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2 Brønsted–Lowry Acids and Bases

2 Brønsted–Lowry Acids and Bases

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End of Chapter Problems   235

*6.104The serum concentration of glucose is expected to be



in the range 80–100 mg/dL. What is the part per million concentration of glucose in serum that contains

95 mg of glucose per deciliter?

6.105The normal serum concentration of potassium ion

(K+) is 3.5–4.9 mEq/L. Convert this concentration

range into mmol/L.

6.106The normal serum concentration of phosphate ion

(PO4 3-) is 2.8–4.5 mEq/L. Convert this concentration range into mmol/L.

6.107For adults, a 1.5 ng/dL serum concentration of the

hormone thyroxine (776.87 g/mol) is in the normal

range. Convert this concentration into



a.picograms per milliliter



b.nanograms per liter



c.picomoles per liter

6.108For men below 45 years of age, a serum concentration of prostate specific antigen (PSA) of 2.5 mg/L is

normal. Convert this concentration into



a. nanograms per milliliter



b. milligrams per deciliter

6.109The normal serum concentration of chloride ion

(Cl-) is 95–107 mmol/L. Convert this concentration

range into mEq/L.

6.110The normal serum concentration of magnesium ion

(Mg2+) is 1.2–2.5 mmol/L. Convert this concentration range into mEq/L.

6.111How many milliequivalents of bicarbonate (HCO3 -)

are present in a 75.0 mL blood serum sample with a

concentration of 25 mEq/L HCO3 -?

6.112How many equivalents of calcium ion (Ca2+) are

present in a 25.0 mL blood serum sample with a Ca2+

concentration of 5.2 mEq/L?

6.113How many moles of sodium ion (Na+) are present in

a 10.0 mL blood serum sample with a Na+ concentration of 132 mEq/L?

6.114How many grams of vitamin B12 are present in a

100.0 mL blood serum sample with a vitamin B12

concentration of 168 ng/L?



6.9Dilution

6.115If 15.0 mL of 3.0 M HCl are diluted to a final volume

of 100.0 mL, what is the new concentration?

6.116If 3.0 mL of 0.15 M HCl are diluted to a final volume

of 250.0 mL, what is the new concentration?

6.117A 10.0% (w/v) solution of ethyl alcohol is diluted

from 50.0 mL to 200.0 mL. What is the new weight/

volume percent?

6.118A 5.0% (w/v) solution of ethyl alcohol is diluted from

50.0 mL to 75.0 mL. What is the new weight/volume

percent?

6.119How many milliliters of 2.00 M NaOH are needed to

prepare 300.0 mL of 1.50 M NaOH?



6.120How many milliliters of 1.50 M NaOH are needed to

prepare 225.0 mL of 0.150 M NaOH?

6.121Calculate the final volume required to prepare each

solution.



a.Starting with 100.0 mL of 1.00 M KBr, prepare

0.500 M KBr.



b.Starting with 50.0 mL of 0.250 M alanine (an

amino acid), prepare 0.110 M alanine.

6.122Calculate the final volume required to prepare each

solution.



a.Starting with 10 mL of 2.0% (w/v) KI, prepare

0.020% (w/v) KI.



b.Starting with 1.0 mL of 60 ppm Cu2+, prepare

18 ppm Cu2+.



6.10Colloids and Suspensions

6.123How is a colloid different from a suspension?

6.124How is a colloid different from a solution?

6.125What do you end up with if you pour dirt into water

and stir—a solution, a colloid, or a suspension?

6.126Give an example of each.



a. a solution

c.  a colloid



b. a suspension



6.11 Diffusion and Osmosis

6.127 ­a. Define the term diffusion.



b.A process called active transport moves certain ions

and compounds across cell membranes from areas of

lower concentration to areas of higher concentration.

Does active transport involve diffusion? Explain.

6.128Define the terms osmosis and osmotic pressure.

6.129To make pickles, you soak cucumbers in a concentrated salt solution called brine. Describe how this process

is related to osmosis.

6.130Soaking in a bath of dissolved Epsom salts is a home

remedy that can help to reduce the swelling and

discomfort of hemorrhoids. Why is this hypertonic

solution of MgSO4 able to reduce the swelling of hemorrhoidal tissue, while pure water is not?



HealthLink | Blood Pressure

*6.131During severe bleeding, ADH (a hormone released by



the hypothalamus) causes vasoconstriction (shrinking of the blood vessels) to take place. What effect

does a decrease in blood vessel volume have on blood

pressure?

6.132Ethyl alcohol, the alcohol present in alcoholic beverages, inhibits the release of ADH (see the previous

problem). What effect does drinking alcohol have on

blood pressure?



236   Chapter 6  Gases, Solutions, Colloids, and Suspensions



Learning Group Problems



HealthLink | Breathing

6.133The chest compressions given during cardiopulmonary resuscitation (CPR) cause the injured person to

exhale. Explain why, in terms of Boyle’s law.

6.134Sometimes an accident victim will be kept alive by

use of a bag valve mask. This mask moves air into the

lungs when its bag is squeezed. In terms of the ideal

gas law, explain how the bag forces air into an accident

victim’s lungs.



HealthLink | Prodrugs

6.135Why is the prodrug chloramphenicol palmitate



(Figure 6.25) less soluble in water than

chloramphenicol?

*6.136Phenacetin is a prodrug that can be prepared from

acetaminophen, a commonly used pain and fever

reducer. While available in many parts of the world,

phenacetin is not sold in the United States due to concerns about its toxicity. What is the theoretical yield

of phenacetin (in grams) upon reaction of 325 mg of

acetaminophen? Assume that acetaminophen is the

limiting reactant.

O



NH¬CCH3

+ K2CO3 + CH3CH2I

HO

Acetaminophen



O



NH¬CCH3

+ KHCO3 + KI

CH3CH2O

Phenacetin



HealthLink | Saliva

6.137Saliva has some characteristics of a solution. Explain.



6.138Saliva has some characteristics of a colloid. Explain.



HealthLink | Diffusion and the Kidneys

6.139Why is it important that dialyzing solution be isotonic

with blood?

6.140Kidney disease can lead to nephrotic syndrome,

which is characterized, in part, by a low blood serum

concentration of albumin (a protein). Explain why

low serum albumin levels result in edema, swelling

caused by the movement of fluid from the blood

into tissue.



6.141a. A 1.50 L gas cylinder contains a mixture of nitrogen



gas (N2) and hydrogen gas (H2) at a temperature of

35.0oC and a pressure of 895 torr. What is the pressure in atmospheres (atm)?



b.What is the pressure in pounds per square inch (psi)?



c.If the partial pressure of N2 is 615 torr, what is the

partial pressure of H2?



d.If the temperature is increased to 70.0oC, what is

the new partial pressure of each gas and what is the

new total pressure?



e.How many moles of N2 and moles of H2 are

present in the cylinder?



f.How many grams of N2 and grams of H2 are

present in the cylinder?



g.How many molecules of N2 and molecules of H2

are present in the cylinder?



h.Nitrogen gas and hydrogen gas react according to

the equation:



N2(g) + 3H2(g) ¡ 2NH3(g)

Assuming that conditions in the cylinder allow the

reaction to take place, which is the limiting reactant,

N2 or H2?



i.For the reaction in part h, what is the theoretical

yield (in grams) of NH3?

6.142 a . Write the balanced equation for the reaction of

lead(II) nitrate with sodium iodide to form sodium

nitrate and lead(II) iodide.



b.Using Table 6.3, label the reactants and products of

this reaction as being aqueous or solid.



c.Write the ionic equation for the reaction in part a.



d.Write the net ionic equation for this reaction.



e.An aqueous solution is prepared that contains

5.72 g of lead(II) nitrate in 125 mL of solution.

What is the weight/volume percent of the solution?



f. What is the molarity of the solution in part e?



g.If 10.00 mL of the solution in part e is diluted to a

final volume of 75.00 mL, what is the weight/volume percent of the new solution?



h.An aqueous solution is prepared that contains 9.88 g

of sodium iodide in 175 mL of solution. What is the

concentration of the solution in parts per million?



i. What is the molarity of the solution in part h?



j.If 5.00 mL of the solution in part h is diluted to a

give a final volume of 25.0 mL, what is the molarity

of the new solution?



k.Once the reaction in part a has taken place, what is

obtained: a homogeneous mixture or a heterogeneous mixture?

SOLUTIONS TO PRACTICE PROBLEMS



6.1 The mass of the atmosphere produces atmospheric

pressure. There is less atmosphere above a mountain

top than at a lower elevation.



Solutions to Practice Problems   237

6.2



a. 1.81 * 103 torr;   b. 0.99 atm







6.3



17.54 torr







6.4



0.8 atm



6.5 It will be 308 cm3 smaller.

6.6



a. 5.6 L;   b. 690 torr



6.7



42 torr



Pb2+(aq) + 2Cl-(aq) ¡ PbCl2(s)

Net Ionic equation



6.11 As the diver breathes, any excess dissolved gases in the

blood can be exhaled.

6.12

O



6.8 All three molecules can form hydrogen bonds and CH CH CH CH CH CH CH CH CH CH CH CH O S O– Na±

3

2

2

2

2

2

2

2

2

2

2

2

contain polar covalent bonds (can interact through

O

dipole–dipole forces). A significant part of the

­structure of acetaminophen and butyl alcohol consists

hydrophobic

hydrophilic

of nonpolar carbon–carbon and carbon–hydrogen

-5

6.13 a. 0.72 ppm; b. 7.2 * 10 % (w/v)

bonds. London forces are an important noncovalent

6.14 2.81 M

interaction between these two molecules.

6.15 0.0011 g

6.9 a. BaSO4 (insoluble); b. LiNO3 (soluble); c. Na2CO3

(soluble)

6.16 3 pg/mL; 3 3 10-6 mg/mL

6.10 Pb(NO3)2(aq) + 2NH4Cl(aq) ¡

6.17 0.13 M

PbCl2(s) + 2NH4NO3(aq)

6.18 a. solution (clear, does not settle); b. colloid (cloudy,

does not settle); c. colloid (cloudy, does not settle) or



Balanced equation

a suspension (if pulp settles to the bottom or floats to

2+

+

Pb (aq) + 2NO3 (aq) + 2NH4 (aq) + 2Cl (aq) ¡

the top); d. solution (clear—hopefully!)



PbCl2(s) + 2NH4+(aq) + 2NO3 -(aq)

6.19 Adding ethyl alcohol to blood serum will cause the



Ionic equation

total solute concentration to rise.



Media Bakery



At the library

One evening you run into a classmate who has put off studying for the upcoming

chemistry exam. Between the stress of trying to learn the material at the last

minute and drinking way too much coffee, she complains of an upset stomach.

She reaches into her book bag, pulls out a roll of antacids, and takes a few.

Seeing this reminds you of an advertisement that talked about how antacids consume excess stomach acid.



7



A



c ids, Bases,

and Equilibrium



About this Chapter

Acids and bases are related classes of compounds that are important to much of the

­chemistry that goes on around us every day. Vinegar is an acid, as is the citric acid found

in oranges, lemons, grapefruit, and other citrus fruits. Bases commonly found around the

home include baking soda, baking powder, ammonia, antacids, and drain cleaners. Your

body also produces and uses acids and bases, and maintaining a proper ­balance of them

is essential for maintaining good health.



Chapter 7 objectives

After completing this chapter, you should be able to:

1 List the common characteristics of acids and bases.

2 Describe Brønsted–Lowry acids and bases and explain how they differ from their c­ onjugates.

3 Write the equilibrium constant expression for a reversible reaction and use Le Châtelier’s principle to

explain how an equilibrium responds to being disturbed.

4 Use H3O+ concentration and pH to identify a solution as being acidic, basic, or neutral.

5 Use Ka and pKa values to assess the relative strength of acids and state the relationship between the

strength of an acid and the strength of its conjugate base.

6 Describe the processes of neutralization and titration.

7 Explain how the pH of a solution can affect the relative concentrations of an acid and its 

conjugate base and describe buffers.

8 Describe the role of buffers, respiration, and the kidneys in maintaining a stable blood serum pH.



239



240   Chapter 7  Acids, Bases, and Equilibrium



7.1



A c i d s a n d Ba s e s

Some compounds are acids and others are bases. Determining whether a particular

­compound belongs in one category or the other can sometimes be as simple as making

a direct observation. Acids have a sour taste (such as that from the citric acid in citrus

fruits), will dissolve some metals, and will turn the plant pigment called litmus to a pink

color (Figure 7.1). Bases, on the other hand, have a bitter taste (such as that of caffeine or

antihistamines), feel slippery or soapy, and cause litmus to turn blue.

In an earlier era, tasting was a routine way of identifying chemicals. This technique is

no longer practiced because the potentially harmful side effects of doing so are well understood. This is especially true of acids and bases—many will cause extensive tissue damage

and can be fatal, even in small doses.

Table 7.1 lists a number of common acids and bases and their uses. Several of the acids

in this table are named differently than might be expected, based on the rules for naming

binary compounds that were presented in Section 3.5. For example, Table 7.1 identifies HCl as hydrochloric acid, not hydrogen chloride. Although both are correct official

names, when acidity is being emphasized, the acid name is used instead of the binary

compound name.

Table | 7.1   Common Acids and Bases

Name



Formula



Uses



Acids

Hydroiodic acid



HI



Disinfectant



Hydrobromic acid



HBr



Veterinary sedative



Hydrochloric acid

HCl

(muriatic acid)



Household cleaning products, swimming

pool maintenance, and metal cleaning



Sulfuric acid



H2SO4



Fertilizers, explosives, dyes, and glues



Nitric acid



HNO3



Fertilizers, explosives, and dyes



Sodium hydroxide



NaOH



Drain cleaners, soap manufacture



Potassium hydroxide



KOH



Paint and varnish removers



Ammonia



NH3



Fertilizers, cleaning



7.1



Some properties of acids

and bases (a) Some metals



react with acids. Here, zinc reacts

with hydrochloric acid and forms

hydrogen (H2) gas. (b) In the

presence of an acid (lemon juice)

litmus, a pH indicator, is pink.

In the presence of a base (baking

soda), it is blue.



Charles D. Winters/Photo Researchers, Inc.



■■  Figure



Martyn F. Chillmaid/Photo Researchers, Inc.



Bases



(a)



(b)



7.2  Brønsted–Lowry Acids and Bases   241



7.2



B r ø n s t e d – l o w r y A c i d s a n d Ba s e s



One of the early challenges that chemists faced was developing a theory that would connect the structure of a compound to its acid or base characteristics. In the 1880s the

Swedish chemist Svante Arrhenius identified acids as compounds that produce H + in

water and bases as compounds that produce OH - . By this definition HCl is an Arrhenius

acid and NaOH is an Arrhenius base.

HO



2

HCl ¡

H+ + Cl-



HO



2

NaOH ¡

Na+ + OH-



In the 1920s, Johannes Brønsted, a Danish chemist, and Thomas Lowry, an English chemist, proposed a more general definition, one which described acids and bases in terms of the

transfer of H + . In the Brønsted–Lowry definition, acids release H + and bases accept H + .

Consider the equation for the reaction that takes place between HCN and water:



n



A Brønsted–Lowry acid releases

H + and a Brønsted–Lowry base

accepts H + .



n



When released in water, H +

rapidly associates with H2O to

form a hydronium ion (H3O + ).

H3O + and H + are often used

interchangeably.



HCN + H2O N CN- + H3O+





Acid



Base



Base



Acid



The double arrows indicate that the reaction is reversible—goes from left to right and

from right to left. In the forward direction of this reaction HCN is the acid (it releases H +

to become CN - ) and H2O is the base (it accepts H + to become H3O + ). An acid and a

base can also be identified for the reverse of this reaction: H3O + is the acid (it releases H +

to become H2O) and CN - is the base (it accepts H + to become HCN).

The household ammonia used for cleaning purposes is an aqueous solution of NH3.

In this solution NH3 acts as a base (it accepts H + to become NH4 +) and H2O serves as

the acid (it releases H + to become OH - ). For the reverse reaction, NH4 + is the acid and

OH - is the base.

NH3 + H2O N NH4 + + OH



Base



Acid



Acid



Base



Sample Problem  7.1



Identifying acids and bases

For the forward and reverse directions of the reaction, identify the Brønsted–Lowry acids

and bases.

HCN + CO3 2- N HCO3 - + CNStrategy



To solve this problem you must look across the reaction arrow to see how a particular reactant changes. CO3 2-, for example, gains H + to become HCO3 -.

Solution



HCN + CO3 2- N HCO3 - + CN



Acid



Base



+



Acid



Base



In the forward reaction, H moves from HCN to CO3 . In the reverse reaction, H +

moves from HCO3 - to CN - .

2-



Practice Problem  7.1



For the forward and reverse directions of each reaction, identify the Brønsted–Lowry acids

and bases.

a.HSO4 - + HPO4 2- N H2SO4 + PO4 3b.HPO4 2- + OH- N H2O + PO4 3-



242   Chapter 7  Acids, Bases, and Equilibrium



n



An acid and its conjugate base

differ by the presence of H+.



Compounds, such as HCN and CN - or NH3 and NH4 +, which differ only in the

­presence or absence of H + , are called conjugates. For the reaction of acetic acid (CH3CO2H)

in water, CH3CO2H and CH3CO2 - are conjugates, as are H3O + and H2O.

Conjugates





CH3CO2H + H2O ∆ CH3CO2 + H3O+

Acid



Base



Base



Acid



Conjugates

Note that an acid (CH3CO2H, for example) and its conjugate base (CH3CO2 -) are

always on opposite sides of the equilibrium reaction arrows.

Water plays two different roles in the reactions described above. With HCN and

CH3CO2H, water acts as a base, while in the reaction with NH3, it is an acid. Compounds

that can act as acids or as bases are called amphoteric. Other amphoteric compounds

include hydrogen carbonate (HCO3 -), dihydrogen phosphate (H2PO4 -), and the amino

acids used to make proteins.



SAMPLE PROBLEM  7.2



Amphoteric compounds

Hydrogen carbonate (HCO3 -) is amphoteric.

a.Complete the chemical equation for HCO3 - reacting as an acid with OH-. Label the

acids and bases on each side of the reaction equation and identify the conjugate pairs.

OH- + HCO3 - N

b.Complete the chemical equation for HCO3 - reacting as a base with HNO3. Label the

acids and bases on each side of the reaction equation and identify the conjugate pairs.

HNO3 + HCO3 - N

Strategy



Based on what makes something an acid or a base, decide which reactant donates H + and

which accepts it. Conjugates differ by the presence or absence of H + .

a.OH- + HCO3 - N H2O + CO3 2 Base



Acid



Acid



Base



-



OH and H2O are conjugates. HCO3 - and CO3 2- are conjugates.

b.HNO3 + HCO3 - N NO3 - + H2CO3

Acid



Base



Base



Acid



HNO3 and NO3 - are conjugates. HCO3 - and H2CO3 are conjugates.

PRACTICE PROBLEM  7.2



Dihydrogen phosphate (H2PO4 -) is amphoteric.

a.Write the chemical equation for hydrogen phosphate reacting as a base with HSO4 -.

Label the acids and bases on each side of the reaction equation and identify the conjugate pairs.

b.Write the chemical equation for hydrogen phosphate reacting as an acid with CN-. Label

the acids and bases on each side of the reaction equation and identify the conjugate pairs.



7.3  Equilibrium   243



7.3



Equilibrium



The acid–base reactions introduced above are reversible, which means that reactants can be

converted into products and products can be converted back into reactants. Let us use the

decomposition of dinitrogen tetroxide (N2O4) to form nitrogen dioxide (NO2), a reversible

reaction that does not involve acids and bases, to introduce the concept of equilibrium.

N2O4( g ) N 2NO2( g )

Use of the terms reactant and product when describing this reversible ­reaction is

somewhat unclear, since N2O4 is the reactant of the forward reaction and the product of the reverse reaction, while NO2 is the product of the forward reaction and the

reactant of the reverse ­reaction. When dealing with reversible reactions, we will define

reactants as the compounds that appear to the left of the reaction equation arrows and

products as the compounds that appear to the right.

If N2O4 is placed in a sealed tube and warmed, a decomposition reaction begins to take

place. Early in the reaction, the concentration of N2O4 is at its highest and the concentration

of NO2 is at its lowest (Figure 7.2). Since, in this case, reactant concentration has a direct effect

on reaction rate (Section 5.7), when the reaction is started the rate of the forward reaction is

at its greatest and the rate of the reverse reaction is at its lowest. As more and more N2O4 is

converted into NO2, the rate of the forward reaction slows (there is less N2O4 present) and the

rate of the reverse reaction increases (there is more NO2 present). At the point where the rate of

the forward reaction and the rate of the reverse reaction are equal, equilibrium has been reached.

At equilibrium reactants and products are produced as rapidly as they are consumed and there

is no change in their concentrations.

Equilibrium can be reached from any set of starting conditions. For the reaction

described above, this includes only N2O4 being present at the beginning of the reaction,

only NO2 being present, or any combination of N2O4 and NO2 being present. Being at

equilibrium does not mean that the concentrations of reactants and products are equal—

each reversible reaction has its own characteristic set of equilibrium concentrations of

reactants and products.



n



At equilibrium the rate of the

forward and reverse reactions

is the same and concentrations

of reactants and products do

not change.



n



Equilibrium constant values vary

with temperature. Examples

used in this chapter are for a

temperature of 25°C.



Equilibrium Constants

If N2O4 and NO2 are allowed to reach equilibrium and the equilibrium concentration of

each is measured, then the following is true:

[NO2]2

= 4.6 * 10-3

Keq =

[N2O4]

where the square brackets stand for concentration in units of molarity. Regardless of the

starting concentrations of N2O4 and NO2, the equilibrium NO2 concentration squared,

divided by the equilibrium concentration of N2O4, will always equal 4.6 * 10 - 3. This

value is the equilibrium constant (Keq) for the reaction.



■■  Figure



7.2



(a)



©1990 Richard Megna/Fundamental

Photographs, NYC



©1990 Richard Megna/Fundamental

Photographs, NYC



©1990 Richard Megna/Fundamental

Photographs, NYC



Reaching equilibrium 



(b)



(c)



(a) N2O4 is a colorless gas. (b) As

N2O4 is warmed it is converted

into brown NO2 gas, as represented by the chemical equation

N2O4( g ) N 2NO2( g ). (c) At

equilibrium, the color in the tube

stops changing because the forward and reverse reactions occur

at the same rate and the concentrations of N2O4(g) and NO2(g)

no longer vary.



244   Chapter 7  Acids, Bases, and Equilibrium



While we will not be doing equilibrium calculations, it will be useful to see how

these expressions are derived. For the generalized reversible reaction

aA + bB N cC + dD

where A and B are reactants, C and D are products, and a, b, c, and d are coefficients, the

equilibrium constant expression takes the form:

Keq =



[C]c[D]d



[A]a[B]b

In this equation, the numerator is obtained by multiplying the concentrations of products together, with each raised to a power equal to its coefficient (if the coefficient is 2,

the concentration is squared; if it is 3, the concentration is cubed). The denominator is

obtained in the same way, using reactant concentrations.

Consider, for example, the reaction that takes place between nitrogen gas (N2) and

hydrogen gas (H2) to produce ammonia (NH3).

n



Some of the reactions introduced

in previous chapters were

reversible. To simplify those

earlier discussions, reversibility

was largely ignored.



N2( g ) + 3H2( g ) N 2NH3( g )

The equilibrium constant for this reversible reaction has a value of 6.9 * 105 and the

equilibrium constant expression is written:

Keq =



[NH3]2



[N2][H2]3



= 6.9 * 105



Sample Problem  7.3



Writing equilibrium constant expressions

Balance the reaction equation and then write the corresponding equilibrium constant

expression.

CO( g ) + O2( g ) N CO2( g )

Strategy



Reaction equations are balanced by changing the coefficients on reactants and/or products.

Using the balanced reaction equation, the equilibrium constant expression is written with

product concentrations in the numerator and reactant concentrations in the denominator,

each raised to a power equal to its coefficient.

Solution



2CO( g ) + O2( g ) N 2CO2( g )



Andy Washnik/Wiley Archive



Keq =



[CO2]2



[CO]2[O2]



Practice Problem  7.3



Write the equilibrium constant expression for each reaction.

a.H2( g ) + CO2( g ) N CO( g ) + H2O( g )

b.2N2O5( g ) N 4NO2( g ) + O2( g )

■■  Figure



7.3



An equilibrium involving a

solid  Solid PbI2, which has a



yellow color, is in equilibrium

with colorless Pb2 + (aq) and

I-(aq).



In some reversible reactions, reactants and products are present in different phases.

For example, when lead(II) iodide, a solid, is dissolved in water, aqueous lead(II) ion and

aqueous iodide ion are produced (Figure 7.3).

PbI2(s) N Pb2+(aq) + 2I-(aq)



Keq = [Pb2+][I-]2 = 7.1 * 10-9



7.3  Equilibrium   245



It may seem that there is an error in the equilibrium constant expression just written—

the concentration of PbI2 does not appear in the denominator. There is no mistake,

however, because when an equilibrium constant expression for a reaction is written, only

concentrations that can change are included. The concentration of a solid is the number of

moles present in a given volume, a value that never changes at a given temperature, so

the concentration of solid reactants and products is omitted. Solvents are also left out of

equilibrium constant expressions because their concentration is very high and does not

change significantly during a reaction. In the reaction of hydrogen cyanide (HCN) and

water to produce cyanide ion (CN - ) and hydronium ion (H3O + ), for example, water is

the solvent and does not appear in the equilibrium constant expression.

HCN(aq) + H2O( l ) N CN-(aq) + H3O+(aq)

Keq =



[CN-][H3O+]

= 4.9 * 10-10

[HCN]



The size of equilibrium constants can vary greatly from one reaction to the next. For

the reaction just described, Keq has a value of 4.9 * 10 - 10. (Because this is an acid–base

­reaction, the equilibrium constant, Keq, is also known as the acidity constant, Ka.) In

terms of the math involved, the only way that Keq can have a value as small as 4.9 * 10 - 10

is if the denominator of the expression is much larger than the numerator. In this particular

case it means that the concentration of HCN is much higher than the concentration of the

two products. It is always true that when an equilibrium constant has a value of less than 1, the

denominator (related to reactant concentrations) of the equilibrium constant expression is

greater than the numerator (related to product concentrations) (Table 7.2).

For another acid–base reaction, that of hydrochloric acid (HCl) and water, the equilibrium constant has a value of 1.0 * 107.

HCl(aq) + H2O( l ) N Cl-(aq) + H3O+(aq)



Keq =



[Cl-][H3O+]

= 1.0 * 107

[HCl]



When an equilibrium constant has a value greater than 1, the numerator of the equilibrium

constant expression is greater than the denominator, which means that at equilibrium there

are more products than reactants. In this example, the product concentrations (Cl- and

H3O + ) far outweigh the reactant concentration (HCl).



Table | 7.2   Interpreting Equilibrium Constants (K eq )

Value of K eq



C omposition of Reaction Mixture



[Reactants] 6 [Products]



Keq 7 1



[Reactants] 7 [Products]



Keq 6 1



[Reactants] L [Products]



Keq L 1



Sample Problem  7.4



Using Keq to predict relative concentrations

In which reaction below are ­equilibrium product concentrations greater than reactant

concentrations?

HI + H2O N I- + H3O+

HF + H2O N F- + H3O+



[I-][H3O+]

= 2.5 * 1010

[HI]

[F-][H3O+]

=

= 6.6 * 10-4

[HF]



Keq =

Keq



n



Solvents and solids are not

included in equilibrium

constant equations.



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2 Brønsted–Lowry Acids and Bases

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