6.1 DEFINING A SOLUTION - Quia · 6.1 DEFINING A SOLUTION PRACTICE ... a substance that dissolves in water to form a conducting solution ... N 2(g) or P 4(s)); compounds consisting

Copyright © 2002 Nelson Thomson Learning Chapter 6 The Nature and Properties of Solutions 165

3. The dissolving of both salt and sugar involves the solid separating into particles too small to see. The salt solutioncontains ions of sodium and chlorine and will conduct a current, while the sugar dissolves to release sugar molecules,so its solution will not conduct electricity.

Try This Activity: Substances in Water

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(a) The potassium permanganate, sugar, and ethanol dissolve.(b) Sugar and ethanol are certainly soluble, as they disappear completely. The solubility of potassium permanganate is

less certain, as some of it remains undissolved.Note: Students may be uncertain about any substance that does not “disappear” completely upon dissolving, becausethey rarely encounter this. Expect discussion about potassum permanganate, if they had some remain in solid state.Students may be uncertain if they speculate about whether some calcium carbonate or vegetable oil dissolves, eventhough there is no visible reduction of the original phase. They have only visible evidence of sample “shrinking” togo on, where no colour change is involved.

(c) The calcium carbonate and vegetable oil do not dissolve.(d) We cannot be entirely certain, as a small amount may have dissolved.(e) Properties are different: solutions are visibly homogeneous. Some other properties that might differ include electrical

conductivity, acidity, melting/freezing points, viscosity, and so forth. (f) Acidity could be tested with pH paper or conductivity with a multimeter.

Note: Tests listed by students should be consistent with their answers to (e).

6.1 DEFINING A SOLUTION

PRACTICE

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Understanding Concepts1. (a) Heterogeneous: different substances are visible.

(b) Homogeneous: only one phase is visible.(c) Homogeneous if it has been decanted; if not, there may be sediment in the bottle and the red wine would then be

considered heterogeneous.(d) Heterogeneous if corroded; if clean, bronze appears homogeneous.(e) Homogeneous: the metal looks all the same throughout.(f) Heterogeneous if corroded; otherwise it is homogeneous.(g) Humid air is usually homogeneous; however, when cloud, fog, or rain forms, the solution is heterogeneous.(h) Heterogeneous: the suspended droplets of water make it opaque.(i) Heterogeneous: the water is not clear.

2. The solutions are (b), (d), (h) and (i).(a), (c), (e), (f) and (g) are not solutions.

3. Solutions may be classified by type of solvent, by electrical conductivity, by acidity, by colour, or by physical state atroom conditions. Even categories such as viscosity, volatility, etc., can be used to classify substances.

4. (a) An aqueous solution is one in which the solvent is water.(b) Aqueous solutions found around the home will be substances such as shampoo, vinegar, syrup, clear fruit juices,

tea, bleach, drain cleaners.5. Methanol is a nonelectrolyte (it is a nonacidic molecular substance); sodium chloride is an electrolyte (it dissolves to

release ions); hydrochloric acid is an electrolyte (acids are the only molecular substances to conduct electricity); andpotassium hydroxide is an electrolyte (it is ionic).

6. (a) Electrolyte solutes include soluble ionic compounds (including ionic hydroxides) and acids.(b) electrolyte: a substance that dissolves in water to form a conducting solution

7. (a) Acidic solutions have acid solutes.(b) Basic soutions have ionic hydroxide solutes. (c) Neutral solutions have molecular solutes (other than acids) or ionic solutes (other than ionic hydroxides).

8. (a) Electrolytes: citric acid, salt (assume sodium chloride), sodium citrate, and monosodium phosphate (4 of the 11substances listed).

166 Unit 3 Solutions and Solubility Copyright © 2002 Nelson Thomson Learning

Nonelectrolytes: water, liquid sugar, glucose-fructose, and brominated vetgetable oil (4 of the 11 substanceslisted).The remaining ingredients (natural flavour, colour, and ester gum) cannot be classified at this stage. Although there are the same number of ingredients that are electrolytes and nonelectrolytes, there is a greaterquantity of nonelectrolytes than electrolytes.

(b) In this product salt (sodium chloride) and sodium citrate contain sodium ions; only the monopotassium phosphatecontains potassium ions. The drink therefore contains more sodium ions than potassium ions.

(c) All of the food energy in the drink comes from carbohydrates—essentially from sugars.(d) The drink attempts to replace water, energy, and alkali metal ions that are lost from the body during physical

activity.

SECTION 6.1 QUESTIONS

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Understanding Concepts1. (a) An appropriate table might be:

Solute

Solvent Solid Liquid Gas

Solid brass or solder Hg amalgams lead in air

Liquid sugar syrup vinegar soda pop

Gas oxygen in ice humid air air

(b) Solutions of gaseous, liquid, and solid solutes in liquid solventsNote: Some students might correctly suggest gas in gas as being a very common solution.

2.

3. Examples include gasoline, lubricating oils, Varsol, mineral spirits, and turpentine.4. The solvent in alkyd paints is a mixture of oils. Any thinner/cleaner for alkyd paints must be an oil, such as mineral

oil or Varsol. Nonpolar alkyd paint components dissolve in nonpolar solvents. Water is used as a thinner/cleaner(solvent) or diluent for latex paints because water is a component of the paint mixture. Polar latex paint componentsdissolve in polar water molecules. Since water and oil will not dissolve in each other (polar does not dissolve innonpolar), it is critical not to mix the two types of solvent.

5. (a) electrolyte(b) nonelectrolyte(c) electrolyte(d) nonelectrolyte

6. (a) HCl(aq): acidic solution; blue litmus turns pink.(b) NaOH(aq): basic solution; pink litmus turns blue.

liquid

aqueoussolution

homogenousmixture

watersolute

solid

solutions

neutral

solvent

gas

acidbase

nonelectrolyteelectrolyte


(c) Methanol: neutral solution; neither colour of litmus changes from its original colour.(d) Sodium hydrogen carbonate: neutral solution; neither colour of litmus changes from its original colour.

Applying Inquiry Skills7. (a)

Predicting Properties of Compounds

Substance Acidic/Basic/Neutral Electrolyte/Nonelectrolyte

C3H7OH(l) (a rubbing alcohol) neutral nonelectrolyte

calcium hydroxide (slaked lime) basic electrolyte

H3PO4(aq) (for manufacturing fertilizer) acidic electrolyte

glucose (a product of photosynthesis) neutral nonelectrolyte

sodium fluoride (in toothpaste) neutral electrolyte

(b) Each compound is dissolved in pure water, and the resulting solutions are tested with litmus. They are also testedfor electrical conductivity.

Making Connections8. Using gasoline as a cleaner in a basement is unsafe because the liquid is very volatile (evaporates very readily), and

the vapours are very flammable. Gasoline should only be used in a very well-ventilated area, with no sources of igni-tion anywhere nearby — preferably outdoors.

Reflecting9. If water really were a universal solvent, everything (and everyone) on Earth would be part of a huge unchanging

sphere of homogeneous solution.

6.2 EXPLAINING SOLUTIONS

PRACTICE

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Understanding Concepts1. Most molecules are polar. The four categories of polar molecules are:

AB (e.g., HCl(g), CO(g))

NxAy (e.g., NH3(g), NF3(g))

OxAy (e.g., H2O(l), OCl2(g))

CxAyBz (e.g., CHCl3(l), C2H5OH(l))

2. Nonpolar molecules are those in which no part of the molecule is significantly more (or less) electronegative than anyother part. The two categories of nonpolar molecules aremolecular elements (e.g., N2(g) or P4(s));compounds consisting only of carbon and one other type of atom, with a general formula CxAy (e.g., CH4(g), CO2(g),and C8H18(l)).

PRACTICE

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Understanding Concepts3. Intramolecular forces act between atoms within a molecule; intermolecular forces act between molecules.4. (a) No, gasoline will not dissolve in water. The film floating on puddles of water at gas stations is evidence that

supports this statement.(b) the “like dissolves like” rule(c) Since gasoline is nonpolar and water is highly polar, the two liquids would not be miscible.


5. (a) The —OH groups of both methanol and water molecules allow hydrogen bonds to form among them, so theseliquids will dissolve well in each other.

(b)

(c) The more hydrogen bonds that can form, the higher the solubility should be. This occurs because the moleculeswill attract each other better if they can form many intermolecular bonds.

PRACTICE

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Understanding Concepts6. (a) Observations might include: the solids disappear; the change of refraction of water near each solid as it dissolves;

and the solutions are colourless. Depending on the type of salt used, the rate of dissolving will vary. If table saltis used, the solution remains cloudy because virtually all brands of table salt have insoluble additives that help tokeep the crystals from bonding together.

(b) In both cases, the solids are separated into particles too small to see, theoretically, by the attraction of the watermolecules.

(c) In theory, the solutions are different because the particles in the sugar solution are neutral molecules, and thosein the salt solution are ions.

(d) Theoretically, the high polarity of water molecules explains why they strongly attract ions in ionic compounds,and the presence of many —OH groups on a sugar molecule allows a lot of hydrogen bonding (hence high solu-bility) with water.

7. (a) NaF(s) → Na+(aq) + F (aq)

–

(b) Na3PO4(s) → 3 Na+(aq) + PO3–

4(aq)

(c) KNO3(s) → K+(aq) + NO–

3(aq)

(d) Al2(SO4)3(s) → 2 Al3+(aq) + 3 SO2–

4(aq)

(e) (NH4)2HPO4(s) → 2 NH+4(aq) + HPO2–

4(aq)

(f) CoCl2�6H2O(s) → Co2+(aq) + 2 Cl–(aq)

PRACTICE

(Page 279)

Understanding Concepts8. Answers will vary but may include the following.

Empirical properties: melting and boiling points; heat capacity; colour and clarity; excellence as a solvent; expansionupon solidification, formation of hexagonal crystals from vapour state; necessity for life; other observable,measurable properties. Theoretical properties: formation of hydrogen bonds; dissociation of dissolved ionic compounds, explaining theexpansion of ice and hexagonal-form snowflakes by theorizing V-shaped molecules that hydrogen bond to each otherin 6-sided patterns; other imagined, explanatory properties.

9. A nonpolar solvent such as mineral spirits (Varsol or turpentine) should dissolve grease, according to the “likedissolves like” rule, because grease is a nonpolar substance.

10. (a) C6H12O6(s), because it is polar.

O

H O

H

O

— C — O

H

H

H

H H

H

H


(b) C2H5OH(l), because it has hydrogen bonding.(c) Na2CO3(s), because it is an ionic compound containing sodium ions.

11. (a) CH3Cl(l), CH2Cl2(l), and CHCl3(l) are most likely to be water soluble because they are polar molecules. (b) CH4(g) and CCl4(l) are most likely to be soluble in nonpolar solvents because they are nonpolar.

12. A logical hypothesis would be that butanol has more C and H atoms, which reduce its polarity, making it more solublein nonpolar solvents.

Applying Inquiry Skills13. Explanations are accepted by the scientific community only if they are logical, relatively simple, and are consistent

with evidence and other related explanations.

Making Connections14. Dry cleaning solvents are chosen for their ability to dissolve nonpolar “dirt and grease” from clothing and to evapo-

rate completely from the fabric. Health regulations apply to all such chemicals — most are toxic/noxious substances.One such solvent that was used for years both domestically and commercially is carbon tetrachloride, CCl4(l). Carbontetrachloride was subsequently found to cause liver damage. Its first substitute, trichloroethene, C2HCl3(l), was alsofound to produce liver damage. Eventually trichloroethane, C2H3Cl3(l), was found to be a safe replacement that neithercaused liver damage nor was carcinogenic.

GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.


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Understanding Concepts1. Water is an effective solvent because its molecules are very polar. This polarity enables it to separate the particles of

the solute, surrounding each of them with water molecules. The polar —OH bonds result in polar molecules thatstrongly attract any charged ion or other polar molecules.A second reason that water is an effective solvent for a few solutes is hydrogen bonding. Although the number of solutesthat can hydrogen bond to water in solution is limited, solutes that have an unpaired electron on a N, O, or F atom, orhave a hydrogen bonded to an N, O, or F atom, have much higher than expected (from polarity) solubility in water.

2. (a) Soluble combinations are (iii), (iv), and (v).(b) Empirical: answers based on observations are by definition empirical.

3. (a) Not soluble: one is polar and the other is nonpolar.(b) Soluble: both are polar.(c) Not soluble: one is nonpolar and the other is polar.(d) Soluble: both have hydrogen bonding.(e) Not soluble: one is nonpolar and the other is polar.

4. Methane, methanol, ammonia. Methane is nonpolar, methanol has hydrogen bonding from one —OH bond, andammonia has hydrogen bonding from three —NH bonds.

Applying Inquiry Skills5. The experimental design is unacceptable because it involves too many variables; the presence of water will make it

impossible to decide anything about the mutual solubility of the other two substances.

6. QuestionWhich of C6H6(l) and C6H5OH(l) is more soluble in water?

PredictionC6H5OH(l) is predicted to be more soluble.

Experimental DesignEach liquid will be added, in 1-mL increments, to 100 mL of water (in separate beakers) while stirring, until no morewill dissolve.

Materials• C6H6(l) and C6H5OH(l)

• two 250-mL beakers

• pure water


• 10-mL graduated cylinder

• 100-mL graduated cylinder

• stirring rod

Making Connections7. Children’s glue must be nontoxic and water washable. This means the chemical substance(s) must be unreactive and

must have very polar molecules.

6.3 SOLUTION CONCENTRATION

PRACTICE

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Understanding Concepts1. W/W (weight to weight), W/V (weight to volume), or V/V (volume to volume) ratios2. v ethanol � 4.1 L

v gasohol � 55 L

c ethanol � �45.51

LL

� � 100%

c ethanol �7.5% V/V

The ethanol concentration in fuel solution is 7.5% V/V (by volume).

3. m zinc chloride � 16 g

v solution � 50 mL

c zinc chloride � �50

16mg

L� � 100%

c zinc chloride � 32% W/V

The concentration of zinc chloride in the flux solution is 32% W/V.

4. m zinc � 1.7 g

m brass � 35.0 g

c zinc � �315.7.0

gg

� � 100%

c zinc � 4.9% W/W

The zinc concentration in the brass is 4.9% W/W (by mass).

5. 8 ppm � 8 mg/L

m oxygen � �81mL�g

� � 1 L�

m oxygen � 8 mg

The mass of oxygen in each litre of water is 8 mg.6. m formaldehyde � 3.2 mg

m air � 0.59 kg

c formaldehyde � �03..529mkgg

�

c formaldehyde � 5.4 mg/kg � 5.4 ppm

The concentration of formaldehyde in air is 5.4 ppm.

7. (a) 1 ppb is 1/1000 of 1 ppm, or 0.001 ppm.

(b) 1 ppb � 1 mg/106 mL (1 mg/kL)

� 1 mg/1000 L (1 mg/m3)

� 1 �g/L

� 1 �g/kg

� 1 ng/g

(c) 30 ppb � 30 �g/kg

8. n CaCl2� 0.11 mol

v CaCl2� 60 mL � 0.060 L

C CaCl2� �

00..10160

mLol

�

C CaCl2� 1.8 mol/L

The molar concentration of calcium chloride is 1.8 mol/L.

Making Connections9. LD50 refers to a toxic substance dosage in mass per kilogram of the recipient (e.g., milligrams per kilogram of body

mass) that will prove lethal to half of the organisms in a test sample. Descriptors of toxicity vary, so students may wellfind differing definitions of “extremely” toxic and “slightly” toxic in their research. However, a common definitionfor extremely toxic is less than 5 mg/kg, while slightly toxic is 5–15 g/kg. Above 15 g/kg is considered to be almostnontoxic.

The quantity of toxin ingested depends on the concentration of the chemical to which one may be exposed.However, the LD50 levels express the concentration of toxin in the body (e.g., 15 mg/kg). Cumulative toxins may beingested at very low concentrations for long periods of time and become toxic over time. Examples of cumulativetoxins include mercury and lead.


Reflecting10. A report mark is a ratio of marks achieved/possible marks. Other common ratios are prices of bulk goods ($/kg),

speeds (km/h), nutritional information (energy per serving, in kJ/g) and cooking times (min/kg).

PRACTICE

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Understanding Concepts11. c C3H7OH � 70.0% V/V � 70.0 mL/100 mL

v C3H7OH � 500 mL

v C3H7OH � 500 mL� � �7100.00

mm

L�L

�

v C3H7OH � 350 mL

The volume of rubbing alcohol present is 350 mL.

12. c H2O2 � 3.0% W/V � 3.0 g/100 mL � 3.0 g/0.100 L

v H2O2 � 1000 bottles� � �

01.2b5o0ttl

Le�

� � 250 L

m H2O2 � 250 L� � �

0.31.000gL�

�

m H2O2 � 7.5 kg

The mass of hydrogen peroxide needed is 7.5 kg.

13. c F– � 1.5 ppm � 1.5 mg/L

v F– � 0.250 L


m F– � 0.250 L� � �1.

15

L�mg�

m F– � 0.38 mg

The mass of fluoride ions present in the glass of water is 0.38 mg.14. C MgCl2

� 0.055 mol/L

v MgCl2 � 75 L

n MgCl2� 75 L� � �

0.0515L�mol�

n MgCl2� 4.1 mol

The amount of magnesium chloride present is 4.1 mol.

15. C HCl � 5.0 mol/L

v HCl � 50 mL � 0.050 L

n HCl � 0.050 L� � �5.0

1mL�

ol�

n HCl � 0.25 mol

The amount of hydrogen chloride in the beaker is 0.25 mol.

16. C NH3 � 1.24 mol/L

n NH3� 0.500 mol

v NH3 � 0.500 mol��

1.214

Lmol��

v NH3 � 0.403 L

The volume of aqueous ammonia solution needed is 403 mL.

17. C Na2SO4 � 2.6 mol/L

n Na2SO4� 0.14 mol

v Na2SO4 � 0.14 mol��

2.61

mL

ol��

v Na2SO4 � 0.054 L � 54 mL

The volume of aqueous sodium sulfate solution needed is 54 mL.

Making Connections18. Prediluted commercial solutions are usually more expensive for equal quantities of solute. If cost is the concern, you

would calculate cost/quantity for the solute. On the other hand, prediluted solutions are both more convenient to useand less dangerous to store.

PRACTICE

(Page 290)

Understanding Concepts19. C NaOH � 0.125 mol/L

v NaOH � 3.00 L

M NaOH � 40.00 g/mol

n NaOH � 3.00 L� � �0.12

15L�mol�

� 0.375 mol

m NaOH � 0.375 mol� � �410.

m00

ol�g

�

m NaOH � 15.0 g


or

m NaOH � 3.00 L� � �0.12

15L�mol��

410.

m00

ol�g

�

m NaOH � 15.0 g

The mass of solid sodium hydroxide required is 15.0 g.

20. C NaCl � 0.56 mol/L

v NaCl � 5.0 L

M NaCl � 58.44 g/mol

n NaCl � 5.0 L� � �0.5

16

L�mol�

� 2.8 mol

m NaCl � 2.8 mol� � �518.

m44

ol�g

�

m NaCl � 0.16 kg

or

m NaCl � 5.0 L� � �0.5

16

L�mol��

518.

m44

ol�g

�

m NaCl � 0.16 kg

The mass of sodium chloride in the sample is 0.16 kg.

21. (a) c CO2 � 355 ppm � 355 mg/L

The mass of carbon dioxide present in a litre of acid rain is 355 mg.

(b) MCO2 � 44.01 g/mol

mCO2 � 0.355 g

vCO2 � 1.00 L

nCO2 � 0.355 g� � �

414.

m01

olg�

�

� 8.07 mmol

CCO2 � �

8.017.0

m0

mL

ol�

CCO2 � 8.07 mmol/L

The molar concentration of aqueous carbon dioxide is 8.07 mmol/L.

22. (a) m NaCl � 235 gv NaCl � 3.00 L � 3.00 � 103 mL

c NaCl ��3.00

2�

3510

g3 mL

�� 100%

c NaCl � 7.83% W/V

The percent concentration of sodium chloride in aqueous solution is 7.83% W/V.

(b) M NaCl � 58.44 g/mol

m NaCl � 235 g

v NaCl � 3.00 L

nNaCl � 235 g� � �518.

m44

olg�

�

� 4.02 mol

CNaCl � �43.0.020mLol

�

CNaCl � 1.34 mol/L

The molar concentration of aqueous sodium chloride is 1.34 mol/L.



(Page 290)

Understanding Concepts1. Consumer products often have percentage concentrations given on the label because these units are more familiar to

most consumers.2. (a) cdextrose � 5.0% W/V � 5.0 g/100 mL

vsolution � 500.0 mL

mdextrose � 500.0 mL� � �10

50.0

mgL�

�

mdextrose � 25 g

The mass of dextrose in the bag is 25 g.

(b) cdextrose � 5.0 g/100 mL � 50 g/L � 5.0 � 104 ppm

The concentration of dextrose in the solution is 5.0 � 104 ppm.

3. c NaCl � 31.6 g/100 mL

v NaCl � 250 mL

m NaCl � 250 mL� � �13010.6m

gL�

�

m NaCl � 79.0 g

The mass of sodium chloride that will dissolve is 79.0 g.

4. The nandrolone concentration of 1000 times 2 mg/L is 2 � 103 mg/L, so the test result concentration is 2 � 103 ppmor 2 ppt.

5. c PCBs � 18.9 mg/kg

m chick � 0.60 kg

m PCBs � 0.60 kg� � �18

1.9

kmg�

g�

m PCBs � 11 mg

The chick would contain 11 mg of PCBs.

6. c PCBs � 4.00 ppm � 4.00 mg/kg

m person � 64.0 kg

m PCBs � 64.0 kg� � �4.0

10kmg�

g�

m PCBs � 256 mg

The mass of PCBs in the person is 256 mg.

7. Assuming the 5 mL volume is a theoretical (exact) value, then the substance concentrations in g/L are, respectively

ammonium carbonate �1553

m�m�L

g� � 30.6 g/L

potassium bicarbonate �2567

m�m�L

g� � 53.4 g/L

menthol �252m�m�Lg

� � 4.4 g/L

camphor �25.2

m�m�Lg

� � 0.44 g/L


8. c each sol’n � 0.10 mol/Lv each sol’n � 100 mL � 0.100 L

n each compound � 0.100 L� � �0.1

10

L�mol� � 0.010 mol


M KCl � 74.55 g/mol

M CaCl2 � 110.98 g/mol

(a) m NaCl � 0.010 mol�� 518.

m44

ol�g

�

m NaCl � 0.58 g

orm NaCl � 0.100 L� � �

0.110

L�mol��

518.

m44

ol�g

�

m NaCl � 0.58 g

The mass of sodium chloride required is 0.58 g.

(b) m KCl � 0.010 mol�� 714.

m55

ol�g

�

m KCl � 0.75 g

orm KCl � 0.100 L� � �

0.110

L�mol��

714.

m55

ol�g

�

m KCl � 0.75 g

The mass of potassium chloride required is 0.75 g.

(c) m CaCl2 � 0.010 mol��

1110m.9

o8l�g

�

m CaCl2 � 1.1 g

orm CaCl2

� 0.100 L� � �0.1

10

L�mol��

1110m.9

o8l�g

�

m CaCl2 � 1.1 g

The mass of calcium chloride required is 1.1 g.

9. c NaCl � 25 g/100 mL � 25 g/0.100 L


C NaCl � �0.

21500

g�L

� � �518.

m44

olg�

�

C NaCl � 4.3 mol/L

The molar concentration of sodium chloride in the brine is 4.3 mol/L.

10. m C6H12O6 � 2.0 g

M C6H12O6 � 180.18 g/mol

C C6H12O6 � 0.055 mol/L

n C6H12O6 � 2.0 g� � �

1810m.1

o8lg�

�

n C6H12O6 � 0.011 mol

v C6H12O6 � 0.011 mol� � �

0.0515Lmol��

v C6H12O6 � 0.20 L

or

v C6H12O6 � 2.0 g� � �

1810m.1

o8l�g�

� � �0.05

15Lmol��

v C6H12O6 � 0.20 L

The volume of solution that contains 2.0 g of glucose is 0.20 L.


Making Connections11. Points in favour: patients would have to swallow less of the medicine; the medicines would be more compact and

could be sold in smaller bottles, which might result in a lower cost (although the saving is likely to be minimal).

Points against: selling medicines in highly concentrated solutions would almost certainly increase the frequency ofpatients accidentally taking the wrong dose (an extra 5 mL would contain considerably more of the active ingredientin a concentrated solution than in a dilute solution); more precise equipment would be required for measuring theappropriate quantity for the prescription and the dose.

12. A common system of communication is crucial for clarity, so there is no confusion about concentrations of blood testresults or drug dosages. There are currently several systems for communicating the concentration of medicines,although all are metric. That all systems are metric is important for international communication and for labelling ofbottles of medicine. • Percent composition is very common for medicines dispensed in solution (e.g., salicylic acid in acne preparations

and D5W intravenous (5% dextrose in water)).

• Percent concentration may also be used for blood-test results such as percent of alcohol (e.g., a maximum limitof 0.080% for drinking and driving).

• Molar concentrations are used for blood sugar analysis where units of millimoles per litre (e.g., 5.1 mmol/L) areemployed.

• Parts per million and/or milligrams per kilogram of body mass are used when very small concentrations areinvolved (e.g., testing for toxins or drug-enhanced sports performances). The effective concentration of medicinesare determined by extensive research with animals and humans. Once the concentration is determined throughresearch, the effective mass of medicine that needs to be ingested every so-many hours is calculated. Taking onecapsule of a medicine every 8 h or at every meal maintains an effective concentration of the chemical in the bodysystems. Time-release medication and patches are newer technologies that release the medication slowly in orderto keep the concentration of the medication fairly constant over time. Sometimes, such as in chemotherapy, theconcentration of chemicals is kept near the toxic level for the patient. Very careful monitoring of the concentra-tions of the chemicals in the blood system is required.

Miscommunication of concentration levels can, of course, result in death. Conventions of communication are not justconvenient and money saving; they also save lives.


6.4 DRINKING WATER

PRACTICE

(Page 294)

Understanding Concepts1. Contaminants are biological (e.g., viruses, bacteria, protozoa), chemical (e.g., mercury, lead, PCBs) and physical (e.g.,

sand, mud, suspended particles of organic matter).2. Some ways contaminants enter water are by leaching from landfills; from inadequate septic/sewage systems; from

overuse of fertilizers, pesticides and herbicides; from livestock wastes; and from road salt runoff.3. Leaking sewer pipes are environmental hazards because untreated sewage contains nitrates and phosphates (which can

act as fertilizers in the environment) and bacteria (which can cause a variety of infections resulting in illness). Rawsewage also has a very unpleasant odour.

4. Water leaching from a landfill site may seep through the surrounding earth and rock, and into wells or streams, thusending up in a drinking water supply. Such leachate could contain heavy metal ions (e.g., mercury, lead, andcadmium); bacteria; acids; and organic compounds (e.g., benzene and tetrachloromethylene). If not removed, thesepollutants could have a negative influence on health: heavy metal ions interfere with brain and nerve development;bacteria can cause infections; acids can damage pipes; and organic compounds may be poisonous or carcinogenic.


5.

6. c tetrachloroethylene � 0.03 ppm � 0.03 mg/Lv water � 250 L

m tetrachloroethylene � 250 L� � �0.0

13L�mg�

m tetrachloroethylene � 8 mg

The mass of tetrachloroethylene in the bath water is 8 mg.7. c cadmium � 0.005 ppm � 0.005 mg/L

c lead � 0.010 ppm � 0.010 mg/L

c mercury � 0.001 ppm � 0.001 mg/L

v water � �11.5

d�L

� � �365

1.2a5 d�

�

� 5.5 � 102 L/a

m cadmium � �5.5 �

11a02 L�

� � �0.0

105

L�mg

�

m cadmium � 3 mg/a

The mass of cadmium consumed per year would be 3 mg.

m lead � �5.5 �

11a02 L�

� � �0.0

110

L�mg

�

m lead � 5.5 mg/a

The mass of lead consumed per year would be 5.5 mg/a.

m mercury � �5.5 �

11a02 L�

� � �0.0

101

L�mg

�

m mercury � 0.5 mg/a

The mass of mercury consumed per year would be 0.5 mg.

8. v water � 10.00 mL � 0.01000 L

m NO3– � 5.4 mg

c NO3– � �

0.50.140m00

gL

�

c NO3– � 0.54 g/L

This concentration of nitrate ion, 0.54 g/L, is much higher than the MAC level, about 540/45, or 12 times the acceptablemaximum.

Making Connections9. Student reports will vary widely. Suitable starting points for research include the Ontario Groundwater Management

program, the Ontario Ministry of Agriculture, Food and Rural Affairs, and reports from the Walkerton Inquiry. Locallyrelevant issues should also be investigated.


forest orcrop spraying

groundwater

leaky tanksor pipelines

gasolinepesticide

landfillleachate

maycomefrom

maycomefrom

maycomefrom

lead

and flow into


10. Student reports will vary. The chosen source of contamination could be any of those listed in Table 1, page 293 of thetext. A proposed solution could involve preventing the contamination at source, preventing the contaminant fromreaching the well, relocating the well, or removing the contaminant from well water before it reaches our taps.

Try This Activity: Simulated Water Treatment

(Page 297)

(a) In glass 1, a control, adding a teaspoon (5 mL) of ammonia to the alum solution causes a white flocculent precipitateto form. The nature of the precipitate is observed. The precipitate gradually settles to the bottom of the glass.

In glass 2, the components are added in the order water, soil, alum, and ammonia, and the contents clear as the floc-culent precipitate settles. The removal of the suspended silt is obvious. The soil/silt adheres to the precipitate. Theprocess is much more complete and quicker than in glass 3.

In glass 3, a control with water and soil only, the soil gradually settles but leaves some of the silt in suspension for alonger period of time than in glass 2.

(b) The precipitate, being gelatinous, will quickly clog a filter paper, making separation by filtration impractical. Allowingthe precipitate to fall to the bottom of the container is slower, but simple and effective.

PRACTICE

(Page 299)

Understanding Concepts11. Disinfection — killing disease-causing organisms — is the most important step in water treatment.12. Usual (all areas) Optional (some areas)

collection aerationcoagulation/sedimentation softeningfiltration fluoridationdisinfectionpostchlorinationammoniation

Making Connections13. Physical treatments might include a variety of filters, such as ceramic, ultrafine, and carbon, plus reverse osmosis. The

pore sizes in the most common filters sold today are 0.1-4 µm. Cysts and bacteria are removed by the filter, but virusesare too small. Boiling the water is effective for disinfecting the water completely.

Chemical treatments might include the use of iodine crystals, iodine-complex tablets, chlorine bleach, calciumhypochlorite crystals, and halazone tablets. The iodine, for example, is added as a measured volume of a saturatedsolution. The dilution procedure adds 4 mg of iodine to 1 L of water to produce a 4 ppm solution that is treated for20 – 30 min before drinking.

Note that information on purifying water while hiking is also readily available at outdoor activity stores.


14. In favour of bottled water: we can be fairly certain that it is free of microorganisms and toxic chemicals; distilled orreverse osmosis-treated water is much more pure; bottled water is convenient, if tap water is contaminated.

Against bottled water: most bottled water is not significantly different from that available through most municipalsystems; bottled water may, in fact, be less safe than local municipally treated water; it is more expensive than tapwater, and distilled or reverse osmosis-treated water is particularly expensive; it is less convenient (e.g. it has to bemoved from storage to display areas, and it is heavy).

The note to the school cafeteria should outline a position and give reasons. It could also request some action on thepart of the cafeteria staff, such as making free tap water available to all students.


15. Water testing, at anything other than a very basic level, is really a branch of analytical chemistry. A person trained asa laboratory technologist can analyze water for chemicals that may be present. However, the analysis also needs toinclude screening for possible biological components. The water must be tested for bacteria (e.g., E-coli), intestinalparasites (e.g., Giardia cysts and cryptosporidium), and viruses. Special equipment and training are required for thebioanalysis. The water-treatment plant operators must also be trained in the interpretation of analytical reports and inhow to respond to these reports (how to adjust the process to remedy any problems).


Explore an Issue

Take a Stand: Safe to Drink?

(Page 299)

(a) Aspects chosen may include: the choice of ground or surface water as a source; the location of wells or intake pipes;the age and condition of the pipes; the treatment the water receives; water metering; or the privatization of waterdelivery. The letter should clearly state the current situation (found through research) and make reasonable sugges-tions for improvements.



(Page 300)

Understanding Concepts1. A typical list might be taken from Section 6.4; Tables 1, 2, and 3:

Pollutant Classificationacid chemicalbacteria biologicalcysts biologicalviruses biologicallead chemicalmercury chemicalcadmium chemicalmineral solids physicalnitrates chemicalphosphates chemicalorganic compounds chemicalbenzene chemicalgasoline chemicalpesticides chemicalsalt chemical

2. The first action should be to notify the community to stop using the water. Next, the water supply system must bedisinfected and flushed.

3. Exceding the MAC for chemicals in drinking water is a potential health hazard. For example, high lead levels cancause brain damage; organic compounds may be carcinogenic; salt makes water taste bad and may contribute to circu-latory problems.

4. Contaminants may be any of those listed in Tables 2 or 3. If benzene is chosen, the presentation should include itssource (e.g., leaked gasoline, industrial effluent, or landfill leachate); its MAC (0.005 ppm); and its effects (it is asuspected carcinogen and, because it floats on water, it interferes with water’s ability to exchange gases with the air).

Research information is plentiful on water contaminants. Students might be encouraged to report on a contaminantthat poses a problem locally.

5. A lead concentration of 0.01 g/L is 1000 times higher than the MAC level of 0.010 mg/L (0.000010 g/L).


Making Connections6. Water treatment is designed to remove physical, biological, and chemical contaminants. Physical contaminants are

removed through coagulation, flocculation, sedimentation, and filtration; biological contaminants are removedthrough disinfection and postchlorination; chemical contaminants are removed through aeration and softening.

Water treatment on a large scale is usually a continuous process (rather than a batch process) in order to provide acontinuous flow of treated water into the water system. Continuous-flow designs are much more difficult to create andmonitor because they must be timed correctly. The size of the container, the mixing of the fluids, and the time that asample of water remains in the container must all be pre-engineered in order for the process to be effective. In a batchprocess, there are fewer variables to control and manipulate — the water is relatively static (still) while being treated.

7. In support of the statement: Almost any contamination problem eventually becomes a human (personal) problembecause of the interconnectedness of the entire ecosystem, which, of course, includes us. Most ground water contam-ination is caused by people, so it is our responsibility to clean it up.In opposition to the statement: Contamination is a problem regardless of whether it affects people and should beavoided if at all possible. Humans are not the only organisms damaged by contamination.Presentations should include suppporting information and reasoned arguments.

8. Precise equipment is required for water testing because “safe” levels of toxic and noxious materials are so low thatthey are hard to measure with imprecise equipment.

6.5 SOLUTION PREPARATION

PRACTICE

(Page 302)

Understanding Concepts1. In solid form, ammonium oxalate is a monohydrate, so the calculation of molar mass must take this into account.

C (NH4)2C2O4 � 0.250 mol/L

v (NH4)2C2O4 � 100.0 mL � 0.1000 L

M (NH4)2C2O4�H2O � 142.14 g/mol

n (NH4)2C2O4� 0.1000 L� � �

0.2510L�mol�

� 0.0250 mol

m (NH4)2C2O4�H2O � 0.0250 mol� � �14

12m.1

o4l�g

�

m (NH4)2C2O4�H2O � 3.55 g

or

m (NH4)2C2O4�H2O � 0.1000 L� � �0.2

150

L�mol��

1412m.1

o4l�g

�

m (NH4)2C2O4�H2O � 3.55 g

The mass of ammonium oxalate monohydrate required is 3.55 g.

2. C NaOH � 10.0 mol/L

v NaOH � 500 mL � 0.500 L


n NaOH � 0.500 L� � �10.

10

L�mol�

� 5.00 mol

m NaOH � 5.00 mol� � �410.

m00

ol�g

�

m NaOH � 200 g

or

m NaOH � 0.500 L� � �10.

10

L�mol��

410.

m00

ol�g

�


Making Connections6. Water treatment is designed to remove physical, biological, and chemical contaminants. Physical contaminants are

removed through coagulation, flocculation, sedimentation, and filtration; biological contaminants are removedthrough disinfection and postchlorination; chemical contaminants are removed through aeration and softening.

Water treatment on a large scale is usually a continuous process (rather than a batch process) in order to provide acontinuous flow of treated water into the water system. Continuous-flow designs are much more difficult to create andmonitor because they must be timed correctly. The size of the container, the mixing of the fluids, and the time that asample of water remains in the container must all be pre-engineered in order for the process to be effective. In a batchprocess, there are fewer variables to control and manipulate — the water is relatively static (still) while being treated.

7. In support of the statement: Almost any contamination problem eventually becomes a human (personal) problembecause of the interconnectedness of the entire ecosystem, which, of course, includes us. Most ground water contam-ination is caused by people, so it is our responsibility to clean it up.In opposition to the statement: Contamination is a problem regardless of whether it affects people and should beavoided if at all possible. Humans are not the only organisms damaged by contamination.Presentations should include suppporting information and reasoned arguments.

8. Precise equipment is required for water testing because “safe” levels of toxic and noxious materials are so low thatthey are hard to measure with imprecise equipment.

6.5 SOLUTION PREPARATION

PRACTICE

(Page 302)

Understanding Concepts1. In solid form, ammonium oxalate is a monohydrate, so the calculation of molar mass must take this into account.

C (NH4)2C2O4 � 0.250 mol/L

v (NH4)2C2O4 � 100.0 mL � 0.1000 L

M (NH4)2C2O4�H2O � 142.14 g/mol

n (NH4)2C2O4� 0.1000 L� � �

0.2510L�mol�

� 0.0250 mol

m (NH4)2C2O4�H2O � 0.0250 mol� � �14

12m.1

o4l�g

�

m (NH4)2C2O4�H2O � 3.55 g

or

m (NH4)2C2O4�H2O � 0.1000 L� � �0.2

150

L�mol��

1412m.1

o4l�g

�

m (NH4)2C2O4�H2O � 3.55 g

The mass of ammonium oxalate monohydrate required is 3.55 g.

2. C NaOH � 10.0 mol/L

v NaOH � 500 mL � 0.500 L


n NaOH � 0.500 L� � �10.

10

L�mol�

� 5.00 mol

m NaOH � 5.00 mol� � �410.

m00

ol�g

�

m NaOH � 200 g

or

m NaOH � 0.500 L� � �10.

10

L�mol��

410.

m00

ol�g

�


m NaOH � 200 g

The mass of sodium hydroxide required is 200 g.

3. Typical answers may include sugar (in coffee, tea, lemonade), salt (in cooking water), fruit drink crystals, and soapsolutions.

Applying Inquiry Skills4. (a) In solid form cobalt (II) chloride is a dihydrate, so the calculation of molar mass must take this into account.

C CoCl2 � 0.100 mol/L

v CoCl2 � 2.00 L

M CoCl2�2H2O � 168.57 g/mol

n CoCl2� 2.00 L� � �

0.1010L�mol�

� 0.200 mol

m CoCl2�2H2O � 0.200 mol� � �16

15m.8

o7l�g

�

m CoCl2�2H2O � 33.2 g

or

m CoCl2�2H2O � 2.00 L� � �0.10

10L�mol��

1615m.8

o7l�g

�

m CoCl2�2H2O � 33.2 g

The mass of cobalt (II) chloride dihydrate required is 33.2 g.

(b) Procedure1. Wear eye protection and a laboratory apron.2. Calculate the mass of CoCl2�2H2O(s) required to prepare 2.00 L of 0.100 mol/L solution.3. Obtain the calculated mass of CoCl2�2H2O(s) in a clean dry 400-mL beaker.4. Dissolve the solid in about 200 mL of water.5. Transfer the solution into a clean 2-L volumetric flask, making sure to rinse the beaker and funnel

several times. Transfer the rinsings into the flask.6. Add pure water to make the final volume 2.00 L.7. Stopper the flask and mix the contents thoroughly by inverting the flask repeatedly.

5. (a) C KMnO4 � 75.0 mmol/L � 0.0750 mol/L

v KMnO4 � 500.0 mL � 0.5000 L

M KMnO4� 158.04 g/mol

n KMnO4� 0.5000 L� � �

0.07150

L�mol

�

� 0.0375 mol

mKMnO4� 0.0375 mol� � �

1518m.0

o4l�g

�

mKMnO4� 5.93 g

or

mKMnO4 � 0.5000 L� � �

0.07150

L�mol�

� � �15

18m.0

o4l�g

�

mKMnO4 � 5.93 g

The mass of potassium permanganate required is 5.93 g.

(b) Procedure1. Wear eye protection and a laboratory apron.

2. Calculate the mass of KMnO4(s) required to prepare 500.0 mL of 75.0 mmol/L solution.3. Obtain the calculated mass of KMnO4(s) in a clean, dry 250-mL beaker.4. Dissolve the solid in about 100 mL of water.


5. Transfer the solution into a clean 500-mL volumetric flask, being sure to rinse the beaker and funnel severaltimes. Transfer the rinsings into the flask.

6. Add pure water to make the final volume 500.0 mL.7. Stopper the flask and mix the contents thoroughly by inverting the flask repeatedly.

ACTIVITY 6.5.2: A STANDARD SOLUTION BY DILUTION

(Page 305)

Analysis(a) Some water is always placed in the volumetric flask initially for safety reasons, so that when concentrated acids are

diluted their heat of solution will be dissipated through this water, and will not cause the flask contents to boil andspurt out. The rule is "Add concentrated acid to water; NEVER add water to concentrated acid.” When the concen-trated solution is not an acid, the rule is followed anyway, for consistency and to build the habit.

(b) If 100 mL of water had initially been placed in the flask, adding the concentrated solution and rinse water duringtransfer would make the final volume too great.

PRACTICE

(Page 306)

Understanding Concepts6. vi � ?

vf � 2.00 L

Ci � 17.8 mol/L

Cf � 0.200 mol/L

viCi � vfCf

vi � �v

CfC

i

f�

vi �

vf � 0.0225 L � 22.5 mL

or

vi � 2.00 L � �01.270.80

mm

ool�l�//LL

�

vi � 0.0225 L � 22.5 mL

The initial volume of 17.8 mol/L hydrochloric acid required is 22.5 mL.

7. (a) vi � 5.00 mL

vf � 100.0 mL

Ci � 0.05000 mol/L

Cf � ?

viCi � vfCf

Cf � �v

viC

f

i�

�

Cf � 0.00250 mol/L � 2.50 mmol/L

5.00 mL� � 0.05000 mol/L��

2.00 L � 0.200 mol/L��

17.8 mol/L�


or

Cf � �0.050

L00 mol� � �

150.000.0

mm

L�L�

�

Cf � 0.00250 mol/L � 2.50 mmol/L

The final concentration of copper(II) sulfate solution is 2.50 mmol/L.

(b) C CuSO4 � 2.50 mmol/L � 0.00250 mol/L

v CuSO4 � 10.0 mL � 0.0100 L

M CuSO4 � 159.61 g/mol

n CuSO4� 0.0100 L� � �

2.501

mL�

mol�

� 0.0250 mmol

m CuSO4 � 0.0250 mmol� � �

1519m.6

o1l�g

�

m CuSO4� 3.99 mg

or

m CuSO4 � 0.0100 L� � �

2.501

mL�

mol��

1519m.6

o1l�g

�

m CuSO4 � 3.99 mg

The mass of copper(II) sulfate in the solution sample is 3.99 mg.

(c) This final dilute solution would be quite difficult to prepare directly, since the entire 100.0 mL volume wouldonly contain 39.9 mg of solute. This mass could only be measured accurately by using a balance with a precision of at least tenths of milligrams (ten-thousandths of grams), which is on the order of a hundred timesmore precise than standard school lab (centigram) balances. Even more difficult to measure would be the 3.99mg needed to prepare the 10.0 mL of the solution. The technique of dilution solves this problem.

Furthermore, mass measurements this precise must be done with the balance pan in an enclosed space inorder to prevent error caused by air currents. The sample must be enclosed in a solid container so that a correc-tion for the buoyant force of air will not be required.

8. (a) The volume increase is 250.0 mL/10.00 mL or 25.00 times.Concentration decrease is thus 1/25.00 or 0.04000 times or 4.000%.(b) The reacting volume would now be 25.00 times as much.(c) We would expect the speed of the reaction to be slower with the diluted solution, since the particles that react

would be spread out more in the solvent and would not collide as often.9. Typical examples of dilutions may include liquid soap solutions (hand cleaner), fruit juice concentrates, chocolate

syrup in milk, and bleach in laundry loads.

Reflecting10. The statement will not be true if the instrument available for measuring mass is very precise, and the equipment

available for measuring liquid volumes is not very precise.


(Page 306)

Understanding Concepts1. Scientists make solutions for chemical analysis, as well as for precise control of reactions. Solutions make substances

easy to handle and measure and safer to use, and many reactions only occur in solution.2. (a) Solutions may be made by dissolving a known mass of solute up to a known volume, or by diluting an existing

solution of higher concentration.(b) The method used normally depends on the form (pure substance or concentrated solution) in which the desired

reagent is available. A solution is prepared from a solid solute by measuring the mass of the solute and addingwater (e.g., the preparation of a sodium carbonate solution). A solution is prepared from an available (moreconcentrated) solution by diluting the available solution (e.g., hydrochloric acid solutions are prepared by dilutingconcentrated hydrochloric acid).


3. C Ba(NO3)2 � 0.125 mol/L

v Ba(NO3)2 � 100 mL � 0.100 L

M Ba(NO3)2 � 261.35 g/mol

n Ba(NO3)2 � 0.100 L� � �

0.1215L�mol�

� 0.0125 mol

m Ba(NO3)2 � 0.0125 mol� � �

2611m.3

o5l�g

�

m Ba(NO3)2 � 3.27 g

or

m Ba(NO3)2 � 0.100 L� � �

0.1215L�mol� � �

2611m.3

o5l�g

�

m Ba(NO3)2 � 3.27 g

The mass of pure barium nitrate required is 3.27 g.

4. vi � 1.00 L

vf � ?

Ci � 17.4 mol/L

Cf � 0.400 mol/L

viCi � vfCf

vf � �vCiC

f

i�

vf �

vf � 43.5 L

or

vf � 1.00 L � �01.74.040

mm

ool�l�//LL

�

vf � 43.5 L

The final volume of diluted acetic acid is 43.5 L

5. vi � 10.00 mL

vf � 250.0 mL

ci � ?

cf � 0.274 g/L

vici � vfcf

ci � �vvfc

i

f�

ci �

ci � 6.85 g/L

or

ci � �0.2

17L4 g� � �

21500.0.00

mm

LL

�

250.0 mL� � 0.274 g/L��

10.00 mL�

1.00 L � 17.4 mol/L��

0.400 mol/L�


ci � 6.85 g/L

The initial concentration of the solution was 6.85 g/L.

Applying Inquiry Skills6. C KSCN � 0.155 mol/L

v KSCN � 100 mL � 0.100 L

M KSCN � 97.18 g/mol

n KSCN � 0.100 L� � �0.15

15L�mol�

� 0.0155 mol

m KSCN � 0.0155 mol� � �917.

m18

ol�g

�

m KSCN � 1.51 g

or

m KSCN � 0.100 L� � �0.15

15L�mol��

917.

m18

ol�g

�

m KSCN � 1.51 g

The mass of potassium thiocyanate required is 1.51 g.

Procedure1. Wear eye protection and a laboratory apron.2. Calculate the mass of KSCN(s) (1.51 g) required to prepare 100 mL of 0.155 mol/L solution.3. Obtain the calculated mass of KSCN(s) in a clean, dry, 250-mL beaker.4. Dissolve the solid in about 50 mL of water.5. Transfer the solution into a clean 100-mL volumetric flask, being sure to rinse the beaker and funnel several times.

Transfer the rinsings into the flask.6. Add pure water to make the final volume 100.0 mL.7. Stopper the flask and mix the contents thoroughly by inverting the flask repeatedly.

7. vi � ?

vf � 1.00 L

Ci � 5.00 mol/L

Cf � 0.125 mol/L

viCi � vfCf

vi � �vCfC

i

f�

vi �

vf � 0.0250 L � 25.0 mL

orvi � 1.00 L � �

05.1.0205

mm

ool�l�//LL

�

vi � 0.0250 L � 25.0 mL

The initial volume of 5.00 mol/L sulfuric acid required is 25.0 mL.

Procedure1. Wear eye protection and a laboratory apron.2. Calculate the initial volume of 5.00 mol/L H2SO4(aq) (25.0 mL) that is required to prepare 1.00 L of 0.125 mol/L

solution.

1.00 L � 0.125 mol/L� ��

5.00 mol/L�


3. Add 400–500 mL of pure water to a 1-L volumetric flask.4. Use a 25-mL volumetric pipet to transfer 25.00 mL of 5.00 mol/L acid to the flask.5. Add pure water to make the final volume 1.000 L (to the mark on the flask).6. Stopper the flask and mix the contents thoroughly by inverting the flask repeatedly.7. This solution may be safely disposed of down the sink, provided it is washed down with plenty of water.8. (a) In solid form, the sample of cobalt (II) chloride is a hexahydrate, so the calculation of molar mass must take this

into account.C CoCl2

� 0.100 mol/L

v CoCl2 � 100.0 mL � 0.1000 L

M CoCl2�6H2O � 237.95 g/mol

n CoCl2� 0.1000 L� � �

0.1010L�mol�

� 0.0100 mol

m CoCl2�6H2O � 0.0100 mol� � �23

17m.9

o5l�g

�

m CoCl2�6H2O � 2.38 g

or

m CoCl2�6H2O � 0.1000 L� � �0.10

10L�mol��

2317m.9

o5l�g

�

m CoCl2�6H2O � 2.38 g

The mass of solid cobalt (II) chloride hexahydrate required is 2.38 g.

(b) vi � ?

vf � 100.0 mL

Ci � 0.100 mol/L

Cf � 0.0100 mol/L

viCi � vfCf

vi � �v

CfC

i

f�

vi � 100.0 mL � �00.0.110000

mm

ool�l�//LL

�

vi � 10.0 mL

The initial volume of 0.100 mol/L CoCl2(aq) required is 10.0 mL.

(c) Materials• lab apron• eye protection• CoCl2�6H2O(s)• wash bottle with pure water• centigram balance• 250-mL beaker• stirring rod• funnel• 2 100-mL volumetric flasks with stoppers• 10-mL volumetric pipet with pipet bulb• medicine dropper• meniscus finder Procedure

1. Wear eye protection and a laboratory apron.2. Calculate the mass of CoCl2�6H2O(s) required to prepare 100.0 mL of 0.100 mol/L solution.3. Obtain the calculated mass (2.38 g) of CoCl2�6H2O(s) in a clean, dry 250-mL beaker.


4. Dissolve the solid in about 50 mL of water.5. Transfer the solution into a clean 100-mL volumetric flask, being sure to rinse the beaker and funnel several times.

Transfer the rinsings into the flask.6. Add pure water to make the final volume 100.0 mL.7. Stopper the flask and mix the contents thoroughly by inverting the flask repeatedly. 8. Calculate the volume of 0.100 mol/L CoCl2(aq) required to make 100.0 mL of 0.0100 mol/L solution by dilution

(according to the above calculations: 10.0 mL).9. Use a 10-mL volumetric pipet to obtain and transfer 10.0 mL of 0.100 mol/L CoCl2(aq) solution to a clean 100-mL

volumetric flask. 10. Add pure water to the flask until the final 100.0 mL volume is reached.11. Stopper and invert the flask to thoroughly mix the contents.12. Solutions may be disposed of down the sink with plenty of water.

Making Connections9. Arguments for the position: There will be a fuel saving from transporting concentrated reagents (less mass and

volume); fewer trucks will be required for their transportation, resulting in less traffic on the roads; loading andunloading will be quicker.Arguments against the position: Any spills or careless handling would be much more dangerous; the recipient willlikely have the inconvenience of diluting the product upon its arrival.

10. The common system that concentrates pollutants (bioaccumulation) is the “web of life” or “food chain.” Organismsthat ingest pollutants can concentrate them in their bodies and pass this concentration on to predators. An example ofsuch a chain is aquatic microorganisms → plankton → fish → seabird chicks → eagles.

Reflecting11. The procedure should show serial dilution, that is, begin by diluting stock HCl(aq) solution and then use samples of

each new dilution to produce a further dilution of, say, one-tenth the concentration each time. These solutions can theneasily be compared according to how rapidly they react with equal samples of zinc.

Make a Summary

(Page 308)


molecularsubstance

does notconduct electricity

molecularsubstance

moleculesin solution

conductselectricity

determinemass

determinevolume

agriculturerunoff

industrialwaste

ioniccompound

preparedfrom solid

solution graduatedcylinder

volumetricflaskprepared

by dilution

ions inwater

refinedwater

humanwaste

ioniccompound

acidcompound

concentratedacid aquifer

acids

lake

river

water

pipet

nonelectrolytes

electrolytes

pollution

volumeof solution

or solute (v)

otherconcentrated

solution

waterpurification

process

landfillleachate

concentrationof solution

amount ofsolution (n)

amount = mass ×

(n = m /M)

1molar mass

amount = volume × concentration(n = v × c)

CHAPTER 6 REVIEW

(Page 309)

Understanding Concepts1. All concentration units express the ratio of the quantity of solute to the quantity of solution.2. (a) Highly polar solvents

(b) Polar solvents(c) Nonpolar solvents

3. (a) Water dissolves very many substances because it is liquid over a wide temperature range and because it is madeup of highly polar molecules that strongly attract (electrostatically) both ions and polar molecules. Water alsohydrogen bonds to substances with N, O and F atoms.

(b) Substances are dissolved in water to make them easier to store, easier to manipulate, and to control when and howrapidly they react. Examples:• Ammonia gas is so corrosive and toxic in pure form that it is very dangerous to handle, but when dissolved

in water can be stored, shipped, and easily handled in ordinary plastic bottles, such as in window cleaner solu-tions, or “household” ammonia cleaner.

• Baking soda (sodium hydrogen carbonate) is an example of a substance that doesn’t react (in cooking) untildissolved. Solid plant fertilizers are another good example of this point.

• All-purpose cleansers are examples of solutions that react more rapidly and to greater extent when they aremade with a higher solute/solvent ratio (more concentrated).

(c) The ability of water to dissolve so many substances is a problem when it comes to the purification of water fordrinking, because often a large number of toxic or noxious solutes must be removed.

4. (a) solute: sodium chloride solvent: water(b) solute: acetic acid solvent: water(c) solute: sodium carbonate decahydrate solvent: water(d) solute: sucrose solvent: water(e) solute: ethanol solvent: water

5. c MF � 2 g/100 mL

v milk � 250 mL

mfat � 250 mL� � �10

20.0

mgL�

�

mfat � 5.0 g

The mass of fat in one glass of milk is 5.0 g.6. cMF � 5.9%, 2.0%, and 1.2% , therefore,

cMF � 5.9 g/100 mL, 2.0 g/100 mL, and 1.2 g/100 mL , respectively.m fat � 3.0 g maximum in each serving

mserving � 3.0 g fat� � �5.

190g0

fgat�

�

mserving � 51 g or 0.051 kg

mserving � 3.0 g fat� � �100

2.g0

sgefravt�ing

�

mserving � 0.15 kg

mserving � 3.0 g fat� � �100

1.g2

sgefravt�ing

�

mserving � 0.25 kg

The serving sizes containing 3.0 g of fat for the three choices of yogurt are as follows: for 5.9% MF yogurt, a 51 g or0.051 kg serving; for 2.0% MF yogurt, a 0.15 kg serving; and for 1.2% MF yogurt, a 0.25 kg serving.


7. (White) vinegar is most commonly sold in stores as a 5% (by volume) solution of acetic acid, HC2H3O2(aq), byvolume. Assume a minimum of 5 mL of acetic acid in 100 mL of solution. (Commercial labelling is always a guar-anteed legal minimum. Significant digits do not apply to this value, because the 5% is not a measurement.)cHC2H3O2

� 5% � 5 mL/100 mL

vHC2H3O2 � 15 mL

vHC2H3O2 � 15 mL � �

1050mmL�

L��

vHC2H3O2 � 0.30 L

The volume of vinegar containing 15 mL of acetic acid is 0.30 L.

8. c nitrate � 55 ppm � 55 mg/L

v water � 200 mL � 0.200 L

m nitrate � 0.200 L� � �55

1mL�

g�

m nitrate � 11 mg

The mass of nitrate ion in the water is 11 mg.

9. (a) n Cu(NO3)2= 0.35 mol

v Cu(NO3)2 = 500 mL � 0.500 L

C Cu(NO3)2 = �

00..35500

mLol

�

C Cu(NO3)2 = 0.70 mol/L

The molar concentration of the solution is 0.70 mol/L.(b) M NaOH = 40.00 g/mol

m NaOH = 10.0 gv NaOH = 2.00 L

n NaOH = 10.0 g� � �410.

m00

olg�

�

= 0.250 mol

C NaOH = �0.

22.5000

mL

ol�

C NaOH = 0.125 mol/L

orC NaOH = 10.0 g� � �

410.

m00

olg�

� � �2.0

10 L�

C NaOH = 0.125 mol/L

The molar concentration of the solution is 0.125 mol/L.

(c) vi � 25 mL

vf � 145 mL

Ci � 11.6 mol/L

Cf � ?

viCi � vfCf

Cf � �v

viC

f

i�

Cf �25 mL� � 11.6 mol/L��

145 mL�


Cf = 2.0 mol/L

or

Cf = 25 mL� � �11

1.465mmoL�l/L

�

Cf = 2.0 mol/L

The final molar concentration of the solution is 2.0 mol/L.

(d) 16 ppm of Mg2+(aq) = 16 mg/L

MMg2+ = 24.31 g/mol

mMg2+ = 16 mg

v = 1.00 L (assume one litre of solution, with a certainty that does not limit the certainty of the answer)

n Mg2+ = 16 mg� � �214.

m31

olg�

�

= 0.66 mmol

CMg2+ = �0.6

16.0

m0

mL

ol�

CMg2+ = 0.66 mmol/L

orCMg2+ = 16 mg� � �

211m.43

og�l

� � �1.0

10L�

CMg2+ = 0.66 mmol/L

The molar concentration of the solution is 0.66 mmol/L.

10. (a) The labelled compounds are (by formula and class of compound):water H2O(l) neutral molecularglucose C6H6O11(l) neutral molecularcitric acid C3H4OH(COOH)3(s) acid molecularpotassium citrate K3C3H4OH(COO)3(s) ionicsodium chloride NaCl(s) ionicpotassium phosphate K3PO4(s) ionic

(b) Dissolved glucose would make the drink taste sweet, and dissolved citric acid would make it taste tangy.(c) Note that the ion mass values in this question are from a commercial label. They thus represent required legalminimum content, and are not measured values. Significant digits can not be assigned to such values, so the certaintyfor this calculation must be taken from (related to) the given serving volume value, 400 mL (3 significant digits).mNa+ � 50 mg

mK+ � 55 mg

v solution � 400 mL = 0.400 L

cNa+ = �05.400m0

gL

�

cNa+ = 125 mg/L = 125 ppm

The concentration of sodium ions is 125 ppm.

cK+ = �05.450m0

gL

�

cK+ = 138 mg/L = 138 ppm

The concentration of potassium ions is 138 ppm.


11. M Na2C2O4= 134.00 g/mol

v Na2C2O4= 250.0 mL � 0.2500 L

C Na2C2O4= 0.375 mol/L

n Na2C2O4� 0.2500 L� � �

0.3715L�mol�

� 0.0938 mol

m Na2C2O4� 0.0938 mol� � �

1314m.0

o0l�g

�

m Na2C2O4� 12.6 g

orm Na2C2O4

� 0.2500 L� � �0.37

15L�mol��

1314m.0

o0l�g

�

m Na2C2O4� 12.6 g

The mass of sodium oxalate required is 12.6 g.

12. vi � ?

vf � 500 mL

Ci � 14.6 mol/L

Cf � 1.25 mol/L

viCi � vfCf

vi � �vCfC

i

f�

vi =

vi = 42.8 mL

orvi = 500 mL � �

11.42.56

mm

ool�l�//LL

�

vi = 42.8 mL

The required volume of the initial phosphoric acid solution is 42.8 mL.

13. c HCl � 36% W/V = 0.36 kg/L

M HCl = 36.46 g/mol

CHCl � �0.3

16Lkg�

� � �316.

m46

olg�

�

� 0.0099 kmol/L � 9.9 mol/L

vi � ?

vf � 5.00 L

Ci � 9.9 mol/L

Cf � 0.12 mol/L

viCi � vfCf

vi � �v

CfC

i

f�

500 mL � 1.25 mol/L��

14.6 mol/L�


vi =

vi = 0.061 L � 61 mL

orvi = 5.00 L � �

09.1.92

mm

ool�l�//LL

�

vi = 0.061 L � 61 mL

The required initial volume of the hydrochloric acid is 61 mL.

Applying Inquiry Skills14. Standard solutions are prepared either by dissolving a measured mass of a solid solute to make a known volume of

solution; or by diluting an existing (more concentrated) solution of known concentration to decrease its concentration. 15. A standard solution is one for which the concentration is accurately known. It is necessary for accurate chemical

analysis or for precise control of chemical reactions. 16. (a) Electrolytes can be distinguished from nonelectrolytes by testing solutions of the compound with an ohmmeter,

or conductivity meter. Electrolytes form conducting solutions; nonelectrolytes form nonconducting solutions.(b) Acids, bases, and neutral compounds can be distinguished by testing a solution of each compound with blue and

pink litmus. In an acidic solution, blue litmus will change to pink (but pink litmus will be unchanged); in a basicsolution, pink litmus will change to blue (but blue litmus will be unchanged); and in a neutral solution, neitherpink nor blue litmus will change colour.

17. (a) M KHC4H4O6= 188.19 g/mol

v KHC4H4O6= 100.0 mL � 0.1000 L

C KHC4H4O6= 0.150 mol/L

n KHC4H4O6� 0.1000 L� � �

0.1510L�mol�

� 0.0150 mol

m KHC4H4O6� 0.0150 mol� � �

1818m.1

o9l�g

�

m KHC4H4O6� 2.82 g

orm KHC4H4O6

� 0.1000 L� � �0.15

10L�mol��

1818m.1

o9l�g

�

m KHC4H4O6� 2.82 g

The mass of potassium hydrogen tartrate measured is 2.82 g.

(b) Procedure

1. Put on eye protection and a lab apron.2. Calculate the mass of solid potassium hydrogen tartrate needed to prepare 100.0 mL of a 0.150 mol/L solution. (2.82

g, as shown above.)3. Obtain the calculated mass of potassium hydrogen tartrate in a clean, dry 150 mL beaker.4. Dissolve the solid in 40 to 50 mL of pure water.5. Transfer the solution into a 100 mL volumetric flask. Rinse the beaker and funnel two or three times with small

quantities of pure water, transferring the rinsings into the volumetric flask.6. Add pure water to the flask until the volume is 100.0 mL.7. Stopper the flask and mix the contents thoroughly by repeatedly inverting the flask.

18. (a) vi � ?

vf � 100.0 mL

Ci � 0.400 mol/L

Cf � 0.100 mol/L

5.00 L � 0.12 mol/L��

9.9 mol/L�


viCi � vfCf

vi � �v

CfC

i

f�

vi =

vi = 25.0 mL

or

vi = 100.0 mL � �00..140000

mm

ooll//L�L�

�

vi = 25.0 mL

The required initial volume of the stock solution is 25.0 mL.

(b) Procedure

1. Wear eye protection and a lab apron.2. Calculate the volume of a 0.400 mol/L stock solution (25.0 mL, as shown above) that will be required to prepare

100.0 mL of a 0.100 mol/L solution by dilution.3. Add 40 to 50 mL of pure water to a 100-mL volumetric flask.4. Use a 25-mL volumetric pipet to transfer 25.00 mL of the stock solution into the 100-mL volumetric flask.5. Add pure water until the final volume of 100.0 mL is reached.6. Stopper the flask and mix the solution thoroughly.

19. Analysis

Solution A shows acidic properties, turning blue litmus red and conducting current; so according to our current knowl-edge it must be the sulfuric acid.Solution B shows neutral molecular properties, not changing litmus and not conducting current; so according to ourcurrent knowledge it must be the glucose.Solution C shows basic properties, turning red litmus blue and conducting current; so so according to our currentknowledge it must be the calcium hydroxide.Solution D shows neutral ionic properties, not changing litmus but conducting current; so according to our currentknowledge it must be the potassium chloride.

Making Connections20. (a) Oil and water will not dissolve in each other because water is polar and oils are not. Essentially, water molecules

attract each other so strongly with hydrogen bonding that oil molecules cannot intermix with them. Dependingon the density of the oil, it will either float on top of the water in a thin layer, or sink to the bottom in discreteglobs.

(b) Oil on the surface of the water prevents the exchange of oxygen and carbon dioxide between the air and the water.It also clogs the fine filaments of organisms, such as fishes’ gills and invertebrates’ tentacles, as well as coatingthe feathers and fur of marine birds and mammals, thus destroying their insulation and waterproofing. Together,these problems result in the death of much marine life, as well as making the area unsightly.

(c) Oil is cleaned up with steam, solvents, absorbent materials, and physical collectors (skimmers and scrapers).Sometimes it is also treated to make it sink to the bottom, but this does not necessarily solve the problem.

(d) Risks of transportation by large oil tanker: oil leakage; introduction of alien species in ballast water; control is inthe hands of relatively few extremely large companies.

Benefits of transportation by large oil tanker: relatively inexpensive; relatively safe (if ships are well built and wellhandled); the system is already established, including fleets of ships, ports, loading and off-loading facilities.Alternatives: no transoceanic transportation of oil; transportation in smaller ships; air transportation.

Stopping tanker oil transport would mean a major change in our civilization, which largely runs on fairly cheapenergy from fossil fuels. But continuing this shipping of oil inevitably means environmental damage will occur. Arisk–benefit analysis should focus on minimizing the risks, in this case, for example, by requiring that tankers bedouble-hulled for increased safety, or that tankers not be allowed to travel near sensitive areas such as the GalapagosIslands.


100.0 mL � 0.100 mol/L��

0.400 mol/L�


21. Solutions collected might include examples such as:• rubbing alcohol (isopropanol): 70% (V/V) CH3CHOHCH3(l) in water (a liquid solute in a liquid solvent); not

to be taken internally; toxic by ingestion;

• decongestant (oxymetazoline hydrochloride): 0.05% in water (a solid solute in a liquid solvent); limit two usesper nostril per day; not to be used by children; causes increase in nasal congestion when use is discontinued;

• bleach (sodium hypochlorite): 5.25% NaOCl(s) in water (a solid solute in a liquid solvent); corrosive; toxic fumes;reacts with acids and bases to produce toxic products;

• window cleaner (household ammonia): NH3(g) in water (a gaseous solute in a liquid solvent); irritant to eyes, noseand throat; concentrated fumes (not in the product mentioned) are toxic and may be fatal;

• metal ornaments (pewter): 4-7% Sb, 1-2%Cu in tin, Sn(s) (solid solutes in a solid solvent); no safety precautions;

• air: about 20% oxygen in nitrogen (gasesous solute in a gaseous solvent); no safety precautions.

22. Research should turn up examples such as• glucose or dextrose (5%); to provide nourishment to a patient by placing sugar directly in the bloodstream for use

by body cells;

• multivitamin infusion (1 vial vitamin concentrate in 1000 mL of IV fluid); to eliminate risk of vitamin deficiencyin patients receiving nutrition primarily through IV feeding;

• antibacterial agent such as gatifloxacin (200 mg (100 ml) or 400 mg (200 ml) of gatifloxacin (2 mg/ml) in 5%dextrose); to treat bacterial infections;

• heart drug, esmolol HCl (10 mg/mL infusion by adding one 2500 mg ampule to a 250 mL container of a compat-ible intravenous solution); to control heartbeat in patients recovering from surgery;

• lidocaine (2% (20 mg/mL)); a local anaesthetic, particularly recommended for use while setting broken bones.

23. A typical chart would show examples as follows:(a) Animal waste washed by heavy rain into a stream that feeds a river, from which water is extracted for a town or city’s

water supply.(b) Human waste migrates underground from an outdoor toilet or an underground septic tank to a well drilled downhill

from the waste.(c) Industrial waste is released into a river system from which water is extracted for domestic purposes farther down river.(d) Municipal waste leaches into the ground and into the aquifer from which well water is extracted for domestic use.(e) Natural waste from nondomestic plants and animals is washed by flooding into a river from which water is removed

for municipal use.

Note: Student research is required for this answer. Other controversial chemicals include alcohol, asbestos, benzene,CFCs, chlorine, cigarette smoke, DDT, lead, mercury, nitrites, nitrogen oxides, ozone, phosphates, radon, and vinylchloride. See Controversial Chemicals: A Citizen's Guide by P. Kruus and I.M. Valeriote (ISBN: 0-9198868-22-3).

24. See a summary of Guidelines for Canadian Drinking Water Quality, published by Health Canada, for acceptable waterorganism and impurity levels.(a) One potentially dangerous contaminant is cryptosporidium, a microorganism and parasite. There is no legal

requirement to test for cryptosporidium, and thus there is no MAC for this contaminant. Many town watersupplies are not tested for cryptosporidium.

metal refining

sulfur dioxide

airborne

respiratoryproblems

source

pollutant

route

potentialhealtheffect

debilitatesplants

(uncertain) nausea, headache,skin irritation, death

burning sulfurouscoal

acid rain

airborne

leakingtransformers

PCBs

ground leaching

pulp-mills

dioxins

rivers


(b) In April/May 2001, there was an outbreak of illnesses in North Battleford, Saskatchewan, that were caused bycryptosporidium.

(c) The symptoms are flulike with severe stomach cramps, nausea, and diarrhea. This outbreak resulted in at leastthree deaths from the cryptosporidium in the drinking water. Only people with immune deficiencies died or were seriously threatened by the contaminant. Cryptosporidium effects are not generally as severe as E. coli effects,such as those experienced in Walkerton, Ontario, in 2000, where seven people died.

(d) A water advisory and then a boil-water order was issued. It was discovered that the drinking-water filtrationsystem malfunctioned after it was shut down for semi-annual maintenance. Provincial authorities consideredlegislation to require testing for cryptosporidium. The counterargument was that the testing is imprecise and thatthe cryptosporidium is in such low concentrations that its presence may not even be detected by the testing. Thewater management design, with the drinking water intake pipe downstream from the used-water treatment plant,was also a concern.

Exploring

25. The report will be in the form of a sales brochure containing information such as the pressure needed to extract purewater through a reverse osmosis membrane from sea water (theoretically about 2500 kPa minimum, empirically about60 000 kPa – 90 000 kPa in practice). There are many commercially operating plants in Israel and the United Statesusing sea water, and one in Yuma, Arizona using saline river water. Reverse osmosis used to be relatively expensivebut is rapidly decreasing in cost with more widespread use. A plant near Tampa Bay in Florida, scheduled for comple-tion in 2002, will treat 45 million U.S. gallons (170 million litres) of sea water each day and produce enough freshpure water to supply about 10% of the needs of that area.


26. Chlorine treatment is used to kill harmful organisms (protozoa, bacteria, and viruses) in drinking water. For example,one Redi Chlor tablet will make one gallon of water suitable for drinking. One system, Pristine, by a Canadiancompany, Advanced Chemicals, mixes solution A (inactivated chlorine dioxide) with solution B (5% phosphoric acid)to produce activated chlorine dioxide, which is then mixed with the water to purify it. Information on ingredients andeffects can also be obtained from a package label.



Documents

6.1 DEFINING A SOLUTION - Quia · 6.1 DEFINING A SOLUTION PRACTICE ... a substance that dissolves in water to form a conducting solution ... N 2(g) or P 4(s)); compounds consisting