CHAPTER 7
SOLUBILITY AND REACTIONS
Reflect on your Learning
(Page 312)
1. Most of the liquids that we encounter in the home are solutions. This could be tested by cooling, evaporating, or
smelling — anything that would encourage the components of a solution to separate.
2. Dissolving rate is usually speeded up by stirring, warming the solution, or finely dividing (grinding) the solute. If
excess solid remains after prolonged stirring, probably no more solute will dissolve.
3. Evidence for a reaction might be colour change, formation of a precipitate, formation of a gas, or a noticeable temperature change. The dissolving of both salt and sugar involves the solid separating into particles too small to see. This
is a physical change. Like most physical changes, it is reversible: the water can be removed, leaving the solute in its
original state.
Try This Activity: Measuring the Dissolving Process
(Page 313)
(a) The slightly cloudy mixture of table salt in water indicates that some substance(s) did not dissolve. Because we know
that sodium chloride is soluble, table salt must be a mixture containing at least one low-solubility solid. The solution
of pickling salt was completely clear.
(b) The ingredients are salt, calcium silicate, potassium iodide, sodium thiosulfate (for Sifto brand). Calcium silicate is
likely the cause of the cloudiness because silicates (such as sand) are not soluble in water.
(c) About 5 to 6 teaspoons of pickling salt appeared to dissolve.
(d) The final volume is a little more than 100 mL, based on the approximate markings on the side of the Erlenmeyer flask.
(e) The volume would likely be around 100 mL but not 120 mL, because solids generally dissolve in water without
increasing the volume. This can be tested by measuring 20.0 mL of sodium chloride in a graduated cylinder and
adding this to 100.0 mL of water measured in another graduated cylinder.
Note: The answer is about 106 mL.
7.1 SOLUBILITY
PRACTICE
(Page 316)
Understanding Concepts
1. (a) About 20 g of K2SO4(s) will dissolve in 100 mL of water at 70°C.
(b) KNO3(s) and KCl(s) have equal solubilities at about 22°C.
(c) To calculate molar concentration we would need to know the volume of the solution. The graph only gives the
volume of water used.
(d) The only substance shown for which 100 g will dissolve in 100 mL of water is KNO3(s), at about 56°C.
2. (a) NaCl would precipitate first at temperatures above 31°C.
(b) KCl would precipitate first at temperatures below 31°C.
Applying Inquiry Skills
3. The graph created from Investigation 7.1.1 will normally differ somewhat from the one in Figure 2. The difference is
primarily experimental error, due to a less than precise design for the Investigation.
Try This Activity: Gas Solubility
(Page 317)
Some suggested answers are given.
(a) The lit match is extinguished. This suggests the presence of carbon dioxide, which does not support combustion.
(b) (Observations will vary depending on the relative humidity.) The air near the outside of the cold glass cools. As the
temperature of the air decreases, so does the solubility of water vapour in the air. Some water vapour condenses to
liquid water. A qualitative test for water is to use cobalt(II) chloride paper.
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Chapter 7 Solubility and Reactions
197
(c) The small gas bubbles on the inside wall of the glass are likely due to air coming out of solution as the temperature
of the water increases to room temperature. As the temperature increases in the cold water glass, the volume of gas
also increases. In the hot water, the temperature is decreasing so the bubbles would stay small or decrease in volume.
PRACTICE
(Page 318)
Understanding Concepts
4. A room temperature can of pop is more likely to spray: as gases are less soluble in warm liquids than in cold, the gases
will tend to come out of solution, so the pressure in the can is higher.
Reflecting
5. (a) Red blood cells contain hemoglobin molecules, that bind chemically to oxygen molecules, making the oxygen
appear more soluble.
(b) If blood held the same amount of oxygen as water, there would be only enough oxygen in our blood to keep our
cells alive for a few seconds, i.e., our lives would be different in the event of such a change by becoming dramatically shorter. To compensate, we would require completely different circulatory systems, for example, one that
supplied a much greater volume of blood to our organs.
PRACTICE
(Page 319)
Understanding Concepts
6. Temperature must always be stated when reporting a solubility.
7.
Solubility
Gases
Solids
Temperature
8. (a) CO2(g) H2O(l) → H2CO3(aq)
(b) H2CO3(aq) Ca(OH)2(aq) → CaCO3(s) 2 H2O(l)
(c) The second reaction can only occur if carbon dioxide is present for the first reaction, so the test is diagnostic.
(d) If the calcium hydroxide solution were too dilute, the calcium carbonate that forms might be below its solubility,
and stay in solution rather than forming a precipitate. Thus, the test would not work.
(e)
Solubility Curve for Calcium Hydroxide
Solubility (g/100 mL)
0.20
0.15
0.10
0.05
0
20
40
60
80
100
Temperature (°C)
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Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
(f) From the solubility curve drawn in (e), the solubility of Ca(OH)2(s) at 22°C is 0.17 g/100 mL, or 0.17 g/0.100 L.
v Ca(OH) 1.0 L
2
c Ca(OH)
2
0.17 g/100 mL 0.17 g/0.100 L
m Ca(OH)
2
m Ca(OH)
2
0.17 g
1.0 L
0.100 L
1.7 g
The minimum mass of calcium hydroxide required to make up 1 L of saturated solution will be about 1.7 g. The actual
mass one would use should be much more than this — say, 15 to 20 g, to ensure a large excess of solute, which in
turn ensures saturation.
Note: In this particular instance the volume of the solute added is very small compared to the volume of the
solvent; so the difference between solvent volume and solution volume may be considered negligible.
(g) The solubility of calcium hydroxide is anomalous — unlike most solids it decreases in solubility with increasing
temperature. The generalization we use is still valid and useful for most soluble solids. It just needs to be understood that generalizations are exactly that — statements that generally (not invariably) describe events correctly.
Applying Inquiry Skills
9. Prediction
(a) The solubility of potassium chlorate should increase with an increase in temperature, according to the generalization for the solubility of solids, and also according to Figure 2 on page 316 of the text.
Analysis
(b)
Solubility Curve for Potassium Chlorate
60
Solubility (g/100 mL)
50
40
30
20
10
0
20
40
60
80
100
Temperature (°C)
(c) According to the Evidence, the solubility of potassium chlorate increases with increasing temperature, for solution temperatures from 0°C to 100°C.
Making Connections
10. Gases such as oxygen are less soluble in warm water, so active fish that require high oxygen levels may not thrive in
water that is warmed by an outside source, such as a power plant. In addition, power plants may alter the ecology of
a body of water by preventing it from freezing over in the winter. This would attract waterfowl that would feed on
aquatic organisms normally under the ice at this time.
PRACTICE
(Page 325)
Understanding Concepts
11. (a) The high/low solubility cutoff point is 0.10 mol/L at SATP.
(b) The 0.10 mol/L cutoff is useful because most laboratory solutions are within an order of magnitude of this
concentration, and because most ionic solids are either markedly more or less soluble than this value.
(c) The table need only be used to determine the solubility of an ionic solid if the solid does not contain a group I
ion, an ammonium ion, or a nitrate ion — because such compounds are always highly soluble. To determine the
solubility of any other ionic solid, find the anion in its column heading, and look down the column to find the row
Copyright © 2002 Nelson Thomson Learning
Chapter 7 Solubility and Reactions
199
(d)
12. (a)
(b)
(c)
(d)
(e)
13. (a)
(b)
(c)
containing the cation. The row where the cation is found will identify the compound as having low or high solubility.
In chemistry class, the word soluble usually refers to compounds that the table classes as “high solubility.” The
word insoluble usually refers to the compounds that the table classes as “low solubility,” provided their solubility
is so low that no effect is noticed from dissolving.
NaOH(s)
high solubility
MnCl2(s)
high solubility
Al(OH)3(s)
low solubility
Ca3(PO4)2(s)
low solubility
CuSO45H2O(s) high solubility
Large crystals take more time to “grow,” so the sugar probably crystallized slowly from a saturated solution.
Solubility is not affected by crystal size, so the specialty sugar would have the same solubility as regular white
sugar.
The dissolving rate is slower for large crystals.
Applying Inquiry Skills
14. Experimental Design
Small quantities of a solid may be dissolved sequentially in a measured sample of a solute until no more will dissolve,
to determine the solubility; or a saturated solution of known volume can be evaporated to dryness, and the solubility
calculated from the measured mass of solid solute remaining.
Making Connections
15. Pollutants in natural water can enter the water cycle through runoff from agricultural areas or landfills, or industrial
tailings ponds. Some pollutants are toxic and/or noxious at extremely low concentrations, so they may be dangerous
even if they have very low solubility.
PRACTICE
(Page 326)
Understanding Concepts
16.
Solubility Curves
Solubility
NH4HCO3
NaHCO3
Temperature
Making Connections
17. In the Solvay process, the scientific knowledge of solubility effects made possible a new technique for producing a
commercial chemical much more efficiently — a classic example of science leading technology.
SECTION 7.1 QUESTIONS
(Page 326)
Understanding Concepts
1. (a) The presence of the solid is evidence that sodium bromide is precipitating, probably because the solvent is evaporating (or maybe because of a drop in temperature). The solution is therefore at the limit of its concentration for
this temperature.
(b) The concentration of sodium bromide in the remaining solution is at the maximum possible value. Such a solution is said to be saturated.
200
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
(c) The mixture could be converted to a single phase solution by adding more solvent, or by warming. Either process
would dissolve the solid present.
2. Solids generally become more soluble as temperature increases, while gases generally become less soluble as temperature increases.
Applying Inquiry Skills
3. Prediction
(a) Barium sulfate should become more soluble as temperature increases, according to the generalization about the
solubility of ionic solids.
Experimental Design
(b) The independent variable is the solution temperature. The dependent variable is the mass of barium sulfate. The
most important controlled variable is the volume of saturated solution taken each time.
Analysis
(c)
Solubility of Barium Sulfate
Mass of BaSO4(s) (mg/100 mL)
0.50
0.40
0.30
0.20
0.10
0
10
20
30
40
50
Temperature (°C)
(d) According to the evidence from this investigation, barium sulfate does become more soluble as temperature
increases.
Evaluation
(e) The experiment could be improved by just making and using a single saturated solution, and taking samples from
it at different temperatures.
(f) The prediction was verified, according to the evidence gathered.
(g) The solubility generalization is supported by the results of this experiment. One test of one compound is certainly
not enough evidence to justify stating such a generalization. Many repeated tests of many compounds would be
required for scientific acceptance and confidence.
Making Connections
4. If a non-aqueous solvent is used, it is probably because the dirt and grease on clothing dissolves better in such
solvents. Also, some clothing fabrics are damaged by water, but are not affected by non-aqueous solvents.
5. (a) Cold, fast-flowing streams will have a higher concentration of dissolved oxygen, because more air will be mixed
with turbulent water, and because gases dissolve better in colder water.
(b) Trout probably require more oxygen than carp.
(c) Thermal pollution would probably affect trout seriously, because warming their water will reduce its oxygen
concentration.
7.2 HARD WATER TREATMENT
PRACTICE
(Page 329)
Understanding Concepts
1. (a) Scale in kettles and soap scum are both evidence of hard water.
(b) Ground water is hard in areas where soluble calcium and magnesium minerals exist in the ground.
Copyright © 2002 Nelson Thomson Learning
Chapter 7 Solubility and Reactions
201
(c) The mixture could be converted to a single phase solution by adding more solvent, or by warming. Either process
would dissolve the solid present.
2. Solids generally become more soluble as temperature increases, while gases generally become less soluble as temperature increases.
Applying Inquiry Skills
3. Prediction
(a) Barium sulfate should become more soluble as temperature increases, according to the generalization about the
solubility of ionic solids.
Experimental Design
(b) The independent variable is the solution temperature. The dependent variable is the mass of barium sulfate. The
most important controlled variable is the volume of saturated solution taken each time.
Analysis
(c)
Solubility of Barium Sulfate
Mass of BaSO4(s) (mg/100 mL)
0.50
0.40
0.30
0.20
0.10
0
10
20
30
40
50
Temperature (°C)
(d) According to the evidence from this investigation, barium sulfate does become more soluble as temperature
increases.
Evaluation
(e) The experiment could be improved by just making and using a single saturated solution, and taking samples from
it at different temperatures.
(f) The prediction was verified, according to the evidence gathered.
(g) The solubility generalization is supported by the results of this experiment. One test of one compound is certainly
not enough evidence to justify stating such a generalization. Many repeated tests of many compounds would be
required for scientific acceptance and confidence.
Making Connections
4. If a non-aqueous solvent is used, it is probably because the dirt and grease on clothing dissolves better in such
solvents. Also, some clothing fabrics are damaged by water, but are not affected by non-aqueous solvents.
5. (a) Cold, fast-flowing streams will have a higher concentration of dissolved oxygen, because more air will be mixed
with turbulent water, and because gases dissolve better in colder water.
(b) Trout probably require more oxygen than carp.
(c) Thermal pollution would probably affect trout seriously, because warming their water will reduce its oxygen
concentration.
7.2 HARD WATER TREATMENT
PRACTICE
(Page 329)
Understanding Concepts
1. (a) Scale in kettles and soap scum are both evidence of hard water.
(b) Ground water is hard in areas where soluble calcium and magnesium minerals exist in the ground.
Copyright © 2002 Nelson Thomson Learning
Chapter 7 Solubility and Reactions
201
2. (a) Na2CO3(s) and Ca(OH)2(s)
(b) The added carbonate ions cause calcium and magnesium ions in the hard water to precipitate, because CaCO3
and MgCO3 have very low solubility.
3. (a) C CaCO
3
7.1 10-5 mol/L
MCaCO 100.09 g/mol
3
0.000071 mol
100.09 g
c CaCO 3
L
mol
c CaCO 0.0071 g/L 7.1 mg/L
3
The concentration of calcium carbonate in the treated water will be equal to the solubility; 7.1 mg/L.
(b) 7.1 mg/L is equal to 7.1 ppm.
(c) The treated water is classified as soft, according to Table 1.
4. A home water softener unit has to be regenerated because the resin in it has a finite capacity to attract hard water ions,
to replace sodium ions. The sodium ions go into the bath and washing machine, but no precipitate (bathtub ring)
2+ and Mg2+ ions can attach to the resin, the regeneration cycle replaces the hard water ions
forms. When no more Ca(aq)
(aq)
with fresh sodium ions. The hard water ions are flushed down the drain.
Making Connections
5. On a large scale, the amount of resin required for softening would be extremely expensive, and the volume of concentrated brine needed for regeneration would be enormous, and thus a problem to make and store. The quantity of
sodium and chloride ions in the water might pose a health problem for some people.
6. Detergents clean well in harder water, producing no “soap scum,” which was a huge benefit. However, they increased
some pollution problems. Early detergents were not very biodegradable, and phosphates from detergents created (and
sometimes still do create) eutrophication problems in lakes and rivers.
7. Home-softened water (from ion-exchange softeners) is often not routed to toilets (for economy), and sometimes not
to a kitchen tap that provides drinking water (for possible health reasons). The incidence of heart disease is statistically slightly lower in areas with hard water than in areas with naturally soft water. People with heart concerns, particularly those on low-sodium diets, may not wish to drink home-softened water (which is often high in sodium).
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
SECTION 7.2 QUESTIONS
(Page 330)
Understanding Concepts
1. Water is naturally hard in areas where ground water comes in contact with minerals (such as limestone) that contain
slightly soluble calcium and/or magnesium compounds.
2+ and Mg2+ are the ions mostly responsible for water hardness.
2. Ca(aq)
(aq)
3. (a) Laundry “scum” and kettle scale are the noticeable effects of hard water.
(b) One serious effect of hard water that is not readily noticeable is the formation of scale in pipes. If it is allowed to
build up unchecked, the scale can significantly reduce the flow of water through the pipe, and come close to
blocking it completely.
4. Na2CO3(aq) Ca(HCO3)2(aq) → 2 NaHCO3(aq) CaCO3(s)
5. (a) Calcium ions are more readily attracted than sodium ions to negatively charged sulfonate sites on a resin molecule. This is mostly because a calcium ion has a charge of 2+, twice as much as the charge of a sodium ion.
(b) During regeneration, there are very many more sodium ions present (at maximum concentration) than calcium
ions, so the chance that sodium ions will collide with and attach to the sulfonate sites becomes very much greater,
overcoming the fact that calcium ions are attracted more strongly than sodium ions.
Applying Inquiry Skills
6. Question
(a) Is water softened by passing through a pipe to which a magnet is attached?
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Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
Experimental Design
Hard water is run through a pipe to a tap, and samples are tested for hardness, with and without a magnet being
attached to the pipe.
(b) Evaluation
The original design is obviously useless in light of the new understanding of the claim made for the magnet; it is
not testing the claimed result.
New Experimental Design
Hard water is run through identical piping that has and has not magnets attached. After several months, the piping is
cut and the ease of removing the scaling from the inside of the pipe is determined.
Making Connections
7. high sodium ion
concentration
WATER
SOFTENING
cleaner laundry;
no scale formation
REGENERATION
high sodium
ion concentration
high calcium ion and
saltwater discharge to
the environment
7.3 REACTIONS IN SOLUTION
PRACTICE
(Page 332)
Understanding Concepts
1. (a) silver sulfide: low solubility
(b) magnesium nitrate: high solubility
(c) zinc carbonate: low solubility
2. (a) Precipitate forms: SrSO4(s)
(b) No precipitate forms
(c) Precipitate forms: CuSO3(s)
PRACTICE
(Page 335)
Understanding Concepts
3. Sr(NO3)2(aq) Na2CO3(aq) → SrCO3(s) 2 NaNO3(aq)
2+ 2 NO–
+
2–
+
–
Sr(aq)
3(aq) 2 Na(aq) CO3(aq) → SrCO3(s) 2 Na(aq) 2 NO3(aq)
2+ CO2–
Sr(aq)
3(aq) → SrCO3(s)
4. (a) Compounds could be any four of:
copper(II) nitrate, copper(II) sulfate, copper(II) acetate, copper(II) chloride, copper(II) bromide,
or copper(II) iodide.
Choose copper(II) nitrate as the example:
(b) 2 Al(s) 3 Cu(NO3)2(aq) → 3 Cu(s) 2 Al(NO3)3(aq)
2+ 6 NO–
3+
–
(c) 2 Al(s) 3 Cu(aq)
3(aq) → 3 Cu(s) 2 Al(aq) 6 NO3(aq)
2+ → 3 Cu 2 Al3+
(d) 2 Al(s) 3 Cu(aq)
(s)
(aq)
5. Cl2(g) 2 Br–(aq) → Br2(l) 2 Cl–(aq)
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Chapter 7 Solubility and Reactions
203
2+ OOCCOO2– → CaOOCCOO
6. Ca(aq)
(s)
(aq)
or (if the oxalate ion formula is written in condensed form)
2+ C O2–
Ca(aq)
2 4(aq) → CaC2O4(s)
+
2+ 2 Ag
7. Pb(s) 2 Ag(aq)
→ Pb(aq)
(s)
Making Connections
3+ 3 OH–
8. (a) Fe(aq)
(aq) → Fe(OH)3(s)
(b) Filtration to remove the precipitate is the most likely process.
Explore an Issue
Debate: Producing Photographs
(Page 336)
(a) At first glance, it seems that digital cameras are more environmentally friendly than film cameras: they don’t require
film, film canisters, developing paper, or processing chemicals. However, this conclusion is based upon the assumption that only the final products affect the environment.
(b) To argue against the proposition you need to recognize that the whole story of a product must be considered, including
everything from the extracting of resources to manufacturing through disposal. For example, are there components of
the digital camera that during manufacture cause the emission of toxins into the environment, or, when the camera
must be disposed of, are there environmental concerns?
(c) Recognize that the resolution requires an environmental perspective only; you need not consider, for example,
economic, social, and technological arguments. In your investigation, you could look at subtopics such as raw materials, manufacturing, the developing process, the developing technology (and its environmental impact back to its
origins), and disposal.
(d) Consider the logic of the presentation and the quantity of evidence used to support the position taken. Were all stages
from pre-manufacturing through post-disposal considered?
SECTION 7.3 QUESTIONS
(Page 336)
Understanding Concepts
2+ SiO2–
1. Pb(aq)
3(aq) → PbSiO3(s)
2+ 2 PO3–
2. 3 Ca(aq)
4(aq) → Ca3(PO4)2(s)
+
2+ 2 Ag
3. Cu(s) 2 Ag(aq)
→ Cu(aq)
(s)
Making Connections
4. Student answers will vary widely, as they will be specific to the regulations controlling local hazardous waste
facilities.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
7.4 WASTE WATER TREATMENT
PRACTICE
(Page 340)
Understanding Concepts
1. Problems from the release of untreated sewage include: transmission of disease, lowering of oxygen levels, and rapid
growth of aquatic plant life.
2. A high BOD reading is an indication that bacteria are using up oxygen to decompose organic material in the water.
This is a problem for any oxygen-using life form in the water.
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Copyright © 2002 Nelson Thomson Learning
3.
Primary Treatment
Secondary Treatment
Tertiary Treatment
• screening
• flotation
• settling
• filtration
• aeration
• precipitation
• chlorination
• reverse osmosis
• distillation
• precipitation
• other...
4. Material that flows into a rural septic system is generally removed only by bacterial decomposition. Ideally, only water
and biodegradable substances should be flushed. Homeowners should be careful not to overwhelm the decomposition
process by flushing large quantities of any pollutants.
Reflecting
5. What goes down our drains is likely to end up in the ground or surface water. Being careful about waste disposal at
home and at school can make a cumulative positive difference in the quality of our environment.
SECTION 7.4 QUESTIONS
(Page 340)
Understanding Concepts
1.
deliver to user
ammoniation
postchlorination
fluoridation
primary treatment:
screening, flotation,
settling, and filtration
softening
aeration
disinfection
secondary treatment:
aeration and chlorination
filtration
coagulation,
flocculation, and
sedimentation
tertiary treatment (discretional):
e.g., reverse osmosis,
steam distillation.
chemical precipitation
collection
ground or
surface
water source
Drinking Water Treatment
Waste Water Treatment
Applying Inquiry Skills
2. (a) Hypotheses
(1) The fish kill may be due to a lack of oxygen caused by sewage discharge of organic matter into the river.
or
(2) The fish kill may be due to discharge of toxic material or a disease-causing organism into the river.
(b) Prediction (1)
The fish kill is caused by a high BOD, due to excess discharge of organic matter in sewage upstream.
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Chapter 7 Solubility and Reactions
205
Experimental Design (1)
The water is tested for oxygen and for organic matter, every 500 m upstream from the fish-kill area to the industrial
town.
Prediction (2)
The fish kill is caused by toxic chemicals or disease organisms in the sewage discharge upstream.
Experimental Design (2)
The water is tested for chemicals and for organisms causing common diseases in fish, every 500 m upstream from the
fish-kill area to the industrial town.
Note: This is an example of a correlational study (see Appendix A1, pages 608 – 9). Technically, correlational
studies are inductive investigations without a hypothesis and a prediction. Note above that when forced, a hypothesis
and a prediction are similar, indicating that the investigation is inductive and that neither a hypothesis nor a prediction should be used.
Making Connections
3. Garbage disposal units decrease the amount of solid bagged waste from a household, which cuts costs and extends the
usefulness of landfills. However, the increase of organic matter in the sewage places more demand on the local waste
water treatment system. Most of this food waste could be diverted to a composting system.
4. Answers will depend on the regulations controlling local hazardous waste facilities. Unless the local area has a tertiary
treatment facility, there will automatically be an argument for improvement. The only logical long-term human goal
is to eventually have all waste water returned to the cycle in a form that puts no stress on the environment.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
7.5 QUALITATIVE CHEMICAL ANALYSIS
PRACTICE
(Page 342)
Understanding Concepts
1. (a) colourless
(b) blue
(c) yellow-brown
(d) orange
(e) colourless
(f) green
2. (a) yellow-red
(b) blue
(c) yellow
(d) violet
(e) colourless
3. (a) yellow-red
(b) light blue-grey
(c) bright red
(d) green
Applying Inquiry Skills
4. (a) Analysis
According to Table 2, Colours of Flames:
solution A contains K+ ions,
solution B contains Cu2+ ions,
solution C contains Na+ ions,
solution D contains Ca2+ ions, and
solution E contains Li+ ions and/or Sr2+ ions.
(b) Evaluation
The design of this experiment is too limited. It only identifies those positive ions (cations) in the solutions that
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Copyright © 2002 Nelson Thomson Learning
happen to produce coloured flames. Other possible cations, not to mention the anions in each of these solutions,
cannot be identified this way. As well, solution E may contain either or both of two cations, since they both
produce the same result.
PRACTICE
(Page 346)
Understanding Concepts
5. Qualitative analysis determines what is in a sample, and quantitative analysis determines how much is present.
6. A diagnostic test statement always includes procedure, evidence, and analysis steps. For example: If a gas is bubbled
through limewater, and a white precipitate forms, then the gas is likely to contain carbon dioxide.
7. Precipitates could be formed with the listed ions by adding:
(a) OH– , CO32– , PO43– , or SO32– aqueous ions.
(b) SO42– , CO32– , PO43– , or SO32– aqueous ions.
(c) Ag+ , Pb2+ , Tl+ , Hg22+ , or Cu+ aqueous ions.
(d) Ag+ , Pb2+ , Ca2+ , Ba2+ , Sr2+ , or Ra2+ aqueous ions.
Applying Inquiry Skills
8. (a) To precipitate carbonate ions from a sample, without at the same time precipitating sulfide ions, add a compound
that supplies any Group II cation, and an anion that is always soluble in combination (e.g., calcium nitrate, barium
nitrate, or magnesium nitrate).
2+ CO2– → CaCO
(b) One example: Ca(aq)
3(aq)
3(s)
9. Experimental Design
Calcium nitrate solution is added to the test solution sample. If a precipitate forms it is filtered, and silver nitrate solution is added to the filtrate (or to the original sample, if no precipitate formed in the initial test).
Making Connections
10. There are innumerable examples of qualitative analysis in society. Common examples include environmental tests for
lead (or other heavy metals) in water supply systems; continuous monitoring by household detectors (carbon
monoxide or natural gas in houses, propane in trailers); simple swimming pool or aquarium water-testing kits
(including quantitative analysis); or even home pregnancy tests. There are also hundreds of industrial and commercial
examples of qualitative analysis. There are many career opportunities as an analyst.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
SECTION 7.5 QUESTIONS
(Page 346)
Understanding Concepts
1. The precipitation reactions are:
2+ 2 Cl–
(a) Pb(aq)
(aq) → PbCl2(s)
2+ S2– → ZnS
(b) Zn(aq)
(aq)
(s)
+
(d) Ag(aq)
C2H3O–2(aq) → AgC2H3O2(s)
2+ + 2 PO3(e) 3 Ba(aq)
4(aq) → Ba3(PO4)2(s)
2+ 2 OH–
(f) Ca(aq)
(aq) → Ca(OH)2(s)
2. (a)
(b)
(c)
(d)
(e)
3. (a)
yellow-brown
colourless
blue
green
colourless
Calcium (yellow-red flame test) can be distinguished from the other ions by a flame test, but lithium and strontium both give bright red flame tests, and so cannot be distinguished from each other in this way.
Copyright © 2002 Nelson Thomson Learning
Chapter 7 Solubility and Reactions
207
(b) Adding sulfate (or carbonate, phosphate, or sulfite) ions to the two unidentified test solutions (aqueous sodium
sulfate, for instance) would precipitate the strontium ions, but not the lithium ions.
Applying Inquiry Skills
4. (a) Experimental Design
The solution is flame tested.
(b) Evaluation
The experimental design is seriously flawed, because both ions produce coloured flames. If the initial flame test
is bright red, there is no way to know whether potassium ions are present, because the violet colour will be hidden
by the strontium ion colour.
Alternative answer:
(a) Experimental Design
Sodium sulfate is added to the solution. Any precipitate is filtered and then the filtrate or original solution is flame
tested.
(b) Evaluation
The experimental design is not valid. Although you will know with certainty whether strontium ions are present
or not, based upon whether a precipitate forms or not, you will not be able to determine whether potassium ions
are present, due to the masking of the potassium flame colour by the sodium flame colour. If cobalt-blue glass is
employed in the materials and the procedure, then the design would be valid. (The cobalt-blue glass filters the
yellow sodium colour from the flame and allows one to determine whether potassium is present or not.)
Alternately, the cation for the sodium solution must be chosen to have a colourless flame test, e.g., hydrogen.
(Unfortunately, a list of colourless flame-test ions is not provided in the text and would have to be researched.)
5. Experimental Design
Sodium chloride solution is added to the sample. If a precipitate forms, it is filtered. Sodium hydroxide solution is
added to the filtrate (or sample, if no precipitate forms). If a precipitate forms, it is filtered. Sodium carbonate (or
sulfate) solution is added to the filtrate (or sample, if no precipitate forms).
Note: A precipitate in the initial step (when you use any soluble halide ion compound) indicates that thallium ions
are present. A precipitate in the second step (when you use any soluble hydroxide compound) indicates that calcium
ions are present. A precipitate in the third step (when you use any soluble sulfate, carbonate, phosphate, or sulfite
compound) indicates that barium ions are present. The sequence of steps is very important: sodium carbonate cannot
be added first, for example.
6. (a) The Experimental Design is satisfactory. The solution colour test can confirm the presence of copper(II) ions, but
not calcium ions; and any red in the flame test will confirm the presence of calcium ions.
(b) Alternative Experimental Design
Sodium sulfate solution is added to the sample solution. If a precipitate forms, it is filtered. Sodium carbonate
solution is added to the filtrate (or sample, if no precipitate forms).
Note: A precipitate in the initial step (when you use any soluble sulfate compound) indicates that calcium ions
are present. A precipitate in the second step (when you use any soluble sulfide, carbonate, phosphate, or sulfite
compound) indicates that copper(II) ions are present.
7. Any carbonated beverage is a home solution with a gaseous solute. A diagnostic test is: If a gas is bubbled through
limewater, and a white precipitate forms, then the gas contains carbon dioxide.
Another example would be household ammonia, or any of several spray window cleaners containing ammonia. The
diagnostic test would be the characteristic odour of ammonia.
8. Sodium carbonate is a typical home solution with a solid solute; e.g., as a water softener (washing soda) in laundry
detergents. Diagnostic tests for this example would be flame testing for sodium ions, and precipitation using calcium
chloride solution, for carbonate ions.
9. Experimental Design
Oxalic acid is added to sample solutions of nitrate or chloride compounds of as many metal cations as can be found
in the school laboratory supplies. Any precipitate formation is recorded.
10. Experimental Design
A solution of the product is first flame tested; then a solution of calcium chloride is added.
Note: A yellow flame test indicates sodium; a precipitate indicates carbonate ions.
208
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
Making Connections
11. Forensic chemists analyze tissue for many things. A typical test would be for the presence and amount of arsenic.
Quantity must be measured precisely to determine if a substance is present in a natural amount, or in an amount much
greater, which could perhaps indicate foul play.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
7.6 QUANTITATIVE ANALYSIS
PRACTICE
(Page 353)
Understanding Concepts
1. 2 NH3(g) H2SO4(aq) → (NH4)2SO4(aq)
24.4 mL
50.0 mL
2.20 mol/L
C
2.20 mol
0.0537 mol
nNH 0.0244 L
3
1L
1
nH SO
0.0537 mol 0.0268 mol
2
4
2
nH SO
0.0268 mol
2
4
CH SO
2
4
CH SO
4
2
0.0268 mol
0.0500 L
0.537 mol/L
or
CH SO
2
CH SO
2
2.20 mol NH
1 mol H2SO4
1
0.0244 L
NH3 3 1L
NH3
2
mol N
H3
0.0500 L
0.537 mol/L
4
4
The concentration of sulfuric acid at this stage is 0.537 mol/L.
2. 3 Ca(OH)2(aq) Al2(SO4)3(aq) → 3 CaSO4(s) 2 Al(OH)3(s)
v
25.0 mL
0.0250 mol/L
0.125 mol/L
nAl (SO )
2
4 3
nCa(OH)
2
vCa(OH)
2
vCa(OH)
2
vCa(OH)
2
vCa(OH)
2
or
0.125 mol
25.0 mL
3.13 mmol
1L
3
3.13 mmol 9.38 mmol
1
1L
9.38 mmol
0.0250 mol
375 mL
OH)2
3
mol Ca(
1 L Ca (OH)2
0.125 mol Al2(SO
4)3
25.0 mL
Al2
(SO4)3 SO4)3
1
mol Al2(
375 mL
The volume of calcium hydroxide solution reacted is 375 mL.
3. (a) 2 FeCl3(aq) 3 Na2CO3(aq) → Fe2(CO3)3(s) 6 NaCl(aq)
75.0 mL
v
Copyright © 2002 Nelson Thomson Learning
Chapter 7 Solubility and Reactions
209
0.200 mol/L
nFeCl
3
nFeCl
3
nNa CO
2
3
nNa CO
3
2
vNa CO
2
3
vNa CO
3
2
or
0.250 mol/L
0.200 mol
75.0 mL
1L
15.0 mmol
3
15.0 mmol 2
22.5 mmol
1L
22.5 mmol
0.250 mol
90.0 mL
eCl3
0.200 m
ol F
3
mol Na2
CO3
1 L Na2CO3
vNa CO 75.0 mL
FeCl
3 2
3
2m
ol FeCl3
0.250 m
ol Na2
CO3
1
FeCl3
L
vNa CO 90.0 mL
2
3
The minimum volume of sodium carbonate solution required for complete reaction is 90.0 mL.
(b) A reasonable volume of sodium carbonate solution would be at least 100 mL, thus providing about a 10%
excess to ensure a complete reaction.
Applying Inquiry Skills
4. (a) Prediction
2 NaOH(aq) ZnCl2(aq) → Zn(OH)2(s) 2 NaCl(aq)
20.0 mL
m
2.50 mol/L
nNaOH
nNaOH
nZn(OH)
2
nZn(OH)
2
mZn(OH)
2
mZn(OH)
2
99.40 g/mol
2.50 mol
20.0 mL
1L
50.0 mmol
1
50.0 mmol 2
25.0 mmol
99.40 g
25.0 mmol
1
mol
2.49 103 mg
mZn(OH) = 2.49 g
2
or
mol Zn(OH)
99.40 g Zn(OH)2
2.50 mol N
aOH 1 mZn(OH) 20.0 mL
NaOH
2 2
1L
N
aOH
2m
ol N
a(OH)
1
mol Zn(OH)2
mZn(OH) 2.49 103 mg
2
mZn(OH) = 2.49 g
2
According to the stoichiometric method, the mass of zinc hydroxide produced is predicted to be 2.49 g.
Note: The calculated answer of 2.485 g can be rounded to 2.48 g or 2.49, depending on the rounding rule used
in the classroom.
(b) Analysis
mZn(OH) 3.30 g 0.91 g 2.39 g
2
According to the evidence, the mass of zinc hydroxide that is actually produced is 2.39 g.
(c) Evaluation
difference 2.39 g 2.48 g 0.10 g
0.10 g
% difference 100% 3.8%
2.48 g
Note: The unrounded value of 2.485 g was used in the calculation. If the rounded value is used, the difference
is 0.09 g and the percentage difference is 3.6%.
210
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
The prediction was 3.8% higher than the value obtained, and so is judged to be verified by the experimental results.
A prediction within 5% is considered acceptably accurate (95% accurate) for this kind of lab work, with any difference probably just due to normal experimental error.
(d) The stoichiometric concept is supported by the results of this investigation, and judged to be acceptable because
the prediction was verified. There is good confidence in this judgment, and no need is seen to modify the concept.
SECTION 7.6 QUESTIONS
(Page 355)
Applying Inquiry Skills
1. Precipitating all the lead(II) ions will require adding an excess of a solution containing an anion (in this case, sulfate)
that forms a low-solubility compound with lead(II). Making a sodium sulfate solution is a logical choice, since sodium
sulfate is soluble, inexpensive, and easy to obtain.
First, the mass of sodium sulfate required must be calculated.
Pb(NO3)2(aq) Na2SO4(aq) → PbSO4(s) 2 NaNO3(aq)
2.0 L
m
0.34 mol/L
142.04 g/mol
n Pb(NO )
3 2
n Na SO
2
4
m Na SO
4
m Na SO
4
m Na SO
4
m Na SO
4
2
2
= 97 g
or
2
2
0.34 mol
= 2.0 L
0.68 mol
1L
1
= 0.68 mol 0.68 mol
1
142.04 g
= 0.68 mol 1
mol
0.34 m
ol Pb
(NO )
1m
ol N
a2SO4
142.04 g Na2SO4
= 3 2 1L
Pb
(NO3)2
1m
ol Pb(N
O3)2
1
mol N
a2SO4
= 97 g
A minimum mass of 97 g of sodium sulfate must be used.
To ensure a complete precipitation of lead(II) ions, using an excess of sodium sulfate will be necessary. Commonly
this amount should be about 10% more than the minimum required — say, 105 g of sodium sulfate, in this case.
The simplest process would be to obtain 105 g of sodium sulfate and dissolve it to make, say, 1.0 L of reacting solution. It is possible that not all of the solute will dissolve, depending on the water temperature. This problem could be
solved by increasing the solvent volume.
2. The experimental design is judged to be inadequate, because the mass of solid measured includes not only the precipitate but also the excess reactant and the second (soluble) product. The latter two chemicals crystallize out of the solution when the water is boiled away. The mass of solid remaining will be of more than one substance, with no way to
calculate amounts from the value. The precipitate should have been separated by filtration and then dried.
If we take the reaction in Question 3 as an example, precipitating and then crystallizing would result in a mixture of
the solids, aluminum nitrate (the excess reactant), aluminum sulfide (the precipitate), and sodium nitrate (the second
(but soluble) product).
3. (a) Prediction
3 Na2S(aq) 2 Al(NO3)3(aq) → Al2S3(s) 6 NaNO3(aq)
20.0 mL
m
0.210 mol/L
n Na S
2
n Na S
2
150.14 g/mol
0.210 mol
20.0 mL
L
4.20 mmol
Copyright © 2002 Nelson Thomson Learning
Chapter 7 Solubility and Reactions
211
n Al S
2 3
n Al S
2 3
m Al S
2 3
m Al S
2 3
or
1
4.20 mmol 3
1.40 mmol
150.14 g
1.40 mmol
1
mol
210 mg = 0.210 g
m Al S
0.210 mol N
a2S
1
mol A
l2S3
150.14 g Al2S3
20.0 mL
Na
2S 1L
N
a2S
3m
ol N
a2S
m
ol A
l 2S3
m Al S
210 mg = 0.210 g
2 3
2 3
According to the stoichiometric method, the mass of aluminum sulfide produced is predicted to be 0.210 g.
(b) Analysis
m Al S 1.17 g 0.97 g 0.20 g
2 3
According to the evidence, the mass of aluminum sulfide that is actually produced is 0.20 g.
(c) Evaluation
The design of the experiment is judged to be adequate, with no obvious flaws. It allowed the question to be
answered easily with simple materials, concepts, and procedures.
difference 0.20 g 0.210 g 0.01 g
0.01 g
% difference 100% 5%
0.210 g
The prediction was 5% higher than the value obtained, and is judged to be verified by the experimental results. A
prediction within 5% is considered acceptably accurate (95% + ...) for this kind of lab work, with any difference probably just due to normal experimental error.
The stoichiometric concept is the authority for this investigation. It is supported by the results of this investigation,
and judged to be acceptable because the prediction was verified. There is good confidence in this judgment, and there
is no need to modify the concept.
4. (a) Analysis
v AgNO = 100 mL = 0.100 L
3
m AgNO = 6.74 g 1.27 g 5.47 g
3
M AgNO = 169.88 g/mol
3
1 mol
nAgNO = 5.47 g 3
169.88 g
0.0322 mol
0.0322 mol
CAgNO = 3
0.100 L
CAgNO = 0.322 mol/L
3
The molar concentration of silver nitrate in solution is 0.322 mol/L.
Making Connections
5. The most common way to check the concentration of antifreeze is to measure its density, since density and concentration are proportional. Battery acid is measured the same way (to determine battery charge). The higher the concentration of solute, the denser the solution will be, and the higher in the solution a buoyant object will float. A calibrated
device used to measure liquid density in this way is called a hydrometer. Common density units for solutions are g/mL
and kg/L, but concentrations of automotive fluids are usually expressed in terms of what they are intended to do:
lowest working temperature (for antifreeze), or charge condition (for battery acid). It is also common to state solution
densities as a ratio with the density of pure water — giving a numerical value greater or less than 1 — called the
“specific gravity” of the solution.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
212
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
CHAPTER 7
REVIEW
(Page 358)
Understanding Concepts
1. Combinations (a), (b), (c), (d), and (f) will react, but only (a), (c), (d), and (f) will form precipitates. The reaction equations are:
2+ + 2 OH– → Cu(OH)
(a) Cu(aq)
2(s)
(aq)
+ + H– → H O
(b) H(aq)
2 (l)
(aq)
2+ + 2 PO3–
(c) 3 Ca(aq)
4(aq) → Ca3(PO4)2(s)
+ + Cl– → AgCl
(d) Ag(aq)
(s)
(aq)
+ Cl– → CuCl
(f) Cu(aq)
(s)
(aq)
2+ + CO2– → ZnCO
2. (a) Zn(aq)
3(s)
3(aq)
2+ + CO2– → PbCO
(b) Pb(aq)
3(s)
3(aq)
3+ + 3 CO2– → Fe (CO )
(c) 2 Fe(aq)
2
3 3(s)
3(aq)
2+ + CO2– → CuCO
(d) Cu(aq)
3(aq)
3(s)
+ + CO2– → Ag CO
(e) Ag(aq)
3(aq)
2
3(s)
2+ + CO2– → NiCO
(f) Ni(aq)
3(aq)
3(s)
(g) The choice of sodium carbonate is good because carbonate ions form low soluble compounds with most metallic
ions and the compound is soluble, common, and inexpensive.
3+ + 3 OH– → Al(OH)
3. (a) Al(aq)
(aq)
3(s)
and
2+ + SO2– → CaSO
Ca(aq)
4(s)
4(aq)
or
3+ + 3 SO2– + 3 Ca2+ + 6 OH– → 2 Al(OH)
2 Al(aq)
(aq)
(aq)
3(s) + 3 CaSO4(s)
4(aq)
Note: The effective precipitate for clarifying the water is the flocculent precipitate, Al(OH)3(s).
2+ + 2 PO3– → Ca (PO )
(b) 3 Ca(aq)
3
4 2(s)
4(aq)
2+ + 2 OH– → Mg(OH)
(c) Mg(aq)
(aq)
2(s)
3+ + 3 OH– → Fe(OH)
(d) Fe(aq)
(aq)
3(s)
2+
4. Cu(aq)
5. Ions of alkali metals, as well as hydrogen, ammonium, and nitrate ions, form compounds with high solubility.
+ could indicate any anion on the solubility chart
6. A violet flame indicates potassium ions. A precipitate with Hg(aq)
except sulfate, nitrate, or acetate. The compound in solution might be KCl, KBr, K2S, K2SO4, KOH, K3PO4, or ...
7. In aqueous solution:
(a) Cu+ is green, Cu2+ is blue.
(b) Fe2+ is pale green, Fe3+ is yellow-brown.
(c) CrO42– is yellow, Cr2O72– is orange.
2+ and Mg2+
8. Ca(aq)
(aq)
9. (a) Na2CO3(aq) CuSO4(aq) → Na2SO4(aq) CuCO3(s)
v
4.54 L
1.25 mol/L
nCuSO
214
4
0.0875 mol/L
0.0875 mol
4.54 L
0.397 mol
1L
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
nNa CO
2
or
3
vNa CO
2
3
vNa CO
2
3
vNa CO
2
3
vNa CO
3
2
1
0.397 mol 0.397 mol
1
1L
0.397 mol 1.25 mol
0.318 L
0.0875 mol C
uSO4
1
mol Na
1 L Na2CO3
2CO3
4.54L
CuSO
4 1
L C
uSO4
1m
ol Cu
SO4
1.25 m
ol Na
2CO3
0.318 L
The minimum volume of sodium carbonate solution required is 0.318 L.
(b) A suitable volume would be about 350 mL. (Assume an excess of 10%.)
10. The mass of zinc reacted (24.89 g 21.62 g) 3.27 g
Zn(s) 2 HCl(aq) → ZnCl2(aq) H2(g)
3.27 g
350 mL
350 mL
65.38 g/mol
C
1 mol
3.27 g 65.38 g
nZn
0.0500 mol
nZn
nZnCl
2
CZnCl
2
CZnCl
2
or
1
0.0500 mol 0.0500 mol
1
0.0500 mol
0.350 L
0.143 mol/L
1
mol Z
n
1 mol ZnCl
1
CZnCl 3.27 g Zn 2 2
65.38 g Z
n
1
mol Zn
0.350 L
CZnCl 0.143 mol/L
2
The molar concentration of zinc chloride solution is 0.143 mol/L.
Applying Inquiry Skills
11. The precipitated anion could be SO42–, CO32–, PO43–, or SO32–.
Since most sulfates are soluble, and most sulfites, carbonates, and phosphates are only slightly soluble, the original
solution could be tested with Zn(NO3)2(aq), Cu(NO3)2(aq), or Ni(NO3)2(aq), etc. If no precipitate forms, the anion must
be SO42–.
12. Experimental Design
The solution is tested with TlNO3(aq) (or Hg2(NO3)2(aq) or CuNO3(aq)) for the presence of halide ions. If a precipitate
forms it is filtered, and the filtrate (or original solution if no precipitate forms) is tested with Ca(NO3)2(aq) (or barium,
strontium, or radium nitrate) for the presence of sulfate ions.
13. Experimental Design
The solutions are tested with litmus to identify the acid and hydroxide (basic) compounds. The remaining solutions
are then tested for conductivity (to identify the ionic compound). To confirm that the final solution contains nitrogen
gas, it could be heated slightly. The formation of tiny bubbles would confirm the presence of a dissolved gas.
Note: The least confidence is for the nitrogen test. Nitrogen has one-half the solubility of oxygen gas in water —
0.00175 g/100 mL.
14. (a) Materials
• deep-seawater sample solution
•
1.00 mol/L Pb(NO3)2 (aq) stock solution
•
KI (aq) test solution
•
medicine dropper
•
50-mL pipet and bulb
Copyright © 2002 Nelson Thomson Learning
Chapter 7 Solubility and Reactions
215
•
250-mL beaker
•
400-mL beaker
•
filtration apparatus
•
filter paper
•
wash bottle of pure water
•
centigram balance
(b) Analysis
mass of PbCl2 (s) precipitate (4.58 g – 0.91 g) = 3.67 g
Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2 (s) 2 NaNO3(aq)
50.0 mL
C
n PbCl
2
n NaCl
3.67 g
278.10 g/mol
1 mol
= 3.67 g = 0.0132 mol
278.10 g
2
= 0.0132 mol = 0.0264 mol
1
0.0264 mol
C NaCl = 0.0500 L
C NaCl = 0.528 mol/L
or
1
mol Pb
Cl2
2 mol NaCl
1
C NaCl 3.67 g PbCl
2 278.10 g P
bCl2
1
mol P
bCl2
0.0500 L
C NaCl = 0.528 mol/L
According to the evidence and the stoichiometric concept, the concentration of sodium chloride in the seawater sample is
0.528 mol/L.
(c) Evaluation
The design of this experiment is judged to be adequate because it allowed the question to be answered easily, and
with confidence in the result. There are no apparent flaws and the equipment is simple and easy to use.
15. (a) Materials
• CuSO4(aq) sample solution
•
0.750 mol/L NaOH (aq) stock solution
•
medicine dropper
•
25-mL pipet and bulb
•
250-mL beaker
•
400-mL beaker
•
filtration apparatus
•
filter paper
•
wash bottle of pure water
•
centigram balance
(b) Analysis
mass of Cu(OH)2 (s) precipitate
(2.83 g – 0.88 g) = 1.95 g
CuSO4(aq) + 2 NaOH(aq) → Cu(OH)2 (s) 2 Na2SO4(aq)
216
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
25.0 mL
C
nCu(OH)
2
nCuSO
4
CCuSO
4
CCuSO
4
CCuSO
4
CCuSO
4
or
1.95 g
97.57 g/mol
1 mol
= 1.95 g = 0.0200 mol
97.57 g
1
= 0.0200 mol = 0.0200 mol
1
0.0200 mol
= 0.02500 L
= 0.799 mol/L
1
mol C
u(OH)2
1 mol CuSO4
1.95 g Cu(OH)
2 97.57 g C
u(OH)2
1
mol C
u(OH)2
= 0.799 mol/L
According to the evidence and the stoichiometric concept, the concentration of copper(II) sulfate in the sample is
0.799 mol/L.
(c) Evaluation
The design of this experiment is judged to be adequate because it allowed the question to be answered easily, and
with confidence in the result. There are no apparent flaws and the equipment is simple and easy to use.
Making Connections
16. For example, the hardness may be 250 ppm (250 mg/L). The soda-lime process could be used by adding washing soda,
Na2CO3(aq), and lime, Ca(OH)2(aq)/(s), to the water to precipitate the hard-water ions as carbonates; e.g., CaCO3(s).
Note: Answers will be specific to school/community location.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
17. Answers will vary, but might express concern about an old septic system at a cottage because of a danger of leakage,
which might release disease-causing organisms into ground water.
Note: Answers will be specific to school/community location. Student discussion should use concepts from this
chapter.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
18. (a) Recall that ppm = mg/L
O2(aq) (maximum) in 50 L is:
14.7 mg
50 L
735 mg at 0°C
L
and
8.7 mg
435 mg at 20°C
50 L
L
735 435 mg 300 mg
The difference in mass of oxygen that can be dissolved in 50 L of water at the two temperatures is 300 mg.
(b) Fish require O2 for respiration, so they might prefer 0°C water, in which O2 solubility is higher and obtaining
sufficient oxygen would be easier. On the other hand, the temperature of the surrounding water affects the activity
level of cold-blooded animals. The fish would not be able to move as fast at the colder temperature.
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Chapter 7 Solubility and Reactions
217