Chapter 9 Review, pages 456–461
Knowledge
1. (b)
2. (b)
3. (c)
4. (b)
5. (a)
6. (a)
7. (b)
8. (a)
9. (c)
10. False. Net ionic equations are not limited to precipitation reactions.
11. True
12. False. Qualitative analysis involves the identification of a substance.
13. True
14. False. Some of the drugs found in water supplies are removed by standard water treatment
processes.
15. True
16. False. A compound is a good electrolyte if it completely dissociates into ions as it dissolves
in water.
17. False. In stoichiometric problems, volume should be in units of litres.
18. True
19. (a) (vi)
(b) (iii)
(c) (ii)
(d) (i)
(e) (iv)
(f) (v)
20. A spectator ion is an ion that is present in a solution, but does not participate in the reaction.
21. Both total ionic equations and net ionic equations show the ions that are involved in the
reaction, as well as other reactants or products. Total ionic equations also include any spectator
ions.
22. The two types of chemical reactions that can be represented by total ionic and net ionic
equations are single-displacement and double-displacement reactions.
23. Most precipitation reactions are double-displacement reactions. During a doubledisplacement reaction, two compounds exchange anions (or cations) and one of the products is
only slightly soluble (a precipitate).
24. Physical contaminants are removed first during the water treatment processes.
25. Sanitation is the safe disposal of human waste so that it does not contaminate drinking water.
Higher levels of sanitation produce cleaner drinking water.
26. Hard water contains relatively high concentrations of magnesium, calcium, and iron ions.
Soft water has low concentrations of these ions. Hard water does not lather and leaves behind a
soap scum, which is a precipitate that forms when the cations in the water combine with the soap
ions. Hard water also forms precipitates that clog pipes and the heating elements of water heaters
and kettles.
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Chapter 9: Solutions and Their Reactions
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27. (a) Leachate is fluid that is contaminated with dissolved or suspended substances picked up
as the fluid passes through solid waste material at mines or landfills.
(b) One way that leachate can pollute nearby water supplies is by carrying heavy metals from
mines into the water. Another way is that leachate from landfills can carry contaminants from
discarded nickel-cadmium batteries into nearby water sources.
28. To determine whether hydrogen gas is present in a distant star, a scientist might compare the
spectrum of the light from the star to the known spectrum of hydrogen gas.
29. Both qualitative analysis and quantitative analysis involve investigating a substance. In
qualitative analysis, the aim is to identify a substance or establish whether a certain substance is
present. Quantitative analysis involves determining how much of a substance is present.
30. (a) The formation of a reddish, scaly compound on the surface of a metal object indicates
that the object contains iron is qualitative analysis.
(b) Two moles of hydrogen gas react with one mole of oxygen gas to produce two moles of
water is quantitative analysis.
31. (a) A filtrate is the clear liquid collected after a mixture has been filtered to remove the
solids.
(b) A filtrate contains the solvent and any substances dissolved in the solvent.
32. During a flame test, a small volume of solution is placed in a flame. If the solution contains a
cation that produces a characteristic colour when heated, the resulting colour of the flame is used
to identify the cation.
33. To reduce the quantities of drugs present in the potable water supply, communities can make
the public aware of the dangers of improper disposal of medications. They can also collect
unused medications and dispose of them properly.
34. Removing drugs from water sources might cause additional environmental problems because
many measures used to clean water require large amounts of energy, which increases greenhouse
emissions.
35. Three advantages of conducting reactions in aqueous solutions are that water is cheap and
easily accessible, water dissolves many different substances, and the reaction rate can be better
controlled in solution.
36. Write the formula equation for the reaction with the number of moles below.
(NH4)3PO4(aq) → 3 NH4+(aq) + PO43–(aq)
1 mol
3 mol
1 mol
The ratio of ammonium ions to phosphate ions in the solution is 3:1.
37. The amount of chemical can be determined by multiplying the amount concentration by the
volume. The equation is n = cV .
Understanding
38. (a) Step 1. Write the formula equation for the reaction.
2 NaI(aq) + Pb(ClO3)2(aq) → 2 NaClO3(aq) + PbI2(s)
Step 2. Write the total ionic equation
2 Na+(aq) + 2 I–(aq) + Pb2+(aq) + 2 ClO3–(aq) → 2 Na+(aq) + 2 ClO3–(aq) + PbI2(s)
(b) Cancel the spectator ions to obtain the net ionic equation.
2 Na+(aq) + 2 I–(aq) + Pb2+(aq) + 2 ClO3–(aq) → 2 Na+(aq) + 2 ClO3–(aq) + PbI2(s)
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Chapter 9: Solutions and Their Reactions
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39. (a) You cannot write a net ionic equation for the reaction of copper metal and iron(III)
chloride because copper is below iron in the activity series so no reaction occurs.
(b) You cannot write a net ionic equation for the decomposition of carbonic acid because no ions
are involved. During the decomposition of carbonic acid into water and carbon dioxide, one
molecular compound breaks down into two molecular compounds.
40. (a) AgNO3 is very soluble at room temperature.
(b) Na3PO4 is very soluble at room temperature.
(c) (NH4)2CO3 is very soluble at room temperature.
(d) PbSO4 is slightly soluble at room temperature.
(e) HI is very soluble at room temperature.
(f) CuCl is slightly soluble at room temperature.
(g) BaS is very soluble at room temperature.
(h) Ca(NO3)2 is very soluble at room temperature.
41. (a) Step 1. Write the formula equation for the reaction.
K2CO3(aq) + 2 AgNO3(aq) → 2 KNO3(aq) + Ag2CO3(s)
Step 2. Write the total ionic equation.
2 K+(aq) + CO32–(aq) + 2 Ag+(aq) + 2 NO3–(aq) → 2 K+(aq) + 2 NO3–(aq) + Ag2CO3(s)
Step 3. Cancel the spectator ions and write the net ionic equation.
CO32–(aq) + 2 Ag+(aq) → Ag2CO3(s)
(b) Step 1. Write the formula equation for the reaction.
CaCl2(aq) + Pb(NO3)2(aq) → PbCl2(s) + Ca(NO3)2(aq)
Step 2. Write the total ionic equation.
Ca2+(aq) + 2 Cl–(aq) + Pb2+(aq) + 2 NO3–(aq) → PbCl2(s) + Ca2+(aq) + 2 NO3–(aq)
Step 3. Cancel the spectator ions and write the net ionic equation.
2 Cl–(aq) + Pb2+(aq) → PbCl2(s)
(c) Step 1. Write the formula equation for the reaction.
2 H3PO4(aq) + 3 CaCl2(aq) → Ca3(PO4)2(s) + 6 HCl(aq)
Step 2. Write the total ionic equation.
6 H+(aq) + 2 PO43–(aq) + 3 Ca2+(aq) + 6 Cl–(aq) → Ca3(PO4)2(s) + 6 H+(aq) + 6 Cl–(aq)
Step 3. Cancel the spectator ions and write the net ionic equation.
2 PO43–(aq) + 3 Ca2+(aq) → Ca3(PO4)2(s)
(d) Step 1. Write the formula equation for the reaction.
Hg2(NO3)2(aq) + 2 NaOH(aq) → Hg2(OH)2(s) + 2 NaNO3(aq)
Step 2. Write the total ionic equation.
Hg22+(aq) + 2 NO3–(aq) + 2 Na+(aq) + 2 OH–(aq) → Hg2(OH)2(s) + 2 Na+(aq) + 2 NO3–(aq)
Step 3. Cancel the spectator ions and write the net ionic equation.
Hg22+(aq) + 2 OH–(aq) → Hg2(OH)2(s)
(e) Step 1. Write the formula equation for the reaction.
6 HCl(aq) + 2 Fe(s) → 2 FeCl3(aq) + 3 H2(g)
Step 2. Write the total ionic equation.
6 H+(aq) + 6 Cl–(aq) + 2 Fe(s) → 2 Fe3+(aq) + 6 Cl–(aq) + 3 H2(g)
Step 3. Cancel the spectator ions and write the net ionic equation.
6 H+(aq) + 2 Fe(s) → 2 Fe3+(aq) + 3 H2(g)
42. The cations NH4+, K+, and Na+ are always spectator ions when they are in a chemical
reaction, as is the anion NO3–.
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Chapter 9: Solutions and Their Reactions
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43. (a) Step 1. Write the formula equation for the reaction.
Hg2(NO3)2(aq) + 2 KCl(aq) → Hg2Cl2(s) + 2 KNO3(aq)
Step 2. Write the total ionic equation.
Hg22+(aq) + 2 NO3–(aq) + 2 K+(aq) + 2 Cl–(aq) → Hg2Cl2(s) + 2 K+(aq) + 2 NO3–(aq)
Step 3. Cancel the spectator ions and write the net ionic equation.
Hg22+(aq) + 2 Cl–(aq) → Hg2Cl2(s)
(b) Step 1. Write the formula equation for the reaction.
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
Step 2. Write the total ionic equation.
Mg(s) + 2 H+(aq) + 2 Cl–(aq) → Mg2+(aq) + 2 Cl–(aq) + H2(g)
Step 3. Cancel the spectator ions and write the net ionic equation.
Mg(s) + 2 H+(aq) → Mg2+(aq) + H2(g)
(c) Step 1. Write the formula equation for the reaction.
2 K(s) + MgSO4(aq) → K2SO4(aq) + Mg(s)
Step 2. Write the total ionic equation.
2 K(s) + Mg2+(aq) + 2 SO4–(aq) → 2 K+(aq) + 2 SO4–(aq) + Mg(s)
Step 3. Cancel the spectator ions and write the net ionic equation.
2 K(s) + Mg2+(aq) → 2 K+(aq) + Mg(s)
44. (a) Physical contaminants mixed with water is a heterogeneous mixture.
(b) Biological contaminants mixed with water is a heterogeneous mixture.
(c) Chemical contaminants mixed with water is a homogeneous mixture.
45. (a) Twigs and pieces of plastic are removed during the collection step.
(b) Suspended particles are removed during the coagulation, flocculation, and
sedimentation step.
(c) Sand and clay are removed during the filtration step.
(d) Calcium and magnesium ions are removed during the softening step.
46. (a) Three micro-organisms that may contaminate a water source include bacteria, viruses,
and protozoa.
(b) The parts of the water purification process that kill micro-organisms are the disinfection and
post-chlorination steps.
(c) Filtration actually removes biological impurities from the water.
47. The quantity of each material added during the water treatment process is carefully
controlled because there are strict guidelines to ensure that any substance in the water is at a
level that is safe for human health.
48. Spectroscopy is used to determine the composition of various substances and the
concentration of the different substances.
49. Two steps of the water treatment process that could remove some medications from water
sources are chlorination and filtration.
50. (a) The formula equation for the reaction:
2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)
(b) To find the amount of sodium hydroxide present in the sample of solution, the volume of
sodium hydroxide (in L) is multiplied by the amount concentration (in mol/L). The equation is
nNaOH = cNaOHVNaOH
(c) To find the mole ratio, use the balanced formula equation.
(d) To convert the amount of sodium sulfate produced to the mass of sodium sulfate, multiply
the amount of the compound (in mol) by the molar mass of the compound (in g/mol).
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Chapter 9: Solutions and Their Reactions
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(e) To find the concentration of the final sodium sulfate solution, divide the amount of sodium
n
sulfate with the total volume of the solution. The equation is c = .
V
51. (a) The amounts of each reagent can be calculated from the concentrations and volumes
given. Using the dissociation equation the ratio of moles can be determined for the two reagents
and the amounts can be compared. The smallest amount determines the limiting reagent.
(b) Given: cHCl = 0.10 mol / L
VHCl = 450 mL
cNaOH = 0.10 mol / L
VNaOH = 250 mL
Required: the limiting reagent
Solution:
Step 1. Convert the volumes of the solutions to litres.
1L
VHCl = 450 mL !
1000 mL
VHCl = 0.45 L
VNaOH = 250 mL !
1L
1000 mL
VNaOH = 0.25 L
Step 2. Determine the amounts of the two reagents present.
nHCl = cHCl ! VHCl
0.10 mol
! 0.45 L
1L
= 0.045 mol
nHCl =
nHCl
nNaOH = cNaOH ! VNaOH
0.20 mol
! 0.25 L
1L
nNaOH = 0.050 mol
Step 3. Write the dissociation equation.
HCl(aq) +
NaOH(aq)
→ NaCl(s) + H2O(aq)
nHCl = 0.045 mol
nNaOH = 0.050 mol
Step 4. Since the ratio of amounts given by the dissociation equation is 1:1, the reagent with the
lesser amount is the limiting reagent. This corresponds to hydrochloric acid.
Statement: The limiting reagent is hydrochloric acid.
nNaOH =
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Chapter 9: Solutions and Their Reactions
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(c) Given: The amount of the limiting reagent nHCl = 0.045 mol
Required: the theoretical yield in moles and mass
Solution:
Step 1. Using the dissociation equation given in (b), calculate the amount of sodium chloride
(theoretical yield).
1 mol NaCl
nNaCl = 0.045 mol HCl !
1 mol HCl
nNaCl = 0.045 mol
Step 2. Using the molar mass of NaCl, calculate the mass.
58.44 g
mNaCl = 0.045 mol !
1 mol
mNaCl = 2.6 g
Statement: The theoretical yield of sodium chloride is 0.045 mol or 2.6 g.
52. (a) (NH4)2SO4(aq) → 2 NH4+(aq) + SO42–(aq)
1 mol
2 mol
1 mol
(b) The ratio of ammonium ions to sulfate ions in the solution is 2:1.
(c) Since there are twice as many moles of ammonium ions, the final concentration of
ammonium ions is two times the initial concentration of ammonium sulfate.
(d) The final concentration of sulfate ions is the same as the initial concentration of ammonium
sulfate.
Analysis and Application
53. (a) The net ionic equation for a neutralization reaction is H+(aq) + OH–(aq) → H2O(l).
(b) One major advantage of using the total ionic equation for this reaction compared to using the
net ionic equation is that the total ionic equation gives more information about the reaction.
Since many neutralization reactions have the same net ionic equation, the equation does not give
a lot of information about the reaction. The total ionic equation shows all the entities present and
identifies the reactant compounds.
54. (a) The formula equation:
Fe(s) + Cu(NO3)2(aq) → Cu(s) + Fe(NO3)2(aq)
(b) The total ionic equation:
Fe(s) + Cu2+(aq) + 2 NO3–(aq) → Cu(s) + Fe2+(aq) + 2 NO3–(aq)
(c) The net ionic equation:
Fe(s) + Cu2+(aq) + 2 NO3–(aq) → Cu(s) + Fe2+(aq) + 2 NO3–(aq)
55. (a) The most likely contaminant is fertilizer, which releases nitrates and phosphates into
the water.
(b) The water turned green because the fertilizers promoted the growth of aquatic plants, which
results in algal blooms.
(c) The fish died because there was not enough oxygen in the water. When the plants died,
decomposers used most of the oxygen in the water during the decomposition process. This
reduced the concentration of oxygen in the water, causing the fish to die.
(d) To resolve this problem, minimal amounts of fertilizers could be used in the surrounding area
to minimize the amount of fertilizer runoff in the pond. Proper disposal of lawn clippings and
vegetation would also help limit the amount of fertilizer in the pond.
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Chapter 9: Solutions and Their Reactions
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56.
57. If the goal were to purify water rather than make it potable, the water treatment process
would have to be more rigorous and would involve boiling the water. Potable water contains
some substances in levels that do not pose a threat to human health. Purified water contains no
substances.
58. Answers may vary. Sample answer: From the table, we can see that potassium ions are
always very soluble, so we can assume that the potassium ion remains in solution regardless of
what compound is added. Therefore, we can focus on separating the other three ions. Since the
chloride ion forms a precipitate with silver ions but not the other two metal ions, we could first
add a soluble chloride solution, such as sodium chloride, to the solution. We could then filter out
the silver chloride precipitate. Since calcium is very soluble with sulfide, but iron is not, we
could add a soluble sulfide compound, such as ammonium sulfide to the solution and filter out
the iron precipitate. We could then add either a hydroxide, such as sodium hydroxide, or a
carbonate, such as sodium carbonate, to form a precipitate with the calcium. When this
precipitate is filtered out, the potassium ion is left in solution.
59. Answers may vary. Sample answer: Electroplating metals can be used to remove metal ions
from contaminated water by adding electrons to change the cations to neutral metal atoms. Two
metal electrodes are placed into the water. They are connected by wires and a battery. As electric
current passes through the system the negative electrode supplies electrons to the positive metal
ions in solution and the metal forms on the electrode. This metal can then be reclaimed and used.
60. The only ion listed in Table 1 that gives a colourless flame is the hydrogen ion. Hydrogen is
therefore the cation present in the acid.
61. You need to add excess anions to make sure that all of the cations are precipitated from the
solution. If too few anions are added, some cations will remain in solution.
62. (a) To make a log burn with flames of different colours, ionic compounds containing a
variety of cations could be added to the fuel.
(b) (i) A bright red flame indicates the presence of lithium ions and strontium ions.
(ii) A yellow-green flame indicates the presence of barium ions.
(iii) A violet flame indicates the presence of potassium ions.
(iv) A green flame indicates the presence of copper ions.
63. Given: cKOH = 0.35 mol/L
cCa(NO
3 )2
VCa(NO
3 )2
= 0.20 mol/L
= 400 mL
Required: volume of potassium hydroxide, VKOH
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Chapter 9: Solutions and Their Reactions
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Solution:
Step 1. Convert volume of the solution to litres.
1L
VCa(NO ) = 400 mL !
3 2
1000 mL
VCa(NO ) = 0.4 L
3 2
Step 2. Write a dissociation equation listing the calculated amounts and the required value(s).
Ca(NO3)2(aq)
+ 2 KOH(aq)
→ Ca(OH)2(s) + 2 KNO3(aq)
cCa(NO ) = 0.20 mol/L
c
= 0.35 mol/L
KOH
3 2
VCa(NO ) = 0.4 L
VKOH
3 2
Step 3. Determine the amount of solute.
n
c=
V
nCa(NO ) = cCa(NO ) VCa(NO )
3 2
3 2
3 2
0.20 mol
! 0.4 L
1L
= 0.08 mol
=
nCa(NO
3 )2
Step 4. Determine the amount of potassium hydroxide.
2 mol KOH
nKOH = 0.08 molCa(NO ) !
3 2
1 molCa(NO )
3 2
nKOH = 0.16 mol KOH [1 extra digit carried]
Step 5. Determine the volume of potassium hydroxide by rearranging the equation.
n
VKOH = KOH
cKOH
=
0.16 mol
0.35 mol
L
VKOH = 0.5 L
Statement: The minimum volume of potassium hydroxide needed to precipitate the calcium ions
is 0.5 L.
Evaluation
64. (a) Answers may vary. Sample answer: A formula equation should be shown first, then a
total ionic equation, and finally a net ionic equation. This order represents listing the equations in
order from showing all compounds to showing all ions to showing just the chemical change
between ions.
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Chapter 9: Solutions and Their Reactions
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(b) Answers may vary. Sample answer: A double-displacement reaction that is a precipitation
reaction is the reaction between silver nitrate and sodium chloride.
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
(c) Answers may vary. Sample answer: The student can use the magnetic letter, numbers, and
symbols to make each type of equation. For the reaction between silver nitrate and sodium
chloride, the student would first use the magnets to make a formula equation.
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
The student could then make a total ionic equation showing how the ionic compounds dissociate.
Ag+(aq) + NO3–(aq) + Na+(aq) + Cl–(aq) → AgCl (s) + Na+(aq) + NO3–(aq)
The student can then take away the spectator ions from both sides of the equation to show the net
ionic equation.
Ag+(aq) + Cl–(aq) → AgCl(s)
65. Joaquin adds the chloride solution first, which precipitates the lead(II) ions, but not the
barium or sodium ions. Adding the sulfate solution to the filtrate and then filtering again removes
the barium ions, but not the sodium ions. Kim adds the sulfate solution first, which precipitates
with both the lead(II) ions and the barium ions. Therefore, Joaquin’s procedure will separate all
of the ions.
66. (a) Adding chloride ions to silver ions and lead(II) ions will not separate the ions since both
cations will precipitate with the chloride ions.
(b) Answers may vary. Sample answer: To separate the ions, we need a solution that precipitates
one of the two cations. From Table 1 in Section 9.1, we can see that acetate anion, C2H3O2–,
forms a precipitate with silver, but not lead(II). Using the acetate ion would remove the silver
ions, leaving the lead(II) ions in solution.
67. Answers may vary. Sample answer: To minimize sodium contamination in a flame test,
avoid touching the wire loop or the inside of any glassware. Use only distilled water. Rinse the
glassware with distilled water before adding the cation solution. Heat the wire loop in a flame to
burn off any sodium ions before testing the sample solution.
Reflect on Your Learning
68. Answers may vary. Sample answer: The statement “Even goldfish are slightly soluble” is
correct. When a goldfish dies in water, the body breaks down in the water. Some of the particles
go into solution, which means that some of the goldfish is soluble.
69. (a) Answers may vary. Sample answer: Two compounds that have a similar solubility to that
of barium sulfate are silver sulfate, Ag2SO4, and lead(II) sulfate, PbSO4, which are both listed as
slightly soluble.
(b) Answers may vary. Sample answer: Other factors that would need to be considered when
choosing a compound used in tests of the human GI tract are that it does not react with stomach
acid and that it effectively blocks X-rays.
70. (a) Answers may vary. Sample answer: If the sanitation and water quality decreased, the
spread of disease would increase. People would get sick and would not be able to work or go to
school. The lack of clean water would also adversely affect agriculture, which means that the
quality of food would decrease.
(b) Answers may vary. Sample answer: To ensure the availability of safe water for northern
communities, frequent testing and treatment of water sources will prevent health problems. New
wells could be built in locations less likely to be contaminated. Sources of contamination could
be isolated from sources of drinking water.
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Chapter 9: Solutions and Their Reactions 9-10
71. (a) Answers may vary. Sample answer: My water is hard. The evidence is that soap scum
forms when soap mixes with the water and the soap does not lather well.
(b) The benefit of a home water softener is that it replaces the ions that make the water hard with
sodium ions. This makes the water soft, which has many benefits. It is easier to wash clothes in
soft water because a scum does not form. Clothes and skin will feel softer using soft water. Pipes
and appliances will have less scale build-up and will operate more efficiently.
72. (a) To determine if the tap water in a municipality contains fluoride, check the community’s
website for information on water. For example, in Ontario, the municipal government decides if
fluoride will be added to the water.
(b) Fluoride is added to water supplies to prevent tooth decay and to keep teeth healthy.
73. It is more important that the amount of barium ion be measured exactly when preparing
barium sulfate. Barium ions are toxic, and if the amount of barium is known, excess sulfate can
be added.
74. (a) Answers may vary. Students’ answers should include a statement explaining how
medications are disposed of in their family. Medications could be disposed of in the garbage or
returned to the pharmacy for proper disposal.
(b) Answers may vary. Students’ answers could include taking the medications to the pharmacy
or to the hazardous waste depot.
75. Step 1. Write the dissociation equation for calcium chloride and sodium chloride.
CaCl2(s) → Ca2+(aq) + 2 Cl–(aq)
NaCl(s) → Na+(aq) + Cl–(aq)
Step 2. Convert the concentration of each compound into concentration of cations. Assume there
is 1 mol of each compound
molCaCl
mol NaCl 1 mol +
2 molCa +
2
Na
cNa + = 1
!
cCa + = 1
!
L
L
1 mol NaCl
1 molCaCl
2
cNa + = 1 mol/L
cCa + = 2 mol/L
Statement: The amount concentration of calcium cations is twice the amount concentration of
sodium cations. This means that calcium chloride would be better at preventing the formation of
ice on roads in the winter.
Research
76. (a) Answers may vary. Sample answer: An example of an acid–base reaction is the reaction
between hydrochloric acid and sodium hydroxide. This reaction is a neutralization reaction.
Step 1. Write the formula equation for the reaction.
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
Step 2. Write the total ionic equation.
Na+(aq) + OH–(aq) + H+(aq) + Cl–(aq) → Na+(aq) + Cl–(aq) + H2O(l)
Step 3. Cancel the spectator ions and write the net ionic equation.
Na+(aq) + OH–(aq) + H+(aq) + Cl–(aq) → Na+(aq) + Cl–(aq) + H2O(l)
Step 4. Name the molecular compound that forms.
Water forms during this reaction.
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Chapter 9: Solutions and Their Reactions 9-11
(b) Answers may vary. Sample answer: An example of the reaction of an acid with a carbonate is
the reaction between sodium carbonate and hydrochloric acid.
Step 1. Write the formula equation for the reaction.
Na2CO3(s) + 2 HCl(aq) → 2 NaCl(aq) + CO2(g) + H2O(l)
Step 2. Write the total ionic equation.
2 Na+(aq) + CO3–(aq) + 2 H+(aq) + 2 Cl–(aq) → 2 Na+(aq) + 2 Cl–(aq) + CO2(g) + H2O(l)
Step 3. Cancel the spectator ions and write the net ionic equation.
2 Na+(aq) + CO3–(aq) + 2 H+(aq) + 2 Cl–(aq) → 2 Na+(aq) + 2 Cl–(aq) + CO2(g) + H2O(l)
Step 4. Name the molecular compound that forms.
Carbon dioxide and water forms during this reaction.
(c) Answers may vary. Sample answer: An example of the reaction of an acid with a sulfide is
the reaction between sodium sulfide and hydrochloric acid. In this type of reaction, hydrogen
sulfide gas forms.
Step 1. Write the formula equation for the reaction.
Na2S(aq) + 2 HCl(aq) → 2 NaCl(aq) + H2S(g)
Step 2. Write the total ionic equation.
2 Na+(aq) + S2–(aq) + 2 H+(aq) + 2 Cl–(aq) → 2 Na+(aq) + 2 Cl–(aq) + H2S(g)
Step 3. Cancel the spectator ions and write the net ionic equation.
2 Na+(aq) + S2–(aq) + 2 H+(aq) + 2 Cl–(aq) → 2 Na+(aq) + 2 Cl–(aq) + H2S(g)
Step 4. Name the molecular compound that forms.
Hydrogen sulfide gas forms during this reaction.
77. Answers may vary. Sample answer: The source of water for Ottawa is the Ottawa River. The
river water is soft and has no need for softening. There are two purification plants for Ottawa’s
water at Britannia and Lemieux Island. At these locations, particulates are removed and the water
is sanitized.
78. (a) Materials other than ionic compounds are used to produce colours in fireworks. For
example, silver colour results from the presence of the metals magnesium, titanium, or
aluminum. A gold colour comes from iron with some form of carbon.
(b) Different cations are responsible for different colours in fireworks. Table 1 lists the different
substances and the colours that they produce.
Table 1 Cations and the Colours They Produce
Cation
Colour
strontium
red
lithium
red
calcium
orange
sodium
yellow
barium and chlorine
green
barium and oxygen
white
copper chlorine
blue
strontium and copper
violet
(c) Metals such as magnesium, titanium, and aluminum burn hot enough to increase the
temperature of the fireworks so that the cations can produce their colours.
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Chapter 9: Solutions and Their Reactions 9-12
79. Table 2 Indicators Used in Qualitative Analysis
Substance for Appearance of a
Indicator
which it tests
positive test
• final colour of
solution depends on
amount of precipitate
(a) Benedict’s • reducing
formed
solution
sugars
• increasing amounts
of sugar changes
colour from green to
orange to red to brown
(b) Biuret
solution
• proteins
(c) Sudan III
stain
• fats in liquids
Copyright © 2011 Nelson Education Ltd.
Quantitive
Purpose of results information
yes
• used to detect
glucose in urine
for diabetes
• used to identify
no
• in the presence of
proteins in urine to
protein, blue solution
diagnose kidney
changes to pink-purple
disorders
• used to determine no
the fat content in
• in the presence of
fat, the stain will turn
foods and can be
used to test stool
red
samples for fats
Chapter 9: Solutions and Their Reactions 9-13