Unit 4: Solutions and Solubility
Chapter 8: Water and Solutions
Section 8.1: The Importance of Water, pages 130–132
1. Answers may vary. Sample answer: Without water, the human body would not be able to transport essential
materials, remove waste from the body, or regulate body temperature.
2. Water contains dipole-dipole intermolecular force that is created by an asymmetrical structure, positively
charged hydrogen atoms, and a negatively charged oxygen atom. This dipole-dipole moment makes it possible
for water molecules to have strong attractions to one another. Forces between water molecules are difficult to
overcome, giving water its strong intermolecular bonds.
3. Table 1 Significance of Hydrogen Bond Benefits
Benefit
Significance
store fats and oils
biological
make ice float
physical
maintain body fluids
biological
4. A hydrogen bond is formed between an electronegative oxygen atom in one water molecule and a hydrogen
atom in another molecule.
5. a
6. The strength of a hydrogen bond increases as the size or polarity of the molecule increases.
7.
8. (a) ii; (b) iv; (c) iii; (d) i
9. Much of the remainder of the water in Earth’s atmosphere comes from the process of transpiration, which is
the evaporation of water residing in tiny pores found in plant leaves.
10. Concept maps may vary. Answers should include the following information: Process 1: Evaporation and
condensation. Water evaporates, leaving behind solutes. When it condenses, it is pure. Process 2: Filtration.
Water passing through sand and gravel has suspended matter removed. Process 3: Bacterial action. Soil
bacteria decompose harmful agents in water.
11. (a) Answers may vary. Sample answer: Aquifers are rocks that have cracks and holes filled up with water
while allowing groundwater to flow through.
(b) Answers may vary. Sample answer: You would expect a small rural area to use aquifers. Many rural
communities get their drinking water from wells that are drilled directly into aquifers.
12. (a) Potable water is important to us because we must have water to survive. Potable water is water that has
gone through the purification process and is safe for human consumption.
(b) Answers may vary. Sample answer: Health is the major risk associated with consuming unpurified water.
Dirty water often contains bacteria and other organisms that can lead to diseases and even death, especially in
children.
13. It takes the longest time to replace water in polar ice caps and glaciers. It takes the least amount of time to
replace water in rivers and the atmosphere.
14. a
15. Flowcharts may vary. Answers should include the following information: Population growth: More people
will be sharing finite water resources. Surface water pollution: Water supplies are contaminated by disposal of
chemicals, mining activities, damming rivers for hydroelectric plants, excess use of fertilizers, etc. Increasing
global demand for water: Water is used by individuals, by industry, and by the agricultural sector.
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Chapter 8: Water and Solutions
8-1
Section 8.2: Solutions and Their Characteristics, pages 133–134
1. a
2. If the unknown solution is not transparent, you can determine that it is a heterogeneous mixture.
3. Table 1 Types of Mixtures
Compound
Type of Mixture
concrete
heterogeneous
sugar water
homogeneous
sand in water
heterogeneous
vinegar
homogeneous
4. The solvent is the component of the solution that is present in the greatest quantity.
5. (a) False. The most important characteristic of solutions is that their composition can change.
(b) True
6. Table 2 Solutes and Solvents
Solution
Solute
Solvent
salt water
salt
water
carbonated beverage
carbon dioxide water
(carbon dioxide and water)
sugar water
sugar
water
rubbing alcohol
water
isopropyl alcohol
(60 % isopropyl alcohol)
7. Concentration is the relationship between the quantity of solute versus the overall quantity of solution or
solvent.
8. Yes, metals can form solutions. An alloy is a solution that is composed of two or more metals.
Section 8.3: The Dissolving Process, pages 135–136
1. b
2. NaOH(s) → Na+(aq) + OH–(aq)
3. (a) False. During dissociation, ions in compound separate and become individual ions.
(b) False. Liquids that completely mix with each other when they come in contact are miscible.
4. Water molecules contain polar bonds. If they come in contact with a compound with polar bonds, the polar
molecules attract and they combine to form an aqueous solution.
5. Hydrocarbons are immiscible in water because their molecules are non-polar. As a result, there are almost
no attractive forces between hydrocarbon molecules and water molecules, so they do not mix.
6. Table 1 Behaviour of Solutes and Solvents
Compound or Solute
Solvent
Dissolve (yes or no)
ionic compound
polar solvent
yes
non-polar solute
polar solvent
no
non-polar solute
non-polar solvent
yes
7. Two of the most well known surfactants are soaps and detergents.
8. Soaps are made by reacting animal fats or vegetable oils with a concentrated base.
9. Surfactants act as a cleaner or detergent and break down the hydrogen bonds present in surface water, which
allows the polar compound, water, to mix with the non-polar compound, oil.
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Chapter 8: Water and Solutions
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Section 8.4: Explore an Issue in Solutions: Oil Dispersants—Is the Fix Worse than
the Problem?, page 137
1. During an oil spill, when the oil that has spilled settles, it floats on the surface of the water, thereby making
it easier to see and begin the cleanup.
2. Table 1 Benefits and Risks of Offshore Oil Drilling
Benefits
Risks
Canada and much of the world depend greatly on Oil rigs can be dangerous.
the natural resources of oil, so the continuation of
drilling now and in the future helps ensure that
we have a steady reserve.
Oil companies hire a lot of workers to help keep
Oil spills can leave a lasting negative
the oil rig going.
environmental impact.
3. Dispersants “clean up” oil spills by breaking the oil into smaller pieces that sink to the bottom of the ocean.
This makes it easier for the micro-organisms to break down the oil into less toxic substances.
4. Answers may vary. Sample answers: Have an emergency plan in place to put into action in the event of an
oil spill. Have dispersants available that can be quickly used in the event of an oil spill. Develop technology
that can help add additional protection from an oil spill. Have detailed maintenance plans.
Section 8.5: Solubility and Saturation, pages 138–139
1. c
2. A supersaturated solution is a solution that has been forced to dissolve more solute.
3. False. Supersaturated solutions are unstable.
4. (a) 20 g of salt would dissolve in 100 g of water at 20 °C.
(b) Based on the four data points, you can conclude that the solutions are saturated.
(c) If you add 10 g of salt to 100 g of water at about 30 °C, it would be an unsaturated solution.
5. Answers may vary. Sample answer: Pressure equals the force applied per unit area.
6. You can use a solubility curve to measure the vertical difference between a point below the curve and a
point on the curve for a particular temperature. The difference is the additional grams of solute needed to make
a saturated solution.
7. Cool water contains more dissolved oxygen than warm water.
8. As soon as you open the cap of the bottled pop, some of the carbon dioxide, which is responsible for the
fizz, escapes. Even though you put the cap back on, since the initial pressure has been released, the carbon
dioxide slowly seeps out until eventually your pop becomes flat.
9. In most cases, the solubility of a solid increases with the rise of temperature. For a gas, as the temperature
rises, the solubility decreases.
10. The solubility in solids and liquids are not affected greatly by changes in pressure because compounds in
these states are not compressible. Pressure does affect gases because they can be easily compressed. The
solubility of gas confined in a liquid increases as the pressure increases.
11. Answers may vary. Sample answer: The power plant is probably creating thermal pollution. The warm
water discharged into the lake is causing the water to heat up, which decreases oxygen levels in the water.
Aquatic wildlife begins to die from a lack of oxygen.
12. Carbonated beverages are bottled under high pressure in order to house a large amount of carbon dioxide
inside. When you pour an unrefrigerated beverage onto ice, the coldness of the ice causes a lot of the carbon
dioxide to dissolve rapidly. This creates the fizz that bubbles over.
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Chapter 8: Water and Solutions
8-3
Section 8.6: Concentration, pages 140–141
1. b
2. Your lab partner probably added water directly to the acid in the beaker. This produces an immense amount
of thermal energy between the acid and water, which can cause the mixture to boil and shatter glass. To
prevent this from happening again, your lab partner should carefully add the acid to the water.
3. A concentrated solution contains a large quantity of dissolved solute per unit volume of solution. A dilute
solution contains a relatively small quantity of solute per unit volume of solution.
4. To determine the amount of solute, use the equation c =
n
V
and solve for the variable n.
5. (i) Make sure the equation is rearranged to solve for the needed variable.
(ii) The amount should be expressed in moles (mol).
(iii) The volume should be expressed in litres (L).
6. b
7. Given: VNaOH = 50 L, mNa = 40 g
Required: cNaOH
Analysis: To determine the amount concentration of the NaOH solution, use the amount concentration
equation cNaOH =
nNaOH
VNaOH
.
Solution:
Step 1: Convert mass of sodium to amount of sodium.
nNa = 40 g !
1 mol Na
22.99 g
nNa = 1.740 mol (two extr digits carried)
Step 2: Convert amount of sodium to amount of NaOH.
Since 1 mol of NaOH contains 1 mol of Na.
nNaOH = nNa = 1.740 mol (two extr digits carried)
Step 3: Substitute the values in the equation.
cNaOH =
=
nNaOH
VNaOH
1.740 mol
50 L
cNaOH = 0.035 mol/L
Statement: The amount concentration of the solution is 0.035 mol/L.
Section 8.7: Preparing Dilutions, pages 142–143
1. b
2. Table 1 Types of Volumetric Glassware
Volumetric
When you might use it
Precision
Glassware
volumetric pipette
to deliver a fixed volume of solution
more precise than the graduated pipette
graduated pipette
to deliver a range of volumes of
not as precise as the volumetric pipette
solution
graduated cylinder
to measure and transfer larger
not precise enough for analytical work
volumes of solution
involving small volumes
volumetric flask
to prepare dilute solutions of larger
similar in precision to the graduated
volumes
cylinder
3. False. During a dilution, the total amount of moles in the solution remains the same.
4. Yes, the concentration has changed because the volume of the solution has increased.
5. c
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Chapter 8: Water and Solutions
8-4
6. Given: cc = 10.0 mol/L; cd = 0.250 mol/L; Vd = 400 mL
Required: volume of initial concentrated solution Vc
Analysis: ccVc = cd Vd
Solution: Rearrange the equation in the appropriate form, substitute the values, and solve.
Vc =
cdVd
cc
!
mol "
## 0.250
$ (400 mL)
L $&
%
=
mol
10.0
L
Vc = 10.0 mL
Statement: In order to make 400 mL of 0.250 mol/L solution, 10.0 mL of 10.0 mol/L hydrochloric acid is
needed.
Section 8.8: Concentrations and Consumer Products, pages 144–145
1. c =
quantity of solute
quantity of solution
2. Given: Vsolute = 30 mL; Vsolution = 100 mL
Required: cv/v
Analysis: cv/v =
Vsolute
Vsolution
! 100 %
Solution:
cv/v =
=
cv/v
Vsolute
Vsolution
! 100 %
30 mL
100 mL
= 30 %
! 100 %
Statement: The percentage volume/volume of the solution is 30 %.
3. The density of water is 1 g/mL at 20 °C.
4. Since the density of very dilute aqueous solutions is very similar to pure water, the density is assumed to be
1 g/mL at 20 °C.
5. a
6. Given: mglucose = 200 g; cglucose = 2.0 % W/V, or 2.0 g/100 mL
Required: Vsolution
Analysis: cglucose =
msolute
Vsolution
Solution:
Step 1: Rearrange the equation in the appropriate form, substitute the values, and solve.
Vsolution =
=
msolute
cglucose
200 g
2.0 g
100 mL
Vsolution = 10 000 mL
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Chapter 8: Water and Solutions
8-5
Step 2: Convert the volume of solution to litres.
Vsolution = 100 00 mL !
1L
1000 mL
Vsolution = 10 L
Statement: 200 g of glucose can make 10 L of a 2.0 % W/V glucose solution.
7. Weight is the force of gravity exerted on an object.
8. The weight tells you the mass of the solute (hydrocortisone).
9. Given: cc = 45 % W/V; Vc = 300 mL; Vd = 4.000 L
Required: concentration of the diluted solution, cd
Analysis: ccVc = cdVd
Solution:
Step 1: Convert the volume of solution to litres.
Vc = 300 mL !
1L
1000 mL
Vc = 0.300 L
Step 2: Rearrange the equation in the appropriate form, substitute the values, and solve.
cd =
=
ccVc
Vd
(45 % W/V )(0.300 L)
4.000 L
cd = 3.4 % W/V
Statement: The final concentration of the diluted solution is 3.4 % W/V.
10. Answers may vary. Sample answer: Scientists express very dilute concentrations in parts per million
(ppm, 1:106), parts per billion (ppb, 1:109), or parts per trillion (ppt, 1:1012).
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Chapter 8: Water and Solutions
8-6
Chapter 8 Summary, page 146
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Chapter 8: Water and Solutions
8-7
Chapter 8 Questions, pages 147–148
1. b
2. (a) False. Intermolecular forces increase as the size or polarity of the molecule increases.
(b) False. A solution that contains a high quantity of solute compared to the total volume of solution is a
concentrated solution.
(c) False. Hydration is the process that occurs when ions are surrounded by water molecules.
3. Dry air is a homogeneous mixture because it is made up of nitrogen, oxygen, and other gases. It is also
transparent. When light passes through the air in your classroom, you can see dust particles in the air. This air
is a heterogeneous mixture.
4. Answers may vary. Sample answers: Properties of water that are the result of hydrogen bonding include
high melting and boiling points; volume expansion when cooling from 4 °C to 0 °C; high surface tension;
ability to exchange thermal energy with little change in temperature; inability to mix with non-polar
compounds.
5. An aqueous solution is a solution in which water is the solvent.
6. Table 1 Factors Affecting Solubility of Liquids, Gases, and Solids
Factor
Liquids
Gases
Solids
increase in temperature
increase
decrease
increase
increase in pressure
no change
increase
no change
7. (a) At 20 °C, the maximum amount of glucose that will dissolve in water is 20 g/100 g H2O.
(b) As the temperature rises, the solubility of glucose increases.
(c) 60 g glucose/100 g H2O at 40 °C is a supersaturated solution.
(d) 15 g (20 g – 5 g) of additional glucose would be needed to saturate 5 g of glucose at 20 °C.
8. Answers may vary. Sample answer:
Figure 2
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Chapter 8: Water and Solutions
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