Unit 4 Review, pages 502–509
Knowledge
1. (c)
2. (b)
3. (a)
4. (c)
5. (a)
6. (c)
7. (a)
8. (d)
9. (c)
10. (d)
11. (b)
12. (a)
13. False. Water expands as it cools to 0 °C, causing ice to float on the surface of liquid water.
14. True
15. False. As sodium chloride dissolves in water, each negative chloride ion gets surrounded by
the hydrogen ends of water molecules, completely hydrating the ion.
16. False. The phrase “like dissolves like” refers to the idea that polar solvents will dissolve
polar solutes.
17. True
18. False. When diluting a concentrated acid, always add the acid to the water.
19. True
20. True
21. False. Flame tests can reliably identify cations in solution when a single cation is present.
22. False. Acidic foods soften teeth, so brushing your teeth immediately after eating acidic foods
is not a good idea.
23. False. Bases readily react with carbon dioxide to produce carbonate compounds.
24. True
25. (a) (i)
(b) (ii)
(c) (i)
(d) (ii)
(e) (ii)
(f) (iii)
(g) (i)
26. (a) (iv)
(b) (i)
(c) (v)
(d) (vi)
(e) (ii)
(f) (iii)
27. Hydrogen bonding is responsible for most of the special characteristics of water.
28. The solvent is ethanol because there is a greater volume of ethanol present.
29. When dissolved in water, ionic compounds undergo the process of dissociation.
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Unit 4: Solutions and Solubility
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30. Since the solvent at a lower temperature holds more solute, we can conclude that the solute is
a gas. The solubility of gases decreases as the temperature increases.
31. Two substances that tend not to dissolve in water are non-polar substances and ionic
compounds that are only slightly soluble.
32. The anion that does not form a precipitate with lead(II) ions is nitrate, NO3–.
33. Of the ions listed, silver ions form the least soluble compound when mixed with sulfate ions.
34. The total ionic equation identifies highly soluble compounds written in their dissociated ionic
form.
35. The net ionic equation identifies only the entities that actually react and change in an
equation.
36. The only spectator ion in the equation is SO42–(aq).
37. Detergents are used in shampoos so that precipitates do not form.
38. Potable means that water is safe to drink.
39. Spectroscopy is a sophisticated form of a flame test.
40. The limiting reagent is reagent B.
41. The two ingredients that combined to make soap were a base and an animal fat.
42. In the presence of phenolphthalein indicator, acids are colourless and bases are pink.
43. (a) H2CO3(aq) is carbonic acid.
(b) HF(aq) is hydrofluoric acid.
(c) H3PO4(aq) is phosphoric acid.
(d) H2S(aq) is hydrosulfuric acid.
(e) HClO4(aq) is perchloric acid.
44. (a) The chemical formula for phosphorous acid is H3PO3(aq).
(b) The chemical formula for hydrochloric acid is HCl(aq).
(c) The chemical formula for hypoiodic acid is HI(aq).
(d) The chemical formula for chlorous acid is HClO2(aq).
(e) The chemical formula for chloric acid is HClO3(aq).
45. When a metal and an acid react, you would expect to see bubbles of hydrogen gas.
46. (a) Dissociation is the process during which a compound separates into ions as it dissolves in
water.
(b) Ionization occurs when uncharged particles react with water and become ions.
47. A net ionic equation that applies to all acid–base neutralization reactions is
H+(aq) + OH–(aq) → H2O(l)
48. A strong acid like hydrochloric acid would ionize more completely in water than a weak
acid.
49. A strong acid like sulfuric acid would conduct electricity better than a weak acid.
50. Titration is the analytical process used to find the concentration of an unknown solute in a
solution.
51. The two events that must coincide for a titration to produce accurate results are the endpoint
and the equivalence point.
Understanding
52. Water has a highly polar structure because of its molecular shape. This structure allows for
hydrogen bonding between water molecules. This hydrogen bonding holds water molecules
together resulting in high surface tension, high melting and boiling points, and expansion when
water freezes.
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Unit 4: Solutions and Solubility
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53. Air can be either a heterogeneous or a homogeneous mixture depending on its composition.
Air is heterogeneous if you can see suspended dust or smog particles. Similarly, foggy air is a
heterogeneous mixture of liquid water droplets and air. Dry, clean air can be totally transparent
and, therefore, qualifies as a homogeneous solution.
54. Ice water is a heterogeneous mixture because it has two distinct phases.
55. Since hexane is a covalent compound that is non-polar, it will not dissolve in water. The
polar ends of water molecules are not attracted to hexane, so hexane is unable to break up
water’s hydrogen bond network.
56. Since polar toxins are soluble in water, when they enter the body of an organism, the polar
toxins do not stay in the body for long because they are excreted in the urine. Non-polar toxins
do not dissolve in blood or urine and instead are stored in body fats. There is no way to get rid of
these toxins so they accumulate in body fats and cause damage to the organism.
57. (a) Soaps and detergents act on the surface of a liquid, breaking down surface tension, which
is what a surfactant does.
(b) Soaps and detergents are ionic compounds that break down into cations and anions that have
a charged head and a long non-polar tail.
(c) The non-polar sections of soap and detergent molecules are attracted to non-polar grease and
oils. The polar ends are attracted to water. When agitated, tension between the two ends is able to
pull the non-polar grease away from the surface to be cleaned. Surfactant molecules then
surround the grease droplets, preventing them from reattaching or coming together.
58. (a) To make a supersaturated glucose solution, add enough glucose to saturate the solution at
a high temperature, such as 80 °C. Then slowly cool the solution. If left undisturbed, the solution
will become supersaturated.
(b) Adding a small quantity of solid glucose to the supersaturated solution or disturbing the
supersaturated solution will cause any excess dissolved glucose to come out of the solution and
crystallize. These crystals could be filtered out, leaving a saturated solution.
59. Amount concentration is defined as the amount of solute (in moles) in a total volume of
solution. Since the student added solute to 1.00 L of water, the final solution has a volume
greater than 1.00 L. Therefore, the amount concentration was actually less than 1.00 mol/L.
60. (a) 2 Al(s) + 6 HCl(aq) → 2 AlCl2(aq) + 3 H2(g)
(b) Aluminum metal, Al(s), is the only solid in the equation.
(c) The total ionic equation is
2 Al(s) + 6 H+(aq) + 6 Cl–(aq) → 2 Al3+(aq) + 6 Cl–(aq) + 3 H2(g).
(d) The chloride ion, Cl–(aq), is the only spectator ion in the equation.
(e) The net ionic equation is
2 Al(s) + 6 H+(aq) → 2 Al3+(aq) + 3 H2(g).
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Unit 4: Solutions and Solubility
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61. Write the formula equation for the reaction of iron and silver nitrate.
Fe(s) + 2 AgNO3(aq) → 2 Ag(s) + Fe(NO3)2(aq)
Write the total ionic equation.
Fe(s) + 2Ag+(aq) + 2 NO3–(aq) → 2 Ag(s) + Fe2+(aq) + 2NO3–(aq)
Write the net ionic equation.
Fe(s) + 2 Ag+(aq) → 2 Ag(s) + Fe2+(aq)
62. The sewage contamination depletes the oxygen in the pond. A fountain would help to mix the
water with air, which would allow more oxygen to dissolve in the water. A higher oxygen
concentration can help threatened fish species survive.
63. The steps in the water treatment process are collection, removal of debris, coagulation,
flocculation, and sedimentation, filtration, disinfection, aeration, softening, fluoridation, postchlorination, and ammoniation.
64. Astronomers are able to identify chemicals in distant stars by analyzing the characteristic
light patterns emitted by the stars in a process called spectroscopy.
65. (a) The concentration of iron(III) sulfate is 2.00 mol/L.
(b) Given: cFe (SO ) = 2.00 mol/L
2
4 3
Required: concentration of iron(III) ions, cFe3+
Solution:
Step 1. Write the net ionic equation for iron(III) sulfate.
2 Fe3+ (aq) + 3 SO42–(aq) → Fe2(SO4)3(aq)
Step 2. Determine the concentration of iron(III) ions using the mole ratio.
cFe3+ = 2.00
mol Fe
2 (SO 4 )3
L
!
2 mol Fe3+
1 mol Fe
2 (SO 4 )3
mol
L
Statement: The concentration of iron(III) ions is 4.00 mol/L.
(c) Given: cFe (SO ) = 2.00 mol/L
cFe3+ = 4.00
2
4 3
Required: concentration of sulfate ions, cSO2–
4
Solution:
Step 1. Write the net ionic equation for iron(III) sulfate.
2 Fe3+ (aq) + 3 SO42–(aq) → Fe2(SO4)3(aq)
Step 2. Determine the concentration of sulfate ions using the mole ratio.
cSO2– = 2.00
4
mol Fe
2 (SO 4 )3
L
!
3 molSO2–
4
1 mol Fe
2 (SO 4 )3
mol
4
L
Statement: The concentration of sulfate ions is 6.00 mol/L.
cSO2– = 6.00
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Unit 4: Solutions and Solubility
U4-8
66. (a) A reaction will take place if the metal is higher than hydrogen in the activity series.
(b) If a reaction occurred, the products would be a dissolved metal chloride compound and
hydrogen gas.
(c) Visible evidence that a reaction is taking place would be bubbles of a gas forming around the
metal.
67. The balanced equation for this reaction is
CaO(s) + 2 HCl(aq) → CaCl2(aq) + H2O(l)
68. (a) The products of the reaction are iron(II) chloride, FeCl3(aq), and hydrogen sulfide,
H2S(g).
(b) The balanced equation for the reaction is
FeS(s) + 2 HCl (aq) → FeCl2(aq) + H2S(g)
69. (a) The test tube most likely contained a basic solution.
(b) The property of bases that this example illustrates is that bases react with carbon dioxide to
produce carbonates, which are generally only slightly soluble so they tend to form precipitates.
70. (a) The general undissolved acid formula is HA(aq).
(b) The general equation describing what happens when an acid dissolves in water is
HA → H+(aq) + A–(aq)
(c) Answers may vary. Sample answer: Nitric acid ionizes in water. The balanced equation is
HNO3(aq) → H+(aq) + NO3–(aq)
71. (a) The products of the neutralization reaction are water and an ionic compound.
(b) The balanced formula equation for the neutralization reaction is
H2SO4(aq) + 2 KOH(aq) → 2 H2O(l) + K2SO4(aq)
(c) The total ionic equation for the neutralization reaction is
2 H+(aq) + SO42–(aq) + 2 K+(aq) + 2 OH–(aq) → 2 H2O(l) + SO42–(aq) + 2 K+(aq)
(d) The net ionic equation for the neutralization reaction is
2 H+(aq) + 2 OH–(aq) → 2 H2O(l)
(e) The net ionic equation shows that any acid–base neutralization reaction can be reduced to an
interaction between hydrogen ions and hydroxide ions to form water.
72. The difference between a strong acid and a weak acid is the degree to which they ionize in
water. A strong acid completely ionizes because all of its molecules react to form ions when they
dissolve in water. A weak acid only partially ionizes because some of its molecules remain
intact. Only a small percentage of the molecules form ions.
73. Strong acids conduct electricity better than weak acids because they ionize completely. Weak
acids only partially ionize. This means that a strong acid will have a higher concentration of ions
in the solution. Since ions conduct electricity, the greater the concentration of ions, the better the
substance conducts electricity.
74. The statement is wrong because a pH 10 solution is less acidic than a pH 9 solution. A
solution of pH 10 is 10 times more basic that a solution of pH 9.
Analysis and Application
75. (a) Li2S(aq) → 2 Li+(aq) + NO3– (aq)
(b) MgCl2(aq) → Mg2+(aq) + 2 Cl– (aq)
(c) Al2(SO4)3(aq) → 2 Al3+(aq) + 3 SO42– (aq)
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Unit 4: Solutions and Solubility
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76. (a) The solution is saturated because the point that corresponds to 50 g of potassium chloride
in 100 mL of water at 60 °C is above the KCl curve. Less than 50 g of the compound dissolves at
that temperature. The solution might actually be supersaturated or include some undissolved
crystals of potassium chloride.
(b) The solution would become unsaturated at about 78 °C. At that point, the solubility value
would be below the curve.
77. (a) Potassium chloride is more soluble at 15 °C than potassium nitrate. At this temperature,
the potassium chloride curve in Figure 1 is above the potassium nitrate curve, which means that a
greater mass of potassium chloride than potassium nitrate will dissolve in 100 mL of water.
(b) Potassium chloride and potassium nitrate would be equally soluble at approximately 22 °C.
(c) Potassium nitrate is more soluble than potassium chloride at any temperature above 22 °C.
78. Given: cNaCl = 0.25 mol/L
mNaCl = 50.00 g
Required: volume of sodium chloride solution, VNaCl
Analysis: cNaCl =
nNaCl
VNaCl
Solution:
Step 1. Convert the mass of solute to the amount of solute.
1 mol
nNaOH = 125 g !
40.00 g
nNaOH = 3.1250 mol [2 extra digits carried]
Step 2. Rearrange the equation and substitute the values.
n
VNaCl = NaOH
cNaOH
= 0.8556 mol !
1L
0.25 mol
VNaCl = 3.42 L
Statement: The volume of a 0.25 mol/l of sodium chloride solution that contains 50.00 g of
sodium chloride is 3.42 L.
79. Given: cc = 2.500 mol/L
Vc = 0.500 L
Vd = 4.250 L
Required: concentration of diluted solution, cd
Analysis: ccVc = cdVd
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Unit 4: Solutions and Solubility
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Solution:
Step 1. Rearrange the equation in the appropriate form, substitute the values, and solve.
cV
cd = c c
Vd
!
mol $
#" 3.0 L &% 4.5 L
cd =
20.0 L
mol
cd = 0.68
L
Statement: The resulting concentration of the solution is 0.294 mol/L.
80. Formula equation:
Cu(s) + 2 AgNO3(aq) → Cu(NO3)2(aq) + 2 Ag(s)
total ionic equation:
Cu(s) + 2 Ag+(aq) + 2 NO3–(aq) → 2 Ag+(aq) + Cu(s) + 2 NO3–(aq)
net ionic equation:
Cu(s) + 2 Ag+(aq) → 2 Ag+(aq) + Cu(s)
spectator ions: NO3–(aq)
81. (a) Physical contaminants are removed during collection, coagulation, flocculation,
sedimentation, and filtration.
(b) Biological contaminants are removed during filtration, disinfection, aeration, and postchlorination.
(c) Chemical contaminants are removed during filtration, aeration, and softening.
82. (a) The influx of fecal material into the pond rapidly increases the growth of
microorganisms. When these organisms die, the decomposition process depletes the oxygen in
the water. In addition, fecal matter may contain harmful micro-organisms, which can cause
serious illness.
(b) The nutrients in the fecal matter create a short-term boom in aquatic plants and algae. When
the nutrients are used up, these organisms die. Decomposers then use up much of the pond’s
oxygen to decompose the biomass. This reduces the concentration of oxygen in the water,
making it difficult for aquatic animals to absorb sufficient oxygen from the water. Since there is
a lack of oxygen, fish and other species with high oxygen requirements die off.
(c) Any organisms that do not require oxygen, such as anaerobic bacteria, can survive in an
oxygen-depleted pond.
83. Insoluble calcium ions in the hard water react with sodium carbonate to form calcium
carbonate, which is only slightly soluble. The calcium carbonate precipitates out, leaving sodium
ions in the water.
Ca2+(aq) + 2 Na2CO3(aq) → 2 Na+(aq) + CaCO3(s)
84. (a) If the solution contains lead, lead(II) ions and iodide ions will react to form a yellow
precipitate.
(b) If the solution contains barium, barium ions and sulfate ions will react to form a white
precipitate.
(c) An insufficient quantity of potassium iodide was added. Without an excess, some lead can
remain after treatment with potassium iodide. So the lead showed up in the later analysis.
(
)
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Unit 4: Solutions and Solubility
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85. (a) Answers may vary. Sample answer: A qualitative analysis for identifying the ions would
involve a series of steps.
Step 1. Add a hydroxide solution of sodium hydroxide, NaOH, to precipitate manganese ions but
not strontium ions. Be sure to add an excess of sodium hydroxide to make sure all of the
manganese precipitates. If a precipitate forms, the solution contained manganese ions. If a
precipitate does not form, there were no manganese ions.
Step 2. To precipitate any strontium ions, add sodium sulfate. Strontium sulfate will precipitate
out of solution.
(b) The flow chart for this qualitative analysis is shown below.
86. (a) Na2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2 NaCl(aq)
(b) Given: cNa CO = 0.25 mol/L
2
3
VCaCl = 200 mL
2
cCaCl = 0.20 mol/L
2
Required: amount of calcium chloride, nCaCl
2
Solution:
Step 1. Convert the volume to litres.
1L
VCaCl = 200 mL !
2
1000 mL
VCaCl = 0.200 L
2
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Unit 4: Solutions and Solubility
U4-12
Step 2. Write a dissociation equation listing the calculated amounts and the required value(s).
Na2CO3(aq)
+
CaCl2(aq)
→
CaCO3(s) + 2 NaCl(aq)
cNa CO = 0.25 mol/L cCaCl = 0.25 mol / L
2
3
VNa CO
2
2
VCaCl = 0.150 L
3
2
Step 3. Determine the amount of calcium chloride by rearranging the equation.
n
c=
V
nSrCl = cSrCl VSrCl
2
2
2
0.25 mol
! 0.15 L
1L
= 0.0375 mol [2 extra digits carried]
=
nSrCl
2
Statement: The amount of calcium chloride is 0.04 mol.
(c) The mole ratio of sodium carbonate to calcium chloride is 1:1.
(d) Given: cNa CO = 0.25 mol/L
2
3
VCaCl = 0.200 L
2
cCaCl = 0.20 mol/L
2
nCaCl = 0.04 mol
2
Required: amount of sodium carbonate, nNa CO
2
3
Solution:
Step 1. Determine the amount of calcium chloride.
1 mol Na CO
2
3
nNa CO = 0.04 molCaCl !
2
3
2
1 molCaCl
2
nNa CO = 0.04 mol
2
3
Statement: The amount of sodium carbonate required is 0.04 mol.
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Unit 4: Solutions and Solubility
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(e) Given: cNa CO = 0.25 mol/L
2
3
nNa CO = 0.04 mol
2
3
Required: volume of sodium carbonate, VNa CO
2
3
Solution:
Step 1. Determine the volume of sodium carbonate by rearranging the equation.
nNa CO
2
3
VNa CO =
2
3
cNa CO
2
=
3
0.04 mol
mol
L
= 0.16 L
0.25
VNa CO
2
3
Statement: The volume of 0.25 mol/L sodium carbonate required is 0.16 L.
87. Jeff’s identification of the limiting reagent could be correct. The limiting reagent is the
substance that is required in the smallest amount to be completely reacted, so the mass of
substances cannot be used for comparison until they are converted to amounts.
88. (a) 2 FeCl3(aq) + 3 H2S(aq) → 6 HCl(aq) + Fe2S3(s)
(b) Given: VFeCl = 0.25 L
3
mH S = 37.50 g
2
cFeCl = 2.20 mol/L
3
Required: amount of FeCl3, nFeCl
3
amount of H2S, nH S
2
Solution:
Step 1. Write the dissociation equation listing the calculated amounts and the required value(s).
2 FeCl3(aq)
+
3 H2S(aq)
→
6 HCl(aq) + Fe2S3(s)
VFeCl = 0.25 L
3
mH S = 37.50 g
2
cFeCl = 2.20 mol/L
3
Step 2. Determine the amount of FeCl3by rearranging the equation.
n
c=
V
nFeCl = cFeCl VFeCl
3
3
3
2.20 mol
! 0.25 L
1L
= 0.55 mol
=
nFeCl
3
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Unit 4: Solutions and Solubility
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Step 3. Convert the mass of H2S to the amount.
1 mol
nH S = 37.50 g !
2
34.09 g
nH S = 1.10 mol
2
Statement: The amount of FeCl3is 0.55 mol and the amount of H2S is 1.10 mol.
(c) The mole ratio of H2S to FeCl3 is 3:2.
(d) The amount ratio of H2S to FeCl3 is 2:1.
(e) The limiting reagent is FeCl3.
(f) Given: nFeCl = 0.55 mol
3
Required: mass of Fe2S3, mFe S
2 3
Step 1. Determine the amount of Fe2S3.
1 mol Fe S
2 3
nFe S = 0.55 mol FeCl !
2 3
3
2 mol FeCl
3
nFe S = 0.275 mol Fe S
2 3
2 3
Step 2. Determine the mass of Fe2S3.
207.91 g
mFe S = 0.275 mol !
2 3
1 mol
mFe S = 57 g
2 3
Statement: The mass of Fe2S3 produced in this reaction is 57 g.
89. The ionization of a strong acid and a weak acid are shown below.
90. (a) The solution with pH 2 is more acidic than a solution with pH 5.
(b) The solution with pH 2 is 1000 times (10 × 10 × 10) more acidic than the solution with pH 5.
(c) The solution with pH 2 has a higher concentration of H+ ions.
91. (a) HCl(aq) → H+(aq) + Cl‒(aq)
(b) The concentration of hydrogen ions in the original solution is the same as the concentration
of hydrochloric acid, which is 1.0 mol/L
(c) Given: cc = 1.0 mol/L
Vc = 1.0 mL
Vd = 1.0 L
Required: concentration of diluted solution, cd
Analysis: ccVc = cdVd
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Unit 4: Solutions and Solubility
U4-15
Solution:
Step 1. Convert the volume to litres.
1L
Vc = 1.0 mL !
1000 mL
Vc = 0.001 L
Step 2. Rearrange the equation in the appropriate form, substitute the values, and solve.
cV
cd = c c
Vd
!
mol $
#" 1.0 L &% 0.001 L
cd =
1.0 L
mol
cd = 0.001
L
Statement: The concentration of H+ ions in the diluted solution is 0.001 mol/L.
(d) The diluted solution is 1000 times less acidic than the original sample.
(e) The acidity decreased a thousand-fold and the concentration of hydrogen ions decreased by
the same factor (1.0 × 103). It appears that acidity is directly related to the concentration of
hydrogen ions.
92. (a) A title of the graph could be “Titration Curve for a Strong Base Titrated with a Strong
Acid.” The x-axis label could be “Volume of acid (mL)” and the y-axis label could be “pH.”
(b) The line starts high on the y-axis because there is a large excess of base in the flask and the
pH is high. The pH decreases slowly as the titration approaches the equivalence point. At the
equivalence point, the pH drops quickly. The base is neutralized, but because more acid is added
past the equivalence point, the pH continues to decrease slowly.
93. (a) Given: volumes of potassium hydroxide solution, VKOH = 15.40 mL
(
)
VKOH = 15.36 mL
VKOH = 15.44 mL
Required: average volume of potassium hydroxide used, VKOH(average)
Solution:
Step 1. Calculate the average volume of titrant used.
15.40 mL + 15.36 mL + 15.44 mL
VKOH(average) =
3
VKOH(average) = 15.40 mL
Step 2. Convert the volume to litres.
1L
VKOH(average) = 15.40 mL !
1000 mL
VKOH(average) = 0.015 40 L
Statement: The average volume of potassium hydroxide used is 0.01540 L.
(b) H2SO4(aq) + 2 KOH(aq) → K2SO4(aq) + 2 H2O(l)
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Unit 4: Solutions and Solubility
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(c) Given: cKOH = 0.20 mol/L
VKOH(average) = 0.015 40 L
Required: amount of potassium hydroxide, nKOH
n
V
Analysis: c =
Solution:
Step 1. Use the concentration equation to determine the amount of acid.
nKOH = cKOHVKOH(average)
0.20 mol
! 0.015 40 L
1L
nKOH = 0.003 08 mol
Statement: The amount of potassium hydroxide used to neutralize the acid is 0.003 08 mol.
(d) Given: amount of potassium hydroxide, nKOH = 0.003 08 mol
=
Required: amount of sulfuric acid, nH SO
2
4
Solution:
Step 1. Use the amount of base and the mole ratio in the balanced equation to determine the
amount of acid.
1 mol H SO
2
4
nH SO = 0.003 08 mol KOH !
2
4
2 mol KOH
nH SO = 1.54 ! 10 –3 mol H SO
2
4
2
4
Statement: The amount of sulfuric acid is 1.54 × 10–3 mol.
(e) Given: nH SO = 1.54 ! 10 –3 mol
2
4
VH SO = 0.0100 L
2
4
Required: concentration of sulfuric acid, nH SO
2
4
Solution:
Step 1. Use the concentration equation to determine the amount of acid.
nH SO
cH SO = 2 4
2
4
VH SO
2
4
1.54 ! 10 –3 mol
0.0100 L
= 0.154 mol/L
=
cH SO
2
4
Statement: The concentration of sulfuric acid is 1.54 × 10–1 mol/L.
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Unit 4: Solutions and Solubility
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94. Given: mKHP = 0.600 g
VNaOH = 16.50 mL
Required: concentration of sodium hydroxide solution, cNaOH
Analysis: c =
n
V
Solution:
Step 1. Convert each volume to litres.
1L
VNaOH = 16.50 mL !
1000 mL
VNaOH = 0.016 50 L
Step 2. Convert the mass of KHP to the amount.
1 mol
nKHP = 0.600 g !
204.20 g
nKHP = 2.94 ! 10 "3 mol
Step 3. Write the balanced equation.
KHP(aq) + NaOH(aq) → H2O(l) + KNaP(aq)
Step 4. Use the amount and the mole ratio in the balanced equation to determine the amount of
the sodium hydroxide.
1 mol NaOH
nNaOH = 2.94 ! 10 "3 mol KHP !
1 mol KHP
nNaOH = 2.94 ! 10 "3 mol NaOH
Step 5. Use the concentration equation to determine the concentration of sodium hydroxide.
n
cNaOH = NaOH
VNaOH
2.94 ! 10 "3 mol
0.016 50 L
mol
cNaOH = 0.178
L
Statement: The concentration of the sodium hydroxide solution is 0.178 mol/L.
=
Evaluation
95. Water’s unique properties make it ideal for the development of life on Earth. Water’s polar
structure and strong hydrogen bonding result in high melting and boiling points, allowing living
things to retain liquids over a wide range of temperatures. Water is able to dissolve a very wide
range of substances, facilitating chemical reactions. Water’s high specific heat capacity
moderates temperatures to prevent temperature extremes from threatening life forms. Since ice is
less dense than liquid water, organisms are able to live in liquid water below ice.
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96. (a) Answers may vary. Sample answer: The global demand for fresh water has increased by a
factor of six, which is twice that of global population growth. This increase indicates that as the
population increases, supply will be less and less able to keep up with demand.
(b) Answers may vary. Sample answer: Consumer behaviour can have a limited impact on water
use. Individuals now account for about 10 % of water use. So even if individuals were to cut
their use by 50 %, overall water consumption would decrease by only 5 %.
(c) Answers may vary. Sample answer: Improvements in the industrial and agricultural use of
water need to be made for the world to meet its demand for water. Many farms are now irrigated
from aquifers that are rapidly being emptied. If this activity continues, the aquifers that supply
large parts of the world will be gone, threatening basic water supplies for millions of people.
(d) Answers may vary. Sample answer: Agricultural practices affect water quality because the
chemical fertilizers run off farm fields and flow through watersheds into major rivers. The
fertilizers eventually end up at the mouth of the river. These runoffs have produced an excess of
nutrients, which leads to massive growth of algae. When the algae die, decomposition processes
deplete oxygen from the water and produce a dead zone. In my opinion, farmers will need to
start raising crops without relying so much on chemical fertilizers.
97. After 15 min, bottle A (the colder bottle) is likely to have more fizz. Gases are more soluble
at lower temperatures than at higher temperatures. Therefore, the carbon dioxide in bottle B (the
warmer bottle) will escape faster, leaving it flat. Bottle A will retain its carbon dioxide in
solution, so it will have more fizz.
98. Given: cc = 1.00 mol/L
Vc = 1.00 L
Vd = 5.00 L
Required: concentration of diluted solution, cd
Analysis: ccVc = cdVd
Solution:
Step 1. Rearrange the equation in the appropriate form, substitute the values, and solve.
cV
cd = c c
Vd
!
mol $
#" 1.00 L &% 1.00 L
cd =
5.00 L
mol
cd = 0.200
L
Statement: The concentration of the solution is 0.200 mol/L. Therefore, the student with the
0.25 mol/L solution has the more concentrated solution.
99. Answers may vary. Sample answer: I think most students chose the net ionic equation as the
equation that best shows the reaction. The net ionic equation shows parts of the reaction in which
substances actually change and leaves out the spectator ions.
(
)
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Unit 4: Solutions and Solubility
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100. Student A is correct because, if lead ions were present, the precipitate lead chloride,
PbCl2(s), would form when the chloride ions were added. The mistake Student B made was that
he assumed that the sulfate precipitate was lead(II) sulfate. Cations such as Ca2+ and Ba2+ form a
precipitate with the sulfate ion, but not with the chloride ion. Any of these cations could be in the
solution instead of lead. Students B’s test does not conclusively determine that the precipitate
was caused by lead(II) ions.
101. (a) Concept maps may vary. Sample concept map is shown below.
(b) The formula for sulfurous acid is H2SO3(aq).
102. To test for the presence of calcium carbonate, the lab assistant is correct to treat the sample
with acid. Carbonates react with acids but not bases. A positive test would result in the
production of bubbles of carbon dioxide gas.
103. (a) To test to see if ions formed, you can pass an electric current through the solution. If it
conducts a current, it contains ions.
(b) Answers may vary. Sample answer: To determine whether dissociation or ionization
occurred, you can test the properties of the undissolved compound. If it has ionic properties (e.g.,
solid, crystal structure, high melting temperature), it will dissociate. If it has non-ionic properties
(e.g., gas or liquid, relatively low melting point), it will ionize. Dissociation occurs when water
molecules pull the ions of a soluble ionic compound apart. Ionization involves the formation of
ions from uncharged molecules.
104. (a) Ammonia is sometimes classified as a base because it has basic properties when
dissolved in water. It has a high pH and can neutralize an acid.
(b) Ammonia is different from other bases because it does not have a hydroxide in its chemical
formula.
(c) We can conclude that substances do not need to contain a hydroxide group to behave as
bases.
105. Methyl red, bromothymol blue, and phenolphthalein are good all candidates for indicators
for acid–base titrations. Bromothymol blue is usually the best choice because its colour change
occurs right at the pH 7.0 in the centre of the steep part of the pH curve.
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Reflect on Your Learning
106. Answers may vary. Sample answer: Detergents are very long molecules that have a
positively charged head and a long, non-polar tail. A new detergent molecule could have a longer
tail and a more polar head. The larger tail would be more strongly attracted to grease. At the
same time, the more polar head would cling more tightly to water. This combination might
produce a stronger tugging force on oil and grease, resulting in more effective cleaning.
107. Copper(I) chloride, CuCl, is slightly soluble in water because the copper and chloride ions
are more tightly attracted to one another. Copper(I) sulfate, Cu2SO4 is highly soluble, which
means that it completely dissociates in solution. The diagram is shown below.
108. Answers may vary. Sample answer: Efficient water conservation can help Canada solve its
energy problems by reducing the amount of energy needed for the water treatment process.
Treating water to ensure high-quality tap water is a process that requires a great deal of energy. If
citizens used water more efficiently, they will require less water, which means that less energy
will be needed to provide water. This will result in overall energy savings, and a reduction in the
carbon footprint and greenhouse gases.
109. Answers may vary. Sample answer: The strategy I used when planning a qualitative
analysis to determine the presence of two cations is to first select a solution that contains an
anion that produces a soluble compound with one of the cations, but not with both. If a
precipitate forms, I know that the cation that produces a slightly soluble compound is present in
the solution. I add an excess of this first reagent to make sure that all of the cation precipitates
out. I filter out the precipitate. Next, I add a second reagent that forms a slightly soluble
compound with the remaining unprecipitated cation. If I see a precipitate, I know that the second
cation is present.
110. (a) Assuming that every car requires four tires, the number of cars would be the limiting
reagent. There are not enough cars to use up all 30 tires. Since seven cars require a total of
28 tires, two tires would be in excess.
(b) Answers may vary. Sample answer: The tires would be the limiting reagent if you had
10 cars and 36 tires. There would be enough tires for only nine cars so that one car would be in
excess.
(c) Answers may vary. Sample answer: The analogy of cars and tires is a good analogy to the
chemistry of limiting reagents. The analogy shows that whichever item (cars or tires) is in short
supply, that item will be the limiting reagent.
111. (a) Answers may vary. Sample answer: Starting the formula for strong acids with an “H”
and putting the “H” at the end of the formula of a weak acid would allow people to tell at a
glance whether an acid is strong and completely ionizes, or an acid is weak and only partially
ionizes. Being able to tell the strength of an acid at a glance would be useful. You would not
need to consult a reference guide to learn whether the acid was strong or weak.
(b) Answers may vary. Sample answer: The disadvantage in changing the formulas of acids is
that it would be confusing to generations of people who have grown up using the old system. Just
by looking at the first letter of the formula, people would not be able to identify an acid.
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Unit 4: Solutions and Solubility
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(c) Answers may vary. Sample answer: Yes, I think overall this would be a good change. There
would be confusion in the beginning, much as there has been with the change from other
traditional conventions to modern conventions. People would gradually adjust to the change.
112. (a) Acid rain in the area has acidified the lakes and ponds. Ionization of acids in the water
causes the lakes and ponds to be much more conductive than pure water, causing any lightning
strikes to be more deadly than usual.
(b) Answers may vary. Sample answer: To test this explanation, I would test the pH and
conductivity of the lakes and ponds to see if the water is at a lower pH and to see if water at a
lower pH conducts electricity significantly better than water at pH 7.
113. Answers may vary. Sample answer: A sample procedure for performing a titration is as
follows.
Step 1. Rinse the burette with water and then with the titrant solution several times.
Step 2. Prepare the burette with the titrant solution.
Step 3. Pipette the required volume of unknown solution into a clean flask.
Step 4. Add an appropriate indicator to the flask. Place a sheet of white paper under the flask to
make colour changes more visible.
Step 5. Record the initial burette volume.
Step 6. Add titrant to the flask quickly at first. As you near the endpoint, add the titrant drop by
drop, swirling the flask after every drop.
Step 7. Use a wash bottle to rinse the sides of the flask to make sure that all titrant has reacted.
Step 8. Stop titrating when you see a permanent colour change. This is the endpoint. Record the
final burette volume.
Step 9. Repeat steps 2 to 8 until you have at least three trials with results that are similar.
Research
114. (a) Answers may vary. Sample answer: There are hydrogen bonds between DNA base pairs
that are essential for determining the structure of DNA. There are four bases: adenine (A),
thymine (T), cytosine (C), and guanine (G). Base A always pairs with T, while C pairs with G.
The base pairs are held together by hydrogen bonds that result from interactions between polar
molecules. This structure allows the DNA molecule to zip and unzip itself, making
complementary copies of its code. Note that if the bases were linked by covalent bonds, it would
require much more energy to break the bonds to zip and unzip the DNA molecule.
(b) Answers may vary. Sample answer: The identification of the base pairing led to the cracking
of the genetic code. Watson and Crick knew that A and T and C and G existed in equal molar
quantities in DNA. What they did not know was how these bases fit together in the DNA
molecule. The two scientists were looking for covalent connections between bases when Watson
realized that it was hydrogen bonds that connected A to T and C to G. Once that fact was
known, the mechanism of DNA was quickly determined. Rather than requiring chemical
reactions to zip and unzip the molecule for copying and expressing, breaking and forming
hydrogen bonds was a much simpler undertaking.
115. (a) Answers may vary. Sample answer: The two most urgent concerns in global water use is
drinking water and proper sanitation. About 1.1 billion people do not have access to clean
drinking water on a daily basis. About 2.6 billion people, which is half of the world’s population,
lack proper sanitation facilities, so the water they do use is threatened. This results in the death of
2 million people a year from diseases caused by drinking contaminated water. Every day,
3900 children die of water-borne diseases.
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Unit 4: Solutions and Solubility
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(b) Answers may vary. Sample answer: The situation in India, China, and neighbouring
Southeast Asian nations is typical of water problems. Many of these nations share common
watersheds, and there is simply not enough water to go around. As industrialization and
population bases increase, demand for water increases even more. Most of the water consumed
in this region is required for agriculture of one type or another. In India, 90 % of water demand is
for farming. In Vietnam and Thailand, much of the water is used for inland fisheries. As water
supplies diminish, countries accuse one another of misuse and other abuses. China, for example,
during a recent drought was accused of hoarding water. Overall, these water tensions could
ultimately result in political conflicts and even warfare if better efficiency in water use is not
achieved.
116. (a) Answers may vary. Sample answer: Extracting soybean oil, for example, traditionally
requires dissolving the beans in a non-polar solvent, hexane, then distilling, or boiling off the
hexane, to recover the pure soybean oil. The distillation process requires a great deal of energy
and time.
(b) Answers may vary. Sample answer: Switchable solvents allow a single solvent to be used and
reused without distillation. The solvent first dissolves the soybean oil in its non-polar stage. Once
all oil is dissolved, carbon dioxide is bubbled through the system, and the solvent changes into a
polar form. Being polar, it separates from the oil and can easily be recovered and reused without
any heating or distillation. This would likely result in considerable energy savings.
(c) Answers may vary. Sample diagram is shown below.
117. (a) Answers may vary. Sample answer: GreenCentre Canada is a chemical company that is
devoted to using chemistry to solve commercial and environmental problems. The philosophy of
GreenCentre Canada is summed up by restating several of the company’s Twelve Principles:
• It is better to prevent waste than to treat or clean up waste after it is formed.
• Wherever practicable, processes should use and generate substances that possess little or no
toxicity to human health and the environment.
• Chemical products should be designed to be effective but less toxic.
• Chemical products should be designed so that at the end of their function they break down into
harmless products and do not persist in the environment.
(b) Answers may vary. Sample answer: GreenCentre Canada is focussing on green products that
are biodegradable, non-toxic, and non-persistent in the environment. These include surfactants
and switchable solvents.
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118. (a) Answers may vary. Sample answer: In Kingston, before the 1950s, sewage flowed
directly from homes and businesses into the harbour and nearby rivers. In the 1950s, a trunk
sewer was built to bring sewage to the Ravensview sewage treatment plant. This system worked
well in normal situations. However, during major rainstorms and other disturbances, the sewers
tended to overflow and dump raw sewage into such places as Richardson Beach, making them
unsafe to swim or draw water from. In the early 1990s, a plan was developed that outline how
the sewer infrastructure was to be improved and the Ravensview sewage treatment plant was to
be updated to prevent overflows even in the most severe storms.
(b) Answers may vary. Sample answer: Kingston’s current water treatment system functions
well. Kingston’s Ravensview plant has been called a “state of the art” facility. During a recent
upgrade, one of the largest biological aeration equipment systems in the world was installed.
This upgrade will should allow Ravensview to become one of Canada’s leaders in reusing
wastewater as irrigation water.
(c) Answers may vary. Sample answer: In Kingston, although great improvements have already
been made, the plan for improved infrastructure has still not been fully implemented because of
high costs and conflicting goals. Currently, efforts are being made to complete the plan as well as
to ensure that Kingston is in compliance with the Ontario Ministry of Environment’s Directive
F-5-5, which calls for strict standards for water quality across all Canadian communities.
119. (a) Answers may vary. Sample answer: The solitary proton, H+, does not exist in nature.
When hydrogen ions form in solution, they quickly become hydrated by nearby water molecules
and form hydronium ions, H3O+. Hydronium ions are identical in behavior to H+ ions. Some
scientists insist that the best representation for a hydrogen ion would be as a complex with a
water molecule: [H:H2O]+. This complex could include more than one water molecule
surrounding the proton.
(b) Answers may vary. Sample answer: The advantage to using the H+ representation is that it is
traditional and easy to understand. The generic acid equation, HA → H+ + A– using the H+ form
of the ion is logical and easy to use. The hydronium ion, H3O+, is a more accurate way of
representing the actual ion. It is also more useful when actually considering the structural
mechanisms of reactions that take place with acids and bases. However, the hydronium ion is not
used so often because it is more clumsy to write.
(c) Answers may vary. Sample answer: No, I think the H+ form of the ion works well because it
is easier to understand and is less confusing than the hydronium ion.
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Unit 4: Solutions and Solubility
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120. Answers may vary. Sample answer: In paper chromatography, a strip of absorbent paper is
extended into a solvent as shown in the following diagram. A spot of ink or concentrated solute
is applied to the paper as a dot. Then the solvent is allowed to travel up the paper by capillary
action. The result is that the various components of the solute travel up the paper with the solute
at different speeds that are based on the polarity of each component. The speed and colour of
each trace on the paper effectively separates the components of the solute. Components can also
be identified using this technique by comparing traces to known samples.
121. (a) Answers may vary. Sample answer: The main sources of drug contamination are
incomplete metabolism of drugs and improper disposal of unused medications. Frequently, drugs
that people take are not completely metabolized by the body. Drugs unused by the body are
eliminated from the body in urine and released into the environment through sewage. Outdated
drugs can also cause contamination when they are disposed of improperly and end up in landfills
or water supplies.
(b) Answers may vary. Sample answer: Many scientists are not concerned about drugs in water
supplies because the concentrations of these medications are very low. Also, few or no negative
effects of these drugs have been observed.
(c) Answers may vary. Sample answer: Drug manufacturers can adjust doses of medications so
that all of the drugs taken are metabolized by the body, and none of the drug is eliminated
through urine or feces. Communities or drug companies can provide ways in which people can
properly dispose of unused drugs rather than just throwing them away.
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Unit 4: Solutions and Solubility
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