upward.

10. (a) Prior to the 19th century, paper was handmade from linen or rags. To meet the skyrocketing demand for paper inthe 19th century, chemists found that large quantities of paper could be made economically from wood pulp.During this process, alum (aluminum sulfate) was added to the paper to fill the microscopic holes in the paper.This procedure is known as sizing. Sizing prevents ink from bleeding across the paper. The acidity of thealuminum ion catalyzes the degradation of cellulose strands of the paper. Over a long period of time, acidic paperbecomes extremely brittle. Acid-free paper is manufactured often with alkaline sizing agents such as alkyl ketenedimers (AKD).

(b) Acid-free paper is used for important documents that must last a long time, such as birth certificates, marriagelicences, photograph album pages, archival records, etc.

(c) Acid-free paper lasts considerably longer and is stronger than regular acidic paper. It also requires less energy andfresh water to manufacture. Paper made from this process can be recycled more easily. One disadvantage of acid-free paper is that the sizing agents tend to form sticky deposits on the paper-making equipment, and sometimeson the paper itself.

8.2 WEAK ACIDS AND BASES

PRACTICE

(Page 554)

Understanding Concepts1. [H�

(aq)] � 10–2.54

[H�(aq)] � 2.88 � 10–3 mol/L

[H�(aq)] � �

1p00� � [HC2H3O2]

p � �[H

[

C

H

2

�(

Haq

3

)

O

]

2]� � 100%

��2.8

[80�

.461m0–

o

3

l/mL

o]l/L

�� 100%

p � 0.63%

2. [H�(aq)] � 10–2.00

[H�(aq)] � 1.0 � 10–2 mol/L

[H�(aq)] � �

1p00� � [HF(aq)]


p � �[

[

H

H

F

�(

(

a

a

q

q

)

)

]

]� � 100%

��1.0

[0�

.1510

m

–2

olm/L

o]l/L

�� 100%

p � 6.7%

PRACTICE

(Page 556)

Understanding Concepts3. 5.8%

HNO2(aq) e H�(aq) � NO2

–(aq)

Ka � �[H

[

�(

Haq

N)][

O

N

2

O

(aq

2

)

–(

]aq)]

�

x � 200 mol/L � 0.058

x � 0.0116 mol/L (extra digits carried)

Ka � ��[H

[

�(

Haq

N)][

O

N

2

O

(aq

2

)

–(

]aq)]

�

� �(0

0.0.118186)2

�

Ka � 7.1 � 10–4

4. (a) 7.8%

HF(aq) e H�(aq) � F–

(aq)

Ka � �[H

[

�(

Haq

F)]

(

[

a

F

q)

–(a

]q)]

�

x � 0.100 mol/L � 0.078

x � 0.0078 mol/L

[H�(aq) ] � 7.8 � 10–3 mol/L

(b) 0.0078%

HCN(aq) e H�(aq) � CN–

(aq)

Ka � �[H

[

�(

Haq

C)][

N

C

(a

N

q)

–(

]aq)]

�

x � 0.100 mol/L � 7.8 � 10–4

x � 7.8 � 10–6 mol/L

[H�(aq)] � 7.8 � 10–6 mol/L

(c) The hydrofluoric solution (a) is more acidic.


ICE Table for the Ionization of HNO2(aq)

HNO2(aq) e H+(aq) + NO2

–(aq)

Initial concentration (mol/L) 0.200

Change in concentration (mol/L) – 0.0116 +0.0116 +0.0116

Equilibrium concentration (mol/L) 0.188 mol/L 0.0116 0.0116

5. [H�(aq)] � �

1p00� � [HA(aq)]

p � �[H

[H

A

�(a

(

q

aq

)]

)]� � 100%

��1.1

[60.

�

10010

m

–3

olm/L

o]l/L

�� 100%

p � 1.16%

The percent ionization of propanoic acid is 1.16%.

PRACTICE

(Page 563)

Understanding Concepts6. (a) Kw � 1.0 � 10–14

Ka � 2.9 � 10–8

KaKb � Kw

Kb � �K

Kw

a�

� �12..09

�

�

1100

–

–

1

8

4�

Kb � 3.4 � 10–7

The base dissociation constant for the hypochlorite ion is 3.4 � 10–7.

(b) Kw � 1.0 � 10–14

Ka � 7.2 � 10–4

KaKb � Kw

Kb � �K

Kw

a�

� �17..02

�

�

1100

–

–

1

4

4�

Kb � 1.4 � 10–11

The base dissociation constant for the nitrite ion is 1.4 � 10–11.

(c) Kw � 1.0 � 10–14

Ka � 6.3 � 10–5

KaKb � Kw

Kb � �K

Kw

a�

� �16..03

�

�

1100

–

–

1

5

4�

Kb � 1.6 � 10–10

The base dissociation constant for the benzoate ion is 1.6 � 10–10.


PRACTICE

(Page 568)

Understanding Concepts7. HC3H5O3(aq) e H�

(aq) � C3H5O3–(aq)

Ka ��[H

[

�(

Haq

C)][

3

C

H3

5

H

O5

3

O

(aq

3

)

–(a

]q)]

�

Ka � 1.4 � 10–4

Predict whether a simplifying assumption is justified:

�[HA

K

]

a

initial� � �

0.10.041�

0 m10

o–l4/L

�

�[HA

K

]

a

initial� � 7.14

Since 7.14 < 100, we may not assume that 0.0010 – x � 0.0010.

�(0.00

x1

2

0 – x)� � 1.4 � 10–4

x2 � 1.4 � 10–4(0.0010 – x)

x �

x � 3.1 � 10–4

pH � –log [H�(aq)]

� –log [3.1 � 10–4]

pH � 3.51

8. HCO2H(aq) e H�(aq) � HCO2

–(aq)

Ka ��[H

[

�(

Haq

C)][

O

H

2

C

H

O

(aq

2

)

–(a

]q)]

�

Ka � 1.8 � 10–4


�[HA

K

]

a

initial� � �

01.1.850

�

m1o0l–/4L

�

�[HA

K

]

a

initial� � 830

Since 830 > 100, we may assume that 0.150 – x � 0.150.

–1.4x � 10–4� �(1.4x �� 10–4)2� – 4(–1�.4 � 1�0–7)��

2


ICE Table for the Ionization of HC3H5O3(aq)

HC3H5O3(aq) e H+(aq) + C3H5O3

–(aq)

Initial concentration (mol/L) 0.0010 0.0000 0.0000

Change in concentration (mol/L) –x +x +x

Equilibrium concentration (mol/L) 0.0010 – x x x

The equilibrium expression becomes

�0.

x1

2

50� � 1.8 � 10–4

which yields

x2 � 2.7 � 10–5

x � 5.2 � 10–3

Validation of the approximation:In general, the approximation is valid if

�[HA

x]initial� � 100% ≤ 5%

[HA]initial � 0.15 mol/L

�[HA

x]initial� � 100% � �

5.20�

.1510–3� � 100%

�[HA

x]initial� � 3.5%

Since 3.5% < 5%, the assumption is valid.

Therefore,

[H�(aq) ] � 5.2 � 10–3 mol/L

pH � –log 5.2 � 10–3

pH � 2.28

PRACTICE

(Page 570)

Understanding Concepts9. pH � 2.40

[H�(aq)] � 10–pH

� 10–2.40

[H�(aq)] � 3.98 � 10–3 mol/L

HC7H5O2(aq) e H�(aq) � C7H5O2

–(aq)

Ka �

[H�(aq)] � [C7H5O2

–(aq)]

[H�(aq)] � 3.98 � 10–3 mol/L

Ka � �(3.98

0�

.2510–3)2�

Ka � 6.3 � 10–5

The Ka for benzoic acid is 6.3 � 10–5.

10. pH � 2.40

[H�(aq)] � 10–pH

� 10–2.40

[H�(aq)] � 3.98 � 10–3 mol/L

[H�(aq)][C7H5O2

–(aq)]

��[HC7H5O2(aq)]



–(aq)

Ka �

[H�(aq)] � [C6H7O6

–(aq)]

[H�(aq)] � 3.98 � 10–3 mol/L

Ka � �(3.98

0�

.2010–3)2�

Ka � 7.9 � 10–5

The Ka for ascorbic acid is 7.9 � 10–5.

Applying Inquiry Skills11. Experimental Design

1. Calibrate the pH following the manufacturer’s instructions.2. Add 25 mL of acetic acid to a 100-mL beaker.3. Measure the pH of the solution using the pH meter.4. Rinse the pH meter as directed by your teacher.5. Discard the acetic acid solution as directed by your teacher.

PRACTICE

(Page 574)

Understanding Concepts12. HC4H7O2(aq) e H�

(aq) � C4H7O2–(aq)

Ka �

Ka � 1.8 � 10–4


�[HA

K

]

a

initial� � �

01.1.850

�

m1o0l–/4L

�

�[HA

K

]

a

initial� � 830

Since 830 > 100, we may assume that 0.150 – x � 0.150.The equilibrium expression becomes

�0.

x1

2

50� � 1.8 � 10–4

which yields

x2 � 2.7 � 10–5

x � 5.2 � 10–3


�[HA

x]initial� � 100% ≤ 5%


�[HA

x]initial� � 100% � �

5.20�

.1510–3� � 100%

[H�(aq)][C4H7O2

–(aq)]

��[HC4H7O2(aq)]

[H�(aq)][C6H7O6

–(aq)]

��[HC6H7O6(aq)]


�[HA

x]initial� � 3.5%

Since 3.5% < 5%, the assumption is valid.Therefore,

[H�(aq)] � 5.2 � 10–3 mol/L

pH � –log 5.2 � 10–3

pH � 2.28

13. C21H22N2O2(aq) � H2O(l) e OH–(aq) � HC21H22N2O2

�(aq)

Kb �

Kb � 1.0 � 10–6

Kb �

Kb � 1.0 � 10–6

�(0.00

x1

2

– x)� � 1.0 � 10–6

Predicting the validity of the assumption …

�1.0

0�

.00110–6� � 1000

Since 1000 > 100, we may assume that 0.001 – x � 0.001.The equilibrium expression becomes

�0.

x0

2

01� � 1.0 � 10–6

which yields

x2 � 1.0 � 10–9

x � 3.2 � 10–5

Justifying the simplification assumption …

�3.2

0�

.00110–5

� � 100% � 3.2%

Since 3.2% < 5%, the assumption is valid.

Therefore,

[OH–(aq)] � 3.2 � 10–5 mol/L

pOH � log 3.2 � 10–5

pOH � 4.5

pH � 14 – pOH

� 14.0 – 4.5

[HC21H22N2O2�(aq)][OH–

(aq)]��

[C21H22N2O2(aq)]

[HC21H22N2O2�(aq)][OH–

(aq)]��

[C21H22N2O2(aq)]


ICE Table for the Ionization of C21H22N2O2(aq)

C21H22N2O2(aq) + H2O(l) e OH –(aq) + HC21H22N2O2

+(aq)


Change in concentration (mol/L) x +x +x


pH � 9.5

The pH of a 0.001 mol/L solution of strychnine is 9.5.

PRACTICE

(Page 578)

Understanding Concepts14. (a) Since H2SO4(aq) is a strong acid, the hydrogen ion concentration from Ka1 is 1.00 mol/L.

H2SO4(aq) e H�(aq) � HSO4

–(aq) Ka1 very large

HSO4–(aq) e H�

(aq) � SO42–(aq) Ka2 � 1.0 �10–2

Ka ��[H�

(a

[q

S)]

O

[H

42

S–

(aq

O

)]4–(aq)]

�

�1.0

x0

2

– x� � 1.0 � 10–2

Predicting whether 1.00 – x � 1.00 …

�[HA

K

]

a

initial� � �

11..000�

m1o0l/–L2�

�[HA

K

]

a

initial� � 100

Since 100 � 100, we can assume that 1.00 – x � 1.00.

�(1.0

x0

2

– x)� � 1.0 � 10–2

x2 � 1.0 � 10–2

x � 1.0 � 10–1

x � 0.10

Validating the assumption …

�1.0

1�

.0010–2� � 100% � 1.0%

Since 1.0% < 5.0%, the assumption is valid.

[H�(aq)] � 0.10 � 1.00

[H�(aq)] � 1.1 mol/L

pH � –log[H�(aq)]

� –log[1.10]

pH � –0.0414

The pH of a 0.001 mol/L solution of sulfuric acid is –0.0414.


ICE Table for the Ionization of HSO4–(aq)

HSO4–(aq) e H+

(aq) + SO42–(aq)




Ka � �[H�

[(a

Hq)

2

]

S

[H

(aq

S

)

–(

]aq)]

�

�0.01

x0

2

– x� � 1.1 � 10–7


�[HA

K

]

a

initial� � �

01.0.110

�

m1o0l–/7L

�

�[HA

K

]

a

initial� � 9.0 � 105

Since 9.0 � 105 � 100, we can assume that 0.010 – x � 0.010.

�(0.

x0

2

10)� � 1.0 � 10–2

x2 � 1.0 � 10–9

x � 3.3 � 10–5


�3.3

0�

.01100–5

� � 100% � 0.33%


[H�(aq)] � 3.3 � 10–5 mol/L


� –log[3.3 � 10–5]

pH � 4.48

The pH of a 0.010 mol/L solution of hydrosulfuric acid is 4.48.

SECTION 8.2 QUESTIONS

(Page 579)

Understanding Concepts1. A dilute solution of a strong acid and a solution of a weak acid may have similar pH values but differ in the degree of

ionization that occurs. Strong acids are fully ionized while weak acids are only partially ionized.2. CN–

(aq) � H2O(l) e HCN(aq) � OH–(aq)

Kb � �[CN

[H

–(aq

C)]

N

[O

(a

H

q)]

–(aq)]

�


ICE Table for the Ionization of H2S(aq)

H2S(aq) e H+(aq) + HS–

(aq)




Kb � �1.0 �

Ka

10–14�

� �16..02

�

�

1100

–

–

1

1

4

0�

Kb � 1.6 � 10–5

�0.1

x8

2

– x� � 1.6 � 10–5


�1.6

0�

.1810–5� � 1.1 � 104

Since 1.1 � 104 � 100, we assume that 0.18 – x � 0.18.

�(0

x.1

2

8)� � 1.6 � 10–5

x2 � 2.9 � 10–6

x � 1.7 � 10–3


�1.7

0�

.1810–3� � 100% � 0.94%


[OH–(aq)] � 1.7 � 10–3 mol/L

pOH � –log[OH–(aq)]

� –log[1.7 � 10–3]

pOH � 2.77

pH � 14 – pOH

� 14 – 2.77

pH � 11.23

The pH of a 0.18 mol/L cyanide solution is 11.23.

4. 2.5%C6H5COOH(aq) e H�

(aq) � C6H5COO–(aq)

Ka �

x � 0.100 mol/L � 0.0025

x � 2.5 � 10–3

[H�(aq)][C6H5COO–

(aq)]��

[C6H5COOH(aq)]


ICE Table for the Ionization of CN–(aq)

CN–(aq) + H2O(l) e HCN(aq) + OH–

(aq)

Initial concentration (mol/L) 0.18 – 0 0

Change in concentration (mol/L) –x – +x +x

Equilibrium concentration (mol/L) 0.18 – x – x x

Ka �

� �(2.5

0�

.09180–3)2

�

Ka � 6.4 � 10–5

The Ka of benzoic acid is 6.4 � 10–5.

5. pH � 2.1

[H�(aq)] � 10–pH

� 10–2.1

[H�(aq)] � 8 � 10–3 mol/L

HNO2(aq) e H�(aq) � NO2

–(aq)

Ka � �[H

[

�(

Haq

N)][

O

N

2

O

(aq

2

)

–(

]aq)]

�

Ka � ��[H

[

�(

Haq

N)][

O

N

2

O

(aq

2

)

–(

]aq)]

�

� �(8 �

0.0190–3)2�

Ka � 7 � 10–4

The Ka for nitrous acid is 7 � 10–4.

6. HBrO(aq) e H�(aq) � BrO–

(aq)

Ka � �[H

[

�(

Haq

B)][

r

B

O

r

(

O

aq)

–(a

]q)]

�

Ka � 2.5 � 10–9

Since 2.5 � 10–9 � 100, we can assume that 0.200 – x � 0.200.

[H�(aq)][C6H5COO–

(aq)]��

[C6H5COOH(aq)]


ICE Table for the Ionization of C6H5COOH(aq)

C6H5COOH(aq) e H+(aq) + C6H5COO–

(aq)


Change in concentration (mol/L) –2.5 � 10–3 �2.5 � 10–3 �2.5 � 10–3

Equilibrium concentration (mol/L) 0.098 2.5 � 10–3 2.5 � 10–3

ICE Table for the Ionization of HNO2(aq)

HNO2(aq) e H+(aq) + NO2

–(aq)

Initial concentration (mol/L) 0.100 0 0

Change in concentration (mol/L) –8 � 10–3 +8 � 10–3 +8 � 10–3

Equilibrium concentration (mol/L) 0.09 8 � 10–3 8 � 10–3


�[HA

K

]

a

initial� � �

2.50�

.20100–9�

�[HA

K

]

a

initial� � 1.3 � 109


�0.

x2

2

00� � 2.5 � 10–9

which yields

x2 � 5.0 � 10–10

x � 2.2 � 10–5


�[HA

x]initial� � 100% 5%


�[HA

x]initial� � 100% � �

2.20�

.20100–5

� � 100%

�[HA

x]initial� � 100% � 0.011%


[H�(aq)] � 2.2 � 10–5 mol/L

pH � –log 2.2 � 10–5

� 2.28

pH � 4.65

The pH of the hypobromous acid solution is 4.65.

7. (a) HCN(aq), HCO2H(aq), HF(aq), HNO3(aq)

(b) Approximate pH of 1.0 mol/L solutions:

8. (a) (strongest acid) H3PO4(aq), HNO2(aq), HC2H3O2(aq), H2S(aq), NH4�(aq) (weakest acid)

(b) (highest pH) NH4�(aq), H2S(aq), HC2H3O2(aq), HNO2(aq), H3PO4(aq) (lowest pH)

(c) (strongest base) NH3(aq), HS–(aq), C2H3O2

–(aq), NO2

–(aq), H2PO4

–(aq) (weakest base)

9. (a) CN–(aq) � H2O(l) e OH–

(aq) � HCN(aq)

Kb ��[HCN

[C(a

Nq)]

–(

[

a

O

q)]

H–(aq)]

�


Acid Approximate pH

HCN(aq) 5

HCO2H(aq) 4

HF(aq) 2

HNO3(aq) 0

(b) SO42–(aq) � H2O(l) e OH–

(aq) � HSO4–(aq)

Kb ��[HSO

[4

S

–(

Oaq

4

)]2

[–

(aq

O

)]

H–(aq)]

�

(c) NO3–(aq) � H2O(l) e OH–

(aq) � HNO3(aq)

Kb ��[HNO

[N3(

Oaq

3

)]–(a

[

q

O

)]

H–(aq)]

�

(d) F–(aq) � H2O(l) e OH–

(aq) � HF(aq)

Kb � �[HF(a

[q

F)]–(a

[

q

O

)]

H–(aq)]

�

10. (a) (weakest base) morphine, erythromycin, atropine (strongest base)(b) pH of morphine …

B(aq) � H2O(l) e OH–(aq) � BH�

(aq)

Kb ��[BH�

(a

[q

B)]

(a

[

q

O

)]

H–(aq)]

�

Kb � 7.9 � 10–7

Since 0.3% < 5%, the assumption is valid.Therefore,[OH–

(aq) ] � 2.8 � 10–4 mol/L

pOH � –log 2.8 � 10–4

pOH � 3.6

pH � 14 – pOH

� 14.0 – 3.6

pH � 10.4

The pH of a 0.1 mol/L solution of morphine is 10.4.

pH of erythromycin …


(aq)

Kb ��[BH�

(a

[q

B)]

(a

[

q

O

)]

H–(aq)]

�

Kb � 6.3 � 10–6


ICE Table for the Ionization of the Base Atropine B(aq)

B(aq)+ H2O(l) e OH–(aq) + BH+

(aq)

Initial concentration (mol/L) 0.1 – 0 0



ICE Table for the Ionization of the Base Erythromycin B(aq)


(aq)

Initial concentration (mol/L) 0.1 – 0.0 0.0



Since 0.8% < 5%, the assumption is valid.Therefore,[OH–

(aq)] � 8.0 � 10–4 mol/L

pOH � –log 8.0 � 10–4

pOH � 3.1

pH � 14 – pOH

� 14.0 – 3.1

pH � 10.9

The pH of a 0.1 mol/L solution of erythromycin is 10.9.

pH of atropine …


(aq)

Kb ��[BH�

(a

[q

B)]

(a

[

q

O

)]

H–(aq)]

�

Kb � 3.2 � 10–5

Since 2% < 5%, the assumption is valid.Therefore,[OH–

(aq)] � 2 � 10–3 mol/L

pOH � –log 2 � 10–3

pOH � 2.7

pH � 14 – pOH

� 14.0 – 2.7

pH � 11.3

The pH of a 0.1 mol/L solution of atropine is 11.3.

11. C3H5O2–(aq) � H2O(l) e OH–

(aq) � HC3H5O2(aq)

Kb �

Kb �[HC3H5O2(aq)][OH–

(aq)]��

[C3H5O2–(aq)]

[HC3H5O2(aq)][OH–(aq)]

��[C3H5O2

–(aq)]


ICE Table for the Ionization of the Base Atropine B(aq)


(aq)




ICE Table for the Ionization of C3H5O2–(aq)

C3H5O2–(aq) + H2O(l) e OH–

(aq) + HC3H5O2(aq)


Change in concentration (mol/L) –1.1 � 10–5 – �1.1 � 10–5 �1.1 � 10–5

Equilibrium concentration (mol/L) 0.157 – 1.1 � 10–5 1.1 � 10–5

� �(1.1

0�

.15170–5)2

�

Kb � 7.7 � 10–10

The base ionization constant of the propanoate ion is 7.7 � 10–10.

12. Kw � 1.0 � 10–14

Ka � 7.2 � 10–4

KaKb � Kw

Kb � �K

Kw

a�

� �17..02

�

�

1100

–

–

1

4

4�

Kb � 1.4 � 10–11

The base dissociation constant for the nitrite ion is 1.4 � 10–11.13. pH of codeine …

Cod(aq) � H2O(l) e OH–(aq) � HCod�

(aq)

Kb �

Kb � 1.73 � 10–6


[OH–(aq) ] � 1.9 � 10–4 mol/L

pOH � –1.9 � 10–4

pOH � 3.73

pH � 14 – pOH

� 14.0 – 3.73

pH � 10.27

The pH of a 0.020 mol/L solution of codeine is 10.27.

14. (a) NH3(aq) � H2O(l) e OH–(aq) � NH4

�(aq)

Kb �

(b) NH4�(aq) e H�

(aq) � NH3(aq)

Ka �[H�

(aq)][NH3(aq)]��

[NH4�(aq)]

[NH4�(aq)][OH–

(aq)]��

[NH3(aq)]

[HCod�(aq)][OH–

(aq)]��

[Cod(aq)]


ICE Table for the Base Cod(aq)

Cod(aq) + H2O(l) e OH –(aq) + HCod+

(aq)




(c) �[H�

(

[a

Nq)

H

][N

4�(

H

aq

3

)](aq)]

� � � [H�(aq)][OH–

(aq)]� Kw

Numerically…

[5.80 � 10–10][1.72 � 10–5] � 1.00 � 10–14

15. (a) NH4�(aq), H2S(aq), SO4

2–(aq)

(b) NH3 : Kb � 1.7 � 10–5

HS–(aq) : Kb � 9.1 � 10–8

SO42–(aq) : Kb � 1.0 � 10–12

16. pH � 10.10

pOH � 14 – pH

� 14.0 – 10.10

pOH � 3.90

[OH–(aq)] � 10–pOH

� 10–3.90

[OH–(aq)] � 1.26 � 10–4 mol/L

Mor(aq) � H2O(l) e OH–(aq) � HMor�

(aq)

Kb ��[HMo

[M

r�(a

oq

r)]

(a

[

q

O

)]

H–(aq)]

�

Kb ��[HMo

[M

r�(a

oq

r)]

(a

[

q

O

)]

H–(aq)]

�

� �(1

9..2867

�

�

1100

–

–

4

3)2

�

Kb � 1.6 � 10–6

The Kb for morphine is 1.6 � 10–6.

17. NH3(aq) � H2O(l) e OH–(aq) � NH4

�(aq)

Kb ��[NH

[4

N

�(a

Hq)

3

][

(a

O

q)

H

]

–(aq)]

�

Kb � 1.77 � 10–5

[NH4�

(aq)][OH–(aq)]

��[NH3(aq)]


ICE Table for the Ionization of Morphine Mor(aq)

Mor(aq) +H2O(l) e OH –(aq) + HMor+

(aq)


Change in concentration (mol/L) –1.26 � 10–4 – �1.26 � 10–4 �1.26 � 10–4

Equilibrium concentration (mol/L) 9.87 � 10–3 – 1.26 � 10–4 1.26 � 10–4

ICE Table for the Ionization of Ammonia

NH3(aq) + H2O(l) e OH–(aq) + NH4

+(aq)





[OH–(aq)] � 1.33 � 10–3 mol/L

pOH � –log 1.33 � 10–3

pOH � 2.877

pH � 14 – pOH

� 14.0 – 2.877

pH � 11.123

The pH of a 0.100 mol/L solution of ammonia is 11.123.

18. C2O42–(aq) � H2O(l) e OH–

(aq) � HC2O4–(aq)

Kb �

Kb � 1.7 � 10–10

Since 5.83 � 10–3 % < 5%, the assumption is valid.Therefore,[OH–

(aq)] � 2.9 � 10–6 mol/L

pOH � – log 2.9 � 10–6

pOH � 5.54

pH � 14 – pOH

� 14.0 – 5.54

pH � 8.46

The pH of a 0.0500 mol/L sodium oxalate solution is 8.460.

Applying Inquiry Skills 19. Strong bases have a higher pH than weaker bases. Weak bases that are molecular produce solutions that do not conduct

electricity well.20. Analysis

Solution 1: HBr(aq)Solution 2: C12H22O11(aq)Solution 3: HC2H3O2(aq)Solution 4: NaCl(aq)

Making Connections21. (a) pH � 10.10

pOH � 14 pH

� 14.0 8.81

pOH � 5.19

[HC2O4–(aq)][OH–

(aq)]��

[C2O42–(aq)]


ICE Table for the Hydrolysis of Oxalate

C2O42–(aq) + H2O(l) e OH–

(aq) + HC2O4–(aq)




�(5.6 �

x1

2

0–3 – x)� � 7.94 � 10–5

Predict whether a simplifying assumption is justified …

�[HA

K

]

a

initial� � �

75..964

�

�

1100

–

–

3

5�

�[HA

K

]

a

initial� � 70

Since 70 � 100, we may not assume that 5.6 � 10–3– x � 5.6 � 10–3.

�(5.6 �

x

1

2

0–3 – x)� � 7.94 � 10–5

x2 � 7.94 � 10–5 (5.6 � 10–3 – x)

x2 � (7.94 � 10–5 x) (4.45 � 10–7) � 0

x �

x � 6.29 � 10–4 mol/L


� –log[6.29 � 10–4]

pH � 3.20

The pH of the lactic acid in the runner’s muscles is 3.20. (b) Lactic acid buildup in muscles causes fatigue, pain, and muscle stiffness.(c) Muscles can oxidize glucose aerobically or anaerobically to release energy. The energy released is stored

temporarily in the molecule ATP that can then be used by muscles to do mechanical work. Anaerobic oxidationof glucose, however, is not very efficient. In a sprint, for example, a great deal of energy is required in a shortperiod of time. Anaerobic oxidation supplies most of this energy but is very inefficient. Short-term energyreserves are depleted quickly and lactic acid accumulates.

Muscles used in long-distance running rely more on aerobic oxidation of glucose for their energy. Whenrunning at a comfortable pace, both systems of oxidation are used but the ratio of anaerobic: aerobic is lowenough to prevent lactic acid from accumulating. As the pace increases, the anaerobic: aerobic ratio increases tothe point where lactic acid begins to accumulate in the blood. This is known as the lactic acid threshold. In orderto improve performance, long-distance runners try to train at the speed at which the lactic acid threshold occurs.This serves to increase the threshold and overall performance.

8.3 ACID–BASE PROPERTIES OF SALT SOLUTIONS

PRACTICE

(Page 588)

Understanding Concepts1. (a) The ammonium ion is a weak acid with Ka� 5.8 � 10–10. The phosphate ion is a base with Kb � 2.4 � 10–2.

Since Kb is larger than Ka, an ammonium phosphate solution is basic.(b) The ammonium ion is a weak acid with Ka� 5.8 � 10–10. The sulfate ion is a base with Kb � 1.0 � 10–12. Since

Ka is larger than Kb, an ammonium sulfate solution is slightly acidic.(c) Magnesium oxide reacts with water to form magnesium hydroxide (a base).(d) MgO(s) � H2O(l) → Mg2�

(aq) � 2 OH–(aq). This makes a solution of magnesium oxide basic.

2. A solution of sodium sulfite will be basic.3. NH4NO3(aq) → NH4

�(aq) � NO3

–(aq)

–7.94 � 10–5 � �(7.94 �� 10–5)–�2 – 4(–�4.45 �� 10–7)��

2


Since NO3–(aq) is the conjugate base of a strong acid, it will not affect the pH of the solution. NH4

�(aq) is the conjugate

acid of a weak base NH3(aq), so it will hydrolyze according to the equation:

NH4�(aq) � H2O(l) e H3O�

(aq) � NH3(aq)

Ka ��[H3O

[

�(

Naq

H)]

4

[�(

N

aq

H

)]3(aq)]

�

Ka � 5.8 � 10–10

Ka ��[H3O

[

�(

Naq

H)]

4

[�(

N

aq

H

)]3(aq)]

�

Ka � 5.8 � 10–10

�0.3

x0

2

– x� � 5.8 � 10–10


�[HA

K

]

a

initial� � �

5.80�

.3100–10�

�[HA

K

]

a

initial� � 5.2 � 108

Since 5.2 � 108 � 100, we may assume that 0.30 – x � 0.30.

�0x.3

2

0� � 5.8 � 10–10

x2 � 1.7 � 10–10

x � 1.3 � 10–5

Since x � [H3O�(aq)]

[H3O�(aq)] � 1.3 � 10–5 mol/L

pH � –log 1.3 � 10–5

pH � 4.88

The pH of a 0.30 mol/L ammonium nitrate solution is 4.88.

4. NH4Br(aq) → NH4�(aq) � Br–

(aq)

Since Br–(aq) is the conjugate base of a strong acid, it will not affect the pH of the solution. NH4

�(aq) hydrolyzes

according to the equation:

NH4�(aq) � H2O(l) e H3O�

(aq) � NH3(aq)

Ka ��[H3O

[

�(

Naq

H)]

4

[�(

N

aq

H

)]3(aq)]

�

Ka � 5.8 � 10–10


ICE Table for the Hydrolysis of NH4+(aq)

NH4+(aq) + H2O(l) e H3O+

(aq) + NH3(aq)


Change in concentration (mol/L) –x – �x �x


Ka ��[H3O

[

�(

Naq

H)]

4

[�(

N

aq

H

)]3(aq)]

�

Ka � 5.8 � 10–10

�0.2

x

5

2

– x� � 5.8 � 10–10


�[HA

K

]

a

initial� � �

5.80�

.2150–10�

�[HA

K

]

a

initial� � 5.2 � 108

Since 5.2 � 108 � 100, we may assume that 0.25 – x � 0.25.

�0x.2

2

5� � 5.8 � 10–10

x2 � 1.5 � 10–10

x � 1.2 � 10–5

Since x � [H3O�(aq)]

[H3O�(aq)] � 1.2 � 10–5 mol/L

pH � –log 1.2 � 10–5

pH � 4.92

The pH of a 0.25 mol/L ammonium bromide solution is 4.92.

5. NH4C2H3O2(aq) → NH4�(aq) � C2H3O2–

(aq). The ammonium ion is a weak acid with Ka � 5.8 � 10–10. The acetate ionis a weak base with Kb � 5.6 � 10–10. Since Kb and Ka are similar, an ammonium acetate solution is essentially neutral.

Making Connections6. Fertilizers containing ammonium compounds, which hydrolyze to produce acidic solutions, are ideal for evergreens.

PRACTICE

(Page 589)

Understanding Concepts 7. HCO3

–(aq), HS–

(aq), HPO42–(aq), H2PO4

–(aq), HSO4

–(aq), HSO3

–(aq), HC2O4

–(aq)

8. (a) acidic(b) basic

PRACTICE

(Page 591)

Understanding Concepts9. (a) Na2O(s) � H2O(l) e 2 Na�

(aq) � 2 OH–(aq)



NH4+(aq) + H2O(l) e H3O+

(aq) + NH3(aq)




(b) SO3(g) � H2O(l) e H3O�(aq) � HSO3

–(aq)

Making Connections10. (a) Carbon dioxide exhaled by the visitors can increase the acidity of the moisture in the caves, according to the

chemical equation:CO2(g) � H2O(l) e H�

(aq) � HCO3–(aq)

The increased acidity can accelerate the dissolving of the cave structures:

H�(aq) � CaCO3(s) e Ca 2�


(b) Possible solutions include providing air circulation systems or large fans that would prevent carbon dioxide gasfrom accumulating.

11. Arguments supporting the use of fluoride:• Some studies of large populations suggest that water fluoridation does prevent tooth decay.

Arguments against the use of fluoride:• Some other studies suggest that the evidence to support that water fluoridation prevents tooth decay is inconclu-

sive.

• The risks associated with long-term exposure to trace amounts of fluoride have not been established.

• The effects of fluoride on aquatic life have not been thoroughly studied.

• Some studies suggest that prolonged exposure to fluoride is associated with a condition known as skeletal fluorosis– a gradual deterioration of bone.

• Fluoride can be administered in more direct ways that do not threaten the environment.

PRACTICE

(Page 594)

Understanding Concepts12. (a) Lewis acid: H�

(aq); Lewis base: OH–(aq)

(b) Lewis acid: H�(aq); Lewis base: NH3(aq)


(Page 594)

Understanding Concepts1. (a) Neutral. The sodium ion, being a member of Group 1, does not hydrolyze. The chloride ion is an extremely weak

base and therefore also does not hydrolyze.(b) Acidic. Chloride does not hydrolyze. Aluminum ions will hydrolyze according to the equation

Al(H2O)63�(aq) e Al(H2O)5(OH)2�

(aq) � H�(aq)

(c) Basic. The sodium ion does not hydrolyze. Carbonate will hydrolyze to release hydroxide ions.CO3

2–(aq) � H2O(l) e OH–


2. The ammonium ion is a weak acid with Ka � 5.8 � 10–10. The carbonate ion is a base with Kb � 2.1 � 10–4. SinceKb is larger than Ka, an ammonium carbonate solution is basic.

3. According to Table C7 in the Appendix, the strongest possible acid is perchloric acid and the strongest possible baseis the hydroxide ion, OH–

(aq) – the base released when the hydroxide ion dissociates. However, for all practicalpurposes, the concentration of oxide in a solution of the hydroxide ion is zero.

4. BeCl2(aq) will turn litmus red because the Be2�(aq) hydrolyzes to release H�

(aq).

5. (a) acidic(b) basic

Applying Inquiry Skills6. Analysis

Na2O(s) basicMgO(s) basic


Ka ��[H

[

�(

Haq

C)][

O

C

2

O

H2

(

H

aq)

–(a

]q)]

�

�0.2

x0

2

– x� � 1.8 � 10–4


�[HA

K

]

a

initial� � �

01..280�

m1o0l/–L4�

� 1100

Since 1100 � 100, we assume that 0.20 – x � 0.20.

�(0

x.2

2

0)� � 1.8 � 10–4

x2 � 3.6 � 10–5

x � 6.0 � 10–3


�6.0

0�

.2010–3� � 100% � 3.0%

Since 3.0% � 5.0%, the assumption is valid.

[H�(aq)] � 6.0 � 10–3 mol/L


� –log [6.0 � 10–3]

pH � 2.22

The pH of the 0.20 mol/L formic acid solution is 2.22.

(b) VHCO2H(aq)� 25.00 mL

CHCO2H(aq)� 0.20 mol/L

nHCO2H(aq)� VHCO2H(aq)

� CHCO2H(aq)

� 25.00 mL � 0.20 mol/L

nHCO2H(aq)� 5.0 mmol

VNaOH(aq)� 10.00 mL

CNaOH(aq)� 0.20 mol/L

nNaOH(aq)� VNaOH(aq)

� CNaOH(aq)

� 10.00 mL � 0.20 mol/L

nNaOH(aq)� 2.0 mmol

nHCO2H(aq)remaining …

nHCO2H(aq)� 5.0 mmol – 2.0 mmol

nHCO2H(aq)� 3.0 mmol

Total volume � 25.00 mL � 10.00 mL

� 35.00 mL

CHCO2H(aq)� �

335.0.0

m0

mm

oLl

�


CHCO2H(aq)� 0.08571 mol/L (extra digits carried)

CCO2H–(aq)

� �325.0.0

m0

mm

oLl

�

CCO2H–(aq)

� 0.05714 mol/L (extra digits carried)

Let x represent the changes in concentration that occur as the system reestablishes equilibrium.

�[H

[

�(

Haq

C)][

O

C

2

O

H2

(

H

aq)

–(a

]q)]

�� Ka

�x(

0

0

.

.

0

0

8

5

5

7

7

1

1

4

–

�

x

x)� � 1.8 � 10–4

Checking whether an approximation is warranted:

�[HCO2H

K(

a

aq)]initial��

1.08.0�

851701–4�

�[HCO2H

K(

a

aq)]initial�� 480

Since 480 > 100, we can assume that

0.08571 – x � 0.08571 and that

0.05714 � x � 0.05714

The equilibrium simplifies to

�x(

00..0085577114)

� � 1.8 � 10–4

x � 2.70 � 10–4

Verifying the assumption with the 5% rule:

�2.70

0�

.2010–4

� � 100% � 0.14%

Since 0.14% < 5%, the simplifying assumption is justified.

[H�(aq)] � 2.70 � 10–4 mol/L

pH � – log[2.70 � 10–4]

pH � 3.569

The pH after the addition of 10.00 mL of NaOH(aq) is 3.57.

(c) At the equivalence point, 25.00 mL of 0.20 mol/L NaOH(aq) were added.Entities remaining in solution: Na�

(aq), CO2H–(aq), H2O(l)

Since Na�(aq) does not hydrolyze, the pH of the solution is determined by the methanoate ion.


ICE Table for the Ionization of HCO2H(aq)

HCO2H(aq) e H+(aq) + CO2H–

(aq)


Change in concentration (mol/L) –x �x �x

Equilibrium concentration (mol/L) 0.08571 – x x 0.05714 �x

CO2H–(aq) � H2O(l) e OH–

(aq) � HCO2H(aq)

� Kb

Ka � 1.8 � 10–4

Kb � �K

Kw

a�

� �11..08

�

�

1100

–

–

1

4

4�

Kb � 5.6 � 10–11

Therefore,

� 5.6 � 10–11

At the equivalence point, the total volume � 50.00 mL.

Since 5.00 mmol of HCO2H(aq) was present initially,

5.00 mmol of CO2H–(aq) is present at the equivalence point.

[CO2H–(aq)] � �

55.00.000

mmm

Lol

�

[CO2H–(aq)] � 0.100 mol/L

� 5.6 � 10–11

�0.10

x0

2

– x� � 5.6 � 10–11


�[CO2H

K

–(aq)]initial� � �

5.60�

.1100–11�

�[CO2H

K

–(aq)]initial� � 1.8 � 109

Since 1.8 � 109 > 100, we can assume that

0.100 – x � 0.100

The equilibrium simplifies to �0.

x1

2

00� � 5.6 � 10–11

x � 2.37 � 10–6


�2.37

0.�

10010–6

� � 100% � 2.4 � 10–3%

[HCO2H(aq)][OH–(aq)]

��[CO2H–

(aq)]


��[CO2H–

(aq)]


��[CO2H–

(aq)]


ICE Table for the Hydrolysis of CO2H–(aq)

CO2H–(aq) + H2O(l) e HCO2H(aq) + OH–

(aq)




Since 2.4 � 10–3% < 5%, the simplifying assumption is justified.

[OH(aq)] � 2.37 � 10–6 mol/L

pOH � – log[2.37 � 10–6]

pOH � 5.625

pH � 14 – pOH

pH � 8.37

The pH at the equivalence point is 8.37.

5. VHOCN(aq)� 30.00 mL

CHOCN(aq)� 0.17 mol/L

nHOCN(aq)� VHOCN(aq)

� CHOCN(aq)

� 30.00 mL � 0.17 mol/L

nHOCN(aq)� 5.1 mmol




� CNaOH(aq)

� 10.00 mL � 0.250 mol/L


nHCON(aq)remaining …

nHOCN(aq)� 5.1 mmol – 2.50 mmol

nHOCN(aq)� 2.6 mmol


� 40.00 mL

CHOCN(aq)� �

420.6.0

m0

mm

oLl

�

CHOCN(aq)� 0.0625 mol/L

COCN–(aq)

� �420.5.0

m0

mm

oLl

�

COCN–(aq)

� 0.0625 mol/L



ICE Table for the Ionization of HOCN(aq)

HOCN(aq) e H +(aq) + OCN–

(aq)



Equilibrium concentration (mol/L) 0.0625 – x x 0.0625 � x

�[H

[

�(

Haq

O)][

C

O

N

C

(a

N

q)

–(a

]q)]

�� Ka

�x(

00..00662255

–�

xx)

� � 3.5 � 10–4


�[HOCN

K(

a

aq)]initial� � �

3.50.

�

062150–4�

��[HOCN

K(

a

aq)]initial� � 180

Since 180 > 100, we can assume that

0.0625 � x � 0.0625 and that 0.0625 – x � 0.0625


�x(

00..00662255)

� � 1.8 � 10–4

x � 1.8 � 10–4


�1.8

0.�

062150–4

� � 100% � 0.29%

Since 0.29% < 5%, the simplifying assumption is justified.

[H�(aq)] � 1.8 � 10–4 mol/L

pH � – log[1.8 � 10–4]

pH � 3.74

The OCN(aq) is such a weak base (Kb � 2.10 � 1012) that any hydrolysis to produce OH–(aq) is insignificant.

The pH of the resulting solution is 3.74.

PRACTICE

(Page 608)

Understanding Concepts6. (a) NH3(aq) � H2O(l) e NH4

�(aq) � OH–

(aq)

Kb ��[HH

[4

N

�(a

Hq)]

3

[

(a

O

q)

H

]

–(aq)]

�

Kb � 1.8 � 10–5


ICE Table for the Ionization of Ammonia

NH3(aq) + H2O(l) e NH4+(aq) + OH–

(aq)




Kb ��[NH4

[N

�(a

Hq)

3

][

(a

O

q)

H

]

–(aq)]

�

Kb � 1.8 � 10–5

�0.1

x5

2

– x� � 1.8 � 10–5

Predicting the validity of the assumption …

�1.8

0�

.1510–5� � 8300

Since 8300 > 100, we may assume that 0.15 – x � 0.15.


�0x.1

2

5� � 1.8 � 10–5

which yields

x2 � 2.7 � 10–6

x � 1.643 � 10–3 (extra digits carried)

Justifying the simplification assumption …

�1.643

0.�

1510–3

� � 100% � 1.1%

Since 1.1% < 5%, the assumption is justified.

Therefore,

[OH(aq)] � 1.643 � 10–3 mol/L

pOH � –log[1.643 � 10–3]


pH � 14 – pOH

� 14.0 – 2.7844

pH � 11.216

The pH before any HI(aq) is added is 11.22.

(b) At the equivalence point, the entities remaining in solution are NH4�(aq), I–

(aq), and H2O(l).

Since I–(aq) does not hydrolyze, the pH of the solution is determined by the ammonium ion.

CNH3(aq)� 0.1500 mol/L

nNH3(aq)� VNH3(aq)

� CNH3(aq)

� 20.0 mL � 0.1500 mol/L

nNH3(aq)� 3.00 mmoL (amount of NH4

�(aq) produced)

Final volume � 40.0 mL

[NH4�(aq)] � �

34.000.0

mmmLol

�

[NH4�(aq)] � 0.075 mol/L

Ka ��[H�

(a

[q

N)]

H

[N

3(

H

aq

4

)]

�(aq)]

�



�0.07

x5

2

– x� � 5.8 � 10–10

Predicting whether 0.075 – x � 0.075:

�[HA

K

]

a

initial� � �

05..087�

5 m10

o–l/1L0�

�[HA

K

]

a

initial� � 1.3 � 109

Since 1.3 � 109 > 100, we assume that 0.075 – x � 0.075.

�(0.

x0

2

75)� � 5.8 � 10–10


[H�(aq)] � 6.595 � 10–6 mol/L


� –log[6.595 � 10–6]

pH � 5.181


PRACTICE

(Page 611)

Understanding Concepts7. Since this is a titration of a weak base with a strong acid, the equivalence point will occur at pH less than 7. Alizarin

yellow’s pH range is 10.1–12.0, beyond the required equivalence point.8. Indicators are weak acids in equilibrium with their conjugate base. At the endpoint some titrant is required to

“neutralize” the indicator. A small volume of indicator solution is used to keep the titrant volume required to producethe colour change insignificant to the results of the titration.

9. Since this is a titration of a weak acid with a strong base, the equivalence point will occur at pH greater than 7.However, methyl red has a pH range of 4.8–6.0, well below the equivalence point. As a result, the endpoint of the titra-tion occurs before the equivalence point is reached.

PRACTICE

(Page 613)

Understanding Concepts10. (a) PO4

3–(aq) � H�

(aq) → HPO42–(aq)

HPO42–(aq) � H�

(aq) → H2PO4–(aq)

H2PO4–(aq) � H�

(aq) → H3PO4(aq)

(b) 25 mL and 50 mL(c) The hydrogen phosphate ion, HPO4

2–(aq) is an extremely weak acid (Ka � 4.2 � 10–13) and does not lose its proton

quantitatively. Only quantitative reactions produce detectable endpoints in an acid–base titration.


(Page 613)

Understanding Concepts1. (a) The endpoint is at pH 9 and the equivalence point is at 26 mL.

(b) Phenolphthalein(c) HC2H3O2(aq) � OH–

(aq) → C2H3O2–(aq) � H2O(l)

2. (a) 7(b) >7

(c) 7(d) <7

3. (a) <7(b) >7(c) �7

4. A pH curve is used to determine the pH of the solution at the equivalence point of a titration, so that a suitable indi-cator that will change colour at that endpoint can be chosen.

5. (a) pink(b) yellow(c) blue(d) red-orange

6. (a)


–(aq)

Ka �

�0.10

x0

2

– x� � 6.3 � 10–5


�[HA

K

]

a

initial� � �

06.1.300

�

m1o0l–/5L

�

�[HA

K

]

a

initial� � 1580

Since 1580 � 100, we assume that 0.100 – x � 0.100.

�(0.

x1

2

00)� � 6.3 � 10–5

x2 � 6.3 � 10–6


The 5% rule justifies the assumption.

[H�(aq)] � 2.51 � 10–3 mol/L


� –log [2.51 � 10–3]

pH � 2.600

The pH of the solution is 2.600.

(b) VHC7H5O2(aq)� 25.00 mL

CHC7H5O2(aq)� 0.100 mol/L

nHC7H5O2(aq)� VHC7H5O2(aq)

� CHC7H5O2(aq)

� 25.00 mL � 0.100 mol/L

nHC7H5O2(aq)� 2.50 mmol

[H�(aq)][C7H5O2

–(aq)]

��[HC7H5O2(aq)]




–(aq)







� CNaOH(aq)

� 10.00 mL � 0.100 mol/L


nHC7H5O2(aq)remaining …

nHC7H5O2(aq)� 2.50 mmol – 1.00 mmol



� 35.00 mL

CHC7H5O2(aq)� �

13.55.000

mmm

Lol

�

CHC7H5O2(aq)� 0.04286 mol/L (extra digits carried)

CC7H5O2–(aq)

� �13.50.000

mmm

Lol

�

CC7H5O2–(aq)

� 0.02857 mol/L (extra digits carried)


� Ka

�x(

0

0

.

.

0

0

4

2

2

8

8

5

6

7

–

�

x

x)� � 6.3 � 10–5


�[HC7H5O

K2

a

(aq)]initial��

6.30�

.10100–5�

� 1600

Since 1600 > 100, we can assume that 0.04286 x � 0.04286 and that 0.02857 � x � 0.02857.


�x(

00..0042288567)

� � 6.3 � 10–5

x � 9.45 � 10–5

The 5% rule validates the assumption:

[H�(aq)] � 9.45 � 10–5 mol/L

pH � – log[9.45 � 10–5]pH � 4.03

[H�(aq)][C7H5O2

–(aq)]

��[HC7H5O2(aq)]




–(aq)



Equilibrium concentration (mol/L) 0.04286 – x x 0.02857 � x

(c) At the equivalence point, 25.00 mL of 0.100 mol/L NaOH(aq) were added.Entities remaining in solution: Na�

(aq), C7H5O2–(aq), H2O(l)

Since Na�(aq) does not hydrolyze, the pH is determined by the benzoate ion.

C7H5O2–(aq) � H2O(l) e OH–


� Kb

Ka � 6.3 � 10–5

Kb � �K

Kw

a�

� �16..03

�

�

1100

–

–

1

5

4�

Kb � 1.587 � 10–10 (extra digits carried)

Therefore,

� 1.587 � 10–10

At the equivalence point, the total volume is 50 mL.Since 2.50 mmol of HC7H5O2(aq) was present initially, 2.50 mmol of C7H5O2

–(aq) is present at the equivalence

point.

[C7H5O2–(aq)] � �

25.05.000

mmm

Lol

�

[C7H5O2–(aq)] � 0.0500 mol/L

� 1.587 � 10–10

�0.05

x0

2

0 – x� � 1.587 � 10–10


�[C7H5O

K2–(aq)]initial��

1.5807.0

�

50100–10�

� 3.2 � 108

Since 3.2 � 108 > 100, we can assume that 0.0500 x � 0.0500.


�0.0

x5

2

00� � 1.587 � 10–10

x � 2.82 � 10–6

(The 5% rule verifies the assumption.)


��[C7H5O2

–(aq)]

[OH–(aq)][HC7H5O2(aq)]

��[C7H5O2

–(aq)]


��[C7H5O2

–(aq)]


ICE Table for the Hydrolysis of C7H5O2–(aq)


(aq) + HC7H5O2(aq)




[OH–(aq)] � 2.82 � 10–6 mol/L

pOH � – log[2.82 � 10–6]

pOH � 5.550

pH � 14 – pOH

pH � 8.450


7. An appropriate indicator for this titration is alizarin yellow.8. (i) HCl(aq) � NH3(aq) → H2O(l) � NH4Cl(aq)

At the equivalence point, nacid � nbase

� CacidVacid � CbaseVbase

VHCl(aq)required ��CNH

3(

Caq)

H

�

Cl(a

V

q)

NH3(aq)

�

�

VHCl(aq)required � 10.0 mL NaOH

Final solution volume � 30.0 mL

Entities in solution at the equivalence point: Cl–(aq), NH4�(aq), H2O(l)

Since Cl–(aq) does not hydrolyze, pH is determined by the ammonium ion.

NH4�(aq) e NH3(aq) � H�

(aq)

Since 2.00 mmol of NH3(aq) was present initially, 2.00 mmol of NH4�(aq) is present at the equivalence point.

[NH4�(aq)] � �

23.000.0

mmmLol

�

[NH4�(aq)] � 0.0667 mol/L

NH4�(aq) e H�

(aq) � NH3(aq)

�[H�

(

[a

Nq)]

H

[N

4�(

H

aq

3

)](aq)]

� � 5.8 � 10–10

�[H�

(

[a

Nq)]

H

[N

4�(

H

aq

3

)](aq)]

� � 5.8 � 10–10

�0.06

x6

2

7 – x� � 5.8 � 10–10


�[NH4

�(

Kaq

a

)]initial� � �

5.80.

�

061607

–10� > 100

0.100 mol/L � 20.0 mL��

0.200 mol/L



NH4+(aq) e H+

(aq) + NH3(aq)





�0.0

x6

2

67� � 5.8 � 10–10

x � 6.22 � 10–6

The 5% rule verifies the assumption.

[H�(aq)] � 6.22 � 10–6 mol/L

pH � – log[6.22 � 10–6]

pH � 5.206


(ii) HC2H3O2(aq) � NaOH(aq) → H2O(l)� NaC2H3O2(aq)

VNaOH required �

�

VNaOH required � 23.3 mL NaOH

Entities remaining in solution at the equivalence point: Na�(aq), C2H3O2

–(aq), H2O(l)

Since Na�(aq) does not hydrolyze, the pH of the solution is determined by the acetate ion.

C2H3O2–(aq) � H2O(l) e OH–


� Kb

Kb � 1.8 � 10–5

Kb � �K

Kw

a�

� �11..08

�

�

1100

–

–

1

5

4�

Kb � 5.555 � 10–10 (extra digits carried)

Therefore,

� 5.555 � 10–10

At the equivalence point, the total volume � 33.33 mL.

Since 3.50 mmol of HC2H3O2(aq) was present initially, 3.50 mmol of C2H3O2–(aq) is present at the equivalence

point.

[C2H3O2–(aq)] � �

33.35.033

mmm

Lol

�

[C2H3O2–(aq)] � 0.105 mol/L


��[C2H3O2

–(aq)]


��[C2H3O2

–(aq)]

0.350 mol/L � 10.0 mL��

0.150 mol/L

CHC2H3O2(aq)� VHC2H3O2(aq)

��CNaOH


� 5.556 � 10–10

�0.10

x5

2

– x� � 5.556 � 10–10


�[C2H3O

K2–(aq)]initial��

5.5560.1

�

0510–10� > 100


�0.

x1

2

05� � 5.556 � 10–10

x � 7.64 � 10–6


[OH–(aq)] � 7.64 � 10–6 mol/L

pOH � – log[7.64 � 10–6]

� 5.117

pH � 14 – pOH

� 14 – 5.117

pH � 8.883


(iii) HBr(aq) � N2H4(aq) → H2O(l) � N2H5Br(aq)

at the equivalence point, nacid � nbase

VHBr(aq) required �

�

VHBr(aq) required � 9.00 mL HBr(aq)

Final solution volume � 24.0 mL

Entities remaining in solution at the equivalence point: Br–(aq), N2H5

�(aq), H2O(l)

Since Br–(aq) does not hydrolyze, the pH of the solution is determined by N2H5

�(aq).

N2H5�(aq) e H�

(aq) � N2H4(aq)

Since 2.25 mmol of N2H4(aq) was present initially, 2.25 mmol of N2H5�(aq) is present at the equivalence point.

0.150 mol/L � 15.0 mL��

0.250 mol/L

CN2H4(aq)� VN2H4(aq)

��CHBr(aq)


��[C2H3O2

–(aq)]


ICE Table for the Hydrolysis of C2H3O2–(aq)


(aq) + HC2H3O2(aq)




[N2H5�(aq)] � �

22.245.0

mmmLol

�

[N2H5�(aq)] � 0.0938 mol/L

N2H5�(aq) e H�

(aq) � N2H4(aq)

�[H�

(

[a

Nq)]

2

[

H

N

5

2�(a

H

q)

4

](aq)]

�� K

Kw

b�

� �19..06

�

�

1100

–

–

1

7

4�

� 1.04 � 10–8 (extra digits carried)

N2H5�(aq) e N2H4(aq) � H�

(aq)

�[H�

(

[a

Nq)]

2

[

H

N

5

2�(a

H

q)

4

](aq)]

�� 1.04 � 10–8

�0.09

x3

2

8 – x� � 1.04 � 10–8


�[N2H5

K

�(a

a

q)]initial� � �

1.004.0�

93180–8� > 100


�0.0

x9

2

38� � 1.04 � 10–8

x � 3.12 � 10–5


[H�(aq)] � 3.12 � 10–5 mol/L

pH � – log[3.12 � 10–5]

pH � 4.505


(b) Appropriate indicators: methyl orange, bromophenol blue.

9. VHC2H3O2(aq)� 45.00 mL

CHC2H3O2(aq)� 0.10 mol/L


ICE Table for the Hydrolysis of N2H5+(aq)

N2H5+(aq) + H2O(l) e OH–

(aq) + N2H4(aq)




nHC2H3O2(aq)� VHC2H3O2(aq)

� CHC2H3O2(aq)

� 45.00 mL � 0.10 mol/L





� CNaOH(aq)

� 25.00 mL � 0.23 mol/L

nNaOH(aq)� 5.75 mmol (extra digit carried)

nNaOH(aq)remaining …

nNaOH(aq)� 5.75 mmol – 4.5 mmol



� 70.00 mL

CNaOH(aq)� �

17.02.500

mmm

Lol

�

CNaOH(aq)� 0.0179 mol/L (extra digits carried)

[OH–(aq)] � 0.0179 mol/L

pOH � –log (0.0179)

pOH � 1.748

pH � 12.25

The pH of the resulting solution is 12.25.10. (a) The pH is between 5.6 and 6.0.

(b) The pH is approximately 6, which corresponds to a [H�(aq)] of 1.0 � 10–6 mol/L.

Applying Inquiry Skills 11.

12. A: between 4.4 and 4.8B: between 6.0 and 6.6C: between 2.8 and 3.2


Volume of NaOH(aq)

pH

Oxalic Acid Reacted with Sodium Hydroxide

�pH � 3.74 – 3.67

�pH � 0.07

The change in pH is 0.07.


(Page 620)

Understanding Concepts1. A buffer is a mixture of a conjugate acid–base pair that maintains a nearly constant pH when diluted or when a strong

acid or base is added.2. Phosphate and carbonate buffers help maintain a normal pH level in the human body.3. (a) H2CO3(aq) → H�


(b) H�(aq) � HCO3

–(aq) → H2CO3(aq)

(c) H2CO3(aq) � OH–(aq) → H2O(l) � HCO3

–(aq)

4. A large excess of strong acid or base can drive the buffer equilibrium to completion. For example, the addition ofHCl(aq) to the carbonate buffer shifts the equilibrium in question 3(a) completely to the left.

5. HC2H3O2(aq) e H�(aq) � C2H3O2

–(aq)

(a) The addition of a small amount of HCl(aq) shifts the equilibrium to the left, consuming some acetate.

(b) NaOH(aq) consumes H�(aq), which causes the equilibrium to shift to the right.

6. (a) base(b) acid(c) neutral(d) base Lowest pH: (b), (c), (d), (a)

7. (a) and (c) will not form effective buffers because they consist of strong acids and bases. Conversely, (b) and (d) willform buffer systems because these mixtures consist of weak acids and their conjugate bases.

8. (a) The pH of nitric acid is much lower than that of nitrous acid. (b) Nitric acid is such a strong acid that it completely donates its hydrogen to water:

100%HNO3(aq) � H2O(l) → H3O�

(aq) � NO3–(aq)

Nitrous acid only partially donates its hydrogen to water. Some molecular nitrous acid remains.

HNO2(aq) � H2O(l) e H3O�(aq) � NO2

–(aq)

9. [C2H3O2–(aq)] � 0.25 mol/L

[HC2H3O2(aq)] � 0.25 mol/L


–(aq)

� 1.8 � 10–5

[H�(aq)] � Ka �

[H

[C

C

2

2

H

H

3

3

O

O

2

2–(a

(

q

aq

)])]

�

� 1.8 � 10–5 � �00..2255

mm

ooll//LL

�

� 1.8 � 10–5

pH � –log[1.8 � 10–5]

pH � 4.74

The addition of hydroxide …

[H�(aq)][C2H3O2

–(aq)]

��


[OH–(aq)]added � 0.15 mol/L

[HC2H3O2(aq)]final � (0.25 0.15) mol/L

[HC2H3O2(aq)]final � 0.10 mol/L

[C2H3O2–(aq)]final � (0.25 � 0.15) mol/L

[C2H3O2–(aq)]final � 0.40 mol/L

[H�(aq)] � Ka �

[H

[C

C

2

2

H

H

3

3

O

O

2

2–(a

(

q

aq

)])]

�

� 1.8 � 10–5 � �00..1400

�

[H�(aq)] � 4.5 � 10–6

pH � 5.35

�pH � 5.35 – 4.74

�pH � 0.61

The change in pH is 0.61.

8.6 CASE STUDY: THE SCIENCE OF ACID DEPOSITION

PRACTICE

(Page 624)

Understanding Concepts1. Acid deposition is suspected as one of the causes of forest decline, particularly in forests at high altitudes and colder

latitudes. (Sample table)

2. (a) Scientific research on catalysis assisted the development of catalytic converter technology.(b) Sensitive detection devices have helped scientific research in the reduction of sulfur oxide emissions.(c) The technology that produces sulfur oxides (smelting and power generating) has a harmful impact on human

health.(d) Society provides the resources, through government funding, that enable scientific research on the causes and

effects of acid deposition.(e) Society affects technology by purchasing its products, such as cars with catalytic converters.


Evidence of acid rain damage Alternative interpretations

The evidence clearly demonstrates that the BlackForest, for example, receives as much as 30 timesmore acid than if the rain fell through clean air.Damage to the trees includes yellowing, prematureneedle loss, and eventual death. Study of the treerings shows that trees grow more slowly in areassubject to acid deposition.

Other scientists question the link between treegrowth and acid deposition. They propose conflictingevidence that seedlings actually grow better in anacidic environment. They counter that the reductionin tree growth rate is more directly due to the reduc-tion in mean annual temperature in the regions inquestion. This hypothesis is supported by empiricaldata. Other researchers indicate that ground-levelozone, rather than acid deposition, is implicated inthe damage to the forests.

10. False: The spontaneity of a reaction depends on enthalpy changes as well as entropy changes. 11. False: The pH of acetic acid is greater than 1.12. False: The hypochlorite ion is a weaker base than ammonia. 13. False: Metal oxides form basic solutions while nonmetal oxides form acidic solutions. 14. False: Potassium sulfate forms a neutral solution. 15. True16. True17. False: The pH at the equivalence point depends on the type of acid and base involved. 18. True19. False: Buffering action occurs during the flat portions of the graph. 20. True21. False: An effective acid–base buffer contains approximately equal amounts of a weak acid and its conjugate base.22. True23. (b)24. (b)25. (e)26. (b)27. (b)28. (c)29. (b)30. (e)31. (c)32. (c)33. (d)34. (c)35. (e)36. (c)37. (b)38. (a)39. (d)40. (d)41. (e)

UNIT 4 REVIEW

(Page 639)

Understanding Concepts

1. �[SO

[2

S(

Og)]

3

2

(g

[O

)]22(g)]

� � K

� �2179�

K � 3.58 � 10–3

The equilibrium constant for the given reaction has a value of 3.58 � 10–3.

2. K � �[N

[

2

N

(g

O

)][(

Og)]

2

2

(g)]�

� �[0.

[603.]1[50].

2

21]�

K � 1.7 � 10–3

The equilibrium constant is 1.7 � 10–3.3. (a) [CO] decreases

(b) [CO] decreases(c) [CO] increases

356 Unit 4 Copyright © 2003 Nelson

(d) [CO] no effect(e) [CO] no effect(f) [CO] increases(g) [CO] decreases

4. (Sample answers) The equilibrium methanol concentration can be increased by:• removing methanol as it is produced,

• increasing the concentration of the reactants,

• increasing the pressure on the system,

• decreasing the volume of the reaction chamber,

• cooling the reaction chamber.

5. (a) An increase in pressure shifts the equilibrium to the right, favouring the production of ammonium carbamate. (b) An increase in temperature and a reduction of pressure favour the production of urea.

6. (a) AgCl(s) e Ag�(aq) � Cl–(aq)

Ksp � [Ag�(aq)][Cl–(aq)]

Ksp � 1.8 � 10–10

[Ag�(aq)] � [Cl–(aq)]

1.8 � 10–10 � [Ag�(aq)]

2

[Ag�(aq)] � 1.8 � 10–5 mol/L

[AgCl(aq)] � [Ag�(aq)]

[AgCl(aq)] � 1.3 � 10–5 mol/L

The solubility of silver chloride is 1.3 � 10–5.(b) AgCl(s) e Ag�

(aq) � Cl–(aq)

Ksp � [Ag�(aq)][Cl–(aq)]

Ksp � 1.8 � 10–10

If [Cl–(aq)] � 0.015 mol/L

1.8 � 10–10 � [Ag�(aq)][0.015]

[Ag�(aq)] � 1.2 � 10–8 mol/L

[AgCl(aq)] � [Ag�(aq)]

The solubility of silver chloride in 0.015 mol/L NaCl(aq) is 1.2 � 10–8.7. The synthesis of ammonia is exothermic. An increase in temperature results in a decrease in the value of the equilib-

rium constant. This observation implies that the concentration of ammonia decreases and the concentrations ofnitrogen and hydrogen increase with a temperature increase. This result can only occur if the energy term is on theright side of the chemical equation.

8. (a) reactants(b) 2 NOCl(g) e 2 NO(g) � Cl2(g)

� 1.60 � 10–5

�[0[N.1

O0]

C

2[l0

(g

.

)

1]02]

� � 1.60 � 10–5

[NOCl(g)] � 7.91 mol/L

The equilibrium concentration of NOCl(g) is 7.91 mol/L.

[NO(g)]2[Cl2(g)]

��[NOCl(g)]

2

Copyright © 2003 Nelson Chemical Systems and Equilibrium 357

9. N2O4(g) e 2 NO2(g)

�[

[

N

N

O

2O2(

4

g

(

)

g

]

)

2

]� � 6.13 � 10–3

Q � �[

[

N

N

O

2O2(

4

g

(

)

g

]

)

2

]�

� �[[48..0000

�

�

1100

–

–

3

4]]

2�

Q � 0.0200

Since Q 6.13 � 10–3, the system is not at equilibrium. The system will become less reddish-brown as it shifts toapproach equilibrium.

13. 2 NH3(g) e N2(g) � 3 H2(g)

�[N

[2

N(g

H)][

3

H

(g

2

)](g2)]

3

� � 1.60 � 10–3

At equilibrium,2 NH3(g) e N2(g) � 3 H2(g)

� 1.60 � 10–3

�[0.

[2x0][

–3x

2]3

x]2� � 1.60 � 10–3

�(0.20

x–

4

2 x)2� � 5.926 � 10–3 (extra digits carried)

��(0.20

x–�4

2 x)2�� 5.926�� 10–3�

�0.20

x–

2

2 x� � 0.770

x2 � 0.154x – 0.00154 � 0

x �

x � 0.3229

[NH3(g)] � 0.20 – 2x

� 0.20 – 2(0.3229)

[NH3(g)] � 0.14 mol/L

[H2(g)] � 3x

� 3(0.3229)

[H2(g)] � 0.097 mol/L

–0.154 � �(0.154�)2 – 4(�1)(–0.0�0154)��

2(l)

[N2(g)][H2(g)]3

��[NH3(g)]

2


ICE Table for the Decomposition of Ammonia

2 NH3(g) e N2(g) + 3 H2(g)


Change in concentration (mol/L) –2x +x �3x

Equilibrium concentration (mol/L) 0.20 – 2x x 3x

[N2(g)] � 2x

[N2(g)] � 0.32 mol/L

The equilibrium concentrations of nitrogen, hydrogen, and ammonia are 0.32 mol/L, 0.097 mol/L, and 0.14 mol/L.

14. H2(g) � I2(g) e 2 HI(g)

K � �[H

[

2

H

(g

I

)](g

[)

I

]

2

2

(g)]�

Initial concentrations are

[H2(g)]initial � �25.0.000mLol

�

[H2(g)]initial � 0.400 mol/L

[I2(g)]initial � �15.0.000mLol

�

[I2(g)]initial � 0.200 mol/L

[HI(g)]initial � 0.0 mol/L

At equilibrium,

K � �[H

[

2

H

(g

I

)](g

[)

I

]

2

2

(g)]�

� 49.7

4x2 � (49.7)(0.400 – x)(0.200 – x)

x2 � (12.425)(0.0800 – 0.6x � x2) (extra digits carried)

x2 – 0.6525x � 0.0870 � 0

x �

x � 0.46565 or 0.18725 (extra digits carried)

The root 0.46565 is rejected as it exceeds the initial amount of hydrogen.

K � �[H

[

2

H

(g

I

)](g

[)

I

]

2

2

(g)]�

[HI(g)] � 2x

� 2 � 0.18725

[HI(g)] � 0.375 mol/L

[H2(g)] � 0.400 – x

� 0.400 – 0.18725

0.6525 � �0.6525�2 – 4(0�.0870)�(1)��

2(l)

(2x)2��(0.400 – x)(0.200 – x)


ICE Table for the Formation of HI(g)

H2(g) + I2(g) e 2 HI(g)


Change in concentration (mol/L) –x –x +2x

Equilibrium concentration (mol/L) 0.400 – x 0.200 – x 2x

[H2(g)] � 0.213 mol/L

[I2(g)] � 0.200 – x

� 0.200 – 0.18725

[I2(g)] � 0.013 mol/L

The equilibrium concentrations of hydrogen, iodine, and hydrogen iodide are 0.213 mol/L, 0.013 mol/L, and 0.375 mol/L, respectively.

15. At equilibrium,

2 SO3(g) e 2 SO2(g) � O2(g)

�[SO

[2

S(

Og)]

3

2

(g

[O

)]22(g)]

� � 6.9 � 10–7

�[0.2

[20x0]2

–[x

2]x]2� � 6.9 � 10–7

�(0.20

40x–

3

2x)2� � 6.9 � 10–7

Assuming 0.200 – 2x � 0.200 …

�(0.

42x0

3

0)2� � 6.9 � 10–7

�04.0x4

3

0� � 6.9 � 10–7

100x3 � 6.9 � 10–7

x3 � 6.9 � 10–5


[SO3(g)] � 0.200 – 2x

� 0.200 – 2(1.904 � 10–3)

[SO3(g)] � 0.20 mol/L

[O2(g)] � 1.9 � 10–3 mol/L

[SO2(g)] � 2(1.904 � 10–3)

[SO2(g)] � 3.8 � 10–3 mol/L

The equilibrium concentrations of sulfur trioxide, oxygen, and sulfur dioxide are 0.20 mol/L, 1.9 � 10–3 mol/L, and 3.8 � 10–3 mol/L, respectively.

16. Ksp � [Ca2�(aq)][SO4

2–(aq)]

Ksp � 7.1 � 10–5

x2 � 7.1 � 10–5

x � 8.4 � 10–3

The molar solubility of calcium sulfate is 8.4 � 10–3 mol/L.


ICE Table for the Dissolving of Calcium Sulfate

CaSO4(s) e Ca2+(aq) + SO4

2–(aq)

Initial concentration (mol/L)

Change in concentration (mol/L) +x +x

Equilibrium concentration (mol/L)

x � 8.4 � 10–3 mol/L

The molar solubility of calcium sulfate is 8.4 � 10–3 mol/L.

17. Ksp � [Pb2�(aq)][Cl–(aq)]

2

Ksp � 1.2 � 10–5

(x)(2x)2 � 1.2 � 10–5

4x3 � 1.2 � 10–5

x � 0.01442 (extra digits carried)

[Cl–(aq)] � 2 � 0.01442

[Cl–(aq)] � 0.029 mol/L

The molar concentration of chloride ions is 0.029 mol/L.

18. Ag�(aq) � Cl–(aq) → AgCl(s)

Before mixing:AgNO3(aq) → Ag�

(aq) � NO3–(aq)

[AgNO3(aq) ] � [Ag�(aq)]

[AgNO3(aq) ] � 0.010 mol/L

[Cl–(aq) ] � 2.2 �10–4 mol/L

After mixing:250.0 mL � 250.0 mL � 500.0 mLConcentrations after mixing:

[Ag+(aq)] � 0.010 mol/L � �

255000..00

mm

LL

�

[Ag+(aq)] � 0.0050 mol/L

[Cl–(aq)] � 2.2 � 10–4 mol/L � �255000..00

mm

LL

�

[Cl–(aq)] � 1.1 � 10–4 mol/L

AgCl(s) e Ag+(aq) � Cl–(aq)

Q � [Ag+(aq)][Cl–(aq)]

� (0.0050)(1.1 � 10–4)

Q � 5.5 � 10–7

Ksp � 1.8 � 10–10

Q is greater than Ksp. Therefore, a precipitate will form.

19. Mg(NO3)2(aq) � 2 KF(aq) → MgF2(s) � 2 KNO3(aq)Before mixing:Mg(NO3)2(aq) → Mg2�

(aq) � 2 NO3–(aq)

[Mg(NO3)2(aq)] � [Mg2�(aq)]

[Mg(NO3)2(aq)] � 0.015 mol/L

[KF(aq)] � [F–(aq)]

[KF(aq)] � 0.10 mol/L

After mixing:300 mL � 100 mL � 400 mL


Concentrations after mixing:

[Mg2�(aq)] � 0.015 mol/L � �

140000

mm

LL

�

[Mg2�(aq)] � 3.75 � 10–3 mol/L (extra digits carried)

[F–(aq)] � 0.10 mol/L � �

340000

mm

LL

�

[F–(aq)] � 0.075 mol/L

MgF2(s) e Mg2�(aq) � 2 F–

(aq)

Q � [Mg2�(aq)][F

–(aq)]

2

� (3.75 � 10–3)(0.075)2

Q � 2.1 � 10–5

Ksp � 6.4 � 10–9

Q is larger than Ksp. Therefore, a precipitate does form.

20. Solubility refers to the amount of a compound that dissolves in a given volume of solution. The solubility product,however, is the product of the concentration of the ions released when a compound dissolves.

21. Compounds that release calcium or sulfate ions will decrease the solubility of calcium sulfate. Some examples arecalcium chloride, calcium nitrate, sodium sulfate, and potassium sulfate.

22. (a) Entropy is positive. The liquid state is more random than the solid state. (b) Entropy is negative. The products are less random due to the formation of the precipitate. (c) Entropy is negative. The formation of two molecules of the same substance HOCl is more ordered than the left

side. (d) Entropy is positive. The gases on the right side of the equation are more randomly arranged than solid ammonium

chloride. 23. 2 NaCl(s) → 2 Na(g) � Cl2(g)

The decomposition of sodium chloride is endothermic (�H > 0). The entropy change is positive, �S > 0, since theproducts are more random than the reactant.

�G � �H – T�S

�G � � – (�)(�)

Since the reaction does not occur spontaneously at room temperature, �G must be positive. Consequently, the temper-ature must be small enough so that the �H term in the equation is greater than T�S.

24. �H° � [�H°f(NH4Cl(aq))] – [�H°f(NH4Cl(s))

]

� [1 mol (– 299.7 kJ/mol)] – [1 mol (–314.4 kJ/mol)]

� [– 299.7 kJ] – [–314.4 kJ]

�H° � �14.7 kJ

�S° � [S°(NH4Cl(aq))] – [S°(NH4Cl(s))

]

� [1 mol (169.9 J/mol•K)] – [1 mol (94.6 J/mol•K)]

� [169.9 J/K] – [94.6 J/K]

�S° � 75.3 J/K

�S° � 75.3 J/K � �1100

k0JJ

�

�S° � 0.0753 kJ/K

�G° � �H° – T�S°

� 14.7 kJ – (298 K)(0.0753 kJ/K)

�G° � – 7.7 kJ


The standard Gibbs free energy change associated with this reaction is –7.7 kJ. Since �G o is negative, the reaction isspontaneous under standard conditions and will proceed to the right as written. A total of –7.7 kJ of free energy ismade available to do useful work for each mole of solid ammonium chloride that reacts.

25. �H° � [�H°f(CO2(g))� 2 �H°f(H2O(g))

] – [�H°f(CH4(g))� 2 �H°f(O2(g))

]

� [1 mol � (– 393.5 kJ/mol) � 2 mol � (–241.8 kJ/mol)] – [(– 74.4 kJ/mol) � 2 mol � (0)]

� [– 877.1 kJ/mol] – [–74.4 kJ]

�H° � 802.7 kJ

�S° � [S°(CO2(g))� 2 S°(H2O(g))

] – [S°(CH4(g))� 2 S°(O2(g))

]

� [1 mol � (213.78 J/mol•K) � 2 mol � (188.84 J/mol•K)] – [(186.3 J/mol•K) � 2 mol � (205.14 J/mol•K)]

� [591.46 J/K] – [596.58 J/K]

�S° � –5.12 J/K

�S° � –5.12 J/K � �1100

k0JJ

�

�S° � –0.00512 kJ/K

�G° � �H° – T�S°

� –802.7 kJ – (298K)(–0.00512 kJ/K)

�G° � – 801.2 kJ

The standard Gibbs free energy change associated with this reaction is –801.2 kJ. Since �G o is negative, the reactionis spontaneous under standard conditions and will proceed to the right as written. A total of –801.2 kJ of free energyis made available to do useful work for each mole of methane that burns.

26. C2H5OH(l) → C2H5OH(g)

T � ��

�

HS°

°�

� [�H°f(C2H5OH(g))– �H°f(C2H5OH(l))

]

� (–235.2 kJ/mol) – (–277.7 kJ/mol)

T � 42.5 kJ

�S° � [S°(C2H5OH(g))– S°(C2H5OH(l))

]

� (�282.70 J/mol•K) – (�160.7 J/mol•K)

�S° � 122 J/mol•K

�S° � 122 J/K � �1100

k0JJ

�

�S° � 0.122 kJ/K

T � ��

�

HS°

°�

� �0.

4122.25

kkJJ/K

�

T � 348 K

t � (T – 273°C)

� (348 K – 273°C)

t � 75°C

The normal condensation point of ethanol is 75°C.

27. (a) HNO2(aq) � SO42–(aq) e HSO4

–(aq) � NO2

–(aq)

(b) HCO3–(aq) � NH4

�(aq) e NH3(g) � H2CO3(aq)

(c) NH3(g) � HS–(aq) e NH4

�(aq) � S2–

(aq)


28. (a) endothermic(b) The increase in Kw with temperature implies that the hydrogen ion concentration also increases. Consequently,

the pH decreases.

29. [H�(aq)]rainwater � 10–5.6

[H�(aq)]rainwater � 2.51 � 10–6 mol/L (extra digits carried)

[H�(aq)]acidic rainwater � 10–4.0

[H�(aq)]acidic rainwater � 1 � 10–4 mol/L

�2.

151

�

�

101

–

0

4

–6m

mol

o/Ll/L

�� 40

The acidic rainwater is 40 times more acidic than normal rainwater.

30. [H�(aq)]cola � 10–2.7

[H�(aq)]cola � 2 � 10–3 mol/L

pOH � 14 – 2.7

pOH � 11.3

[OH–(aq)]cola � 10–11.3

[OH–(aq)]cola � 5 � 10–12 mol/L

The hydrogen and hydroxide ion concentrations in cola are 2 � 10–3 mol/L and 5 � 10–12 mol/L, respectively.

31. 5.9%HOCN(aq) e H�

(aq) � OCN–(aq)

Ka ��[H

[

�(

Haq

O)][

C

O

N

C

(a

N

q)

–(

]aq)]

�

[H�(aq)] � (5.9 � 10–3)(0.100 mol/L)

[H�(aq)] � 5.9 � 10–4

Ka ��[H

[

�(

Haq

O)][

C

O

N

C

(a

N

q)

–(

]aq)]

�

�

Ka � 3.5 � 10–6

The Ka for cyanic acid is 3.5 � 10–6.32.

(5.9 � 10–4)2��0.100 – 5.9 � 10–4


ICE Table for the Ionization of HOCN(aq)

HOCN(aq) e H+(aq) + OCN –

(aq)


Change in concentration (mol/L) –5.9 � 10–4 5.9 � 10–4 5.9 � 10–4

Equilibrium concentration (mol/L) 0.100 – 5.9 � 10–4 5.9 � 10–4 5.9 � 10–4

ICE Table for the Ionization of HC7H4NO3S(aq)

HC7H4NO3S(aq) e H+(aq) + C7H4NO3S–

(aq)


Change in concentration (mol/L) – x +x +x


HC7H4NO3S(aq) eH+(aq) � C7H4NO3S–

(aq)

Ka �

�0.50

x0

2

–x� � 2.1 � 10–12


�[HA

K

]

a

initial� � �

02..110�

0 m10

o–l/1L2�

>100Therefore, we assume that 0.500 – x � 0.500.

�(0.

x5

2

00)� � 2.1 � 10–12

x2 � 1.05 � 10–12 (extra digits carried)

x � 1.1 � 10–6


[H�(aq)] � 1.1 � 10–6 mol/L


� –log[1.1 � 10–6]

pH � 5.99

The pH of the saccharin solution is 5.99.

33. (a) CO2H–(aq) � H2O(l) e OH–

(aq) � HCO2H(aq)

(b) HCO2H(aq) e H�(aq) � CO2H–

(aq)

(c) � � [H�(aq)][OH–

(aq)] � Kw

34.

HOCl(aq) e H�(aq) � OCl–(aq)

Ka �

�0.10

x0

2

– x� � 2.9 � 10–8

[H+(aq)][OCl–(aq)]

��[HOCl(aq)]


��[CO2H–

(aq)]

[H�(aq)][CO2H–

(aq)]��

[HCO2H(aq)]

[H�(aq)][CO2H–

(aq)]��

[HCO2H(aq)]


��[CO2H–

(aq)]

[H+(aq)][C7H4NO3S–

(aq)]��

[HC7H4NO3S(aq)]


ICE Table for the Ionization of HOCl(aq)

HOCl(aq) e H+(aq) + OCl–(aq)





�[HA

K

]

a

initial� � �

02.1.900

�

m1o0l–/8L

�

>100Therefore, we assume that 0.100 – x � 0.100.

�(0.

x1

2

00)� � 2.9 � 10–8

x2 � 2.9 � 10–8



[H�(aq)] � 5.4 � 10–5 mol/L


� –log[5.39 � 10–5]

pH � 4.27

The pH of the hypochlorous acid solution is 4.27.

35. Hydrogen sulfite as an acid: HSO3–(aq) � HS–

(aq) e SO32–(aq) � S2–

(aq)

Hydrogen sulfite as a base: HSO4–(aq) � HSO3

–(aq) e SO4

2–(aq) � H2SO3(aq)

36. Despite having identical molar concentrations, solutions of these compounds differ in pH because the hydrogen sulfiteion, HSO4

–(aq), is a stronger acid than the hydrogen sulfite ion, HSO3

–(aq).

37. (a) Calcium carbonate would raise soil pH because the hydrolysis of the carbonate ion releases hydroxide ions:

CO32–(aq) � H2O(l) e OH–


(b) Alum lowers soil pH due to the hydrolysis of the aluminum ion:

Al(H2O)63�(aq) e H�

(aq) � Al(H2O)5(OH)2�(aq)

38. (a) Sodium phosphate solution is likely to be basic.

(b) PO43–(aq) � H2O(l) e HPO4

2–(aq) � OH–

(aq)

Kb � �K

Kw

a�

� �14..02

�

�

1100

–

–

1

1

4

3�

Kb � 2.3 � 10–2

Kb �[HPO4

2–(aq)][OH–

(aq)]��

[PO43–(aq)]


ICE Table for the Hydrolysis of Phosphate

PO43–(aq) + H2O(l) e HPO4

2–(aq) + OH–

(aq)




Kb � 2.3 � 10–2

�0.1

x0

2

– x� � 2.3 � 10–2

x2 � 2.3 � 10–2 (0.10 – x)

x2 � 2.3 � 10–3 – 2.3 � 10–2x

x2 � 2.3 � 10–2x – 2.3 � 10–3 � 0

x �

x � 0.0378 (the only positive root) (extra digits carried)

[OH–(aq)] � 0.0378 mol/L

pOH � –log[0.0378]

� 1.42

pH � 14 – pOH

� 14.0 – 1.42

pH � 12.58

The pH of the phosphate solution is 12.58.

39. (a) (i) HPO42–(aq) � H2O(l) e H3O�

(aq) � PO43–(aq)

(ii) HPO42–(aq) � H2O(l) e OH–

(aq) � H2PO4–(aq)

(b) HPO42–(aq); Ka � 4.2 � 10–13

HPO42–(aq); Kb � 2.4 � 10–2

Kb for the hydrogen phosphate ion is larger than Ka.

(c) Since Kb > Ka, a solution of Na2HPO4 is basic.

40. (a) <7(b) >7(c) <7(d) >7(e) >7(f) 7

41.

42.

2.3 � 102 � �(2.3 �� 102)�2 4(�2.3 �� 103)��

2


Lewis acid: Lewis base:

(a) HCl (g) NH3(g)

(b) Cu2�(aq) H2O(l)

(c) Al(OH)3 OH–(aq)

Type of titration pH at equivalence point

strong acid/strong base 7

strong acid/weak base 6

weak acid/strong base 9

43.

(a) 7(b) 1(c) VHCl remaining � 5.00 mL

CHCl after addition of NaOH � 0.100 mol/L � �155

mm

LL

�

CHCl after addition of NaOH � 0.0333 mol/L

pH � –log 0.0333

pH � 1.477

The pH after adding 5.00 mL of base is 1.477.

(d) H2O(l), Cl–(aq), Na�(aq)

(e) 7

(f) VNaOH remaining � 5.00 mL

CNaOH � 0.100 mol/L � �255.0.000

mm

LL

�

� 0.0200 mol/L

pOH � –log 0.0200

pOH � 1.699

pH � 12.301



pH

Titration of 0.100 mol/L HCl(aq)with 0.100 mol/L NaOH(aq)

(d) (H2O(l), Na(aq), Cl(aq))+ –

(c) 1.477

(f) 12.301

(b) 1

7(a)(e)

Volume of NaOH(aq) added

44.

(a) 1(b) VHCl remaining � 10.00 mL

CHCl after addition of NaOH � 0.100 mol/L � �1300..0000

mm

LL

�

CHCl after addition of NaOH � 0.0333 mol/L

pH � –log 0.0333

pH � 1.477


(c) VHCl remaining � 0.10 mL

CHCl after addition of NaOH � 0.100 mol/L � �309.1.900

mm

LL

�

CHCl after addition of NaOH � 2.506 � 10–4 mol/L (extra digits carried)

pH � –log 2.506 � 10–4

pH � 3.601


(d) VHCl remaining � 0.01 mL

CHCl after addition of NaOH � 0.01 mol/L � �309.0.919

mm

LL

�

CHCl after addition of NaOH � 2.5006 � 10–6 mol/L (extra digits carried)

pH � –log 2.5006 � 10–6

pH � 4.602


(e) VNaOH remaining � 0.01 mL

CNaOH � 0.100 mol/L � �400.0.011

mm

LL

�

CNaOH � 2.499 � 10–5 mol/L (extra digits carried)

pOH � –log 2.499 � 10–5


pH

Titration of 0.100 mol/L HCl(aq)with 0.100 mol/L NaOH(aq)

equivalence point

halfway toequivalence point

7

10.00 mL 20.00 mL

Volume of NaOH(aq) added

1

pOH � 4.600

pH � 9.400


(f) VNaOH remaining � 5.00 mL

CNaOH � 0.100 mol/L � �455.0.000

mm

LL

�

CNaOH � 11 � 10–2 mol/L

pOH � 1.954

pH � 12.046


45. (a) HF(aq) � NaOH(aq) e H2O(l) � NaF(aq)(b)

(c) F–(aq) � H2O(l) e OH–

(aq) � HF(aq)

Small additions of hydroxide are consumed by the reverse reaction of the equilibrium.

(d) The pH at the equivalence point will be greater than 7: basic.46. (a) The bottom curve represents the titration of a strong acid with sodium hydroxide solution. The strongest acid has

the lowest initial pH. (b) The top curve represents the titration of the acid with the smallest Ka value. The weakest acid will have the highest

initial pH.(c) At the equivalence point, the moles of acid present equal the moles of base added.

47.

48. Thymolphthalein49. The molar concentration of sodium acetate should also be 0.5 mol/L. Dissolve 1.5 g of anhydrous sodium acetate in

enough distilled water to make 50 mL of solution. Combine this solution with 50 mL of 0.5 mol/L acetic acid. 50. (a) Bromothymol blue changes colour during the steep portion of the graph.

(b) The colour change of phenolphthalein is within the steep portion of the graph.51. (a) Equilibrium shifts to the left.

(b) Equilibrium shifts to the right.(c) Equilibrium shifts to the left.


pH

HF(aq) Titrated with NaOH(aq)

Volume of NaOH (mL)

4

25 15 2010

6

25 30 35 40 45 50

8

10

12

14

methyl red orange (red in transition to yellow)

thymolphthalein colourless

bromothymol blue green (blue in transition to yellow)

indigo carmine yellow

52. [CO2H–(aq)] � 0.15 mol/L

[HCO2H(aq)] � 0.25 mol/L

HCO2H(aq) e H�(aq) � CO2H–

(aq)

�[H

[

�(

Haq

C)][

O

C

2

O

H2

(

H

aq)

–(a

]q)]

�� 1.8 � 10–4

[H�(aq)] � Ka �

[H

[C

C

O

O

2

2

H

H–(a

(

q

aq

)])]

�

� 1.8 � 10–4 � �00..2155

mm

ooll//LL

�

[H�(aq)] � 3.0 � 10–4

pH � –log [1.8 � 10–4]

pH � 3.52

The addition of H�(aq) …

[H�(aq)]added � 0.10 mol/L

[HCO2H(aq)]final � (0.25 � 0.10) mol/L

[HCO2H(aq)]final � 0.35 mol/L

[CO2H–(aq)]final � (0.15 – 0.10) mol/L

[CO2H–(aq)]final � 0.05 mol/L

[H�(aq)] � Ka �

[H

[C

C

O

O

2

2

H

H–(a

(

q

aq

)])]

�

� 1.8 � 10–4 � �00..3055

�

[H�(aq)] � 1.26 � 10–3 (extra digits carried)

pH � 2.90

The change in pH is 3.52 – 2.90.

�pH � 0.62

53. (a) nAg2SO4� �

3110.8.205gg/mol

�

nAg2SO4� 8.0 � 10–4 mol

(b) molar solubility of Ag2SO4

� �85.00.

�

001�

0–4

10m

–o3l

�

� 0.016 mol/L

(c) [Ag�(aq)] � 0.032 mol/L

(d) [SO42–(aq)] � 0.016 mol/L

(e) Ksp � [Ag�(aq)]

2[SO42–(aq)]

(f) Ksp � [Ag�(aq)]

2[SO42–(aq)]

� [0.032]2[0.016]Ksp � 1.6 � 10–5


(g) The accepted value for the solubility product of silver sulfate is 1.2 � 10–5. The perfect difference between thecalculated value and the accepted value is 33%. Given the large difference, the calculated value is judged to beunacceptable. (Note: Large errors in Ksp experiments are quite common.)

54. (a) Volume of unreacted acid � 17.40 mL

CHCl � 17.40 mL � 1.00 mol/L

CHCl � 17.4 mmol

The amount of unreacted acid is 17.4 mmol.

(b) 7.60 mL of HCl reacted.

nHCl � 7.60 mL � 1.00 mol/L

nHCl � 7.60 mmol

2:1 ratio, therefore

nCaCO3� 3.80 mmol

The amount of calcium carbonate in the shell was 3.80 mmol.

(c) mCaCO3� 3.80 mmol � 100.09 g/mol

mCaCO3� 0.380 g

The mass of calcium carbonate in the shell was 0.380 g.

%CaCO3 � �00..3485

�

%CaCO3 � 84%

The percent by mass of calcium carbonate in the shell is 84%.

(d) Grinding the shell increases its surface area, allowing the reaction with acid to occur more quickly. (e) The contents of the flask are boiled to remove dissolved carbon dioxide from the solution and to ensure that the

reaction with the acid is complete. (f) The addition of water dilutes the acid but does not alter the moles of acid present. The amount of base added

depends only on the amount of acid.

55. (a) nKH(IO3)2� �

2631.0.010gg/mol

�

nKH(IO3)2� 3.802 � 10–3 mol

nKH(IO3)2� nNaOHused

CNaOH � �32.58.0524

�

�

1100

–

–

3

3mm

ooll

�

CNaOH � 0.185 mol/L

The concentration of the sodium hydroxide solution is 0.185 mol/L.

(b) Since this is a titration of a strong acid with a strong base, the pH change at the equivalence point is large. ThepH range of both indicators falls well within the pH change that occurs.

(c) Boiling the water removes dissolved carbon dioxide that could combine with the hydroxide ions. (d) The test tube prevents atmospheric carbon dioxide from dissolving into the solution.

56. (a) Maintaining a high pressure inside the washing machine forces the equilibrium to shift to the right – the side withfewer gas molecules.

(b) Fresh-air circulation systems must be installed in CO2 dry-cleaning facilities to ensure that workers andcustomers are not exposed to excessive levels of carbon dioxide. Periodic monitoring of carbon dioxide levelsshould occur on a regular basis. In the extreme, a sudden leak of CO2 could suffocate everyone in the facility.

(c) Carbon dioxide is an ideal dry-cleaning solvent. It is a nonpolar molecule that is just as effective in dissolvinggrease and oil as the organic solvents currently used in the dry-cleaning process. It is nonflammable, almostchemically inert, and does not harm the ozone layer as many dry-cleaning solvents do. Once the washing cycleis complete, carbon dioxide can be evaporated, leaving behind the residue of the cleaning process. The “clean”


gas is then collected and can be used for the next cleaning cycle. The clean clothing can be packaged immedi-ately because no drying is required, saving considerable time.

57. (a) Assuming only Au3� is present …

AuCl3(s) e Au3�(aq) � 3 Cl–(aq)

Concentrations after mixing:

[Au3�(aq)] � 5.6 � 10–11 mol/L � �

12

�

[Au3�(aq)] � 2.8 � 10–11 mol/L

[Cl(aq)] � 0.100 mol/L � �12

�

[Cl(aq)] � 5.0 � 10–2 mol/L

Q � [Au3�(aq)][Cl(aq)]

2

� (2.8 � 10–11)(5.0 � 10–2)3

Q � 3.5 � 10–15

Ksp � 3.2 � 10–25

Q is larger than Ksp. Therefore, a precipitate does form.

(b) There is so little gold dissolved in seawater that huge quantities of water would have to be processed to extract ameasurable amount of gold. This would require investment in an extremely large filtration system. Assuming thegold precipitates could be separated from all the other particulates that are found in seawater, we would then haveto extract the gold from the precipitate. Another complication is that the amount of gold that has already precip-itated with the existing chloride ions in seawater is unknown. In summary, the extraction of gold from seawaterusing precipitation is not feasible.

58. (a) nNaOH � �40

2.040.8

g8/mg

ol�

nNaOH � 0.6000 mol

CNaOH(aq)� �

0.06.070500

mL

ol�


pOH � –log 0.800

pOH � 0.0969

pH � 14 – pOH

pH � 13.903

(b) Safety goggles and chemical-resistant gloves and apron should be worn when preparing this solution. 59. (a) The approximately 2-mm-thick enamel layer that protects teeth is a mineral that has a very low Ksp —

Ca5(PO4)3OH(s). The action of bacteria, particularly on the sweets we consume, can produce acetic and lacticacids. If the pH drops below 5.5, hydroxide ions are removed from enamel, resulting in demineralization.

(b) Tooth decay can be prevented by:• regular brushing to remove food particles, which become nutrients for bacteria, from teeth;• flossing to remove trapped food particles and plaque in places that the toothbrush cannot reach;• using toothpastes and fluoride treatments, which add F–

(aq) ions to replace the OH–(aq) ions, to remineralize

teeth with Ca5(PO4)3F(s);• avoiding sticky, sweet foods because they remain on teeth, providing bacteria with a trapped supply of

sugar.• Some foods, such as milk and other dairy products, actually raise the pH of saliva. They are also rich in cal-

cium and phosphorus and can help remineralize teeth.60. (a) A decrease in pH corresponds to an increase in the hydrogen ion concentration in the pool water. In response, the

system shifts to the left in an effort to consume the excess acid. This produces more hypochlorous acid, which isan eye irritant.


(b) Raising the pH removes hydrogen ions from the equilibrium. The system responds by producing more hydrogenhypochlorite ions via the forward reaction. The concentration of hypochlorite increases.

(c) A “shock treatment” involves adding a larger than normal amount of an oxidizing compound like sodiumhypochlorite to the pool. These compounds oxidize organic contaminants or ammonia and other nitrogencompounds and sanitize the water.

(d) A “shock treatment” introduces excess hypochlorite ions to the pool that cause the hypochlorous acid–hypochlo-rite equilibrium to shift to the left, consuming hydrogen ions in the process. The reduction of hydrogen ionsincreases the pH of the water.

61. The addition of calcium hydroxide or “slaked lime” has been useful in temporarily reducing the effects of acid depo-sition in lakes. Calcium hydroxide is relatively inexpensive, readily available in Ontario, nontoxic, and it dissolvesreadily in water. However, liming a lake can only be a temporary solution. If acid deposition continues, it is noteconomically feasible to be continually adding calcium hydroxide to lakes. Also, the already existing metal contami-nants of the lake pose a possible toxic risk to aquatic life. An alternative strategy would be to reduce the amount ofacid deposition contaminating the lakes in the first place.

Extension/Challenge62. (a) The large value of K suggests that the reaction strongly favours the products of this reaction. Consequently, very

little hydrogen sulfide or sulfur dioxide is released.(b) An increase in pressure shifts the equilibrium to the side having fewer gas molecules – the right side. The effi-

ciency of the reaction should increase with increasing pressure. (c) An increase in temperature favours the reverse reaction, which would decrease the efficiency of the reaction.

63. (a) (i) An increase in pressure increases the yield of sulfur trioxide. (ii) The removal of sulfur trioxide increases the yield of sulfur trioxide.(iii) Continually adding SO2(g) and O2(g) increases the yield of sulfur trioxide.

(b) The use of a catalyst does not affect the position of the equilibrium because the catalyst increases the rate of bothforward and reverse reactions. However, a catalyst does allow more SO3 to be produced in a given period of time– an increase in the rate of production.

(c) An increase in temperature shifts the equilibrium to the left. However, it also increases the rate of both reactions,resulting in more SO3 being produced.

64. (a)

Ksp � [Ca2�(aq)][CO3

2–(aq)]

� 5.0 � 10–9

(x)(x) � 5.0 � 10–9

x � 7.1 � 10–5

The molar solubility of calcium carbonate is 7.1 � 10–5 mol/L.

(b) nCaCO3� �

1005..000

g/gmol

�

nCaCO3� 0.050 mol

molar solubility � 7.1 � 10–5 mol/L

Vwater required ��7.1

0�

.01500–5m

mol

ol/L�

Vwater required � 700 L

Since the kettle holds 2.0 L of water, it would have to be filled 350 times to dissolve the calcium carbonate scale.This assumes that more scale deposits do not form and that the existing calcium carbonate does not break off dur-ing this process.


ICE Table for the Solubility of Calcium Carbonate

CaCO3(s) e Ca2+(aq) + CO3

2–(aq)

Initial concentration (mol/L) – 0.00 0.00

Change in concentration (mol/L) – +x +x

Equilibrium concentration (mol/L) – x x

(c) Precipitation of calcium carbonate occurs because water cannot hold as much solute when its temperaturedecreases.

(d) This is a common problem in hot-water pipes and hot-water heaters.


chapter 8 acid–base equilibrium -...

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