Chemistry

CHAPTER REVIEW

USING THE HANDBOOK ALTERNATIVE ASSESSMENT

40. Review the information on alloys in the 43. Make a comparison of the electrolyte concen-
Elements Handbook. tration in various brands of sports drinks. Using
a. Why is aluminum such an important compo- the labeling information for sugar, calculate the
nent of alloys? molarity of sugar in each product or brand.
b. What metals make up bronze? Construct a poster to show the results of your
c. What metals make up brass? analysis of the product labels.
d. What is steel?
e. What is the composition of the mixture 44. Write a set of instructions on how to prepare a
called cast iron? solution that is 1 M CuSO4 using CuSO4•5H2O
as the solute. How do the instructions differ if
41. Table 5A of the Elements Handbook contains the solute is anhydrous CuSO4? Your instruc-
carbon monoxide concentration data expressed tions should include a list of all materials needed.
as parts per million (ppm). The OSHA
(Occupational Safety and Health Graphing Calculator
Administration) limit for worker exposure to Solubility vs. Temperature
CO is 200 ppm for an eight-hour period. Go to go.hrw.com for a graphing calculator
a. At what concentration do harmful effects exercise that asks you to graph the solubility
occur in less than one hour? versus temperatures for NaCl and KNO3.
b. By what factor does the concentration in
item (a) exceed the maximum limit set Keyword: HC6SLNX
by OSHA?

RESEARCH & WRITING

42. Find out about the chemistry of emulsifying
agents. How do these substances affect the dis-
solution of immiscible substances such as oil
and water? As part of your research on this
topic, find out why eggs are an emulsifying
agent for baking mixtures.

S O L U T I O N S 429

Math Tutor CALCULATING SOLUTION CONCENTRATION

You can use the relationship below to calculate the concentration in molarity of any
solution.

molarity of solution (M) = m⎯oles o⎯f solute⎯(mol)
volume of solution (L)

Suppose you dissolve 20.00 g of NaOH in some water and dilute the solution to a vol-
ume of 250.0 mL (0.2500 L). You don't know the molarity of this solution until you
know how many moles of NaOH were dissolved. You know that the number of moles
of a substance can be found by dividing the mass of the substance by the mass of
1 mol (molar mass) of the substance. The molar mass of NaOH is 40.00, so the num-
ber of moles of NaOH dissolved is

20.00 g NaOH × ⎯1 mol ⎯NaOH = 0.5000 mol NaOH
40.00 g NaOH

Now you know that the solution has 0.5000 mol NaOH dissolved in 0.2500 L of solu-
tion, so you can calculate molarity.

molarity of NaOH × m⎯ol N⎯aOH = 0⎯.5000 m⎯ol NaOH = 2.000 mol/L = 2.000 M NaOH
L solution 0.2500 L solution

Problem-Solving S

• Remember that balances measure mass, not moles, so you often have to
convert between mass and moles of solute when making or using solutions.

SAMPLE

A 0.5000 L volume of a solution contains 36.49 g of magnesium chloride, MgCl2. What is the
molarity of the solution?

You know the volume of the solution, but you need to find the number of moles of the

solute MgCl2 by the following conversion.

mass MgCl2 × ⎯1 m⎯ol Mg⎯Cl2 = mol MgCl2
molar mass MgCl2

36.49 g MgCl2 × ⎯1 mo⎯l Mg⎯Cl2 = 0.3833 mol MgCl2
95.20 g MgCl2

Now you can calculate mol MgCl2 per liter of solution (molarity).

0⎯.3833 mo⎯l MgCl2 = 0.7666 M MgCl2
0.5000 L solution

PRACTICE PROBLEMS 2. What is the molarity of a solution that contains

1. What is the molarity of a solution that contains 45.00 g of cadmium nitrate, Cd(NO3)2, dissolved
0.0350 mol of sodium sulfate, Na2SO4, dissolved in 400.0 mL of solution?
in 50.0 mL of solution?

430 C H A P T E R 1 2

Standardized Test Prep

Answer the following items on a separate piece of 8.Which one of the following statements is false?
paper. A. Gases are generally more soluble in water
under high pressures than under low pressures.
MULTIPLE CHOICE B. As temperature increases, the solubilities of
some solids in water increase and the solubil-
1.Water is an excellent solvent because ities of other solids in water decrease.
A. it is a covalent compound. C. Water dissolves many ionic solutes because
B. it is a nonconductor of electricity. of its ability to hydrate ions in solution.
C. its molecules are quite polar. D. Many solids dissolve more quickly in a cold
D. it is a clear, colorless liquid. solvent than in a warm solvent.

2.Two liquids are likely to be immiscible if SHORT ANSWER
A. both have polar molecules.
B. both have nonpolar molecules. 9.Several experiments are carried out to deter-
C. one is polar and the other is nonpolar. mine the solubility of cadmium iodide, CdI2, in
D. one is water and the other is methyl alcohol, water. In each experiment, a measured mass of
CH3OH. CdI2 is added to 100 g of water at 25°C and the
mixture is stirred. Any undissolved CdI2 is then
3.The solubility of a gas in a liquid would be filtered off and dried, and its mass is deter-
increased by an mined. Results for several such experiments are
A. addition of an electrolyte. shown in the table below. What is the solubility
B. addition of an emulsifier. of CdI2 in water at this temperature?
C. agitation of the solution.
D. increase in its partial pressure.

4.Which of the following types of compounds is Mass of CdI2 Mass of undissolved
most likely to be a strong electrolyte? added, g CdI2 recovered, g
A. a polar compound
B. a nonpolar compound 17.9 0.0
C. a covalent compound 38.2 0.0
D. an ionic compound 53.6 0.0
79.3 0.0
5.A saturated solution can become supersaturat- 93.6 7.4
ed under which of the following conditions? 104.3 18.1
A. It contains electrolytes.
B. The solution is heated and then allowed EXTENDED RESPONSE
to cool.
C. More solvent is added. 10.Explain why oil and water do not mix.
D. More solute is added.
11.Write a set of instructions on how to prepare a
6.Molarity is expressed in units of solution that is 0.100 M KBr, using solid KBr
A. moles of solute per liter of solution. (molar mass 119 g/mol) as the solute. Your
B. liters of solution per mole of solute. instructions should include a list of all materials
C. moles of solute per liter of solvent. and equipment needed.
D. liters of solvent per mole of solute.

7.What mass of NaOH is contained in 2.5 L of a

0.010 M solution?

A. 0.010 g C. 2.5 g Allow a few minutes at the end
of the test-taking period to check for careless mistakes,
B. 1.0 g D. 0.40 g such as marking two answers for a single question.

S O L U T I O N S 431

CHAPTER LAB SEE PRE-LAB MICRO

PAPER CHROMATOGRAPHY LAB

Separation of Pen Inks by
Paper Chromatography

OBJECTIVES BACKGROUND
Paper Chromatography
• Demonstrate proficiency in qualitatively Details on this technique can be found in the
separating mixtures using paper Pre-Laboratory Procedure “Paper Chromatography”
chromatography. on page 848.

• Determine the Rf factor(s) for each Writing Inks
component of each tested ink. Most ballpoint pen inks are complex mixtures, con-
taining pigments or dyes that can be separated by
• Explain how the inks are separated paper chromatography, as shown in Figure A.
by paper chromatography.
Black inks can contain three or more colors; the
• Observe the separation of a mixture number of colors depends on the manufacturer.
by the method of paper chromatography. Each ink formulation has a characteristic pattern that
uniquely identifies it.
MATERIALS
In this experiment you will develop radial paper
• 12 cm circular chromatography paper chromatograms for four black ballpoint pen inks,
or filter paper using water as solvent. You will then repeat this
process using isopropanol as the solvent. You will then
• distilled water measure the distance traveled by each of the individ-
ual ink components and the distance traveled by the
• filter paper wick, 2 cm equilateral triangle solvent front. Finally, you will use these measurements
to calculate the Rf factor for each component.
• isopropanol
SAFETY
• numbered pens,
each with a different For review of safety, please see Safety in the
black ink, 4 Chemistry Laboratory in the front of your book.

• pencil PREPARATION
1. Determine the formula, structure, polarity,
• petri dish with lid density, and volatility at room temperature for
water and isopropanol. The following titles are
• scissors sources that provide general information on spe-
cific elements and compounds: CRC Handbook
FIGURE A Paper chromatog- of Chemistry and Physics, McGraw-Hill
raphy reveals the different col- Dictionary of Chemical Terms, and Merck Index.
ored dyes that black ink contains.

432 C H A P T E R 1 2

2. Prepare two data tables, one for the chro- CLEANUP AND DISPOSAL
matogram made with water and one for the 9. The water may be poured down the
chromatogram made with isopropanol. Record sink. Chromatograms and other pieces
the pen number you are testing. In your data of filter paper may be discarded in the
tables, create columns for each color in which to trash. The isopropanol solution should be placed
record the distance and Rf value. Your teacher in the waste disposal container designated by
can provide you with sample data tables. Leave your teacher. Clean up your equipment and
room below each data table to record the dis- lab station. Thoroughly wash your hands after
tance that the solvent reaches. completing the lab session and cleanup.

PROCEDURE ANALYSIS AND INTERPRETATION
1. Evaluating Conclusions: Is the color in each
Part A: Prepare a chromatogram using pen the result of a single dye or multiple dyes?
water as the solvent Justify your answer.

1. Construct an apparatus for paper chromatog- 2. Relating Ideas: What can be said about the
raphy as described in the Pre-Laboratory properties of a component ink that has an Rf
Procedure on page 848. You will make only four value of 0.50?
dots. You will use ballpoint pens rather than
micropipets to spot your paper. 3. Analyzing Methods: Suggest a reason for stop-
ping the process when the solvent front is 1 cm
2. After 15 min or when the water is about 1 cm from the edge of the filter paper rather than
from the outside edge of the paper, remove the when it is even with the edge of the paper.
paper from the Petri dish and allow the chro-
matogram to dry. Record in the data table the 4. Predicting Outcomes: Predict the results of
colors that have separated from each of the four forgetting to remove the chromatogram from
different black inks. the water in the petri dish until the next day.

Part B: Prepare a chromatogram using CONCLUSIONS
isopropanol as the solvent 1. Analyzing Results: Compare the Rf values
for the colors from pen number 2 when water
3. Repeat Procedure steps 1 to 2, replacing the was the solvent and the Rf values obtained
water in the Petri dish with isopropanol. when isopropanol was the solvent. Explain
why they differ.
Part C: Determine Rf values for each component
4. After the chromatogram is dry, use a pencil to 2. Evaluating Methods: Would you consider iso-
mark the point where the solvent front stopped. propanol a better choice for the solvent than
water? Why or why not?
5. With a ruler, measure the distance from the
initial ink spot to your mark, and record this 3. Analyzing Conclusions: Are the properties of
distance on your data table. the component that traveled the farthest in the
water chromatogram likely to be similar to the
6. Make a small dot with your pencil in the center properties of the component that traveled the
of each color band. farthest in the isopropanol chromatogram?
Explain your reasoning.
7. With a ruler, measure the distance from the
initial ink spot to each dot separately, and 4. Inferring Conclusions: What can you conclude
record each distance on your data table. about the composition of the inks in ballpoint
pens from your chromatogram?
8. Divide each value recorded in Procedure step 7
by the value recorded in Procedure step 5. The S O L U T I O N S 433
result is the Rf value for that component.
Record the Rf values in your data table. Tape
or staple the chromatogram to your data table.

CHAPTER 13

Ions in Aqueous
Solutions and
Colligative Properties

These formations were made by the precipitation
of ionic compounds from an aqueous solution.

Cyanotrichite Crystals

Compounds in SECTION 1
Aqueous Solutions
OBJECTIVES
A s you have learned, solid compounds can be ionic or molecular. In
Write equations for the
an ionic solid, a crystal structure is made up of charged particles held dissolution of soluble ionic
together by ionic attractions. In a molecular solid, molecules are com- compounds in water.
posed of covalently bonded atoms. The solid is held together by non-
covalent, intermolecular forces. When they dissolve in water, ionic Predict whether a precipitate
compounds and molecular compounds behave differently. will form when solutions of
soluble ionic compounds are
Dissociation combined, and write net ionic
equations for precipitation
When a compound that is made of ions dissolves in water, the ions sep- reactions.
arate from one another, as shown in Figure 1. This separation of ions
that occurs when an ionic compound dissolves is called dissociation. For Compare dissociation of ionic
example, dissociation of sodium chloride and calcium chloride in water compounds with ionization of
can be represented by the following equations. (As usual, (s) indicates a molecular compounds.
solid species, and (aq) indicates a species in an aqueous solution. Note
that each equation is balanced for charge as well as for atoms.) Draw the structure of the
hydronium ion, and explain
NaCl(s) ⎯H⎯2O→ Na+(aq) + Cl−(aq) why it is used to represent
the hydrogen ion in solution.
CaCl2(s) ⎯H⎯2O→ Ca2+(aq) + 2Cl −(aq)
Distinguish between strong
Notice the number of ions produced per formula unit in the equations electrolytes and weak
above. One formula unit of sodium chloride gives two ions in solution, electrolytes.
whereas one formula unit of calcium chloride gives three ions in solution.

H2O Na+
Cl–

NaCl NaCl FIGURE 1 When NaCl dissolves
in water, the ions separate
as they leave the crystal.

I O N S I N A Q U E O U S S O L U T I O N S A N D C O L L I G A T I V E P R O P E R T I E S 435

Assuming 100% dissociation, a solution that contains 1 mol of so-
dium chloride contains 1 mol of Na+ ions and 1 mol of Cl− ions. In this

book, you can assume 100% dissociation for all soluble ionic com-

pounds. The dissociation of NaCl can be represented as follows.

NaCl(s) ⎯H⎯2O→ Na+(aq) + Cl−(aq)

1 mol 1 mol 1 mol

A solution that contains 1 mol of calcium chloride contains 1 mol of
Ca2+ ions and 2 mol of Cl− ions—a total of 3 mol of ions.

CaCl2(s) ⎯H⎯2O→ Ca2+(aq) + 2Cl −(aq)

1 mol 1 mol 2 mol

SAMPLE PROBLEM A

Write the equation for the dissolution of aluminum sulfate, Al2(SO4)3, in water. How many moles of alu-
minum ions and sulfate ions are produced by dissolving 1 mol of aluminum sulfate? What is the total
number of moles of ions produced by dissolving 1 mol of aluminum sulfate?

SOLUTION

1 ANALYZE Given: amount of solute = 1 mol Al2(SO4)3
solvent identity = water

Unknown: a. moles of aluminum ions and sulfate ions

b. total number of moles of solute ions produced

2 PLAN The coefficients in the balanced dissociation equation will reveal the mole relationships,
so you can use the equation to determine the number of moles of solute ions produced.

Al2(SO4)3(s) ⎯H⎯2O→ 2Al 3+(aq) + 3SO24−(aq)

3 COMPUTE a. 1 mol Al2(SO4)3 → 2 mol Al 3+ + 3 mol SO42−
b. 2 mol Al3+ + 3 mol SO42− = 5 mol of solute ions

4 EVALUATE The equation is correctly balanced. Because one formula unit of Al2(SO4)3 produces 5 ions,
1 mol of Al2(SO4)3 produces 5 mol of ions.

PRACTICE Answers in Appendix E

1. Write the equation for the dissolution of each of the following in Go to go.hrw.com for
water, and then determine the number of moles of each ion pro- more practice problems
duced as well as the total number of moles of ions produced. that ask you to calculate
a. 1 mol ammonium chloride moles of ions produced
b. 1 mol sodium sulfide in dissociation.
c. 0.5 mol barium nitrate
Keyword: HC6IONX

436 C H A P T E R 1 3

FIGURE 2 Ionic compounds

can be soluble or insoluble in

water. NiCl2, KMnO4, CuSO4, and
Pb(NO3)2 are soluble in water. AgCl
and CdS are insoluble in water.

Precipitation Reactions www.scilinks.org
Topic: Precipitation Reactions
Although no ionic compound is completely insoluble, compounds of Code: HC61203
very low solubility can be considered insoluble for most practical pur-
poses. Some examples of ionic compounds that are soluble and insolu-
ble in water are shown in Figure 2. It is difficult to write solubility rules
that cover all possible conditions. However, we can write some general
guidelines to help predict whether a compound made of a certain com-
bination of ions is soluble. These general solubility guidelines are given
in Table 1.

By looking at the table you can tell that most sodium compounds are
soluble. Sodium carbonate, Na2CO3, is soluble because it contains so-
dium. Its dissociation equation is as follows.

Na2CO3(s) ⎯H⎯2O→ 2Na+(aq) + CO23−(aq)

TABLE 1 General Solubility Guidelines

1. Sodium, potassium, and ammonium compounds are soluble
in water.

2. Nitrates, acetates, and chlorates are soluble.

3. Most chlorides are soluble, except those of silver, mercury(I),
and lead. Lead(II) chloride is soluble in hot water.

4. Most sulfates are soluble, except those of barium, strontium,
lead, calcium, and mercury.

5. Most carbonates, phosphates, and silicates are insoluble, except
those of sodium, potassium, and ammonium.

6. Most sulfides are insoluble, except those of calcium, strontium,
sodium, potassium, and ammonium.

I O N S I N A Q U E O U S S O L U T I O N S A N D C O L L I G A T I V E P R O P E R T I E S 437

FIGURE 3 Ammonium sulfide is a Is calcium phosphate, Ca3(PO4)2, soluble or insoluble? According to
Table 1, most phosphates are insoluble. Calcium phosphate is not one of
soluble compound that dissociates in the exceptions listed, so it is insoluble. Dissociation equations cannot be
written for insoluble compounds.
water to form NH + and S 2− ions.
4 The information in Table 1 is also useful in predicting what will hap-
pen if solutions of two different soluble compounds are mixed. If the
Cadmium nitrate is a soluble com- mixing results in a combination of ions that forms an insoluble com-
pound, a double-displacement reaction and precipitation will occur.
Precipitation occurs when the attraction between the ions is greater
than the attraction between the ions and surrounding water molecules.

Will a precipitate form when solutions of ammonium sulfide and
cadmium nitrate are combined? By using the table, you can tell that
cadmium nitrate, Cd(NO3)2 , is soluble because it is a nitrate and all
nitrates are soluble. You can also tell that ammonium sulfide, (NH4)2S,
is soluble. It is one of the sulfides listed in the table as being soluble.
Their dissociation equations are as follows.

(NH4)2S(s) ⎯H⎯2O→ 2NH 4+(aq) + S2−(aq)

Cd(NO3)2(s) ⎯H⎯2O→ Cd2+(aq) + 2NO−3(aq)

pound that dissociates in water to
form NO3− and Cd2+ ions. Precipi-
tation of cadmium sulfide occurs

when the two solutions are mixed.

Before reaction
S2–

After reaction
H2O

NH4+ (NH4)2S(aq)
NO3–
Cd2+

CdS(s) + NH4NO3(aq)

CdS(s)

Cd(NO3)2(aq)

438 C H A P T E R 1 3

The two possible products of a double-displacement reaction between
(NH4)2S and Cd(NO3)2 are ammonium nitrate, NH4NO3, and cadmium
sulfide, CdS. The question marks indicate that the states are unknown.

(NH4)2S(aq) + Cd(NO3)2(aq) ⎯→ 2NH4NO3(?) + CdS(?)

To decide whether a precipitate can form, you must know the solubil-
ities of these two compounds. Consulting Table 1, you can see that
NH4NO3 is soluble in water. However, CdS is insoluble. You can there-
fore predict that when solutions of ammonium sulfide and cadmium
nitrate are combined, ammonium nitrate will not precipitate and cadmi-
um sulfide will. As illustrated in Figure 3, crystals of CdS form when the
solutions are mixed. In the following equation, the designations (aq) and
(s) show that NH4NO3(aq) remains in solution and CdS(s) precipitates.

(NH4)2S(aq) + Cd(NO3)2(aq) ⎯→ 2NH4NO3(aq) + CdS(s)

Net Ionic Equations CROSS-DISCIPLINARY

Reactions of ions in aqueous solution are usually represented by net ionic Go to go.hrw.com for a full-length
equations rather than formula equations. A net ionic equation includes article on minerals.
only those compounds and ions that undergo a chemical change in a reac-
tion in an aqueous solution. To write a net ionic equation, you first con- Keyword: HC6IONX
vert the chemical equation into an overall ionic equation. All soluble
ionic compounds are shown as dissociated ions in solution. The precipi-
tates are shown as solids. The precipitation of cadmium sulfide described
previously can be shown by the following overall ionic equation.

Cd2+(aq) + 2NO3−(aq) + 2NH+4(aq) + S2−(aq) ⎯→
CdS(s) + 2NO−3(aq) + 2NH+4(aq)

Notice that the aqueous ammonium ion, NH+4(aq), and the aqueous
nitrate ion, NO−3(aq), appear on both sides of this equation. Therefore,
they have not undergone any chemical change and are still present in

their original form. Ions that do not take part in a chemical reaction and

are found in solution both before and after the reaction are spectator

ions.

To convert an ionic equation into a net ionic equation, the spectator

ions are canceled on both sides of the equation. Eliminating the NH +
4
and NO3− ions from the overall ionic equation above gives the following

net ionic equation.

Cd2+(aq) + S2−(aq) ⎯→ CdS(s)

This net ionic equation applies not only to the reaction between 439
(NH4)2S and Cd(NO3)2 but also to any reaction in which a precipitate
of cadmium sulfide forms when the ions are combined in solution. For
example, it is also the net ionic equation for the precipitation of CdS
when CdSO4 and H2S react.

IONS IN AQUEOUS SOLUTIONS AND COLLIGATIVE PROPERTIES

SAMPLE PROBLEM B For more help, go to the Math Tutor at the end of Chapter 14.

Identify the precipitate that forms when aqueous solutions of zinc nitrate and ammonium sulfide are com-
bined. Write the equation for the possible double-displacement reaction. Then write the formula equation,
overall ionic equation, and net ionic equation for the reaction.

SOLUTION

1 ANALYZE Given: identity of reactants: zinc nitrate and ammonium sulfide
reaction medium: aqueous solution

Unknown: a. equation for the possible double-displacement reaction b. identity of the
precipitate c. formula equation d. overall ionic equation e. net ionic equation

2 PLAN Write the possible double-displacement reaction between Zn(NO3)2 and (NH4)2S. Use
Table 1 to determine if any of the products are insoluble and will precipitate. Write a for-

mula equation and an overall ionic equation, and then cancel the spectator ions to produce

a net ionic equation.

3 COMPUTE a. The equation for the possible double-displacement reaction is as follows.

Zn(NO3)2(aq) + (NH4)2S(aq) ⎯→ ZnS(?) + 2NH4NO3(?)
b. Table 1 reveals that zinc sulfide is not a soluble sulfide and is therefore a precipitate.

Ammonium nitrate is soluble according to the table.

c. The formula equation is as follows.

Zn(NO3)2(aq) + (NH4)2S(aq) ⎯→ ZnS(s) + 2NH4NO3(aq)
d. The overall ionic equation is as follows.

Zn2+(aq) + 2NO3−(aq) + 2NH+4(aq) + S2−(aq) ⎯→ ZnS(s) + 2NH+4(aq) + 2NO−3(aq)
e. The ammonium and nitrate ions appear on both sides of the equation as spectator ions.

The net ionic equation is as follows.
Zn2+(aq) + S2−(aq) ⎯→ ZnS(s)

PRACTICE Answers in Appendix E

1. Will a precipitate form if solutions of potassium sulfate and bar- Go to go.hrw.com for
ium nitrate are combined? If so, write the net ionic equation for more practice problems
the reaction. that ask you to identify
precipitates that form
2. Will a precipitate form if solutions of potassium nitrate and mag- when solutions are
nesium sulfate are combined? If so, write the net ionic equation combined.
for the reaction.
Keyword: HC6IONX
3. Will a precipitate form if solutions of barium chloride and sodium
sulfate are combined? If so, identify the spectator ions and write
the net ionic equation.

4. Write the net ionic equation for the precipitation of nickel(II)
sulfide.

440 C H A P T E R 1 3

Ionization www.scilinks.org
Topic: Hydronium Ion
Some molecular compounds can also form ions in solution. Usually such Code: HC60778
compounds are polar. Ions are formed from solute molecules by the
action of the solvent in a process called ionization. The more general
meaning of this term is the creation of ions where there were none. Note
that ionization is different from dissociation. When an ionic compound
dissolves, the ions that were already present separate from one another.
When a molecular compound dissolves and ionizes in a polar solvent,
ions are formed where none existed in the undissolved compound. Like
all ions in aqueous solution, the ions formed by such a molecular solute
are hydrated. The energy released as heat during the hydration of the
ions provides the energy needed to break the covalent bonds.

In general, the extent to which a solute ionizes in solution depends
on the strength of the bonds within the molecules of the solute and the
strength of attraction between the solute and solvent molecules. If the
strength of a bond within the solute molecule is weaker than the attrac-
tive forces of the solvent molecules, then the covalent bond of the solute
breaks and the molecule is separated into ions. Hydrogen chloride, HCl,
is a molecular compound that ionizes in aqueous solution. It contains a
highly polar bond. The attraction between a polar HCl molecule and the
polar water molecules is strong enough to break the HCl bond, forming
hydrogen ions and chloride ions.

HCl ⎯H⎯2O→ H+(aq) + Cl −(aq)

The Hydronium Ion

Many molecular compounds contain a hydrogen atom bonded by a
polar covalent bond. Some of these compounds ionize in an aqueous
solution to release H+. The H+ ion attracts other molecules or ions so
strongly that it does not normally exist alone. The ionization of hydro-
gen chloride in water is better described as a chemical reaction in which
a proton is transferred directly from HCl to a water molecule, where it
becomes covalently bonded to oxygen and forms H3O+.

H2O(l) + HCl(g) ⎯→ H3O+(aq) + Cl−(aq)

This process is represented in Figure 4. The H3O+ ion is known as the
hydronium ion. The reaction of the H+ ion to form the hydronium ion
produces much of the energy needed to ionize a molecular solute.

FIGURE 4 When hydrogen chlo-
ride gas dissolves in water, it ionizes
to form an H+ ion and a Cl− ion. The

+ – H+ ion immediately bonds to a water

+ + molecule, forming a hydronium ion.
The aqueous solution of hydrogen
chloride is called hydrochloric acid.

I O N S I N A Q U E O U S S O L U T I O N S A N D C O L L I G A T I V E P R O P E R T I E S 441

Na+ HF
Cl–
H3O+
H2O F–

NaCl(aq) HF(aq)
C12H22O11

FIGURE 5 Strong electrolytes, C12H22O11(aq)
such as NaCl, yield only ions when
they dissolve in aqueous solution. Strong and Weak Electrolytes
Weak electrolytes, such as HF,
exist as both ions and unionized As discussed in Chapter 12, substances that yield ions and conduct
molecules in aqueous solution. an electric current in solution are electrolytes. Substances that do not yield
Nonelectrolytes, such as sucrose, ions and do not conduct an electric current in solution are non-
C12H22O11, do not form any ions in electrolytes. Hydrogen chloride is one of a series of compounds
aqueous solution. composed of hydrogen and the members of Group 17 (known as the halo-
gens). The hydrogen halides are all molecular compounds with single
442 C H A P T E R 1 3 polar-covalent bonds.All are gases, all are very soluble in water, and all are
electrolytes. Hydrogen chloride, hydrogen bromide, and hydrogen iodide
strongly conduct an electric current in an aqueous solution. However,
hydrogen fluoride only weakly conducts an electric current at the same
concentration.The strength with which substances conduct an electric cur-
rent is related to their ability to form ions in solution, as shown in Figure 5.

Strong Electrolytes

Hydrogen chloride, hydrogen bromide, and hydrogen iodide are 100%
ionized in dilute aqueous solution. A strong electrolyte is any com-
pound whose dilute aqueous solutions conduct electricity well; this is due
to the presence of all or almost all of the dissolved compound in the form
of ions. Hydrogen chloride, hydrogen bromide, and hydrogen iodide are
all acids in aqueous solution. These acids, several other acids, and all sol-
uble ionic compounds are strong electrolytes.

The distinguishing feature of strong electrolytes is that to whatever
extent they dissolve in water, they yield only ions. For example, some
strong electrolytes, such as NaCl, may be highly soluble in water and
form ions in solution. Others may not dissolve much, but the amount
that does dissolve exists solely as ions in solution.

Weak Electrolytes

Some molecular compounds form aqueous solutions that contain not
only dissolved ions but also some dissolved molecules that are not ion-
ized. Hydrogen fluoride, HF, dissolves in water to give an acidic solution
known as hydrofluoric acid. However, the hydrogen-fluorine bond is
much stronger than the bonds between hydrogen and the other halo-
gens. When hydrogen fluoride dissolves, some molecules ionize. But the
reverse reaction—the transfer of H+ ions back to F− ions to form hydro-
gen fluoride molecules—also takes place.

HF(aq) + H2O(l) ←⎯⎯→ H3O+(aq) + F−(aq)

Thus, the concentration of dissolved intact HF is much greater than the
concentration of H3O+ and F− ions.

Hydrogen fluoride is an example of a weak electrolyte. A weak
electrolyte is any compound whose dilute aqueous solutions conduct
electricity poorly; this is due to the presence of a small amount of the
dissolved compound in the form of ions. This is in contrast to a non-
electrolyte, such as the molecular compound sucrose, which dissolves
but does not produce any ions in solution. Another example of a weak
electrolyte is acetic acid, CH3COOH. Only a small percentage of the
acetic acid molecules ionize in aqueous solution.

CH3COOH(aq) + H2O(l) ←⎯⎯→ CH3COO−(aq) + H3O+(aq)

The description of an electrolyte as strong or weak must not be con-
fused with the description of a solution as concentrated or dilute. Strong
and weak electrolytes differ in the degree of ionization or dissociation.
Concentrated and dilute solutions differ in the amount of solute dis-
solved in a given quantity of a solvent. Hydrochloric acid is always a
strong electrolyte. This is true even in a solution that is 0.000 01 M—a
very dilute solution. By contrast, acetic acid is always considered a weak
electrolyte, even in a 10 M solution—a very concentrated solution.

SECTION REVIEW 4. Explain why HCl is a strong electrolyte and HF is a
weak electrolyte.
1. Write the equation for the dissolution of Sr(NO3)2
in water. How many moles of strontium ions and Critical Thinking
nitrate ions are produced by dissolving 0.5 mol of 5. PREDICTING OUTCOMES For each of the following
strontium nitrate?
pairs, tell which solution contains the larger total
2. Will a precipitate form if solutions of magnesium concentration of ions.
acetate and strontium chloride are combined? a. 0.10 M HCl and 0.05 M HCl
b. 0.10 M HCl and 0.10 M HF
3. What determines whether a molecular compound c. 0.10 M HCl and 0.10 M CaCl2
will be ionized in a polar solvent?

I O N S I N A Q U E O U S S O L U T I O N S A N D C O L L I G A T I V E P R O P E R T I E S 443

The Riddle of Electrolysis

Historical Perspective position of the solids, with the two products of the
When Michael Faraday performed his electrochemical solids’ breakdown also evolving at opposite poles of the
experiments, little was known about the relationship battery. This electrochemical decomposition was later
between matter and electricity. Chemists were still debat- named electrolysis.
ing the existence of atoms, and the discovery of the elec-
tron was more than 50 years in the future. Combining his The Roots of Electrolytic Theory
talents in electrical and chemical investigation, Faraday The discovery of electrolysis led two pioneering chemists
pointed researchers to the intimate connection between to ponder the connection between chemical forces and
chemical reactions and electricity while setting the stage electricity. One of the chemists was Humphry Davy, who
for the development of a new branch of chemistry. thought that chemical bonding must be driven by the
same forces that drive electrical attractions.
Electrifying Experiments
In 1800, Italian physicist Alessandro Volta introduced his The Swedish chemist Jöns Jacob Berzelius took Davy’s
voltaic pile, better known as the battery. The stack of alter- idea a step further. He postulated that matter consisted of
nating zinc and silver disks provided scientists with a combinations of “electropositive” and “electronegative”
source of electric current for the first time. substances, classifying the parts by the pole at which they
accumulated during electrolysis.
That same year, chemists discovered a new phenome-
non using Volta’s device. They immersed the two poles These ideas inspired two early electrolytic theories, each
of a battery at different locations in a container of of which ultimately proved incorrect but contributed to
water. The current caused the water to decompose into our present understanding of the phenomenon. The con-
its elemental components, with hydrogen evolving at the tact theory proposed that electrolytic current was due
positive pole of the battery and oxygen evolving at the merely to the contact of the battery’s metals with the elec-
negative pole. Similar experiments using solutions of trolytic solution. The chemical theory, on the other hand,
certain solids dissolved in water resulted in the decom- attributed the current to undefined changes in the solu-
tion’s components.

Chlorine being produced by electrolysis.

444 C H A P T E R 1 3

Faraday Provides a Spark Michael Faraday used this instrument in his
Although Michael Faraday is best remembered for his studies of electrolysis.
work in electromagnetism, he began his career as
Humphry Davy’s laboratory assistant at the Royal Questions
Institution in London and went on to be the professor of
chemistry there for over 30 years. In the 1830s, Faraday 1. To which scientist does this feature give the most
devised several ingenious experiments to determine credit for the initial understanding of the basic nature
whether the current in an electrolytic solution is depend- of electrolysis? Briefly summarize his contributions in
ent solely on the contact of the battery’s poles with the this area.
solution. In a typical setup, one of the poles was separated
from the solution, and electricity was permitted to enter 2. How are amounts of hydrogen and oxygen gas made
the solution by way of a spark. In all cases, Faraday from the electrolytic decomposition of water related
observed current in the electrolytic cell despite one or both to the chemical formula of water?
of the poles not being in direct contact with the electrolyt-
ic solution. In 1833, he made the hypothesis that the www.scilinks.org
process of electrolysis was due to the intrinsic properties Topic: Electrolysis
of the metals in solution and the effect of current on these Code: HC60479
properties.

Although the battery’s poles were, in fact, later shown
to play a part in the current, Faraday had established the
active role of the electrolytic solution in electrolysis. And in
realizing that electricity affected the chemical nature of
the solution, he anticipated the ideas of oxidation and
reduction despite that the concepts of electrons and ions
were unknown at the time.

Faraday’s Legacy
Faraday continued to study the role of the electrolytic
solution, or electrolyte, as he named it, in electrolysis. He
also coined most of the other modern terms of electrolysis,
including electrode, ion, anode, cathode, anion, and
cation. These investigations culminated in the discovery of
his basic laws of electrolysis.

Still valid today, these principles put electrolysis on a
quantitative footing, leading to our current understanding
of the phenomenon. They also bolstered the atomic theory,
which was still seriously contested by many chemists at
the time. And perhaps most important, Faraday’s experi-
ments inspired his successors to further clarify the chemi-
cal nature of solutions. This ultimately led to Svante
Arrhenius’s theory of electrolytic dissociation and the evo-
lution of a new division in the chemical field, known today
as physical chemistry.

I O N S I N A Q U E O U S S O L U T I O N S A N D C O L L I G A T I V E P R O P E R T I E S 445

SECTION 2 Colligative Properties
of Solutions
OBJECTIVES
T he presence of solutes affects the properties of the solutions. Some
List four colligative properties,
and explain why they are of these properties are not dependent on the nature of the dissolved
classified as colligative substance but only on how many dissolved particles are present. Prop-
properties. erties that depend on the concentration of solute particles but not on their
identity are called colligative properties. In calculations involving some
Calculate freezing-point colligative properties, the concentration is given in terms of molality, m.
depression, boiling-point
elevation, and solution molal- Vapor-Pressure Lowering
ity of nonelectrolyte solutions.
The boiling point and freezing point of a solution differ from those of
Calculate the expected the pure solvent. The graph in Figure 6 shows that a nonvolatile solute
changes in freezing point and raises the boiling point and lowers the freezing point. A nonvolatile
boiling point of an electrolyte substance is one that has little tendency to become a gas under existing
solution. conditions.

Discuss causes of the differ-
ences between expected and
experimentally observed
colligative properties of
electrolyte solutions.

Vapor Pressure vs. Temperature for a Pure Solvent
and a Solution with a Nonvolatile Solute

Vapor pressure of solvent (atm) 1.0 Liquid ΔPsolvent
Solid PureSosolulvteionnt at bp of
pure solvent

FIGURE 6 Vapor pressure as a Gas
function of temperature is shown for
a pure solvent and for a solution of tf tf tb tb
a nonvolatile solute in that solvent. solution pure pure solution
The vapor pressure of the solution is solvent solvent Δt b
lower than the vapor pressure of the Δtf
pure solvent. This can be seen by
noting the decrease in pressure Temperature (ºC)
between the pure solvent and the
solution at the temperature that is
the boiling point of the pure solvent.
The solute thus reduces the freezing
point and elevates the boiling point.

446 C H A P T E R 1 3

Pure water Aqueous solution of nonvolatile solute

Used to represent C12H22O11, sucrose FIGURE 7 At a given tempera-
Used to represent H2O, water ture, the vapor pressure of water
over pure water is greater than the
To understand why a nonvolatile solute changes the boiling point and vapor pressure of water over an
freezing point, you must consider equilibrium vapor pressure, which aqueous solution containing a non-
was discussed in Chapter 10. Vapor pressure is the pressure caused by volatile solute, such as sucrose.
molecules in the gas phase that are in equilibrium with the liquid phase.
Experiments show that the vapor pressure of a solvent containing a
nonvolatile solute is lower than the vapor pressure of the pure solvent
at the same temperature, as shown in Figure 7. As the number of solute
particles increases in a given volume of solution, the proportion of solvent
(water) molecules decreases. Fewer water molecules will be available to
escape from the liquid. As a result, the tendency of water molecules to
leave the solution and enter the vapor phase decreases. Thus, the vapor
pressure of the solution is less than the vapor pressure of pure water.

Nonelectrolyte solutions of the same molality have the same concen-
tration of particles. Equally dilute molal solutions of any nonelectrolyte
solutes in the same solvent lower the vapor pressure equally. For exam-
ple, a 1 m aqueous solution of the nonelectrolyte glucose, C6H12O6, low-
ers the vapor pressure of water by 5.5 × 10−4 atm at 25°C. A 1 m solution
of sucrose, C12H22O11, another nonelectrolyte, also lowers the vapor
pressure by 5.5 × 10−4 atm. Because vapor-pressure lowering depends on
the concentration of a nonelectrolyte solute and is independent of
solute identity, it is a colligative property.

Refer to the graph in Figure 6. Because the vapor pressure has been
lowered, the solution remains liquid over a larger temperature range.
This lowers the freezing point and raises the boiling point. It follows
that changes in boiling point and freezing point also depend on the con-
centration of solute and are therefore colligative properties.

I O N S I N A Q U E O U S S O L U T I O N S A N D C O L L I G A T I V E P R O P E R T I E S 447

Freezing-Point Depression

The freezing point of a 1 m solution of any nonelectrolyte solute in
water is found by experiment to be 1.86°C lower than the freezing point
of water. That is, when 1 mol of a nonelectrolyte solute is dissolved in
1 kg of water, the freezing point of the solution is −1.86°C instead of
0.00°C. When 2 mol of a nonelectrolyte solute is dissolved in 1 kg of
water, the freezing point of the solution is −3.72°C. This is 2 × (−1.86°C).
In fact, for any concentration of a nonelectrolyte solute in water, the
decrease in freezing point can be estimated by using the value of
−1.86°C/m. This value, called the molal freezing-point constant (Kf) is
the freezing-point depression of the solvent in a 1-molal solution of a
nonvolatile, nonelectrolyte solute.

Each solvent has its own characteristic molal freezing-point constant.
The values of Kf for some common solvents are given in Table 2. These
values are most accurate for dilute solutions at 1 atmosphere of pres-
sure. Some variations are introduced in the value of Kf at other pres-
sures and with more-concentrated solutions. The table also shows the
values of a related quantity called Kb, which you will study next.

As stated earlier, the freezing point of a solution containing 1 mol of a
nonelectrolyte solute in 1 kg water is 1.86°C lower than the normal freez-
ing point of water. The freezing-point depression, Δtf, is the difference
between the freezing points of the pure solvent and a solution of a non-
electrolyte in that solvent, and it is directly proportional to the molal con-
centration of the solution. As shown by the previous example, if the molal
concentration is doubled, the freezing-point depression is doubled.
Freezing-point depression can be calculated by the following equation.

Δtf = Kf m

Kf is expressed as °C/m, m is expressed in mol solute/kg solvent (molal-
ity), and Δtf is expressed in °C. Sample Problems C and D show how this
relationship can be used to determine the freezing-point depression and
molal concentration of a solution.

TABLE 2 Molal Freezing-Point and Boiling-Point Constants

Solvent Normal f.p. Molal f.p. constant, Normal b.p. Molal b.p. constant,
Acetic acid (°C) Kf (°C/m) (°C) Kb (°C/m)
Camphor −3.90 117.9 3.07
Ether 16.6 −39.7 207.4 5.61
Naphthalene 178.8 −1.79 34.6 2.02
Phenol −116.3 − 6.94 217.7 5.80
Water 80.2 −7.40 181.8 3.60
40.9 −1.86 100.0 0.51

0.00

448 C H A P T E R 1 3

SAMPLE PROBLEM C For more help, go to the Math Tutor at the end of this chapter.

What is the freezing-point depression of water in a solution of 17.1 g of sucrose, C12H22O11, in 200. g of
water? What is the actual freezing point of the solution?

SOLUTION

1 ANALYZE Given: solute mass and chemical formula = 17.1 g C12H22O11
solvent mass and identity = 200. g water

Unknown: a. freezing-point depression

b. freezing point of the solution

2 PLAN Find the molal freezing-point constant, Kf , for water in Table 2. To use the equation for
3 COMPUTE freezing-point depression, Δtf = Kf m, you need to determine the molality of the solution.

mass of solute (g) × 1 mol solute = amount of solute (mol)

molar mass of solute (g)

amount of solute (mol) 1000 g water
× = molality
mass of solvent (g) 1 kg water

Δtf = Kf m
f.p. solution = f.p. solvent + Δtf

17.1 g C12H22O11 × 1 mol C12H22O11 = 0.0500 mol C12H22O11
342.34 g C12H22O11

0.0500 mol C12H22O11 × 1000 g water = 0.250 mol C12H22O11 = 0.250 m
200. g water kg water kg water

a. Δtf = 0.250 m × (−1.86°C/m) = −0.465°C
b. f.p. solution = 0.000°C + (−0.465°C) = −0.465°C

SAMPLE PROBLEM D For more help, go to the Math Tutor at the end of this chapter.

A water solution containing an unknown quantity of a nonelectrolyte solute is found to have a freezing
point of – 0.23°C. What is the molal concentration of the solution?

SOLUTION

1 ANALYZE Given: freezing point of solution = −0.23°C
Unknown: molality of the solution

2 PLAN Water is the solvent, so you will need the value of Kf , the molal-freezing-point constant for
water, from Table 2. The Δtf for this solution is the difference between the f.p. of water and
the f.p. of the solution. Use the equation for freezing-point depression to calculate molality.

Δtf = f.p. of solution − f.p. of pure solvent
Δtf = Kf m Solve for molality, m.

m = Δtf
Kf

I O N S I N A Q U E O U S S O L U T I O N S A N D C O L L I G A T I V E P R O P E R T I E S 449

3 COMPUTE Δtf = −0.23°C − 0.00°C = −0.23°C
4 EVALUATE − 0.23°C

m = = 0.12 m
−1.86°C/m

As shown by the unit cancellation, the answer gives the molality, as desired. The answer is
properly limited to two significant digits.

PRACTICE Answers in Appendix E

1. A solution consists of 10.3 g of the nonelectrolyte glucose, Go to go.hrw.com for
C6H12O6, dissolved in 250. g of water. What is the freezing-point more practice problems
depression of the solution? that ask you to calculate
the freezing-point
2. In a laboratory experiment, the freezing point of an aqueous solu- depression.
tion of glucose is found to be − 0.325°C. What is the molal concen-
tration of this solution? Keyword: HC6IONX

3. If 0.500 mol of a nonelectrolyte solute are dissolved in 500.0 g of
ether, what is the freezing point of the solution?

4. The freezing point of an aqueous solution that contains a non-
electrolyte is −9.0°C.
a. What is the freezing-point depression of the solution?
b. What is the molal concentration of the solution?

Boiling-Point Elevation

As discussed in Chapter 12, the boiling point of a liquid is the tempera-
ture at which the vapor pressure of the liquid is equal to the prevailing
atmospheric pressure. Therefore, a change in the vapor pressure of the
liquid will cause a corresponding change in the boiling point. As stated
earlier, the vapor pressure of a solution containing a nonvolatile solute
is lower than the vapor pressure of the pure solvent. This means that
more energy as heat will be required to raise the vapor pressure of the
solution to equal the atmospheric pressure. Thus, the boiling point of a
solution is higher than the boiling point of the pure solvent.

The molal boiling-point constant (Kb) is the boiling-point elevation of
the solvent in a 1-molal solution of a nonvolatile, nonelectrolyte solute.
The boiling-point elevation of a 1-molal solution of any nonelectrolyte
solute in water has been found by experiment to be 0.51°C. Thus, the
molal boiling-point constant for water is 0.51°C/m.

For different solvents, the molal boiling-point constants have differ-
ent values. Some other values for Kb are included in Table 2. Like the
freezing-point constants, these values are most accurate for dilute
solutions.

The boiling-point elevation, Δtb, is the difference between the boiling
points of the pure solvent and a nonelectrolyte solution of that solvent,

450 C H A P T E R 1 3

and it is directly proportional to the molal concentration of the solution.
Boiling-point elevation can be calculated by the following equation.

Δtb = Kbm

When Kb is expressed in °C/m and m is expressed in mol of solute/kg of
solvent, Δtb is the boiling-point elevation in °C.

SAMPLE PROBLEM E For more help, go to the Math Tutor at the end of this chapter.

What is the boiling-point elevation of a solution made from 20.1 g of a nonelectrolyte solute and 400.0 g
of water? The molar mass of the solute is 62.0 g.

SOLUTION

1 ANALYZE Given: solute mass = 20.1 g
solute molar mass = 62.0 g
solvent mass and identity = 400.0 g of water

Unknown: boiling-point elevation

2 PLAN Find the molal boiling-point constant, Kb, for water in Table 2. To use the equation for
3 COMPUTE boiling-point elevation, Δtb = Kbm, you need to determine the molality of the solution.

mass of solute (g) × 1 mol solute = amount of solute (mol)

molar mass of solute (g)

amount of solute (mol) 1000 g water
× = molality
mass of solvent (g) 1 kg water

Δtb = Kbm

20.1 g of solute × 1 mol solute = 0.324 mol of solute

62.0 g of solute

0.324 mol of solute 1000 g water mol solute
× = 0.810 = 0.810 m
400.0 g water 1 kg water kg water

Δtb = 0.51°C/m × 0.810 m = 0.41°C

PRACTICE Answers in Appendix E

1. A solution contains 50.0 g of sucrose, C12H22O11, a nonelectrolyte,
dissolved in 500.0 g of water. What is the boiling-point elevation?

2. A solution contains 450.0 g of sucrose, C12H22O11, a nonelectrolyte,
dissolved in 250.0 g of water. What is the boiling point of the solution?

3. If the boiling-point elevation of an aqueous solution containing a Go to go.hrw.com for
nonvolatile electrolyte is 1.02°C, what is the molality of the solution? more practice problems
that ask you to calculate
4. The boiling point of an aqueous solution containing a nonvolatile the boiling-point
electrolyte is 100.75°C. elevation.
a. What is the boiling-point elevation?
b. What is the molality of the solution? Keyword: HC6IONX

I O N S I N A Q U E O U S S O L U T I O N S A N D C O L L I G A T I V E P R O P E R T I E S 451

(a) (b)

Pure water

Sucrose solution

Semipermeable membrane

Used to represent C12H22O11, sucrose

Used to represent H2O, water

FIGURE 8 (a) When pure water Osmotic Pressure
and an aqueous sucrose solution are
separated by a semipermeable mem- Figure 8 illustrates another colligative property. In the figure, an aque-
brane, the net movement of water ous sucrose solution is separated from pure water by a semipermeable
molecules through the membrane is membrane. A semipermeable membrane allows the passage of some
from the pure water side into the particles while blocking the passage of others. The level of the sucrose
aqueous solution. (b) The level of solution will rise until a certain height is reached. What causes the level
the solution rises until pressure of the solution to rise?
exerted by the height of the solution
equals the osmotic pressure, at The semipermeable membrane allows water molecules, but not
which point no net movement of sucrose molecules, to pass through. The sucrose molecules on the solu-
water molecules occurs. tion side allow fewer water molecules to strike the membrane than
strike on the pure water side in the same amount of time. Thus, the rate
www.scilinks.org at which water molecules leave the pure water side is greater than the
Topic: Osmosis rate at which they leave the solution. This causes the level of the solu-
Code: HC61090 tion to rise. The level rises until the pressure exerted by the height of the
solution is large enough to force water molecules back through the
membrane from the solution at a rate equal to that at which they enter
from the pure water side.

The movement of solvent through a semipermeable membrane from
the side of lower solute concentration to the side of higher solute concen-
tration is osmosis. Osmosis occurs whenever two solutions of different
concentrations are separated by a semipermeable membrane. Osmotic
pressure is the external pressure that must be applied to stop osmosis. In
the example given above, osmosis caused the level of the solution to rise
until the height of the solution provided the pressure necessary to stop
osmosis. Because osmotic pressure is dependent on the concentration of

452 C H A P T E R 1 3

solute particles and not on the type of solute particles, it is a colligative Water Purification
property. The greater the concentration of a solution, the greater the by Reverse Osmosis
osmotic pressure of the solution.
The process of reverse osmosis is used
Regulation of osmosis is vital to the life of a cell because cell mem- to purify water because it removes many
branes are semipermeable. Cells lose water and shrink when placed in a contaminants at relatively low cost. The
solution of higher concentration. They gain water and swell when placed osmosis process can be reversed by
in a solution of lower concentration. In vertebrates, cells are protected applying a pressure greater than the
from swelling and shrinking by blood and lymph that surround the cells. osmotic pressure to an aqueous solu-
Blood and lymph are equal in concentration to the concentration inside tion. This causes the water, but not the
the cell. dissolved solutes, to flow from the solu-
tion through the semipermeable mem-
Electrolytes and Colligative Properties brane. The largest municipal desalination
plant in the United States is in Sarasota,
Early investigators were puzzled by experiments in which certain sub- Florida. It treats salt water by reverse
stances depressed the freezing point or elevated the boiling point of a osmosis to produce more than 12 mil-
solvent more than expected. For example, a 0.1 m solution of sodium lion gallons of fresh water per day. Other
chloride, NaCl, lowers the freezing point of the solvent nearly twice as applications include the desalination of
much as a 0.1 m solution of sucrose. A 0.1 m solution of calcium chloride, sea water on ocean-going vessels and
CaCl2, lowers the freezing point of the solvent nearly three times as much wastewater purification.
as a 0.1 m solution of sucrose. The effect on boiling points is similar.
FIGURE 9 Compare the
To understand why this is so, contrast the behavior of sucrose with that number of particles produced per
of sodium chloride in aqueous solutions. Sugar is a nonelectrolyte. Each formula unit for these three solutes.
sucrose molecule dissolves to produce only one particle in solution, so Colligative properties depend on
1 mol of sucrose dissolves to produce only 1 mol of particles in solution. the total concentration of particles.
NaCl, however, is a strong electrolyte. Each mole of NaCl dissolves to
produce 2 mol of particles in solution: 1 mol of sodium ions and 1 mol
of chloride ions. Figure 9 compares the production of particles in solu-
tion for three different solutes. As you can see, electrolytes produce
more than 1 mol of solute particles for each mole of compound dissolved.

H2O Na+

C12H22O11 Cl–
Ca2+

Sucrose solution Sodium chloride solution Calcium chloride solution
C12H22O11 ⎯H2→O C12H22O11(aq) NaCl ⎯H2→O Na+(aq) + Cl–(aq) CaCl2 ⎯H2→O Ca2+(aq) + 2Cl–(aq)

I O N S I N A Q U E O U S S O L U T I O N S A N D C O L L I G A T I V E P R O P E R T I E S 453

Calculated Values for Electrolyte Solutions

Remember that colligative properties depend on the total concentra-
tion of solute particles regardless of their identity. The changes in col-
ligative properties caused by electrolytes will be proportional to the
total molality of all dissolved particles, not to formula units. For the
same molal concentrations of sucrose and sodium chloride, you would
expect the effect on colligative properties to be twice as large for sodi-
um chloride as for sucrose. What about barium nitrate, Ba(NO3)2? Each
mole of barium nitrate yields 3 mol of ions in solution.

Ba(NO3)2(s) ⎯→ Ba2+(aq) + 2NO3−(aq)

You would expect a Ba(NO3)2 solution of a given molality to lower the
freezing point of its solvent three times as much as a nonelectrolytic
solution of the same molality.

SAMPLE PROBLEM F For more help, go to the Math Tutor at the end of this chapter.

What is the expected change in the freezing point of water in a solution of 62.5 g of barium nitrate,
Ba(NO3)2, in 1.00 kg of water?

SOLUTION

1 ANALYZE Given: solute mass and formula = 62.5 g Ba(NO3)2
solvent mass and identity = 1.00 kg water

Δtf = Kf m
Unknown: expected freezing-point depression

2 PLAN The molality can be calculated by converting the solute mass to moles and then dividing by
the number of kilograms of solvent. That molality is in terms of formula units of Ba(NO3)2
and must be converted to molality in terms of dissociated ions in solution. It must be multi-
plied by the number of moles of ions produced per mole of formula unit. This adjusted
molality can then be used to calculate the freezing-point depression.

mass of solute (g) 1 mol solute mol
΂ ΃mass of solvent (kg) molar mass of solute (g)
× = molality of solution kg

mol ions
× molality conversion

mol
mol
molality of solution

΂ ΃ ΂ ΃ ΂ ΃kg
× Kf °C • kg H2O
mol ions

= expected freezing-point depression (°C)

This problem is similar to Sample Problem E, except that the solute is ionic rather than a
nonionizing molecular solute. The number of particles in solution will therefore equal the
number of ions of the solute.

3 COMPUTE 62.5 g Ba(NO3)2 × mol Ba(NO3)2 = 0.239 mol Ba(NO3)2

1.00 kg H2O 261.35 g Ba(NO3)2 kg H2O

454 C H A P T E R 1 3

Ba(NO3)2(s) ⎯→ Ba2+(aq) + 2NO3−(aq)

Each formula unit of barium nitrate yields three ions in solution.

0.239 mol Ba(NO3)2 × 3 mol ions × −1.86°C • kg H2O = −1.33°C
kg H2O mol Ba(NO3)2 mol ions

4 EVALUATE The units cancel properly to give the desired answer in °C. The answer is correctly given
to three significant digits. The mass of the solute is approximately one-fourth its molar
mass and would give 0.75 mol of ions in the 1 kg of solvent, so the estimated answer of
0.75 × (−1.86°C) = −1.4°C supports our computation.

PRACTICE Answers in Appendix E

1. What is the expected freezing-point depression for a solution that Go to go.hrw.com for
contains 2.0 mol of magnesium sulfate dissolved in 1.0 kg of water? more practice problems
that ask you to calculate
2. What is the expected boiling-point elevation of water for a solution the freezing-point
that contains 150 g of sodium chloride dissolved in 1.0 kg of water? depression and boiling-
point elevation for
3. The freezing point of an aqueous sodium chloride solution is electrolyte solutions.
−0.20°C. What is the molality of the solution?
Keyword: HC6IONX

Actual Values for Electrolyte Solutions

It is important to remember that the values just calculated are only
expected values. As stated earlier, a 0.1 m solution of sodium chloride
lowers the freezing point nearly twice as much as a 0.1 m solution of
sucrose. The actual values of the colligative properties for all strong elec-
trolytes are almost what would be expected based on the number of
particles they produce in solution. Some specific examples are given in
Table 3. The freezing-point depression of a compound that produces two
ions per formula unit is almost twice that of a nonelectrolytic solution.
The freezing-point depression of a compound that produces three ions
per formula unit is almost three times that of a nonelectrolytic solution.

TABLE 3 Molal Freezing-Point Depressions for Aqueous Solutions of Ionic Solutes

Solute Concentration Δtf, observed Δtf, nonelectrolyte Δtf , observed
KCl (m) (°C) solution (°C) Δtf , nonelectrolyte solution

MgSO4 0.1 − 0.345 − 0.186 1.85
0.01 − 0.0361 − 0.0186 1.94
BaCl2 0.001 − 0.00366 − 0.00186 1.97

0.1 − 0.225 − 0.186 1.21
0.01 − 0.0285 − 0.0186 1.53
0.001 − 0.00338 − 0.00186 1.82

0.1 − 0.470 − 0.186 2.53
0.01 − 0.0503 − 0.0186 2.70
0.001 − 0.00530 − 0.00186 2.84

I O N S I N A Q U E O U S S O L U T I O N S A N D C O L L I G A T I V E P R O P E R T I E S 455

FIGURE 10 The salts applied to Look at the values given for KCl solutions in Table 3. The freezing-
icy roads are electrolytes. They lower point depression of a 0.1 m KCl solution is only 1.85 times greater than
the freezing point of water and melt that of a nonelectrolyte solution. However, as the concentration de-
the ice. creases, the freezing-point depression comes closer to the value that is
twice that of a nonelectrolyte solution.

The differences between the expected and calculated values are
caused by the attractive forces that exist between dissociated ions in
aqueous solution. The attraction between the hydrated ions in the solu-
tion is small compared with those in the crystalline solid. However,
forces of attraction do interfere with the movements of the aqueous
ions. Only in very dilute solutions is the average distance between the
ions large enough and the attraction between ions small enough for the
solute ions to move about almost completely freely.

Peter Debye and Erich Hückel introduced a theory in 1923 to account
for this attraction between ions in dilute aqueous solutions. According to
this theory, the attraction between dissociated ions of ionic solids in dilute
aqueous solutions is caused by an ionic atmosphere that surrounds each
ion. This means that each ion is, on average, surrounded by more ions of
opposite charge than of like charge. A cluster of hydrated ions can act as
a single unit rather than as individual ions. Thus, the effective total con-
centration is less than expected, based on the number of ions known to
be present.

Ions of higher charge attract other ions very strongly. They therefore
cluster more and have lower effective concentrations than ions with
smaller charge. For example, ions formed by MgSO4 have charges of 2+
and 2−. Ions formed by KCl have charges of 1+ and 1−. Note in Table 3
that MgSO4 in a solution does not depress the freezing point as much as
the same concentration of KCl.

SECTION REVIEW level of the more-concentrated solution or the
less-concentrated solution rise?
1. What colligative properties are displayed by each
of the following situations? 4. a. Calculate the expected freezing-point depres-
sion of a 0.200 m KNO3 solution.
a. Antifreeze is added to a car’s cooling system to
prevent freezing when the air temperature is b. Will the value you calculated match the actual
below 0°C. freezing-point depression for this solution?
Why or why not?
b. Ice melts on sidewalks after salt has been
spread on them. Critical Thinking

2. Two moles of a nonelectrolyte solute are dissolved 5. INFERRING RELATIONSHIPS The freezing-point
in 1 kg of an unknown solvent. The solution freezes depressions of aqueous solutions A, B, and C are
at 7.8°C below its normal freezing point. What is −2.3°C, −1.2°C, and −4.1°C, respectively. Predict
the molal freezing-point constant of the unknown the order of the boiling-point elevations of these
solvent? Suggest a possible identity of the solvent. solutions, from lowest to highest. Explain your
ranking.
3. If two solutions of equal amounts in a U-tube are
separated by a semipermeable membrane, will the

456 C H A P T E R 1 3

CHAPTER HIGHLIGHTS

Compounds in Aqueous Solutions

Vocabulary • The separation of ions that occurs when an ionic solid dis-
dissociation
net ionic equation solves is called dissociation.
spectator ions
ionization • When two different ionic solutions are mixed, a precipitate may
hydronium ion
strong electrolyte form if ions from the two solutions react to form an insoluble
weak electrolyte compound.

• A net ionic equation for a reaction in aqueous solution

includes only compounds and ions that change chemically in
the reaction. Spectator ions are ions that do not take part in
such a reaction.

• Formation of ions from molecular compounds is called ioniza-

tion. A molecular compound may ionize in an aqueous solution
if the attraction of the polar water molecules is strong enough
to break the polar-covalent bonds of the solute molecules.

• An H3O+ ion is called a hydronium ion.
• All, or almost all, of a dissolved strong electrolyte exists as ions

in an aqueous solution, whereas a relatively small amount of a
dissolved weak electrolyte exists as ions in an aqueous solution.

Colligative Properties of Solutions

Vocabulary • Colligative properties of solutions depend only on the total num-

colligative properties ber of solute particles present. Boiling-point elevation, freezing-
point depression, vapor-pressure lowering, and osmotic pressure
nonvolatile substance are colligative properties.

molal freezing-point constant, Kf • The molal boiling-point and freezing-point constants are used
freezing-point depression, Δtf
molal boiling-point constant, Kb to calculate boiling-point elevations and freezing-point depres-
boiling-point elevation, Δtb sions of solvents containing nonvolatile solutes.
semipermeable membrane
• Electrolytes have a greater effect on the freezing and boiling
osmosis
points of solvents than nonelectrolytes do.
osmotic pressure
• Except in very dilute solutions, the values of colligative proper-

ties of electrolyte solutions are less than expected because of
the attraction between ions in solution.

I O N S I N A Q U E O U S S O L U T I O N S A N D C O L L I G A T I V E P R O P E R T I E S 457

CHAPTER REVIEW

For more practice, go to the Problem Bank in Appendix D. 10. Write the equation for the dissolution of each
of the following in water, and then indicate the
Compounds in Aqueous total number of moles of solute ions formed.
Solutions a. 0.50 mol strontium nitrate
b. 0.50 mol sodium phosphate
SECTION 1 REVIEW
11. Using Table 1, write the balanced chemical equa-
1. How many moles of ions are contained in 1 L of tion, write the overall ionic equation, identify
a 1 M solution of KCl? of Mg(NO3)2? the spectator ions and possible precipitates, and
write the net ionic equation for each of the fol-
2. Use Table 1 to predict whether each of the follow- lowing reactions. (Hint: See Sample Problem B.)
ing compounds is considered soluble or insoluble: a. mercury(II) chloride (aq) + potassium sul-
a. KCl fide (aq) ⎯→
b. NaNO3 b. sodium carbonate (aq) + calcium chloride
c. AgCl (aq) ⎯→
d. BaSO4 c. copper(II) chloride (aq) + ammonium phos-
e. Ca3(PO4)2 phate (aq) ⎯→
f. Pb(ClO3)2
g. (NH4)2S 12. Identify the spectator ions in the reaction
h. PbCl2 (in cold water) between KCl and AgNO3 in an aqueous solution.
i. FeS
j. Al2(SO4)3 13. Copper(II) chloride and lead(II) nitrate react in
aqueous solutions by double displacement.
3. What is a net ionic equation? Write the balanced chemical equation, the over-
4. a. What is ionization? all ionic equation, and the net ionic equation for
this reaction. If 13.45 g of copper(II) chloride
b. Distinguish between ionization and react, what is the maximum amount of precipi-
dissociation. tate that could be formed?

5. a. Define and distinguish between strong Colligative Properties
electrolytes and weak electrolytes. of Solutions

b. Give two examples of each type. SECTION 2 REVIEW
6. What determines the strength of a solute as an
14. How does the presence of a nonvolatile solute
electrolyte? affect each of the following properties of the
7. Distinguish between the use of the terms strong solvent in which the solute is dissolved?
a. vapor pressure
and weak and the use of the terms dilute and b. freezing point
concentrated when describing electrolyte c. boiling point
solutions. d. osmotic pressure

PRACTICE PROBLEMS 15. Using Figure 6 as a guide, make a graph of
vapor pressure versus temperature that shows
8. Write the equation for the dissolution of each the comparison of pure water, a solution with
x concentration of solute, and a solution with
of the following ionic compounds in water. 2x concentration of solute. What is the relation-
ship between Δtf for the x solution and Δtf for
(Hint: See Sample Problem A.) the 2x solution?

a. KI c. MgCl2

b. NaNO3 d. Na2SO4

9. For the compounds listed in the previous prob-

lem, determine the number of moles of each ion

produced as well as the total number of moles

of ions produced when 1 mol of each compound

dissolves in water.

458 C H A P T E R 1 3

CHAPTER REVIEW

16. a. Why does the level of the more-concentrated 24. If 1.500 g of a solute that has a molar mass of
solution rise when two solutions of different 125.0 g were dissolved in 35.00 g of camphor,
concentrations are separated by a semiper- what would be the resulting freezing point of
meable membrane? the solution?

b. When does the level of the solution stop rising? 25. Determine the boiling-point elevation of H2O
c. When the level stops rising, what is the net in each of the following solutions. (Hint: See
Sample Problem E.)
movement of water molecules across the a. 2.5 m solution of C6H12O6 (glucose) in H2O
membrane? b. 3.20 g C6H12O6 in 1.00 kg H2O
17. a. Compare the effects of nonvolatile elec- c. 20.0 g C12H22O11 (sucrose) in 500. g H2O
trolytes with the effects of nonvolatile non-
electrolytes on the freezing and boiling points 26. Given the following boiling points, determine
of solvents in which they are dissolved. the molality of each water solution.
b. Why are such differences observed? a. 100.25°C
18. Why does the actual freezing-point depression of b. 101.53°C
an electrolytic solution differ from the freezing- c. 102.805°C
point depression calculated on the basis of the
concentration of particles? 27. Given 1.00 m aqueous solutions of each of the
following electrolytic substances, determine the
PRACTICE PROBLEMS expected change in the freezing point of the
solvent. (Hint: See Sample Problem F.)
19. Determine the freezing-point depression of a. KI
H2O in each of the following solutions. (Hint: b. CaCl2
See Sample Problem C.) c. Ba(NO3)2
a. 1.50 m solution of C12H22O11 (sucrose) in H2O
b. 171 g of C12H22O11 in 1.00 kg H2O 28. What is the expected change in the freezing
c. 77.0 g of C12H22O11 in 400. g H2O point of water for a 0.015 m aqueous solution
of AlCl3?
20. Given the following freezing-point depressions,
determine the molality of each solution of an 29. What is the expected freezing point of a solu-
unknown nonelectrolyte in water. (Hint: See tion containing 85.0 g NaCl dissolved in 450. g
Sample Problem D.) of water?
a. −0.930°C
b. −3.72°C 30. Determine the expected boiling point of a solu-
c. −8.37°C tion made by dissolving 25.0 g of barium chlo-
ride in 0.150 kg of water.
21. A solution contains 20.0 g of C6H12O6 (glucose)
in 250. g of water. 31. The change in the boiling point of water for an
a. What is the freezing-point depression of the aqueous solution of potassium iodide is 0.65°C.
solvent? Determine the molal concentration of potas-
b. What is the freezing point of the solution? sium iodide.

22. How many grams of antifreeze, C2H4(OH)2, 32. The freezing point of an aqueous solution of
would be required per 500. g of water to pre- barium nitrate is −2.65°C. Determine the molal
vent the water from freezing at a temperature concentration of barium nitrate.
of −20.0°C?
33. Calculate the expected freezing point of a solu-
23. Pure benzene, C6H6, freezes at 5.45°C. A solu- tion containing 1.00 kg H2O and 0.250 mol
tion containing 7.24 g C2Cl4H2 in 115 g of ben- NaCl.
zene (specific gravity = 0.879) freezes at 3.55°C.
Based on these data, what is the molal freezing- 34. Experimental data for a 1.00 m MgI2 aqueous
point constant for benzene? solution indicate an actual change in the freezing
point of water of −4.78°C. Find the expected
change in the freezing point of water. Suggest a
possible reason for the discrepancy between the
experimental and expected values.

I O N S I N A Q U E O U S S O L U T I O N S A N D C O L L I G A T I V E P R O P E R T I E S 459

CHAPTER REVIEW

MIXED REVIEW 45. Find the expected freezing point of a water
solution that contains 268 g of aluminum
35. Given 0.01 m aqueous solutions of each of the nitrate, Al(NO3)3, in 8.50 kg of water.
following, arrange the solutions in order of
increasing change in the freezing point of the CRITICAL THINKING
solution.
a. NaI 46. Applying Models
b. CaCl2 a. You are conducting a freezing-point determi-
c. K3PO4 nation in the laboratory by using an aqueous
d. C6H12O6 (glucose) solution of KNO3. The observed freezing
point of the solution is −1.15°C. Using a pure
36. What is the molal concentration of an aqueous water sample, you recorded the freezing
calcium chloride solution that freezes at −2.43°C? point of the pure solvent on the same ther-
mometer as 0.25°C. Determine the molal
37. a. Write the balanced formula equation that concentration of KNO3. Assume that there
shows the possible products of a double- are no forces of attraction between ions.
displacement reaction between calcium b. You are not satisfied with the result in part
nitrate and sodium chloride. (a) because you suspect that you should not
ignore the effect of ion interaction. You take
b. Using Table 1, determine whether there is a a 10.00 mL sample of the solution. After
precipitate. carefully evaporating the water from the
solution, you obtain 0.415 g KNO3. Deter-
c. Does this reaction occur? mine the actual molal concentration of
38. Write a balanced equation to show what occurs KNO3 and the percentage difference
between the concentration observed in the
when hydrogen bromide dissolves and reacts freezing-point determination and the actual
with water. Include a hydronium ion in the concentration of KNO3. Assume that the
equation. solution’s density is 1.00 g/mL.
39. Write the equation for the dissolution of each
of the following in water, and then indicate the 47. Analyzing Information The observed freezing-
total number of moles of solute ions formed. point depression for electrolyte solutions is
a. 0.275 mol of potassium sulfide sometimes less than the calculated value. Why
b. 0.15 mol of aluminum sulfate does this occur? Is the difference greater for
40. Calculate the expected change in the boiling concentrated solutions or dilute solutions?
point of water in a solution made up of 131.2 g
of silver nitrate, AgNO3, in 2.00 kg of water. 48. Analyzing Information The osmotic pressure
41. Nitrous acid, HNO2, is a weak electrolyte. Nitric of a dilute solution can be calculated as follows:
acid, HNO3, is a strong electrolyte. Write equa-
tions to represent the ionization of each in p = MRT
water. Include the hydronium ion, and show the p = osmotic pressure
appropriate kind of arrow in each equation. M = concentration in moles per liter
42. Find the boiling point of an aqueous solution con- R = ideal gas constant
taining a nonelectrolyte that freezes at −6.51°C. T = absolute temperature of the solution
43. Write a balanced equation for the dissolution of
sodium carbonate, Na2CO3, in water. Find the How does the osmotic-pressure equation com-
number of moles of each ion produced when pare with the ideal gas law?
0.20 mol of sodium carbonate dissolves. Then,
find the total number of moles of ions.
44. Given the reaction below and the information
in Table 1, write the net ionic equation for the
reaction.
potassium phosphate (aq) + lead (II) nitrate (aq)

460 C H A P T E R 1 3

CHAPTER REVIEW

USING THE HANDBOOK ALTERNATIVE ASSESSMENT

49. Common reactions for Group 13 elements are 53. Performance Determine the freezing point of
found in the Elements Handbook. Review this four different mixtures of water and ethylene
material, and answer the following: glycol (use commercial antifreeze). What mix-
a. Write net ionic equations for each of the ture has the lowest freezing point?
example reactions shown on page 809.
b. Which reactions did not change when written 54. Performance Find the optimum mixture of
in net ionic form? Why? salt and ice for reducing the temperature of the
chilling bath for an ice-cream freezer. Use your
50. Common reactions for Group 14 elements are data to write a set of instructions on how to pre-
found in the Elements Handbook. Review this pare the chilling bath for making ice cream.
material, and answer the following:
a. Write net ionic equations for each of the 55. Performance Using a low-voltage dry cell,
example reactions shown on page 813. assemble a conductivity apparatus. Secure sev-
b. Which reactions did not change when written eral unknown aqueous solutions of equal molal-
in net ionic form? Why? ity from your instructor, and use the apparatus
to distinguish the electrolytes from the non-
RESEARCH & WRITING electrolytes. Among those identified as electro-
lytes, rank their relative strengths as conductors
51. Find out how much salt a large northern city, from good to poor.
such as New York City or Chicago, uses on its
streets in a typical winter. What environmental 56. Performance Using equal volumes of the
problems result from this use of salt? What unknown solutions from the preceding activity,
substitutes for salt are being used to melt ice explain how you could use the freezing-point
and snow? depression concept to distinguish the electro-
lytes from the nonelectrolytes. Explain how you
52. Research the role of electrolytes and electro- could determine the number of ions contained
lytic solutions in your body. Find out how per molecule among the solutes identified as
electrolytes work in the functioning of nerves electrolytes. Design and conduct an experiment
and muscles. What are some of the health prob- to test your theories.
lems that can arise from an imbalance of elec-
trolytes in body fluids?

I O N S I N A Q U E O U S S O L U T I O N S A N D C O L L I G A T I V E P R O P E R T I E S 461

Math Tutor BOILING AND FREEZING POINTS OF SOLUTIONS

The presence of a nonvolatile solute causes the freezing point of a solution to be lower
and the boiling point to be higher than those of the pure solvent. The freezing-point
depression, Δtf, is the amount that the freezing point is lowered. It is calculated by using
the formula Δtf = Kfm. The boiling-point elevation, Δtb, is the amount that the boiling
point is elevated. It is found by using the equation Δtb = Kbm. To determine Δtf or Δtb,
you need to know its molal concentration, m (moles of solute per kilogram of solvent).
You also need to know the molal freezing-point constant, Kf, or the molal boiling-point
constant, Kb. The values for Kf and Kb depend on the solvent and are given in Table 2.
When these equations are used for electrolytes, the molality represents the total number
of moles of ions in solution.

Problem-Solving S

• Make sure that you find the molal concentration, not the molar concentration.
• For electrolytes, calculate the total number of moles of ions in solution.

SAMPLE

What is the theoretical boiling point of a solution of 247 g of potassium chloride, KCl, dis-
solved in 2.90 kg of water? Potassium chloride is a strong electrolyte.

First, determine the molality of the KCl that dissolved.

mass of solute (g) × 1 mol solute = amount of solute (mol)

molar mass of solute (g)

247 g KCl × 1 mol KCl = 3.31 mol KCl

74.55 g KCl

mol solute particles 3.31 mol KCl
molality KCl(m) = = = 1.14 m KCl
mass of solvent (kg) 2.90 kg H2O

Because KCl is an electrolyte, the total moles of ions in solution must now be determined.
The equation KCl(s) ⎯→ K+(aq) + Cl−(aq) shows that 1 mol KCl will yield 2 mol of ions.

To find the total molality of particles in solution, multiply the molality of the solute by

the moles of ions produced by the dissociation: 2 × 1.14 m KCl = 2.28 m.

Next, use the equation Δtb = Kbm, where Kb for water is 0.51°C/m and the value for m is 2.28.
Δtb = (0.51°C/m)(2.28 m) = 1.16°C

The new boiling point is equal to the sum of the boiling point of the solvent and Δtb.
boiling point of solution = 100°C + 1.16°C = 101.16°C

PRACTICE PROBLEMS 2. What is the boiling point of a solution com-

1. What is the freezing point of a solution contain- posed of 850 g of ethylene glycol, C2H6O2,
ing 28.0 g of the strong electrolyte calcium chlo- mixed with 1100 g of water?
ride, CaCl2, dissolved in 295 g of water?

462 C H A P T E R 1 3

Standardized Test Prep

Answer the following items on a separate piece of paper. The table below shows the freezing points of solu-
tions of three nonvolatile solutes, X, Y, and Z, in
MULTIPLE CHOICE water. Refer to the table to answer items 6 and 7.
The Kf for water is −1.86°C/m.
1.Acetic acid is a weak electrolyte because it
A. is miscible with water. Solute Solute Water Freezing
B. forms hydronium and hydroxide ions in X (mol) (g) point (°C)
aqueous solution. Y
C. lowers the freezing point of water. Z 1.00 1000 −5.58
D. ionizes only slightly in aqueous solution. 1.00 1000 −1.86
1.00 1000 −3.72
2.Which of the following solutions would contain
the highest concentration of hydronium ions, 6.Which of the following statements is not true?
H3O+? A. All three solutes are nonelectrolytes.
A. 0.10 M HCl B. Solute X could be Ca(NO3)2.
B. 0.10 M HF C. Solute Y could be sucrose.
C. 0.10 M CH3COOH D. Solute Z could be KCl.
D. 0.10 M NaCl
7.Which of the solutions described in the table
3.Which of the following is the best representa- would have the highest boiling point?
tion of the precipitation reaction that occurs A. the solution of solute X
when aqueous solutions of sodium carbonate B. the solution of solute Y
and calcium chloride are mixed? C. the solution of solute Z
A. Na+(aq) + Cl−(aq) ⎯→ 2NaCl(s) D. All three solutions would have the same
B. 2Na+(aq) + CO23−(aq) + boiling point.
Ca2+(aq) + 2Cl−(aq) ⎯→ CaCO3(s) + 2NaCl(s)
C. Ca2+(aq) + CO23−(aq) ⎯→ CaCO3(s) SHORT ANSWER
D. No precipitation reaction occurs.
8.An aqueous solution of an unknown quantity of
4.Which of the following is not a colligative a nonelectrolyte solute is found to have a freez-
property? ing point of −0.58°C. What is the molal concen-
A. molality tration of the solution?
B. vapor-pressure lowering
C. boiling-point elevation 9.Give the names and formulas of two strong
D. osmotic pressure electrolytes.

5.Solution A contains 0.1 mol of sucrose, EXTENDED RESPONSE
C12H22O11, dissolved in 500 g of water. Solution
B contains 0.1 mol of sodium chloride, NaCl, in 10.Write the formula equation, the overall ionic
500 g of water. Which of the following state- equation, and the net ionic equation for the pre-
ments about these solutions is true? cipitation reaction that occurs when solutions of
A. Both solutions have the same vapor pressure. zinc chloride, ZnCl2, and sodium sulfide, Na2S,
B. Solution A would boil at a higher tempera- are mixed.
ture than solution B would.
C. Solution A would freeze at a higher tempera- 11.Distinguish between dissociation and ionization.
ture than solution B would. Write one chemical equation to illustrate each
D. Both solutions would boil at the same of these terms.
temperature.
Before choosing an answer to a
question, try to answer the question without looking
at the answer choices on the test.

I O N S I N A Q U E O U S S O L U T I O N S A N D C O L L I G A T I V E P R O P E R T I E S 463

CHAPTER LAB MICRO-

LAB

Testing Water for Ions

OBJECTIVES BACKGROUND
The physical and chemical properties of aqueous
• Observe chemical reactions involving aque- solutions are affected by small amounts of dissolved
ous solutions of ions. ions. For example, if a water sample has enough Mg2+
or Ca2+ ions, it does not create lather when soap is
• Relate observations of chemical properties added. This is common in places where there are
to the presence of ions. many minerals in the water (hard water). Other ions,
such as Pb2+ and Co2+, can accumulate in body tis-
• Infer whether an ion is present in a water sues; therefore, solutions of these ions are poisonous.
sample.
Because some sources of water may contain
• Apply concepts concerning aqueous solu- harmful or unwanted substances, it is important to
tions of ions. find out what ions are present. In this experiment,
you will test various water samples for the presence
MATERIALS of four common, nontoxic ions: Fe3+, Ca2+, Cl–, and
SO42−. Some of the samples may contain these ions
• 24-well microplate lid in very small concentrations, so make very careful
observations.
• fine-tipped dropper bulbs, labeled, with
solutions, 10 SAFETY

• overhead projector (optional) For review of safety, please see Safety in the
Chemistry Laboratory in the front of your book.
• paper towels
PREPARATION
• solution 1: reference (all ions) 1. Create a data table in your lab notebook follow-
ing your teacher’s instructions. Record all of
• solution 2: distilled water (no ions) your observations in your data table.
2. Place the 24-well microplate lid in front of you
• solution 3: tap water (may have ions) on a white background. Label the columns and
rows as instructed by your teacher. The coordi-
• solution 4: bottled spring water (may have nates will designate the individual circles. For
ions) example, the circle in the top right corner would
be 1-D.
• solution 5: local river or lake water (may
have ions) PROCEDURE
1. Obtain labeled dropper bulbs containing the six
• solution 6: solution X, prepared by your different solutions from your teacher.
teacher (may have ions)

• solution A: NaSCN solution (test for Fe3+)
• solution B: Na2C2O4 solution (test for Ca2+)
• solution C: AgNO3 solution (test for Cl–)
• solution D: Sr(NO3)2 solution (test for

SO24–)
• white paper

464 C H A P T E R 1 3

2. Place a drop of the solution from bulb 1 into 8. Follow the procedure used for bulb A with bulbs
circles 1-A, 1-B, 1-C, and 1-D (the top row). B, C, and D to test for the other ions. Record
Solution 1 contains all four of the dissolved ions, your observations about the test results. Specify
so these drops will show what a positive test for whether the solutions contained Ca2+, Cl−, or SO24−
each ion looks like. Be careful to keep the solu- and whether the ions seemed to be present at a
tions in the appropriate circles. Any spills will high, moderate, or low concentration. A black
cause poor results. background may be useful for these three tests.

3. Place a drop of the solution from bulb 2 into 9. If some of the results are difficult to discern,
each of the circles in row 2. This solution is dis- place your microplate on an overhead projector.
tilled water and should not contain any of the Examine the drops for signs of cloudiness.
ions. It will show what a negative test looks like. Looking at the drops from the side, keep your
line of vision 10° to 15° above the plane of the
4. Place a drop from bulb 3 into each of the circles lid. Compare each drop tested with the control
in row 3 and a drop from bulb 4 into each of the drops in row 2. If any sign of cloudiness is
circles in row 4. Follow the same procedure for detected in a test sample, it is due to the Tyndall
bulb 5 (into row 5) and bulb 6 (into row 6). effect and is a positive test result. Record your
These solutions may or may not contain ions. results.
The materials list gives contents of each bulb.
CLEANUP AND DISPOSAL
5. Now that each circle contains a solution to be 10. Clean all equipment and your lab sta-
analyzed, use the solutions in bulbs A–D to test
for the presence of the ions. Bulb A contains tion. Return equipment to its proper
NaSCN, sodium thiocyanate, which reacts with place. Dispose of chemicals and solu-
any Fe3+ to form the complex ion Fe(SCN)2+, tions in the containers designated by your
which results in a deep red solution. Bulb B con- teacher. Do not pour any chemicals down the
tains Na2C2O4, sodium oxalate, which reacts with drain or in the trash unless your teacher directs
Ca2+ ions. Bulb C contains AgNO3, silver nitrate, you to do so. Wash your hands thoroughly before
which reacts with Cl− ions. Bulb D contains you leave the lab and after all work is finished.
Sr(NO3)2, strontium nitrate, which reacts with
SO24− ions. The contents of bulbs B–D react with ANALYSIS AND INTERPRETATION
the specified ion to yield insoluble precipitates. 1. Organizing Ideas: Describe what each positive
test looked like. Write the balanced chemical
6. Holding the tip of bulb A 1 to 2 cm above the equations and net ionic equations for each of
drop of water to be tested, add one drop of solu- the positive tests.
tion A to the drop of reference solution in circle
1-A and one drop to the distilled water in circle CONCLUSIONS
2-A. Circle 1-A should show a positive test, and 1. Organizing Conclusions: List the solutions that
circle 2-A should show a negative test. In your you tested and the ions that you found in each
data table, record your observations about what solution. Include notes on whether the concen-
the positive and negative tests look like. tration of each ion was high, moderate, or low
based on your observations.
7. Use the NaSCN solution in bulb A to test the
rest of the water drops in column A to deter- 2. Predicting Outcomes: Using your test results,
mine whether they contain the Fe3+ ion. Record predict which water sample would be the
your observations in your data table. For each of “hardest.” Explain your reasoning.
the tests in which the ion was present, specify
whether it seemed to be at a high, moderate, or
low concentration.

I O N S I N A Q U E O U S S O L U T I O N S A N D C O L L I G A T I V E P R O P E R T I E S 465

CHAPTER 14

Acids and Bases

Acids and bases change the color
of compounds called indicators.

Hydrangeas

Properties of Acids SECTION 1
and Bases
OBJECTIVES
H ow many foods can you think of that are sour? Chances are that
List five general properties
almost all the foods you thought of, like those in Figure 1a, owe their of aqueous acids and bases.
sour taste to an acid. Sour milk contains lactic acid. Vinegar, which can
be produced by fermenting juices, contains acetic acid. Phosphoric acid Name common binary acids
gives a tart flavor to many carbonated beverages. Most fruits contain and oxyacids, given their
some kind of acid. Lemons, oranges, grapefruits, and other citrus fruits chemical formulas.
contain citric acid. Apples contain malic acid, and grape juice contains
tartaric acid. List five acids commonly used
in industry and the laboratory,
Many substances known as bases are commonly found in household and give two properties of
products, such as those in Figure 1b. Household ammonia is an ammo- each.
nia-water solution that is useful for all types of general cleaning. Sodium
hydroxide, NaOH, known by the common name lye, is present in some Define acid and base accord-
commercial cleaners. Milk of magnesia is a suspension in water of mag- ing to Arrhenius’s theory of
nesium hydroxide, Mg(OH)2, which is not very water-soluble. It is used ionization.
as an antacid to relieve discomfort caused by excess hydrochloric acid
in the stomach. Aluminum hydroxide, Al(OH)3, and sodium hydrogen Explain the differences
carbonate, NaHCO3, are also bases commonly found in antacids. between strong and weak
acids and bases.

Benzoic acid, HC7H5O2 Citric acid, NH3(aq) NaOH
Sorbic acid, HC6H7O2 H3C6H5O7
Phosphoric acid, H3PO4 NaHCO3
Carbonic acid, H2CO3 Ascorbic acid,
H2C6H6O6

(a) (b) Al(OH)3

FIGURE 1 (a) Fruits and fruit juices contain acids such as citric acid A C I D S A N D B A S E S 467
and ascorbic acid. Carbonated beverages contain benzoic acid, phosphoric acid,
and carbonic acid. (b) Many household cleaners contain bases such as ammonia
and sodium hydroxide. Antacids contain bases such as aluminum hydroxide.

FIGURE 2 A strip of pH paper Acids
dipped into vinegar turns red, show-
ing that vinegar is an acid. Acids were first recognized as a distinct class of compounds because of
the common properties of their aqueous solutions. These properties are
listed below.
1. Aqueous solutions of acids have a sour taste. Taste, however, should

NEVER be used as a test to evaluate any chemical substance. Many
acids, especially in concentrated solutions, are corrosive; that is, they
destroy body tissue and clothing. Many are also poisons.
2. Acids change the color of acid-base indicators. When pH paper is
used as an indicator, the paper turns certain colors in acidic solution.
This reaction is demonstrated in Figure 2.
3. Some acids react with active metals and release hydrogen gas, H2.
Recall that metals can be ordered in terms of an activity series.
Metals above hydrogen in the series undergo single-displacement
reactions with certain acids. Hydrogen gas is formed as a product, as
shown by the reaction of barium with sulfuric acid.

Ba(s) + H2SO4(aq) ⎯→ BaSO4(s) + H2(g)

4. Acids react with bases to produce salts and water. When chemically
equivalent amounts of acids and bases react, the three properties just
described disappear because the acid is “neutralized.” The reaction
products are water and an ionic compound called a salt.

5. Acids conduct electric current. Some acids completely separate into ions
in water and are strong electrolytes. Other acids are weak electrolytes.

Acid Nomenclature

A binary acid is an acid that contains only two different elements: hydro-
gen and one of the more electronegative elements. Many common inor-
ganic acids are binary acids. The hydrogen halides—HF, HCl, HBr, and
HI—are all binary acids.

The procedure used to name binary acids is illustrated by the exam-
ples given in Table 1. In pure form, each compound listed in the table is
a gas. Aqueous solutions of these compounds are known by the acid
names. From the table you can see that naming binary compounds can
be summarized as follows.

TABLE 1 Names of Binary Acids

Formula Acid name Molecule name
hydrogen fluoride
HF hydrofluoric acid hydrogen chloride
hydrogen bromide
HCl hydrochloric acid hydrogen iodide
hydrogen sulfide
HBr hydrobromic acid

HI hydriodic acid

H2S hydrosulfuric acid

468 C H A P T E R 1 4

H O FIGURE 3 Structures of (a) phos-
O HOS OH phoric acid and (b) sulfuric acid
HO P OH
O O

(a) H3PO4 (b) H2SO4

Binary Acid Nomenclature

1. The name of a binary acid begins with the prefix hydro-.
2. The root of the name of the second element follows this prefix.
3. The name then ends with the suffix -ic.

An oxyacid is an acid that is a compound of hydrogen, oxygen, and a
third element, usually a nonmetal. Nitric acid, HNO3, is an oxyacid. The
structures of two other oxyacids are shown in Figure 3. Oxyacids are
one class of ternary acids, which are acids that contain three different
elements. Usually, the elements in an oxyacid formula are written as one
or more hydrogen atoms followed by a polyatomic anion. But as you
can see from the structures, the H atoms are bonded to O atoms. The
names of oxyacids follow a pattern, and the names of their anions are
based on the names of the acids. Some common oxyacids and their
anions are given in Table 2.

TABLE 2 Names of Common Oxyacids and Oxyanions

Formula Acid name Anion
acetic acid
CH3COOH carbonic acid CH3COO−, acetate
H2CO3 hypochlorous acid CO32−, carbonate
HClO chlorous acid ClO−, hypochlorite
chloric acid
HClO2 perchloric acid ClO−2, chlorite
HClO3 iodic acid ClO−3, chlorate
HClO4 nitrous acid ClO4−, perchlorate
HIO3 nitric acid IO−3, iodate
HNO2 phosphorous acid NO2−, nitrite
HNO3 phosphoric acid NO−3, nitrate
H3PO3 sulfurous acid PO33−, phosphite
H3PO4 sulfuric acid PO34−, phosphate
H2SO3 SO32−, sulfite
H2SO4 SO24−, sulfate

A C I D S A N D B A S E S 469

www.scilinks.org Some Common Industrial Acids
Topic: Acids
Code: HC60012 The properties of acids make them important chemicals both in the lab-
oratory and in industry. Sulfuric acid, nitric acid, phosphoric acid,
hydrochloric acid, and acetic acid are all common industrial acids.

Sulfuric Acid

Sulfuric acid is the most commonly produced industrial chemical in the
world. More than 37 million metric tons of it are made each year in the
United States alone. Sulfuric acid is used in large quantities in petrole-
um refining and metallurgy as well as in the manufacture of fertilizer. It
is also essential to a vast number of industrial processes, including the
production of metals, paper, paint, dyes, detergents, and many chemical
raw materials. Sulfuric acid is the acid used in automobile batteries.

Because it attracts water, concentrated sulfuric acid is an effective
dehydrating (water-removing) agent. It can be used to remove water
from gases with which it does not react. Sugar and certain other organic
compounds are also dehydrated by sulfuric acid. Skin contains organic
compounds that are attacked by concentrated sulfuric acid, which can
cause serious burns.

Nitric Acid

Pure nitric acid is a volatile, unstable liquid. Dissolving the acid in water
makes the acid more stable. Solutions of nitric acid are widely used in
industry. Nitric acid also stains proteins yellow. The feather in Figure 4
was stained by nitric acid.The acid has a suffocating odor, stains skin, and
can cause serious burns. It is used in making explosives, many of which
are nitrogen-containing compounds. It is also used to make rubber,

FIGURE 4 Concentrated nitric
acid stains a feather yellow.

470 C H A P T E R 1 4

plastics, dyes, and pharmaceuticals. Initially, nitric acid solutions are col- www.scilinks.org
orless; however, upon standing, they gradually become yellow because Topic: Bases
of slight decomposition to brown nitrogen dioxide gas. Code: HC60137

Phosphoric Acid

Phosphorus, along with nitrogen and potassium, is an essential element
for plants and animals. The bulk of phosphoric acid produced each year
is used directly for manufacturing fertilizers and animal feed. Dilute
phosphoric acid has a pleasant but sour taste and is not toxic. It is used
as a flavoring agent in beverages and as a cleaning agent for dairy
equipment. Phosphoric acid is also important in the manufacture of
detergents and ceramics.

Hydrochloric Acid

The stomach produces HCl to aid in digestion. Industrially, hydrochloric
acid is important for “pickling” iron and steel. Pickling is the immersion
of metals in acid solutions to remove surface impurities. This acid is also
used in industry as a general cleaning agent, in food processing, in the
activation of oil wells, in the recovery of magnesium from sea water, and
in the production of other chemicals.

Concentrated solutions of hydrochloric acid, commonly referred to as
muriatic acid, can be found in hardware stores. It is used to maintain the
correct acidity in swimming pools and to clean masonry.

Acetic Acid

Pure acetic acid is a clear, colorless, and pungent-smelling liquid known
as glacial acetic acid. This name is derived from the fact that pure acetic
acid has a freezing point of 17°C. It can form crystals in a cold room. The
fermentation of certain plants produces vinegars containing acetic acid.
White vinegar contains 4% to 8% acetic acid.

Acetic acid is important industrially in synthesizing chemicals used
in the manufacture of plastics. It is a raw material in the production of
food supplements—for example, lysine, an essential amino acid. Acetic
acid is also used as a fungicide.

Bases FIGURE 5 pH paper turns blue
in the presence of this solution of
How do bases differ from acids? You can answer this question by com- sodium hydroxide.
paring the following properties of bases with those of acids.
1. Aqueous solutions of bases taste bitter. You may have noticed this fact A C I D S A N D B A S E S 471

if you have ever gotten soap, a basic substance, in your mouth. As
with acids, taste should NEVER be used to test a substance to see if
it is a base. Many bases are caustic; they attack the skin and tissues,
causing severe burns.
2. Bases change the color of acid-base indicators. As Figure 5 shows, an
indicator will be a different color in a basic solution than it would be
in an acidic solution.

Wear safety goggles, gloves and an apron.

Household Acids and Bases Materials

Question • dishwashing liquid, dishwash-
er detergent, laundry deter-
Which of the household sub- gent, laundry stain remover,
stances are acids, and which fabric softener, and bleach
are bases?
• mayonnaise, baking powder,
Procedure 4. Add a drop or two of the red baking soda, white vinegar,
cider vinegar, lemon juice,
Record all your results in a data cabbage juice to the solution soft drinks, mineral water,
table. being tested, and note the and milk
color. The solution will turn
1. To make an acid-base indica- red if it is acidic and green if • fresh red cabbage
it is basic.
tor, extract juice from red cab- • hot plate
bage. First, cut up some red Discussion
cabbage and place it in a • beaker, 500 mL or larger
large beaker. Add enough 1. Are the cleaning products
water so that the beaker is • beakers, 50 mL
half full. Then, bring the mix- acids, bases, or neither?
ture to a boil. Let it cool, and • spatula
then pour off and save the 2. What are acid/base character-
cabbage juice. This solution is • tap water
an acid-base indicator. istics of foods and beverages?
• tongs
2. Assemble foods, beverages, 3. Did you find consumer warn-

and cleaning products to be ing labels on basic or acidic
tested. products?

3. If the substance being tested Red cabbage, which contains an
anthocyanin pigment, can be made
is a liquid, pour about 5 mL into an acid-base indicator.
into a small beaker. If it is a
solid, place a small amount
into a beaker, and moisten it
with about 5 mL of water.

www.scilinks.org 3. Dilute aqueous solutions of bases feel slippery. You encounter this
Topic: Household Acids and property of aqueous bases whenever you wash with soap.

Bases 4. Bases react with acids to produce salts and water. The properties of a
Code: HC60762 base disappear with the addition of an equivalent amount of an acid.
It could also be said that “neutralization” of the base occurs when
472 C H A P T E R 1 4 these two substances react to produce a salt and water.

5. Bases conduct electric current. Like acids, bases form ions in aqueous
solutions and are thus electrolytes.

Arrhenius Acids and Bases –

Svante Arrhenius, a Swedish chemist who lived from 1859 to 1927,
understood that aqueous solutions of acids and bases conducted electric
current. Arrhenius therefore theorized that acids and bases must pro-
duce ions in solution. An Arrhenius acid is a chemical compound that
increases the concentration of hydrogen ions, H+, in aqueous solution. In
other words, an acid will ionize in solution, increasing the number of
hydrogen ions present. An Arrhenius base is a substance that increases
the concentration of hydroxide ions, OH−, in aqueous solution. Some
bases are ionic hydroxides. These bases dissociate in solution to release
hydroxide ions into the solution. Other bases are substances that react
with water to remove a hydrogen ion, leaving hydroxide ions in
the solution.

Aqueous Solutions of Acids

The acids described by Arrhenius are molecular compounds with ioniz-
able hydrogen atoms. Their water solutions are known as aqueous acids.
All aqueous acids are electrolytes.

Because acid molecules are sufficiently polar, water molecules
attract one or more of their hydrogen ions. Negatively charged anions
are left behind. As explained in Chapter 13, the hydrogen ion in aque-
ous solution is best represented as H3O+, the hydronium ion. The ion-
ization of an HNO3 molecule is shown by the following equation.
Figure 6 also shows how the hydronium ion forms when nitric acid
reacts with water.

HNO3(l) + H2O(l) ⎯→ H3O+(aq) + NO−3(aq)

Similarly, ionization of a hydrogen chloride molecule in hydrochloric
acid can be represented in the following way.

HCl(g) + H2O(l) ⎯→ H3O+(aq) + Cl−(aq)

+

++

HNO3 H2O H3O+ NO3–
Nitric acid Water Hydronium ion Nitrate ion

FIGURE 6 Arrhenius’s observations form the basis of a definition of A C I D S A N D B A S E S 473
acids. Arrhenius acids, such as the nitric acid shown here, produce hydroni-
um ions in aqueous solution.

TABLE 3 Common Aqueous Acids

Strong acids Weak acids

HI + H2O ⎯→ H3O+ + I − HSO4− + H2O ←⎯⎯→ H3O+ + SO42−
HClO4 + H2O ⎯→ H3O+ + ClO4−
HBr + H2O ⎯→ H3O+ + Br− H3PO4 + H2O ←⎯⎯→ H3O+ + H2PO4−
HCl + H2O ⎯→ H3O+ + Cl −
H2SO4 + H2O ⎯→ H3O+ + HSO4− HF + H2O ←⎯⎯→ H3O+ + F −
HClO3 + H2O ⎯→ H3O+ + ClO3−
CH3COOH + H2O ←⎯⎯→ H3O+ + CH3COO−

H2CO3 + H2O ←⎯⎯→ H3O+ + HCO3−

H2S + H2O ←⎯⎯→ H3O+ + HS−

HCN + H2O ←⎯⎯→ H3O+ + CN−

HCO3− + H2O ←⎯⎯→ H3O+ + CO23−

Strength of Acids

A strong acid is one that ionizes completely in aqueous solution.A strong
acid is a strong electrolyte. Perchloric acid, HClO4, hydrochloric acid,
HCl, and nitric acid, HNO3, are examples of strong acids. In water,
100% of the acid molecules are ionized. The strength of an acid depends
on the polarity of the bond between hydrogen and the element to which
it is bonded and the ease with which that bond can be broken. Acid
strength increases with increasing polarity and decreasing bond energy.

An acid that releases few hydrogen ions in aqueous solution is a weak
acid. The aqueous solution of a weak acid contains hydronium ions,
anions, and dissolved acid molecules. Hydrocyanic acid is an example of
a weak acid. In aqueous solution, both the ionization of HCN and the
reverse reaction occur simultaneously. In a 1 M solution of HCN there
will be only two H+ ions and two CN− ions out of 100,000 molecules.
The other 99,998 molecules remain as HCN.

HCN(aq) + H2O(l) ←⎯⎯→ H3O+(aq) + CN−(aq)

Common aqueous acids are listed in Table 3. Each strong acid ionizes
completely in aqueous solution to give up one hydrogen ion per mol-
ecule. Notice that the number of hydrogen atoms in the formula does not
indicate acid strength. Molecules with multiple hydrogen atoms may not
readily give up each hydrogen. The fact that phosphoric acid has three
hydrogen atoms per molecule does not mean that it is a strong acid.
None of these ionize completely in solution, so phosphoric acid is weak.

Organic acids, which contain the acidic carboxyl group —COOH, are
generally weak acids. For example, acetic acid, CH3COOH, ionizes
slightly in water to give hydronium ions and acetate ions, CH3COO−.

CH3COOH(aq) + H2O(l) ←⎯⎯→ H3O+(aq) + CH3COO−(aq)

A molecule of acetic acid contains four hydrogen atoms. However, only
one of the hydrogen atoms is ionizable. The hydrogen atom in the

474 C H A P T E R 1 4

carboxyl group in acetic acid is the one that is “acidic” and forms the II HO
hydronium ion. This acidic hydrogen can be seen in the structural dia- JHICICIOIH
gram in Figure 7.
H acidic hydrogen
Aqueous Solutions of Bases FIGURE 7 Acetic acid contains

Most bases are ionic compounds containing metal cations and the four hydrogen atoms, but only one
hydroxide anion, OH−. Because these bases are ionic, they dissociate of them is “acidic” and forms the
when dissolved in water. When a base completely dissociates in water to hydronium ion in solution.
yield aqueous OH− ions, the solution is referred to as strongly basic.
Sodium hydroxide, NaOH, is a common laboratory base. It is water-sol- Chemistry in Action
uble and dissociates as shown by the equation below.
Go to go.hrw.com for a full-length
NaOH(s) ⎯H⎯2O→ Na+(aq) + OH−(aq) article on drain cleaners.

You will remember from Chapter 5 that Group 1 elements are the alkali Keyword: HC6ACDX
metals. This group gets its name from the fact that the hydroxides of Li,
Na, K, Rb, and Cs all form alkaline (basic) solutions.

Not all bases are ionic compounds. A base commonly used in house-
hold cleaners is ammonia, NH3, which is molecular. Ammonia is a base
because it produces hydroxide ions when it reacts with water molecules,
as shown in the equation below.

NH3(aq) + H2O(l) ←⎯⎯→ NH4+ (aq) + OH−(aq)

Strength of Bases

As with acids, the strength of a base also depends on the extent to which
the base dissociates, or adds hydroxide ions to the solution. For exam-
ple, potassium hydroxide, KOH, is a strong base because it completely
dissociates into its ions in dilute aqueous solutions.

KOH(s) ⎯H⎯2O→ K+(aq) + OH −(aq)

Strong bases are strong electrolytes, just as strong acids are strong elec-
trolytes. Table 4 lists some strong bases.

TABLE 4 Common Aqueous Bases

Strong bases Weak bases

Ca(OH)2 ⎯→ Ca2+ + 2OH− NH3 + H2O ←⎯⎯→ NH4+ + OH−
Sr(OH)2 ⎯→ Sr2+ + 2OH− C6H5NH2 + H2O ←⎯⎯→ C6H5NH+3 + OH−
Ba(OH)2 ⎯→ Ba2+ + 2OH−
NaOH ⎯→ Na+ + OH−
KOH ⎯→ K+ + OH−
RbOH ⎯→ Rb+ + OH−
CsOH ⎯→ Cs+ + OH−

A C I D S A N D B A S E S 475

Na+(aq) + OH−(aq) Chloride ion,
Cl−
Chloride ion, Cl− Sodium ion,
Na+

Copper(ΙΙ) ion, Water molecule, Cu(OH)2(s)

Cu2+ Cu2+(aq) + 2OH−(aq) ⎯→ Cu(OH)2(s)

H2O Bases that are not very soluble do not produce a large number of
hydroxide ions when added to water. Some metal hydroxides, such as
FIGURE 8 The hydroxides of Cu(OH)2, are not very soluble in water, as seen in Figure 8. They
cannot produce strongly alkaline solutions. The alkalinity of aqueous
most d-block metals are nearly insol- solutions depends on the concentration of OH− ions in solution. It is unre-
lated to the number of hydroxide ions in the undissolved compound.
uble in water, as is shown by the
Now consider ammonia, which is highly soluble but is a weak elec-
gelatinous precipitate, copper(II) trolyte. The concentration of OH− ions in an ammonia solution is rela-
tively low. Ammonia is therefore a weak base. Many organic compounds
hydroxide, Cu(OH)2, in the beaker that contain nitrogen atoms are also weak bases. For example, codeine,
on the right. C18H21NO3, a pain reliever and common cough suppressant found in
prescription cough medicine, is a weak base.

SECTION REVIEW 4. a. Why are strong acids also strong electrolytes?

1. a. What are five general properties of aqueous b. Is every strong electrolyte also a strong acid?
acids?
Critical Thinking
b. Name some common substances that have
one or more of these properties. 5. RELATING IDEAS A classmate states, “All com-
pounds containing H atoms are acids, and all com-
2. Name the following acids: a. HBrO b. HBrO3. pounds containing OH groups are bases.” Do you
3. a. What are five general properties of aqueous agree? Give examples.

bases?

b. Name some common substances that have
one or more of these properties.

476 C H A P T E R 1 4

Acid Water—A Hidden Menace

Many people are unaware of the pH awhile first could be adding copper 2. In the morning, let your water tap
of the tap water in their home until and lead ions to their tea or coffee. run for about half a minute
they are confronted with such phe- before you fill your kettle or drink
nomena as a blue ring materializing Lead poisoning is of particular the water. If the water is acidic,
around a porcelain sink drain, a concern in young children. The the first flush of water will have
water heater suddenly giving out, or absorption rate of lead in the intesti- the highest concentration of lead
tropical fish that keep dying. Each of nal tract of a child is much higher and copper ions.
these events could be due to acidic than in that of an adult, and lead
water. Acidic water can also cause poisoning can permanently impair a 3. Install an alkali-injection pump, a
the amount of lead in the water child’s rapidly growing nervous sys- low-cost, low-maintenance solu-
to rise. tem. The good news is that lead poi- tion that can save your plumbing
soning and other effects of acidic and lessen the risk of lead poi-
The possibility of lead poisoning water in the home can be easily pre- soning from your own water sup-
from home water supplies is a con- vented by following these tips: ply. The pump injects a small
cern. Many older homes still have 1. Monitor the pH of your water on amount of an alkali (usually
lead pipes in their plumbing, while potassium carbonate or
most modern homes use copper pip- a regular basis, especially if you sodium carbonate) into your
ing. Copper pipe joints, however, are have well water. This can easily water-pressure tank each time
often sealed with lead-containing be done with pH test kits (see your well’s pump starts. This
solder. Highly acidic water can leach photograph) that are sold in effectively neutralizes the acidity
out both the lead from the solder hardware or pet stores—many of your water.
joints and copper from the pipes tropical fish are intolerant of
themselves, which turns the sink water with a pH that is either too Questions
drain blue. In addition, people who high (basic) or too low (acidic).
are in the habit of filling their tea The pH of municipal water sup- 1. What is the source of lead conta-
kettles and coffee pots in the morn- plies is already regulated, but mination in water in the home?
ing without letting the tap run regularly checking your water’s
pH yourself is a good idea. 2. Does the use of copper water
pipes ensure that your household
water is free from lead?

3. Why does lead poisoning affect
children more severely than it
affects adults?

The pH of your home’s www.scilinks.org
water supply can be easily Topic: Acid Water
monitored using a test Code: HC60011
kit, such as the one
shown here.

A C I D S A N D B A S E S 477

SECTION 2 Acid-Base Theories

OBJECTIVES F or most uses, scientists found the Arrhenius definition of acids and

Define and recognize bases to be adequate. However, as scientists further investigated acid-
Brønsted-Lowry acids base behavior, they found that some substances acted as acids or bases
and bases. when they were not in a water solution. Because the Arrhenius defini-
Define a Lewis acid and a tion requires that the substances be aqueous, the definitions of acids
Lewis base. and bases had to be revised.
Name compounds that are
acids under the Lewis defini- Brønsted-Lowry Acids and Bases
tion but are not acids under
the Brønsted-Lowry definition. In 1923, the Danish chemist J. N. Brønsted and the English chemist
T. M. Lowry independently expanded the Arrhenius acid definition. A
478 C H A P T E R 1 4 Brønsted-Lowry acid is a molecule or ion that is a proton donor. Because
H+ is a proton, all acids as defined by Arrhenius donate protons to water
and are Brønsted-Lowry acids as well. Substances other than molecules,
such as certain ions, can also donate protons. Such substances are not
Arrhenius acids but are included in the category of Brønsted-Lowry acids.

Hydrogen chloride acts as a Brønsted-Lowry acid when it reacts with
ammonia. The HCl transfers protons to NH3 much as it does in water.

HCl + NH3 ⎯→ NH+4 + Cl −

A proton is transferred from the hydrogen chloride molecule, HCl, to
the ammonia molecule, NH3. The ammonium ion, NH4+, is formed.
Electron-dot formulas show the similarity of this reaction to the reac-

tion of HCl with water.

H Cl + HO ⎯→ + Cl −

HOH +

HH

H+
H Cl + H N H ⎯→ H N H + Cl −

HH

In both reactions, hydrogen chloride is a Brønsted-Lowry acid.
Water can also act as a Brønsted-Lowry acid. Consider, for example,

the following reaction, in which the water molecule donates a proton to
the ammonia molecule.

H2O(l) + NH3(aq) ←⎯⎯→ NH4+(aq) + OH−(aq)

H+ −
H O + H N H →⎯⎯→ H N H +
O

HH HH