16.1 WATER IONIZATION AND ACID–BASE STRENGTH
Lab Exercise 16A: The Chromate–Dichromate Equilibrium
(Page 712)
Purpose
The scientific purpose of this investigation is to test a Design for varying the acidity of an
equilibrium.
Problem
How does changing the hydrogen ion concentration affect the chromatedichromate equilibrium?
Prediction
2 CrO 4 2– (aq) + 2 H + (aq) p Cr2 O7 2– (aq) + H 2 O(l)
According to Le Châtelier’s principle and the equilibrium equation, as the hydrogen ion
concentration increases, the equilibrium will shift to the right, increasing the dichromate ion
concentration. As the hydrogen ion concentration decreases, the equilibrium will shift to the left,
increasing the chromate ion concentration. In both cases, the equilibrium shifts to oppose the
change: to the right to try to use up the added hydrogen ions, and to the left to try to replace the
hydrogen ions that are removed.
Design
Equal samples of a chromatedichromate equilibrium mixture are placed in three separate test
tubes. One test tube acts as a control. The added acid or base (to remove hydrogen ions) is the
manipulated variable, and the colour of the mixture is the responding variable. Controlled
variables are temperature and volume.
If a few drops of a strong acid are added to one test tube, and the colour becomes more
orange, then the dichromate ion concentration has increased. If a few drops of a strong base are
added to another test tube, and the colour becomes more yellow, then the chromate ion
concentration has increased.
Career Connection: Environmental Engineer
(Page 714)
Becoming an environmental engineer in Alberta requires, at minimum, a degree in environmental
engineering or a related field (such as chemical or civil engineering).
Both the University of Calgary and the University of Alberta offer engineering programs
which lead to B.Sc., M.Eng., M.Sc., and Ph.D. degrees in various engineering disciplines, such as
civil, chemical, and mechanical engineering. Some disciplines permit students to pursue a
specialization in environmental engineering. The University of Calgary, for example, offers a
graduate-level environmental engineering program taught by staff members from four different
engineering disciplines.
Both universities also offer a co-op engineering program which involves alternating
periods of academic study and industrial work experience. The co-op program generally takes
one year longer to complete.
Employment opportunities in this field are expected to improve over the next ten years as
many currently middle-aged engineers approach retirement.
The average salary in this field (in 2006) is $77 400 per year.
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Copyright © 2007 Thomson Nelson
Practice
(Page 716)
[Units are usually omitted for equilibrium constants. If concentrations substituted into the
equilibrium law expression are always in moles per litre, then any calculated concentration
answer must also be in units of moles per litre. Students do not need to show any units in an
equilibrium constant expression until a final amount concentration is reported.]
Kw
1.0 u 10 –14
1.0 u 10 –14
1. ª¬ OH – (aq)º¼
or
4.40 u 10 –3 mol/L
4.40 u 10 –3
ª¬ H 3O + (aq) º¼
= 2.3 u 10 –12 mol/L
Using the KW relationship, the concentration of hydroxide ions is 2.3 u1012 mol/L.
Kw
1.0 u 10 –14
2. ª¬ H 3O + (aq) º¼
= 3.3 u 10 –11 mol/L
–4
–
2.99
u
10
mol/L
ª¬OH (aq) º¼
Using the KW relationship, the concentration of hydronium ions is 3.3 u1011 mol/L.
3. HCl(aq) H 2 O(l) o H 3O + (aq) + Cl – (aq) (strong acid: 100% ionized)
1 mol
nHCl = 0.37 g u
= 1.0 u 10 –2 mol = 10 mmol
36.46 g
10 m mol
= 0.040 mol/L
250 m L
[H 3O + (aq)] = [HCl(aq)] = 0.040 mol/L
[HCl(aq)] =
ª¬ OH – (aq)º¼
Kw
+
(strong acid)
–14
1.0 u 10
= 2.5 u 10 –13 mol/L
4.0 u 10 –2 mol/L
ª¬ H 3O (aq) º¼
Using the KW relationship, the amount concentration of hydroxide ions is 2.5 u1013 mol/L.
4. Ca(OH) 2 (s) p Ca 2+ (aq) + 2 OH – (aq)
2
[OH–(aq)] = 6.9 mmol/L u = 14 mmol/L
1
K
1.0 u 10 –14
w
ª¬ H 3O + (aq) º¼
= 7.2 u 10 –13 mol/L
ª¬OH – (aq) º¼ 1.4 u 10 –2 mol/L
Using the KW relationship, the amount concentration of hydronium ions is 7.2 u1013 mol/L.
5. KOH(s) o K + (aq) + OH – (aq)
1 mol
nKOH = 20.0 g u
= 0.356 mol
56.11 g
0.356 mol
= 0.713 mol/L
0.500 L
[OH–(aq)] = [KOH(aq)] = 0.713 mol/L
(assume complete dissociation)
–14
Kw
1.0 u 10
ª¬ H 3O + (aq) º¼
= 1.4 u 10 –14 mol/L
–
ª¬ OH (aq) º¼ 0.713 mol/L
[KOH(aq)] =
According to the dissociation equation and the KW relationship, the amount concentration of
hydronium ions is 1.4 u1014 mol/L.
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625
6. H2O(l) + H2O(l) p H3O+(aq) + OH–(aq)
cH2 O(l) =
1000 g
1L
u
1 mol
= 55.49 mol/L
18.02 g
1.0 u 10 –7 mol/L
u 100% = 1.8 u 10 –7 %
55.49 mol/L
The percent ionization of water is 1.8 u107 %. (Approximately 2 of every billion molecules
is ionized.)
p
Practice
(Page 718)
7. (a) (Solutions follow.)
Food
oranges
[H3O+(aq)]
(mol/L)
5.5 u 103
[OH–(aq)]
(mol/L)
1.8 u 1012
pH
2.26
pOH
11.74
asparagus
olives
blackberries
4 u 109
5.0 u 104
4.0 u 104
3 u 106
2.0 u 1011
2.5 u 1011
8.4
3.30
3.40
5.6
10.70
10.6
[For Table 1, the solutions may be calculated in different sequences. Only one sequence
of calculations is shown here for each substance.]
Kw
1.0 u 10 –14
ª¬ OH – (aq)º¼
oranges:
= 1.8 u 10 –12 mol/L
ª¬ H 3O + (aq)º¼ 5.5 u 10 –3 mol/L
asparagus:
pH
= log(5.5 u 103) = 2.26
pOH = 14.00 – pH = 14.00 – 2.26 = 11.74
pH = 14.00 – pOH = 14.00 – 5.6 = 8.4
[OH–(aq)] = 10–pOH = 10–5.6 = 3 u 106 mol/L
Kw
1.0 u 10 –14
ª¬ H 3O + (aq) º¼
= 4 u 10 –9 mol/L
ª¬ OH – (aq) º¼ 3 u 10 –6 mol/L
ª¬ H 3O + (aq) º¼
olives:
blackberries
Kw
1.0 u 10 –14
= 5.0 u 10 –4 mol/L
2.0 u 10 –11 mol/L
ª¬ OH – (aq) º¼
pH
= log(5.0 u 104) = 3.30
pOH = 14.00 – pH = 14.00 – 3.30 = 10.70
pH = 14.00 – pOH = 14.00 – 10.60 = 3.40
[OH–(aq)] = 10–pOH = 10–10.60 = 2.5 u 1011 mol/L
Kw
1.0 u 10 –14
ª¬ H 3O + (aq) º¼
= 4.0 u 10 –4 mol/L
–11
–
2.5
10
mol/L
u
ª¬ OH (aq) º¼
(b) On the basis of pH only, orangeswith the lowest pHwould be predicted to taste the
most sour.
8. NaOH(s) o Na + (aq) + OH – (aq)
1 mol
nNaOH = 26 g u
= 0.65 mol
40.00 g
nOH- = nNaOH = 0.65 mol
[OH–(aq)] =
626
0.65 mol
= 4.3 mol/L
0.150 L
Unit 8 Solutions Manual
Copyright © 2007 Thomson Nelson
pOH = -log[OH–(aq)] = -log(4.3) = -0.64
pH = 14.00 – pOH = 14.00 – (-0.64) = 14.64
According to the dissociation equation and the definition of pH, the pH and pOH of the
solution are 14.64 and 0.64.
9. pOH = 14.00 – 11.5 = 2.5
[OH–(aq)] =10–pOH = 10-2.5 mol/L = 3 u 10-3 mol/L
[KOH(aq)] = [OH–(aq)] = 3 u 10-3 mol/L
3 u 10-3 mol
nKOH = 0.500 L u
= 0.002 mol
1 L
56.11 g
mKOH = 0.002 mol u
= 0.09 g
1 mol
The mass of potassium hydroxide is 0.09 g. The certainty of this calculation is limited to one
significant digit, which for any quantity value is really just a crude approximation.
Web Activity: Canadian Achievers—Edgar Steacie
(Page 721)
1. Edgar Steacie’s main area of research was free-radical research, extending later into
photochemistry and chemical kinetics. All of the areas of study which Steacie focused on,
involved detailed descriptions of chemical reaction systems.
2. Steacie was known as a statesman of science for Canada because he was instrumental in
building up university research in this country, and was the architect of enduring programs to
support industrial innovation. He played a leading role in BritishCanadian collaboration in
atomic energy, which led to the construction of the Chalk River reactor, the first to be built
outside the USA.
3. Every year, Natural Sciences and Engineering Research Canada (NSERC) awards up to six
Steacie Fellowships for a two-year period. Successful Fellows are relieved of teaching and
administrative duties, so that they can devote all their time and energy to research. The
Fellowships are held at a Canadian university or affiliated research institution. The
Fellowship normally includes a contribution to the university in the amount of
$90 000 per year toward the Fellow's salary. As part of the Fellowship agreement, the
university is expected to fund a replacement for the Fellow’s teaching and administrative
responsibilities or to enhance the research environment of the Fellow's department.
Section 16.1 Questions
(Page721)
1. (a) If a solution is neutral, the hydronium and hydroxide concentrations are the same.
(b) If a solution is acidic, the hydronium ion concentration is greater than the hydroxide ion
concentration.
(c) If a solution is basic, the hydroxide ion concentration is greater than the hydronium ion
concentration.
2. Conductivity and pH tests can both distinguish a weak acid from a strong acid if the
temperature and the initial acid solute concentrations are the same.
3. According to Arrhenius’ theory, all bases dissolve in water and increase the hydroxide ion
concentration of solutions.
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3
7.8 u10 %

o H 3O + (aq) + CN (aq)
4. (a) HCN(aq) + H 2 O(l) m

p
7.8 u 10 –3 %
+
u [HCN(aq)]initial =
u 0.10 mol/L = 7.8 u 10-6 mol/L
(b) [H 3O (aq)]equil =
100
100
pH = -log [H 3O + (aq)] = -log(7.8 u 10 –6 ) = 5.11
5. [OH – (aq)] =
Kw
1.0 u 10 –14
=
[H 3O + (aq)] 1.3 u 10 –3 mol/L
7.7 u 10 –12 mol/L
6. [H 3O + (aq)] =
Kw
1.0 u 10 –14
=
–
[OH (aq)] 2.5 u 10-7 mol/L
4.0 u 10 –8 mol/L
pH = –log [H3O + (aq)] = –log(4.0 u 10 –8 ) = 7.40
7. [H 3O + (aq)] = 10–pH = 10-5.6 mol/L
3 u 10-6 mol/L
According to the definition of pH, the hydronium ion concentration in the acid rain is
3 u 106 mol/L.
8. The hydronium ion changes (decreases) by a factor of 1000 (103).
9. NaOH(s) o Na + (aq) + OH – (aq)
nNaOH
8.50 g u
1 mol
40.00 g
0.213 mol
0.213 mol
= 0.425 mol/L
0.500 L
[OH - (aq)] = [NaOH(aq)] = 0.425 mol/L
[NaOH(aq)] =
pOH = –log [OH - (aq)] = –log(0.425) = 0.372
10. HCl(aq) + H 2 O(l) o H3O + (aq) + Cl – (aq)
[H 3O + (aq)] 10 –pH
101.57 mol/L = 0.027 mol/L
[HCl(aq)] = [H 3O + (aq)] = 0.027 mol/L.
12. nHCl
0.250 L u
0.027 mol
0.0067 mol
1 L
36.46 g
mHCl 0.0067 mol u
0.25 g
1 mol
p
0.054
11. [H 3O + (aq)] =
u [HA(aq)] =
u 0.10 mol/L = 5.4 u 10-5 mol/L
100
100
pH = –log [H3 O + (aq)] = –log(5.4 u 10-5 ) = 4.27
nHCl
30.5 k g u
1 mol
36.46 g
0.837 kmol = 837 mol
837 mol
= 1.04 mol/L
806 L
[H 3O + (aq)] [HCl(aq)] = 1.04 mol/L
[HCl(aq)] =
pH = –log [H3 O+ (aq)] = –log(1.04) = –0.016
pOH = 14.00 - pH = 14.00 - (–0.016) = 14.02
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We assume 100% percent reaction (ionization) of the dissolved hydrogen chloride because
hydrochloric acid is a strong acid. We also assume that the volume of solutions is equal to the
volume of water.
0.48%

o H3O + (aq) + CH 3COO - (aq)
13. CH 3COOH(aq) + H 2 O(l) m

1 mol
nCH3COOH = 60.0 k g u
= 0.999 kmol
60.06 g
0.999 k mol
= 0.799 mol/L
1.25 k L
p
0.48
[H 3O + (aq)] =
u [CH 3COOH(aq)] =
u 0.799 mol/L = 0.0038 mol/L
100
100
pH = -log [H3O + (aq)] = -log(3.8 u 10-3 ) = 2.42
pOH = 14.00 - pH = 14.00 - 2.42 = 11.58
14. pOH = 14.00 - 10.35 = 3.65
[CH 3COOH(aq)] =
[OH - (aq)] = 10–pOH = 10–3.65 mol/L = 2.2 u 10–4 mol/L
nOH = 2.00 L u
2.2 u 10-4 mol
= 4.5 u 10-4 mol
1 L
nNaOH = nOH - = 4.5 u 10-4 mol
mNaOH = 4.5 u 10-4 mol u
40.00 g
= 0.018 g (or 18 mg)
1 mol
15. Design
The solutions are tested with blue litmus (or pH paper) to determine whether they are acidic
or neutral, and then for conductivity in order to distinguish between the strong and weak
acids, and the ionic and molecular solutes.
Diagnostic test
Strong acid
Weak acid
Conductivity
Blue litmus
high
turns red
low
turns red
Neutral
molecular
none
no change
Neutral ionic
high
no change
16.
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