16.6 16.9 16.10 16 - sri lanka's educational...

16.9 acid–BaSe properTieS of SalT SoluTionS We learn that the ions of a soluble ionic compound can serve as Brønsted–Lowry acids or bases.

16.10 acid–BaSe Behavior and chemical STrucTure We explore the relationship between chemical structure and acid–base behavior.

16.11 lewiS acidS and BaSeS Finally, we see the most general definition of acids and bases, namely the Lewis acid–base definition. A Lewis acid is an electron-pair acceptor and a Lewis base is an electron-pair donor.

16.6 weak acidS We learn that the ionization of a weak acid in water is an equilibrium process with an equilibrium constant Ka that can be used to calculate the pH of a weak acid solution.

16.7 weak BaSeS We learn that the ionization of a weak base in water is an equilibrium process with equilibrium constant Kb that can be used to calculate the pH of a weak base solution.

16.8 relaTionShip BeTween Ka and Kb We see that Ka and Kb are related by the relationship Ka × Kb = Kw. Hence, the stronger an acid, the weaker its conjugate base.

672 cHApTer 16 Acid–Base equilibria

our lives all critically depend on the acidity or basicity of solutions. We will thus explore in this chapter how we measure acidity and how the chemical reactions of acids and bases depend on their concentrations.

We first encountered acids and bases in Sections 2.8 and 4.3, in which we discussed the naming of acids and some simple acid–base reactions, respectively. In this chap-ter we take a closer look at how acids and bases are identified and characterized. In doing so, we consider their behavior both in terms of their structure and bonding and in terms of the chemical equilibria in which they participate.

16.1 | Acids and Bases: A Brief ReviewFrom the earliest days of experimental chemistry, scientists have recognized acids and bases by their characteristic properties. Acids have a sour taste and cause certain dyes to change color, whereas bases have a bitter taste and feel slippery (soap is a good exam-ple). Use of the term base comes from the old English meaning of the word, “to bring low.” (We still use the word debase in this sense, meaning to lower the value of some-thing.) When a base is added to an acid, the base “lowers” the amount of acid. Indeed, when acids and bases are mixed in the right proportions, their characteristic properties seem to disappear altogether. (Section 4.3)

By 1830 it was evident that all acids contain hydrogen but not all hydrogen- containing substances are acids. During the 1880s, the Swedish chemist Svante Arrhenius (1859–1927) defined acids as substances that produce H+ ions in water and bases as sub-stances that produce OH- ions in water. Over time the Arrhenius concept of acids and bases came to be stated in the following way: • Anacid is a substance that, when dissolved in water, increases the concentration of

H+ ions. • Abase is a substance that, when dissolved in water, increases the concentration of

OH- ions.Hydrogen chloride gas, which is highly soluble in water, is an example of an Arrhenius

acid. When it dissolves in water, HCl(g) produces hydrated H+ and Cl- ions:

HCl1g2 ¡H2O H+1aq2 + Cl-1aq2 [16.1]

The aqueous solution of HCl is known as hydrochloric acid. Concentrated hydrochloric acid is about 37% HCl by mass and is 12 M in HCl.

Sodium hydroxide is an Arrhenius base. Because NaOH is an ionic compound, it dissociates into Na+ and OH- ions when it dissolves in water, thereby increasing the concentration of OH- ions in the solution.

give it some thoughtWhich two ions are central to the Arrhenius definitions of acids and bases?

▲ Figure 16.1 two organic acids: tartartic acid, H2C4H4O6, and malic acid, H2C4H4O5.

Tartaric acid Malic acid

O

H OH

CCHO

H OH

O

C OHC

O

H H

CCHO

H OH

O

C OHC

secTiON 16.2 Brønsted–Lowry Acids and Bases 673

16.2 | Brønsted–Lowry Acids and Bases

The Arrhenius concept of acids and bases, while useful, is rather limited. For one thing, it is restricted to aqueous solutions. In 1923 the Danish chemist Johannes Brønsted (1879–1947) and the English chemist Thomas Lowry (1874–1936) independently proposed a more gen-eral definition of acids and bases. Their concept is based on the fact that acid–base reactions involve the transfer of H+ ions from one substance to another. To understand this definition better, we need to examine the behavior of the H+ ion in water more closely.

The H+ Ion in WaterWe might at first imagine that ionization of HCl in water produces just H+ and Cl- . A hydrogen ion is no more than a bare proton—a very small particle with a positive charge. As such, an H+ ion interacts strongly with any source of electron density, such as the nonbonding electron pairs on the oxygen atoms of water molecules. For exam-ple, the interaction of a proton with water forms the hydronium ion, H3O+1aq2:

O

H

HHO

H

HH+

+ +

+

+

+

[16.2]

The behavior of H+ ions in liquid water is complex because hydronium ions interact with additional water molecules via the formation of hydrogen bonds. (Section 11.2) For example, the H3O+ ion bonds to additional H2O molecules to generate such ions as H5O2

+ and H9O4+ (▶ Figure 16.2).

Chemists use the notations H+1aq2 and H3O+1aq2 interchangeably to represent the hydrated proton responsible for the characteristic properties of aqueous solutions of acids. We often use the notation H+1aq2 for simplicity and convenience, as we did in Chapter 4 and Equation 16.1. The notation H3O+1aq2, however, more closely represents reality.

Proton-Transfer ReactionsIn the reaction that occurs when HCl dissolves in water, the HCl molecule transfers an H+ ion (a proton) to a water molecule. Thus, we can represent the reaction as occurring between an HCl molecule and a water molecule to form hydronium and chloride ions:

+−

+ +

O

H

HHO

H

HCl Cl H + ++

HCl(g) Cl−(aq)H2O(l)

Acid Base

H3O+(aq)+ +

−

[16.3]

Notice that the reaction in Equation 16.3 involves a proton donor (HCl) and a pro-ton acceptor 1H2O2. The notion of transfer from a proton donor to a proton acceptor is the key idea in the Brønsted–Lowry definition of acids and bases: • Anacid is a substance (molecule or ion) that donates a proton to another substance. • Abase is a substance that accepts a proton.

O OH H H

HH

OH

HHO

H

O

H

H

OH

H H

H9O4+

+

H5O2+

+

+

+

▲ Figure 16.2 ball-and-stick models and lewis structures for two hydrated hydronium ions.

Go FiGuREWhich type of intermolecular force do the dotted lines in this figure represent?


Thus, when HCl dissolves in water (Equation 16.3), HCl acts as a Brønsted–Lowry acid (it donates a proton to H2O), and H2O acts as a Brønsted–Lowry base (it accepts a proton from HCl). We see that the H2O molecule serves as a proton acceptor by using one of the nonbonding pairs of electrons on the O atom to “attach” the proton.

Because the emphasis in the Brønsted–Lowry concept is on proton transfer, the concept also applies to reactions that do not occur in aqueous solution. In the reaction between gas phase HCl and NH3, for example, a proton is transferred from the acid HCl to the base NH3:

+−

+ +

N

H

HHN

H

H H

HCl ClH + +

+

Acid Base

−

[16.4]

The hazy film that forms on the windows of general chemistry laboratories and on glassware in the laboratory (◀ Figure 16.3) is largely solid NH4Cl formed by the gas-phase reaction between HCl and NH3.

Let’s consider another example that compares the relationship between the Arrhenius and Brønsted–Lowry definitions of acids and bases—an aqueous solution of ammonia, in which we have the equilibrium:

NH31aq2 + H2O1l2 ∆ NH4+1aq2 + OH- 1aq2 [16.5]

Base Acid

Ammonia is a Brønsted–Lowry base because it accepts a proton from H2O. Ammonia is also an Arrhenius base because adding it to water leads to an increase in the concen-tration of OH- 1aq2.

The transfer of a proton always involves both an acid (donor) and a base (acceptor). In other words, a substance can function as an acid only if another substance simulta-neously behaves as a base. To be a Brønsted–Lowry acid, a molecule or ion must have a hydrogen atom it can lose as an H+ ion. To be a Brønsted–Lowry base, a molecule or ion must have a nonbonding pair of electrons it can use to bind the H+ ion.

Some substances can act as an acid in one reaction and as a base in another. For exam-ple, H2O is a Brønsted–Lowry base in Equation 16.3 and a Brønsted–Lowry acid in Equa-tion 16.5. A substance capable of acting as either an acid or a base is called amphiprotic. An amphiprotic substance acts as a base when combined with something more strongly acidic than itself and as an acid when combined with something more strongly basic than itself.

give it some thoughtIn the forward reaction of this equilibrium, which substance acts as the Brønsted–Lowry base?

H2S1aq2 + CH3NH21aq2 ∆ HS- 1aq2 + CH3NH3+1aq2

Conjugate Acid–Base PairsIn any acid–base equilibrium, both the forward reaction (to the right) and the reverse reaction (to the left) involve proton transfer. For example, consider the reaction of an acid HA with water:

HA1aq2 + H2O1l2 ∆ A- 1aq2 + H3O+1aq2 [16.6]

In the forward reaction, HA donates a proton to H2O. Therefore, HA is the Brønsted–Lowry acid and H2O is the Brønsted–Lowry base. In the reverse reaction, the H3O+ ion

▲ Figure 16.3 fog of NH4Cl 1s 2 caused by the reaction of HCl 1g 2 and NH3 1g 2 .


donates a proton to the A- ion, so H3O+ is the acid and A- is the base. When the acid HA donates a proton, it leaves behind a substance, A- , that can act as a base. Likewise, when H2O acts as a base, it generates H3O+, which can act as an acid.

An acid and a base such as HA and A- that differ only in the presence or absence of a proton are called a conjugate acid–base pair.* Every acid has a conjugate base, formed by removing a proton from the acid. For example, OH- is the conjugate base of H2O, and A- is the conjugate base of HA. Every base has a conjugate acid, formed by adding a proton to the base. Thus, H3O+ is the conjugate acid of H2O, and HA is the conjugate acid of A- .

In any acid–base (proton-transfer) reaction, we can identify two sets of conjugate acid–base pairs. For example, consider the reaction between nitrous acid and water:

HNO2(aq) H2O(l) NO2 (aq) H3O (aq)Acid Base Conjugate

baseConjugate

acid

remove H

add H

− +

+

+

++

[16.7]

Likewise, for the reaction between NH3 and H2O (Equation 16.5), we have

NH3(aq) H2O(l) NH4 (aq) OH (aq)AcidBase Conjugate

baseConjugate

acid

remove H

add H

−+ ++

+

+

[16.8]

*The word conjugate means “joined together as a pair.”

(a) What is the conjugate base of HClO4, H2S, PH4+, HCO3

- ?(b) What is the conjugate acid of CN - , SO4

2 - , H2O, HCO3- ?

sAmpLE ExERcisE 16.1 identifying conjugate Acids and Bases

soLuTionAnalyze We are asked to give the conjugate base for several acids and the conjugate acid for several bases.Plan The conjugate base of a substance is simply the parent substance minus one proton, and the conjugate acid of a substance is the parent substance plus one proton.Solve

(a) If we remove a proton from HClO4, we obtain ClO4- , which is

its conjugate base. The other conjugate bases are HS- , PH3, and CO3

2 - .(b) If we add a proton to CN - , we get HCN, its conjugate acid. The

other conjugate acids are HSO4- , H3O+, and H2CO3. Notice that

the hydrogen carbonate ion 1HCO3-2 is amphiprotic. It can act

as either an acid or a base.

practice Exercise 1Consider the following equilibrium reaction:

HSO4- 1aq2 + OH- 1aq2 ∆ SO4

2 - 1aq2 + H2O1l2Which substances are acting as acids in the reaction?(a) HSO4

- and OH- (b) HSO4- and H2O

(c) OH- and SO42 - (d) SO4

2 - and H2O(e) None of the substances are acting as acids in this reaction.

practice Exercise 2Write the formula for the conjugate acid of each of the following: HSO3

- , F - , PO43 - , CO.

Once you become proficient at identifying conjugate acid–base pairs it is not difficult to write equations for reactions involving Brønsted–Lowry acids and bases (proton-transfer reactions).


The hydrogen sulfite ion 1HSO3- 2 is amphiprotic. Write an equation for the reaction of HSO3

- with water (a) in which the ion acts as an acid and (b) in which the ion acts as a base. In both cases identify the conjugate acid–base pairs.

sAmpLE ExERcisE 16.2 Writing Equations for proton-Transfer Reactions

soLuTionAnalyze and Plan We are asked to write two equations representing re-actions between HSO3

- and water, one in which HSO3- should donate

a proton to water, thereby acting as a Brønsted–Lowry acid, and one in which HSO3

- should accept a proton from water, thereby acting as a base. We are also asked to identify the conjugate pairs in each equation.

Solve

(a) HSO3- 1aq2 + H2O1l2 ∆ SO3

2 - 1aq2 + H3O+1aq2

The conjugate pairs in this equation are HSO3- (acid) and SO3

2 - (conjugate base), and H2O (base) and H3O+ (conjugate acid).

(b) HSO3- 1aq2 + H2O1l2 ∆ H2SO31aq2 + OH- 1aq2

The conjugate pairs in this equation are H2O (acid) and OH- (conju-gate base), and HSO3

- (base) and H2SO3 (conjugate acid).

practice Exercise 1The dihydrogen phosphate ion, H2PO4

- , is amphiprotic. In which of the following reactions is this ion serving as a base? (i) H3O+1aq2 + H2PO4

- 1aq2 ∆ H3PO41aq2 + H2O1l2 (ii) H3O+1aq2 + HPO4

2 - 1aq2 ∆ H2PO4- 1aq2 + H2O1l2

(iii) H3PO41aq2 + HPO42 - 1aq2 ∆ 2 H2PO4

- 1aq2(a) i only (b) i and ii (c) i and iii (d) ii and iii(e) i, ii, and iii

practice Exercise 2When lithium oxide 1Li2O2 is dissolved in water, the solution turns basic from the reaction of the oxide ion 1O2-2 with water. Write the equation for this reaction and identify the conjugate acid–base pairs.

Relative Strengths of Acids and BasesSome acids are better proton donors than others, and some bases are better proton accep-tors than others. If we arrange acids in order of their ability to donate a proton, we find that the more easily a substance gives up a proton, the less easily its conjugate base accepts a proton. Similarly, the more easily a base accepts a proton, the less easily its conjugate acid gives up a proton. In other words, the stronger an acid, the weaker its conjugate base, and the stronger a base, the weaker its conjugate acid. Thus, if we know how readily an acid donates protons, we also know something about how readily its conjugate base accepts protons.

The inverse relationship between the strengths of acids and their conjugate bases is illustrated in ▼ Figure 16.4. Here we have grouped acids and bases into three broad categories based on their behavior in water:

BASECl−

HSO4−

NO3−

H2OSO4

2−

H2PO4−

F−

CH3COO−

HCO3−

HS−

HPO42−

NH3CO3

2−

PO43−

OH−

O2−

H−

CH3−

Strong bases

Weak bases

Negligible basicity

Base strength increasing

Aci

d s

tren

gth

incr

easi

ng

ACIDHClH2SO4HNO3H3O+(aq)HSO4

−

H3PO4HFCH3COOHH2CO3H2SH2PO4

−

NH4+

HCO3−

HPO42−

H2OOH−

H2CH4

Strong acids

Weak acids

Negligible acidity

▲ Figure 16.4 relative strengths of select conjugate acid–base pairs. The two members of each pair are listed opposite each other in the two columns.

Go FiGuREIf O2- ions are added to water, what reaction, if any, occurs?


1. A strong acid completely transfers its protons to water, leaving essentially no undis-sociated molecules in solution. (Section 4.3) Its conjugate base has a negligible tendency to accept protons in aqueous solution. (The conjugate base of a strong acid shows negligible basicity.)

2. A weak acid only partially dissociates in aqueous solution and therefore exists in the solution as a mixture of the undissociated acid and its conjugate base. The con-jugate base of a weak acid shows a slight ability to remove protons from water. (The conjugate base of a weak acid is a weak base.)

3. A substance with negligible acidity contains hydrogen but does not demonstrate any acidic behavior in water. Its conjugate base is a strong base, reacting com-pletely with water, to form OH- ions. (The conjugate base of a substance with neg-ligible acidity is a strong base.)

The ions H3O+1aq2 and OH- 1aq2 are, respectively, the strongest possible acid and strongest possible base that can exist at equilibrium in aqueous solution. Stronger acids react with water to produce H3O+1aq2 ions, and stronger bases react with water to pro-duce OH- 1aq2 ions, a phenomenon known as the leveling effect.

give it some thoughtGiven that HClO4 is a strong acid, how would you classify the basicity of ClO4

-?

We can think of proton-transfer reactions as being governed by the relative abili-ties of two bases to abstract protons. For example, consider the proton transfer that occurs when an acid HA dissolves in water:

HA1aq2 + H2O1l2 ∆ H3O+1aq2 + A- 1aq2 [16.9]

If H2O (the base in the forward reaction) is a stronger base than A- (the conjugate base of HA), it is favorable to transfer the proton from HA to H2O, producing H3O+ and A- . As a result, the equilibrium lies to the right. This describes the behavior of a strong acid in water. For example, when HCl dissolves in water, the solution consists almost entirely of H3O+ and Cl- ions with a negligible concentration of HCl molecules:

HCl1g2 + H2O1l2 ¡ H3O+1aq2 + Cl- 1aq2 [16.10]

H2O is a stronger base than Cl- (Figure 16.4), so H2O acquires the proton to become the hydronium ion. Because the reaction lies completely to the right, we write Equation 16.10 with only an arrow to the right rather than using the double arrows for an equilibrium.

When A- is a stronger base than H2O, the equilibrium lies to the left. This situation occurs when HA is a weak acid. For example, an aqueous solution of acetic acid consists mainly of CH3COOH molecules with only a relatively few H3O+ and CH3COO - ions:

CH3COOH1aq2 + H2O1l2 ∆ H3O+1aq2 + CH3COO - 1aq2 [16.11]

The CH3COO - ion is a stronger base than H2O (Figure 16.4) and therefore the reverse reaction is favored more than the forward reaction.

From these examples, we conclude that in every acid–base reaction, equilibrium favors transfer of the proton from the stronger acid to the stronger base to form the weaker acid and the weaker base.

For the following proton-transfer reaction use Figure 16.4 to predict whether the equilibrium lies to the left 1Kc 6 12 or to the right 1Kc 7 12:

HSO4- 1aq2 + CO3

2 - 1aq2 ∆ SO42 - 1aq2 + HCO3

- 1aq2soLuTionAnalyze We are asked to predict whether an equilibrium lies to the right, favoring products, or to the left, favoring reactants.

sAmpLE ExERcisE 16.3 predicting the position of a proton-Transfer

Equilibrium


16.3 | The Autoionization of WaterOne of the most important chemical properties of water is its ability to act as either a Brønsted–Lowry acid or a Brønsted–Lowry base. In the presence of an acid, it acts as a proton acceptor; in the presence of a base, it acts as a proton donor. In fact, one water molecule can donate a proton to another water molecule:

Acid Base

O

H

HHO

H

HO

H

H O

H

+−

+

+

+

+

+

H2O(l) OH−(aq)H2O(l) H3O+(aq)+ +

−

[16.12]

We call this process the autoionization of water.Because the forward and reverse reactions in Equation 16.12 are extremely

rapid, no water molecule remains ionized for long. At room temperature only about two out of every 109 water molecules are ionized at any given instant. Thus, pure water consists almost entirely of H2O molecules and is an extremely poor conductor of electricity. Nevertheless, the autoionization of water is very important, as we will soon see.

Plan This is a proton-transfer reaction, and the position of the equilibrium will favor the proton going to the stronger of two bases. The two bases in the equation are CO3

2 - , the base in the forward reaction, and SO4

2 - , the conjugate base of HSO4- . We can find the relative positions

of these two bases in Figure 16.4 to determine which is the stronger base.Solve The CO3

2 - ion appears lower in the right-hand column in Figure 16.4 and is there-fore a stronger base than SO4

2 - . Therefore, CO32- will get the proton preferentially to become

HCO3- , while SO4

2 - will remain mostly unprotonated. The resulting equilibrium lies to the right, favoring products (that is, Kc 7 1):

HSO4-1aq2 + CO3

2-1aq2 ∆ SO42-1aq2 + HCO3

-1aq2

Kc 7 1Acid Base Conjugate base Conjugate acid

Comment Of the two acids HSO4- and HCO3

-, the stronger one 1HSO4-2 gives up a proton

more readily, and the weaker one 1HCO3-2 tends to retain its proton. Thus, the equilibrium

favors the direction in which the proton moves from the stronger acid and becomes bonded to the stronger base.

practice Exercise 1Based on information in Figure 16.4, place the following equilibria in order from smallest to largest value of Kc: (i) CH3COOH1aq2 + HS- 1aq2 ∆ CH3COO - 1aq2 + H2S1aq2 (ii) F - 1aq2 + NH4

+1aq2 ∆ HF1aq2 + NH31aq2 (iii) H2CO31aq2 + Cl- 1aq2 ∆ HCO3

- 1aq2 + HCl1aq2(a) i 6 ii 6 iii (b) ii 6 i 6 iii (c) iii 6 i 6 ii (d) ii 6 iii 6 i(e) iii 6 ii 6 i

practice Exercise 2For each reaction, use Figure 16.4 to predict whether the equilibrium lies to the left or to the right:(a) HPO4

2 - 1aq2 + H2O1l2 ∆ H2PO4- 1aq2 + OH- 1aq2

(b) NH4+1aq2 + OH- 1aq2 ∆ NH31aq2 + H2O1l2

secTiON 16.3 The Autoionization of Water 679

The Ion Product of WaterThe equilibrium-constant expression for the autoionization of water is Kc = 3H3O+43OH-4 [16.13]The term [H2O] is excluded from the equilibrium-constant expression because we exclude the concentrations of pure solids and liquids. (Section 15.4) Because this expression refers specifically to the autoionization of water, we use the symbol Kw to denote the equilibrium constant, which we call the ion-product constant for water. At 25 °C, Kw equals 1.0 * 10-14. Thus, we have Kw = 3H3O+43OH-4 = 1.0 * 10-14 1at 25 °C2 [16.14]

Because we use H+1aq2 and H3O+1aq2 interchangeably to represent the hydrated proton, the autoionization reaction for water can also be written as H2O1l2 ∆ H+1aq2 + OH-1aq2 [16.15]Likewise, the expression for Kw can be written in terms of either H3O+ or H+, and Kw has the same value in either case: Kw = 3H3O+43OH-4 = 3H+43OH-4 = 1.0 * 10-14 1at 25 °C2 [16.16]This equilibrium-constant expression and the value of Kw at 25 °C are extremely im-portant, and you should commit them to memory.

A solution in which 3H+4 = 3OH-4 is said to be neutral. In most solutions, how-ever, the H+ and OH- concentrations are not equal. As the concentration of one of these ions increases, the concentration of the other must decrease, so that the product of their concentrations always equals 1.0 * 10-14 (▼ Figure 16.5).

Acidic solution[H+] > [OH−]

[H+][OH−] = 1.0 × 10−14

Neutral solution[H+] = [OH−]

[H+][OH−] = 1.0 × 10−14

Basic solution[H+] < [OH−]

[H+][OH−] = 1.0 × 10−14

Hydrochloricacid

HCl(aq)

WaterH2O

SodiumhydroxideNaOH(aq)

▲ Figure 16.5 relative concentrations of H+ and OH− in aqueous solutions at 25 °C.

Go FiGuRESuppose that equal volumes of the middle and right samples in the figure were mixed. Would the resultant solution be acidic, neutral, or basic?

Calculate the values of 3H+4 and 3OH- 4 in a neutral aqueous solution at 25 °C.

sAmpLE ExERcisE 16.4 calculating 3h+4 for pure Water

soLuTionAnalyze We are asked to determine the concentrations of H+ and OH- ions in a neutral solution at 25 °C.Plan We will use Equation 16.16 and the fact that, by definition, 3H+4 = 3OH-4 in a neutral solution.Solve We will represent the concentration of H+ and OH- in neutral solution with x. This gives

3H+43OH-4 = 1x21x2 = 1.0 * 10-14

x2 = 1.0 * 10-14

x = 1.0 * 10-7M = 3H+4 = 3OH-4In an acid solution 3H+4 is greater than 1.0 * 10-7 M; in a basic solu-tion 3H+4 is less than 1.0 * 10-7 M.


What makes Equation 16.16 particularly useful is that it is applicable both to pure water and to any aqueous solution. Although the equilibrium between H+1aq2 and OH- 1aq2 as well as other ionic equilibria are affected somewhat by the presence of additional ions in solution, it is customary to ignore these ionic effects except in work requiring exceptional accuracy. Thus, Equation 16.16 is taken to be valid for any dilute aqueous solution and can be used to calculate either 3H+4 (if 3OH- 4 is known) or 3OH- 4 (if 3H+4 is known).

practice Exercise 1In a certain acidic solution at 25 °C, 3H+4 is 100 times greater than 3OH-4. What is the value for 3OH-4 for the solution?(a) 1.0 * 10-8 M (b) 1.0 * 10-7 M (c) 1.0 * 10-6 M(d) 1.0 * 10-2 M (e) 1.0 * 10-9 M

practice Exercise 2Indicate whether solutions with each of the following ion concen-trations are neutral, acidic, or basic: (a) 3H+4 = 4 * 10-9 M; (b) 3OH-4 = 1 * 10-7 M; (c) 3OH-4 = 1 * 10-13 M.

Calculate the concentration of H+1aq2 in (a) a solution in which 3OH-4 is 0.010 M, (b) a solution in which 3OH-4 is 1.8 * 10-9M. Note: In this problem and all that follow, we assume, unless stated otherwise, that the temperature is 25 °C.

sAmpLE ExERcisE 16.5 calculating 3h+4 from 3oh - 4

soLuTionAnalyze We are asked to calculate the 3H+4 concentration in an aqueous solution where the hydroxide concentration is known.

Plan We can use the equilibrium-constant expression for the auto-ionization of water and the value of Kw to solve for each unknown concentration.

Solve

(a) Using Equation 16.16, we have 3H+43OH-4 = 1.0 * 10-14

3H+4 =11.0 * 10-142

3OH-4 =1.0 * 10-14

0.010= 1.0 * 10-12 M

This solution is basic because 3OH- 4 7 3H+4

(b) In this instance 3H+4 =11.0 * 10 - 142

3OH- 4 =1.0 * 10 - 14

1.8 * 10 - 9 = 5.6 * 10 - 6 M

This solution is acidic because 3H+4 7 3OH- 4

practice Exercise 1A solution has 3OH- 4 = 4.0 * 10 - 8. What is the value of 3H+4 for the solution?(a) 2.5 * 10 - 8 M (b) 4.0 * 10 - 8 M (c) 2.5 * 10 - 7 M(d) 2.5 * 10 - 6 M (e) 4.0 * 10 - 6 M

practice Exercise 2Calculate the concentration of OH- 1aq2 in a solution in which(a) 3H+4 = 2 * 10 - 6 M; (b) 3H+4 = 3OH- 4;(c) 3H+4 = 200 * 3OH- 4.

16.4 | The ph scaleThe molar concentration of H+1aq2 in an aqueous solution is usually very small. For convenience, we therefore usually express 3H+4 in terms of pH, which is the negative logarithm in base 10 of 3H+4:* pH = - log[H+] [16.17]If you need to review the use of logarithms, see Appendix A.

In Sample Exercise 16.4, we saw that 3H+4 = 1.0 * 10-7 M for a neutral aqueous solution at 25 °C. We can now use Equation 16.17 to calculate the pH of a neutral solution at 25 °C:

pH = - log11.0 * 10-72 = -1-7.002 = 7.00

*Because 3H+4 and 3H3O+4 are used interchangeably, you might see pH defined as - log 3H3O+4.

secTiON 16.4 The pH scale 681

Notice that the pH is reported with two decimal places. We do so because only the num-bers to the right of the decimal point are the significant figures in a logarithm. Because our original value for the concentration 11.0 * 10-7 M2 has two significant figures, the corresponding pH has two decimal places (7.00).

What happens to the pH of a solution as we make the solution more acidic, so that 3H+4 increases? Because of the negative sign in the logarithm term of Equation 16.17, the pH decreases as 3H+4 increases. For example, when we add sufficient acid to make 3H+4 = 1.0 * 10-3 M the pH is

pH = - log11.0 * 10-32 = -1-3.002 = 3.00

At 25 °C the pH of an acidic solution is less than 7.00.We can also calculate the pH of a basic solution, one in which 3OH-4 7

1.0 * 10-7 M. Suppose 3OH-4 = 2.0 * 10-3 M. We can use Equation 16.16 to calcu-late 3H+4 for this solution and Equation 16.17 to calculate the pH:

3H+4 =Kw

3OH-4 =1.0 * 10-14

2.0 * 10-3 = 5.0 * 10-12 M

pH = - log15.0 * 10-122 = 11.30

At 25 °C the pH of a basic solution is greater than 7.00. The relationships among 3H+4, 3OH-4, and pH are summarized in ▲ Table 16.1.

give it some thoughtIs it possible for a solution to have a negative pH? If so, would that pH signify a basic or acidic solution?

One might think that when 3H+4 is very small, as is often the case, it would be unim-portant. That reasoning is quite incorrect! Remember that many chemical processes depend on the ratio of changes in concentration. For example, if a kinetic rate law is first order in 3H+4, doubling the H+ concentration doubles the rate even if the change is merely from 1 * 10-7 M to 2 * 10-7 M. (Section 14.3) In biological systems, many reactions involve proton transfers and have rates that depend on 3H+4. Because the speeds of these reactions are crucial, the pH of biological fluids must be maintained within narrow limits. For example, human blood has a normal pH range of 7.35 to 7.45. Illness and even death can result if the pH varies much from this narrow range.

Table 16.1 Relationships among 3h+4 , 3oh−4 , and ph at 25 °c

solution Type 3h+4 1M 2 3oh−4 1M 2 ph

Acidic 7 1.0 * 10-7 61.0 * 10-7 6 7.00

Neutral 1.0 * 10-7 1.0 * 10-7 7.00

Basic 61.0 * 10-7 71.0 * 10-7 7 7.00

soLuTionAnalyze We are asked to determine the pH of aqueous solutions for which we have already calculated 3H+4.Plan We can calculate pH using its defining equation, Equation 16.17.Solve

(a) In the first instance we found 3H+4 to be 1.0 * 10-12 M, so that

pH = - log11.0 * 10-122 = -1-12.002 = 12.00

Because 1.0 * 10-12 has two significant figures, the pH has two decimal places, 12.00.

sAmpLE ExERcisE 16.6 calculating ph from 3h+4Calculate the pH values for the two solutions of Sample Exercise 16.5.

(b) For the second solution, 3H+4 = 5.6 * 10-6 M. Before performing the calculation, it is helpful to estimate the pH. To do so, we note that 3H+4 lies between 1 * 10-6 and 1 * 10-5. Thus, we expect the pH to lie between 6.0 and 5.0. We use Equation 16.17 to calculate the pH:

pH = - log15.6 * 10-62 = 5.25

Check After calculating a pH, it is useful to compare it to your esti-mate. In this case the pH, as we predicted, falls between 6 and 5. Had the calculated pH and the estimate not agreed, we should have recon-sidered our calculation or estimate or both.


pOH and Other “p” ScalesThe negative logarithm is a convenient way of expressing the magnitudes of other small quantities. We use the convention that the negative logarithm of a quantity is labeled “p” (quantity). Thus, we can express the concentration of OH- as pOH: pOH = - log 3OH-4 [16.18]Likewise, pKw equals - log Kw.

By taking the negative logarithm of both sides of the equilibrium-constant expres-sion for water, Kw = 3H+43OH-4, we obtain - log3H+4 + 1- log 3OH-42 = - log Kw [16.19]from which we obtain the useful expression pH + pOH = 14.00 1at 25 °C2 [16.20]The pH and pOH values characteristic of a number of familiar solutions are shown in (▼ Figure 16.6). Notice that a change in 3H+4 by a factor of 10 causes the pH to change by 1. Thus, the concentration of H+1aq2 in a solution of pH 5 is 10 times the H+1aq2 concentra-tion in a solution of pH 6.

practice Exercise 1A solution at 25 °C has 3OH-4 = 6.7 * 10-3. What is the pH of the solution?(a) 0.83 (b) 2.2 (c) 2.17 (d) 11.83 (e) 12

practice Exercise 2(a) In a sample of lemon juice, 3H+4 = 3.8 * 10-4 M. What is the pH?(b) A commonly available window-cleaning solution has 3OH-4 = 1.9 * 10-6 M. What is the pH at 25 °C?

[H+] (M) [OH−] (M)

pH + pOH = 14

[H+][OH−] = 1×10−14

pOHpH

1 (1×100)

1×10−1

1×10−2

1×10−3

1×10−4

1×10−5

1×10−6

1×10−7

1×10−8

1×10−9

1×10−10

1×10−11

1×10−12

1×10−13

1×10−14

1×10−14

1×10−13

1×10−12

1×10−11

1×10−10

1×10−9

1×10−8

1×10−7

1×10−6

1×10−5

1×10−4

1×10−3

1×10−2

1×10−1

1 (1×100)

14.0

13.0

12.0

11.0

10.0

9.0

8.0

7.0

6.0

5.0

4.0

3.0

2.0

1.0

0.0

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

10.0

11.0

12.0

13.0

14.0

Milk

Tomatoes

Black coffee

Cola, vinegarLemon juice

Stomach acid

SalivaRain

Wine

Household ammoniaHousehold bleach

Lime water

Human blood

Seawater

Borax

Increasing base strength

Incr

easi

ng a

cid

str

engt

h

▲ Figure 16.6 Concentrations of H+ and OH- , and pH and poH values of some common substances at 25 °C.

Go FiGuREWhich is more acidic, black coffee or lemon juice?

secTiON 16.4 The pH scale 683

give it some thoughtIf the pOH for a solution is 3.00, what is the pH? Is the solution acidic or basic?

soLuTionAnalyze We need to calculate 3H+4 from pOH.Plan We will first use Equation 16.20, pH + pOH = 14.00, to calcu-late pH from pOH. Then we will use Equation 16.17 to determine the concentration of H+.Solve From Equation 16.20, we have

pH = 14.00 - pOH pH = 14.00 - 10.24 = 3.76

Next we use Equation 16.17:pH = - log3H+4 = 3.76

Thus,log3H+4 = -3.76

To find 3H+4, we need to determine the antilogarithm of -3.76. Your calculator will show this command as 10x or INV log (these functions are usually above the log key). We use this function to perform the calculation:

3H+4 = antilog 1-3.762 = 10-3.76 = 1.7 * 10-4 M

Comment The number of significant figures in 3H+4 is two because the number of decimal places in the pH is two.

sAmpLE ExERcisE 16.7 calculating 3h+ 4 from poh

A sample of freshly pressed apple juice has a pOH of 10.24. Calculate 3H+4.

Check Because the pH is between 3.0 and 4.0, we know that 3H+4 will be between 1.0 * 10-3 M and 1.0 * 10-4 M. Our calculated 3H+4 falls within this estimated range.

practice Exercise 1A solution at 25 °C has pOH = 10.53. Which of the following statements is or are true? (i) The solution is acidic. (ii) The pH of the solution is 14.00 - 10.53.(iii) For this solution, 3OH-4 = 10-10.53M.(a) Only one of the statements is true.(b) Statements (i) and (ii) are true.(c) Statements (i) and (iii) are true.(d) Statements (ii) and (iii) are true.(e) All three statements are true.

practice Exercise 2A solution formed by dissolving an antacid tablet has a pOH of 4.82. Calculate 3H+4.

Measuring pHThe pH of a solution can be measured with a pH meter (▶ Figure 16.7). A complete understanding of how this important device works requires a knowledge of elec-trochemistry, a subject we take up in Chapter 20. In brief, a pH meter consists of a pair of electrodes connected to a meter capable of measuring small voltages, on the order of millivolts. A voltage, which varies with pH, is generated when the elec-trodes are placed in a solution. This voltage is read by the meter, which is calibrated to give pH.

Although less precise, acid–base indicators can be used to measure pH. An acid–base indicator is a colored substance that can exist in either an acid or a base form. The two forms have different colors. Thus, the indicator has one color at lower pH and another at higher pH. If you know the pH at which the indicator turns from one form to the other, you can determine whether a solution has a higher or lower pH than this value. Litmus, for example, changes color in the vicinity of pH 7. The color change, however, is not very sharp. Red litmus indicates a pH of about 5 or lower, and blue lit-mus indicates a pH of about 8 or higher.

Some common indicators are listed in Figure 16.8. The chart tells us, for instance, that methyl red changes color over the pH interval from about 4.5 to 6.0. Below pH 4.5 it is in the acid form, which is red. In the interval between 4.5 and 6.0, it is gradually converted to its basic form, which is yellow. Once the pH rises above 6 the conversion

▲ Figure 16.7 a digital pH meter. The device is a millivoltmeter, and the electrodes immersed in a solution produce a voltage that depends on the pH of the solution.


is complete, and the solution is yellow. This color change, along with that of the indi-cators bromthymol blue and phenolphthalein, is shown in ◀ Figure 16.9. Paper tape impregnated with several indicators is widely used for determining approximate pH values.

16.5 | strong Acids and BasesThe chemistry of an aqueous solution often depends critically on pH. It is therefore im-portant to examine how pH relates to acid and base concentrations. The simplest cases are those involving strong acids and strong bases. Strong acids and bases are strong elec-trolytes, existing in aqueous solution entirely as ions. There are relatively few common strong acids and bases (see Table 4.2).

Strong AcidsThe seven most common strong acids include six monoprotic acids (HCl, HBr, HI, HNO3, HClO3, and HClO4), and one diprotic acid 1H2SO42. Nitric acid 1HNO32 exemplifies the behavior of the monoprotic strong acids. For all practical purposes, an aqueous solution of HNO3 consists entirely of H3O+ and NO3

- ions:

HNO31aq2 + H2O1l2 ¡ H3O+1aq2 + NO3- 1aq2 1complete ionization2 [16.21]

We have not used equilibrium arrows for this equation because the reaction lies entirely to the right. (Section 4.1) As noted in Section 16.3, we use H3O+1aq2 and H+1aq2 interchangeably to represent the hydrated proton in water. Thus, we can simplify this acid ionization equation to

HNO31aq2 ¡ H+1aq2 + NO3- 1aq2

In an aqueous solution of a strong acid, the acid is normally the only significant source of H+ ions.* As a result, calculating the pH of a solution of a strong monoprotic

pH range for color change

Methyl violet

Thymol blue

Methyl orange

Methyl red

Bromthymol blue

Phenolphthalein

Alizarin yellow R

0 2 4 6

Yellow

Yellow

Violet

Red

YellowRed

Yellow

Yellow

Red

Red

PinkColorless

Yellow Blue

Yellow Blue

8 10 12 14

▲ Figure 16.8 pH ranges for common acid–base indicators. Most indicators have a useful range of about 2 pH units.

Go FiGuREIf a colorless solution turns pink when we add phenolphthalein, what can we conclude about the pH of the solution?

Methyl red

Bromthymol blue

Phenolphthalein

▲ Figure 16.9 solutions containing three common acid–base indicators at various pH values.

Go FiGuREWhich of these indicators is best suited to distinguish between a solution that is slightly acidic and one that is slightly basic?

*If the concentration of the acid is 10 - 6 M or less, we also need to consider H+ ions that result from H2O autoionization. Normally, the concentration of H+ from H2O is so small that it can be neglected.

secTiON 16.5 strong Acids and Bases 685

acid is straightforward because 3H+4 equals the original concentration of acid. In a 0.20 M solution of HNO31aq2, for example, 3H+4 = 3NO3

-4 = 0.20 M. The situation with the diprotic acid H2SO4 is somewhat more complex, as we will see in Section 16.6.

soLuTionAnalyze and Plan Because HClO4 is a strong acid, it is completely ionized, giving 3H+4 = 3ClO4

-4 = 0.040 M.

Solve

pH = - log10.0402 = 1.40

Check Because 3H+4 lies between 1 * 10-2 and 1 * 10-1, the pH will be between 2.0 and 1.0. Our calculated pH falls within the estimated range. Furthermore, because the concentration has two significant figures, the pH has two decimal places.

sAmpLE ExERcisE 16.8 calculating the ph of a strong Acid

What is the pH of a 0.040 M solution of HClO4?

practice Exercise 1Order the following three solutions from smallest to largest pH:(i) 0.20 M HClO3 (ii) 0.0030 M HNO3 (iii) 1.50 M HCl(a) i 6 ii 6 iii (b) ii 6 i 6 iii (c) iii 6 i 6 ii (d) ii 6 iii 6 i (e) iii 6 ii 6 i

practice Exercise 2An aqueous solution of HNO3 has a pH of 2.34. What is the concentration of the acid?

Strong BasesThe most common soluble strong bases are the ionic hydroxides of the alkali metals, such as NaOH, KOH, and the ionic hydroxides heavier alkaline earth metals, such as Sr1OH22. These compounds completely dissociate into ions in aqueous solution. Thus, a solution labeled 0.30 M NaOH consists of 0.30 M Na+1aq2 and 0.30 M OH- 1aq2; there is essentially no undissociated NaOH.

give it some thoughtWhich solution has the higher pH, a 0.001 M solution of NaOH or a 0.001 M solution of Ba1OH22?

What is the pH of (a) a 0.028 M solution of NaOH, (b) a 0.0011 M solution of Ca1OH22?

soLuTionAnalyze We are asked to calculate the pH of two solutions of strong bases.Plan We can calculate each pH by either of two equivalent methods. First, we could use Equation 16.16 to calculate 3H+4 and then use Equation 16.17 to calculate the pH. Alternatively, we could use 3OH- 4 to calculate pOH and then use Equation 16.20 to calculate the pH.Solve

(a) NaOH dissociates in water to give one OH- ion per formula unit. Therefore, the OH - con-centration for the solution in (a) equals the stated concentration of NaOH, namely 0.028 M.

Method 1:

3H+4 =1.0 * 10-14

0.028= 3.57 * 10-13 M pH = - log13.57 * 10-132 = 12.45

Method 2:pOH = - log10.0282 = 1.55 pH = 14.00 - pOH = 12.45

(b) Ca1OH22 is a strong base that dissociates in water to give two OH - ions per formula unit. Thus, the concentration of OH -1aq2 for the solution in part (b) is 2 * 10.0011 M2 = 0.0022 M.

Method 1:

3H+4 =1.0 * 10-14

0.0022= 4.55 * 10-12 M pH = - log14.55 * 10-122 = 11.34

Method 2:pOH = - log10.00222 = 2.66 pH = 14.00 - pOH = 11.34

sAmpLE ExERcisE 16.9 calculating the ph of a strong Base


Although all of the alkali metal hydroxides are strong electrolytes, LiOH, RbOH, and CsOH are not commonly encountered in the laboratory. The hydroxides of the heavier alkaline earth metals—Ca1OH22, Sr1OH22, and Ba1OH22:are also strong electrolytes. They have limited solubility, however, so they are used only when high solubility is not critical.

Strongly basic solutions are also created by certain substances that react with water to form OH-1aq2. The most common of these contain the oxide ion. Ionic metal oxides, espe-cially Na2O and CaO, are often used in industry when a strong base is needed. The O2- reacts very exothermically with water to form OH-, leaving virtually no O2- in the solution:

O2-1aq2 + H2O1l2 ¡ 2 OH-1aq2 [16.22]

Thus, a solution formed by dissolving 0.010 mol of Na2O1s2 in enough water to form 1.0 L of solution has 3OH-4 = 0.020 M and a pH of 12.30.

give it some thoughtThe CH3

- ion is the conjugate base of CH4, and CH4 shows no evidence of being an acid in water. Write a balanced equation for the reaction of CH3

- and water.

16.6 | Weak AcidsMost acidic substances are weak acids and therefore only partially ionized in aqueous solution (▶ Figure 16.10). We can use the equilibrium constant for the ionization reac-tion to express the extent to which a weak acid ionizes. If we represent a general weak acid as HA, we can write the equation for its ionization in either of the following ways, depending on whether the hydrated proton is represented as H3O+1aq2 or H+1aq2:

HA1aq2 + H2O1l2 ∆ H3O+1aq2 + A- 1aq2 [16.23]or

HA1aq2 ∆ H+1aq2 + A- 1aq2 [16.24]

These equilibria are in aqueous solution, so we will use equilibrium-constant expres-sions based on concentrations. Because H2O is the solvent, it is omitted from the equilibrium-constant expression. (Section 15.4) Further, we add a subscript a on the equilibrium constant to indicate that it is an equilibrium constant for the ionization of an acid. Thus, we can write the equilibrium-constant expression as either:

Ka =3H3O+43A- 4

3HA4 or Ka =3H+43A- 43HA4 [16.25]

Ka is called the acid-dissociation constant for acid HA.▶ Table 16.2 shows the structural formulas, conjugate bases, and Ka values for a

number of weak acids. Appendix D provides a more complete list. Many weak acids are organic compounds composed entirely of carbon, hydrogen, and oxygen. These compounds usually contain some hydrogen atoms bonded to carbon atoms and some bonded to oxygen atoms. In almost all cases, the hydrogen atoms bonded to carbon do not ionize in water; instead, the acidic behavior of these compounds is due to the hydrogen atoms attached to oxygen atoms.

practice Exercise 1Order the following three solutions from smallest to largest pH:(i) 0.030 M Ba1OH22 (ii) 0.040 M KOH (iii) pure water(a) i 6 ii 6 iii (b) ii 6 i 6 iii (c) iii 6 i 6 ii (d) ii 6 iii 6 i (e) iii 6 ii 6 i

practice Exercise 2What is the concentration of a solution of (a) KOH for which the pH is 11.89, (b) Ca1OH22 for which the pH is 11.68?

secTiON 16.6 Weak Acids 687

The magnitude of Ka indicates the tendency of the acid to ionize in water: The larger the value of Ka, the stronger the acid. Chlorous acid 1HClO22, for example, is the strongest acid in Table 16.2, and phenol 1HOC6H52 is the weakest. For most weak acids Ka values range from 10-2 to 10-10.

give it some thoughtBased on the entries in Table 16.2, which element is most commonly bonded to the acidic hydrogen?

▲ Figure 16.10 species present in a solution of a strong acid and a weak acid.

Strong acidHA molecules

completely dissociate

HA(aq) + H2O(l) A−(aq) + H3O+(aq)

A−A−

H3O+H3O+

HA

Weak acidHA molecules

partially dissociate

HA(aq) + H2O(l) A−(aq) + H3O+(aq)

−

−

−

−

−−

++

+

++ +

+−

Table 16.2 some Weak Acids in Water at 25 °c

Acid structural Formula* conjugate Base Ka

Chlorous 1HClO22 ClO2 - 1.0 * 10-2

Hydrofluoric (HF) F- 6.8 * 10-4

Nitrous 1HNO22 NO2 - 4.5 * 10-4

Benzoic 1C6H5COOH2 C6H5COO- 6.3 * 10-5

Acetic 1CH3COOH2 CH3COO- 1.8 * 10-5

Hypochlorous (HOCl) OCl- 3.0 * 10-8

Hydrocyanic (HCN) CN- 4.9 * 10-10

Phenol 1HOC6H52 C6H5O- 1.3 * 10-10

*The proton that ionizes is shown in red.

H F

O ONH

OH

OH H

H

H

O

C

OH

O

C

OH Cl

O OH Cl

CH N

C


Calculating Ka from pHIn order to calculate either the Ka value for a weak acid or the pH of its solutions, we will use many of the skills for solving equilibrium problems developed in Section 15.5. In many cases the small magnitude of Ka allows us to use approximations to simplify the problem. In doing these calculations, it is important to realize that proton-transfer reactions are generally very rapid. As a result, the measured or calculated pH for a weak acid always represents an equilibrium condition.

A student prepared a 0.10 M solution of formic acid (HCOOH) and found its pH at 25 °C to be 2.38. Calculate Ka for formic acid at this temperature.

soLuTionAnalyze We are given the molar concentration of an aqueous solution of weak acid and the pH of the solution, and we are asked to determine the value of Ka for the acid.Plan Although we are dealing specifically with the ionization of a weak acid, this problem is very similar to the equilibrium problems we encountered in Chapter 15. We can solve this problem using the method first outlined in Sample Exercise 15.8, starting with the chemical reaction and a tabulation of initial and equilibrium concentrations.Solve The first step in solving any equilibrium problem is to write the equation for the equilib-rium reaction. The ionization of formic acid can be written as

HCOOH1aq2 ∆ H+1aq2 + HCOO-1aq2The equilibrium-constant expression is

Ka =3H+43HCOO-43HCOOH4

From the measured pH, we can calculate 3H+4: pH = - log 3H+4 = 2.38

log3H+4 = -2.38

3H+4 = 10-2.38 = 4.2 * 10-3 M

To determine the concentrations of the species involved in the equilibrium, we imagine that the solution is initially 0.10 M in HCOOH molecules. We then consider the ionization of the acid into H+ and HCOO-. For each HCOOH molecule that ionizes, one H+ ion and one HCOO- ion are produced in solution. Because the pH measurement indicates that 3H+4 = 4.2 * 10-3 M at equilibrium, we can construct the following table:

HCOOH1aq2 ∆ H+1aq2 + HCOO-1aq2Initial concentration (M) 0.10 0 0Change in concentration (M) -4.2 * 10-3 +4.2 * 10-3 +4.2 * 10-3

Equilibrium concentration (M) 10.10 - 4.2 * 10-32 4.2 * 10-3 4.2 * 10-3

Notice that we have neglected the very small concentration of H+1aq2 due to H2O autoioniza-tion. Notice also that the amount of HCOOH that ionizes is very small compared with the initial concentration of the acid. To the number of significant figures we are using, the subtraction yields 0.10 M:

10.10 - 4.2 * 10-32 M ≃ 0.10 MWe can now insert the equilibrium concentrations into the expression for Ka:

Ka =14.2 * 10-3214.2 * 10-32

0.10= 1.8 * 10-4

Check The magnitude of our answer is reasonable because Ka for a weak acid is usually between 10-2 and 10-10.

sAmpLE ExERcisE 16.10 calculating Ka from measured ph


Percent IonizationWe have seen that the magnitude of Ka indicates the strength of a weak acid. Another measure of acid strength is percent ionization, defined as

Percent ionization =concentration of ionized HAoriginal concentration of HA

* 100% [16.26]

The stronger the acid, the greater the percent ionization.If we assume that the autoionization of H2O is negligible, the concentration of acid

that ionizes equals the concentration of H+1aq2 that forms. Thus, the percent ioniza-tion for an acid HA can be expressed as

Percent ionization =3H+4equilibrium

3HA4initial* 100% [16.27]

For example, a 0.035 M solution of HNO2 contains 3.7 * 10-3 M H+1aq2 and its per-cent ionization is


3HNO24initial* 100% =

3.7 * 10-3 M0.035 M

* 100% = 11%

practice Exercise 1A 0.50 M solution of an acid HA has pH = 2.24. What is the value of Ka for the acid?(a) 1.7 * 10-12 (b) 3.3 * 10-5 (c) 6.6 * 10-5 (d) 5.8 * 10-3 (e) 1.2 * 10-2

practice Exercise 2Niacin, one of the B vitamins, has the molecular structure

C

O

O H

N

A 0.020 M solution of niacin has a pH of 3.26. What is the acid-dissociation constant for niacin?

As calculated in Sample Exercise 16.10, a 0.10 M solution of formic acid (HCOOH) contains 4.2 * 10-3 M H+1aq2. Calculate the percentage of the acid that is ionized.

soLuTionAnalyze We are given the molar concentration of an aqueous solution of weak acid and the equi-librium concentration of H+1aq2 and asked to determine the percent ionization of the acid.Plan The percent ionization is given by Equation 16.27.Solve


3HCOOH4initial* 100% =

4.2 * 10-3 M0.10 M

* 100% = 4.2

practice Exercise 1A 0.077 M solution of an acid HA has pH = 2.16. What is the percentage of the acid that is ionized?(a) 0.090% (b) 0.69% (c) 0.90% (d) 3.6% (e) 9.0%

practice Exercise 2A 0.020 M solution of niacin has a pH of 3.26. Calculate the percent ionization of the niacin.

sAmpLE ExERcisE 16.11 calculating percent ionization


Using Ka to Calculate pHKnowing the value of Ka and the initial concentration of a weak acid, we can calculate the concentration of H+1aq2 in a solution of the acid. Let’s calculate the pH at 25 °C of a 0.30 M solution of acetic acid 1CH3COOH2, the weak acid responsible for the charac-teristic odor and acidity of vinegar.

1. Our first step is to write the ionization equilibrium:

CH3COOH1aq2 ∆ H+1aq2 + CH3COO-1aq2 [16.28]

Notice that the hydrogen that ionizes is the one attached to an oxygen atom. 2. The second step is to write the equilibrium-constant expression and the value for

the equilibrium constant. Taking Ka = 1.8 * 10-5 from Table 16.2, we write

Ka =3H+43CH3COO-43CH3COOH4 = 1.8 * 10-5 [16.29]

3. The third step is to express the concentrations involved in the equilibrium reac-tion. This can be done with a little accounting, as described in Sample Exercise 16.10. Because we want to find the equilibrium value for 3H+4, let’s call this quan-tity x. The concentration of acetic acid before any of it ionizes is 0.30 M. The chem-ical equation tells us that for each molecule of CH3COOH that ionizes, one H+1aq2 and one CH3COO-1aq2 are formed. Consequently, if x moles per liter of H+1aq2 form at equilibrium, x moles per liter of CH3COO-1aq2 must also form and x moles per liter of CH3COOH must be ionized:

CH3COOH1aq2 ∆ H+1aq2 + CH3COO-1aq2Initial concentration (M)

0.30 0 0

Change in concentration (M)

-x +x +x

Equilibrium concentration (M)

10.30 - x2 x x

4. The fourth step is to substitute the equilibrium concentrations into the equilibrium-constant expression and solve for x:

Ka =3H+43CH3COO-43CH3COOH4 =

1x21x20.30 - x

= 1.8 * 10-5 [16.30]

This expression leads to a quadratic equation in x, which we can solve by using either an equation-solving calculator or the quadratic formula. We can simplify the problem, however, by noting that the value of Ka is quite small. As a result, we anticipate that the equilibrium lies far to the left and that x is much smaller than the initial concentration of acetic acid. Thus, we assume that x is negligible relative to 0.30, so that 0.30 - x is essentially equal to 0.30. We can (and should!) check the validity of this assumption when we finish the problem. By using this assumption, Equation 16.30 becomes

Ka =x2

0.30= 1.8 * 10-5

Solving for x, we have

x2 = 10.30211.8 * 10-52 = 5.4 * 10-6

x = 25.4 * 10-6 = 2.3 * 10-3

3H+4 = x = 2.3 * 10-3 M

pH = - log12.3 * 10-32 = 2.64


Now we check the validity of our simplifying assumption that 0.30 - x ≃ 0.30. The value of x we determined is so small that, for this number of significant figures, the assumption is entirely valid. We are thus satisfied that the assumption was a reasonable one to make. Because x represents the moles per liter of acetic acid that ionize, we see that, in this particular case, less than 1% of the acetic acid molecules ionize:

Percent ionization of CH3COOH =0.0023 M

0.30 M* 100% = 0.77%

As a general rule, if x is more than about 5% of the initial concentration value, it is better to use the quadratic formula. You should always check the validity of any simplifying assumptions after you have finished solving a problem.

We have also made one other assumption, namely that all of the H+ in the solution comes from ionization of CH3COOH. Are we justified in neglecting the autoionization of H2O? The answer is yes—the additional [H+4 due to water, which would be on the order of 10-7 M, is negligible compared to the [H+4 from the acid (which in this case is on the order of 10-3 M). In extremely precise work, or in cases involving very dilute solutions of acids, we would need to consider the autoionization of water more fully.

give it some thoughtWould a 1.0 * 10-8M solution of HCl have pH 6 7, pH = 7, or pH 7 7?

Finally, we can compare the pH value of this weak acid with the pH of a solution of a strong acid of the same concentration. The pH of the 0.30 M acetic acid is 2.64, but the pH of a 0.30 M solution of a strong acid such as HCl is - log10.302 = 0.52. As expected, the pH of a solution of a weak acid is higher than that of a solution of a strong acid of the same molarity. (Remember, the higher the pH value, the less acidic the solution.)

sAmpLE ExERcisE 16.12 using Ka to calculate ph

Calculate the pH of a 0.20 M solution of HCN. (Refer to Table 16.2 or Appendix D for the value of Ka.)

soLuTionAnalyze We are given the molarity of a weak acid and are asked for the pH. From Table 16.2, Ka for HCN is 4.9 * 10-10.

Plan We proceed as in the example just worked in the text, writing the chemical equation and constructing a table of initial and equilibrium concentrations in which the equilibrium concentration of H+ is our unknown.

HCN1aq2 ∆ H+1aq2 + CN-1aq2Initial concentration (M)

0.20 0 0


-x +x +x


10.20 - x2 x x

Solve Writing both the chemical equation for the ionization reaction that forms H+1aq2 and the equilibrium-constant 1Ka2 expression for the reaction:

HCN1aq2 ∆ H +1aq2 + CN-1aq2

Ka =3H+43CN-43HCN4 = 4.9 * 10-10

Next, we tabulate the concentrations of the species involved in the equilibrium reaction, letting x = 3H+4 at equilibrium:

Substituting the equilibrium concentrations into the equilibrium-constant expression yields

Ka =1x21x2

0.20 - x= 4.9 * 10-10

We next make the simplifying approximation that x, the amount of acid that dissociates, is small compared with the initial con-centration of acid, 0.20 - x ≃ 0.20. Thus,

x2

0.20= 4.9 * 10-10


The properties of an acid solution that relate directly to the concentration of H+1aq2, such as electrical conductivity and rate of reaction with an active metal, are less evident for a solution of a weak acid than for a solution of a strong acid of the same concentration. ▼ Figure 16.11 presents an experiment that demonstrates this difference with 1 M CH3COOH and 1 M HCl. The concentration of H+1aq2 in 1 M CH3COOH is only 0.004 M, whereas the 1 M HCl solution contains 1 M H+1aq2. As a result, the reac-tion rate with the metal is much faster in the HCl solution.

As the concentration of a weak acid increases, the equilibrium concentration of H+1aq2 increases, as expected. However, as shown in ▶ Figure 16.12, the percent ioniza-tion decreases as the concentration increases. Thus, the concentration of H+1aq2 is not directly proportional to the concentration of the weak acid. For example, doubling the concentration of a weak acid does not double the concentration of H+1aq2.

practice Exercise 1What is the pH of a 0.40 M solution of benzoic acid, C6H5COOH? (The Ka value for benzoic acid is given in Table 16.2.)(a) 2.30 (b) 2.10 (c) 1.90 (d) 4.20 (e) 4.60

practice Exercise 2The Ka for niacin (Practice Exercise 16.10) is 1.5 * 10-5. What is the pH of a 0.010 M solution of niacin?

▲ Figure 16.11 rates of the same reaction run in a weak acid and a strong acid. The bubbles are H2 gas, which along with metal cations, is produced when a metal is oxidized by an acid.

(Section 4.4)

Reaction proceeds more rapidly in strong acid, leading to formation of larger H2 bubbles and rapid disappearance of metal

Reaction eventually goes to completion in both acids

Reaction complete in strong acid

H2 bubbles show reaction still in progress in weak acid

1 M HCl(aq)[H+] = 1 M

1 M CH3COOH(aq)[H+] = 0.004 M

Solving for x, we have x2 = 10.20214.9 * 10-102 = 0.98 * 10-10

x = 20.98 * 10-10 = 9.9 * 10-6 M = 3H+4A concentration of 9.9 * 10-6 M is much smaller than 5% of 0.20, the initial HCN concentration. Our simplifying approxima-tion is therefore appropriate. We now calculate the pH of the solution: pH = - log3H+4 = - log19.9 * 10-62 = 5.00


0 0.05 0.10 0.15

1.0

2.0

3.0

4.0

5.0

6.0

Acid concentration (M)

Perc

ent i

oniz

ed

As concentration increases, a smaller percentage of CH3COOH molecules dissociates.

▲ Figure 16.12 effect of concentration on percent ionization in an acetic acid solution.

Go FiGuREIs the trend observed in this graph consistent with Le Châtelier’s principle? Explain.

sAmpLE ExERcisE 16.13 using the Quadratic Equation to calculate ph and percent ionization

Calculate the pH and percentage of HF molecules ionized in a 0.10 M HF solution.

soLuTionAnalyze We are asked to calculate the percent ionization of a solution of HF. From Appendix D, we find Ka = 6.8 * 10-4.

Plan We approach this problem as for previous equilibrium problems: We write the chemical equation for the equilibrium and tabulate the known and unknown concentrations of all species. We then substitute the equilibrium concentrations into the equilibrium-constant expres-sion and solve for the unknown concentration of H+.

Ka =3H+43F-43HF4 =

1x21x20.10 - x

= 6.8 * 10-4

Solve The equilibrium reaction and equilibrium concentrations are as follows:

HF1aq2 ∆ H +1aq2 + F -1aq2Initial concentration (M)

0.10 0 0


-x +x +x


10.10 - x2 x x

The equilibrium-constant expression is

When we try solving this equation using the approximation 0.10 - x ≃ 0.10 (that is, by neglecting the concentration of acid that ionizes), we obtain x = 8.2 * 10-3 M

Because this approximation is greater than 5% of 0.10 M, how-ever, we should work the problem in standard quadratic form. Rearranging, we have

x2 = 10.10 - x216.8 * 10-42 = 6.8 * 10-5 - 16.8 * 10-42x

x2 + 16.8 * 10-42x - 6.8 * 10-5 = 0

Substituting these values in the standard quadratic formula givesx =

-6.8 * 10-4 { 216.8 * 10-422 - 41-6.8 * 10-522

=-6.8 * 10-4 { 1.6 * 10-2

2


Polyprotic AcidsAcids that have more than one ionizable H atom are known as polyprotic acids. Sulfu-rous acid 1H2SO32, for example, can undergo two successive ionizations:

H2SO31aq2 ∆ H+1aq2 + HSO3-1aq2 Ka1 = 1.7 * 10-2 [16.31]

HSO3-1aq2 ∆ H+1aq2 + SO3

2-1aq2 Ka2 = 6.4 * 10-8 [16.32]

Note that the acid-dissociation constants are labeled Ka1 and Ka2. The numbers on the constants refer to the particular proton of the acid that is ionizing. Thus, Ka2 always refers to the equilibrium involving removal of the second proton of a polyprotic acid.

We see that Ka2 for sulfurous acid is much smaller than Ka1. Because of electro-static attractions, we would expect a positively charged proton to be lost more readily from the neutral H2SO3 molecule than from the negatively charged HSO3

- ion. This observation is general: It is always easier to remove the first proton from a polyprotic acid than to remove the second. Similarly, for an acid with three ionizable protons, it is easier to remove the second proton than the third. Thus, the Ka values become succes-sively smaller as successive protons are removed.

give it some thoughtWhat is the equilibrium associated with Ka3 for H3PO4?

The acid-dissociation constants for common polyprotic acids are listed in ▼ Table 16.3, and Appendix D provides a more complete list. The structure of citric acid illustrates the presence of multiple ionizable protons ◀ Figure 16.13.

Of the two solutions, only the positive value for x is chemically rea-sonable. From that value, we can determine 3H+4 and hence the pH x = 3H+4 = 3F-4 = 7.9 * 10-3 M, so pH = - log3H+4 = 2.10

From our result, we can calculate the percent of molecules ionized: Percent ionization of HF =concentration ionizedoriginal concentration

* 100%

=7.9 * 10-3 M

0.10 M* 100% = 7.9%

practice Exercise 1What is the pH of a 0.010 M solution of HF?(a) 1.58 (b) 2.10 (c) 2.30 (d) 2.58 (e) 2.64

practice Exercise 2In Practice Exercise 2 for Sample Exercise 16.11, we found that the percent ionization of niacin 1Ka = 1.5 * 10-52 in a 0.020 M solution is 2.7%. Calculate the percentage of niacin molecules ionized in a solution that is (a) 0.010 M, (b) 1.0 * 10-3 M.

Table 16.3 Acid-dissociation constants of some common polyprotic Acids

name Formula Ka1 Ka2 Ka3

Ascorbic H2C6H6O6 8.0 * 10-5 1.6 * 10-12

Carbonic H2CO3 4.3 * 10-7 5.6 * 10-11

Citric H3C6H5O7 7.4 * 10-4 1.7 * 10-5 4.0 * 10-7

Oxalic HOOC ¬ COOH 5.9 * 10-2 6.4 * 10-5

Phosphoric H3PO4 7.5 * 10-3 6.2 * 10-8 4.2 * 10-13

Sulfurous H2SO3 1.7 * 10-2 6.4 * 10-8

Sulfuric H2SO4 Large 1.2 * 10-2

Tartaric C2H2O21COOH22 1.0 * 10-3 4.6 * 10-5

H

O

C O

H

O

O

C O

HC O

C

H2C

H2C

HO

Citric acid

▲ Figure 16.13 the structure of the polyprotic acid, citric acid.

Go FiGuRECitric acid has four hydrogen atoms bonded to oxygen. How does the hydrogen atom that is not an acidic proton differ from the other three?


Notice in Table 16.3 that in most cases the Ka values for successive losses of protons differ by a factor of at least 103. Notice also that the value of Ka1 for sulfuric acid is listed simply as “large.” Sulfuric acid is a strong acid with respect to the removal of the first proton. Thus, the reaction for the first ionization step lies completely to the right:

H2SO41aq2 ¡ H+1aq2 + HSO4-1aq2 1complete ionization2

However, HSO4- is a weak acid for which Ka2 = 1.2 * 10-2.

For many polyprotic acids Ka1 is much larger than subsequent dissociation constants, in which case the H+1aq2 in the solution comes almost entirely from the first ionization reaction. As long as successive Ka values differ by a factor of 103 or more, it is usually possible to obtain a satisfactory estimate of the pH of polyprotic acid solutions by treating the acids as if they were monoprotic, consid-ering only Ka1.

sAmpLE ExERcisE 16.14 calculating the ph of a solution of a polyprotic Acid

The solubility of CO2 in water at 25 °C and 0.1 atm is 0.0037 M. The common practice is to assume that all the dissolved CO2 is in the form of carbonic acid 1H2CO32, which is produced in the reaction

CO21aq2 + H2O1l2 ∆ H2CO31aq2What is the pH of a 0.0037 M solution of H2CO3?

soLuTionAnalyze We are asked to determine the pH of a 0.0037 M solution of a polyprotic acid.

Plan H2CO3 is a diprotic acid; the two acid-dissociation constants, Ka1 and Ka2 (Table 16.3), differ by more than a factor of 103. Conse-quently, the pH can be determined by considering only Ka1, thereby treating the acid as if it were a monoprotic acid.

Ka1 =3H +43HCO3

-43H2CO34 =

1x21x20.0037 - x

= 4.3 * 10-7

Solve Proceeding as in Sample Exercises 16.12 and 16.13, we can write the equilib-rium reaction and equilibrium concentra-tions as

H2CO31aq2 ∆ H +1aq2 + HCO3-1aq2

Initial concentration (M)

0.0037 0 0


-x +x +x


10.0037 - x2 x x

The equilibrium-constant expression is

Solving this quadratic equation, we get x = 4.0 * 10-5 M

Alternatively, because Ka1 is small, we can make the simplifying approximation that x is small, so that 0.0037 - x ≃ 0.0037

Thus, 1x21x20.0037

= 4.3 * 10-7

Solving for x, we have x2 = 10.0037214.3 * 10-72 = 1.6 * 10-9

x = 3H +4 = 3HCO3-4 = 21.6 * 10-9 = 4.0 * 10-5M

Because we get the same value (to 2 signifi-cant figures) our simplifying assumption was justified. The pH is therefore pH = - log3H+4 = - log14.0 * 10-52 = 4.40


16.7 | Weak BasesMany substances behave as weak bases in water. Weak bases react with water, ab-stracting protons from H2O, thereby forming the conjugate acid of the base and OH- ions:

B1aq2 + H2O1l2 ∆ HB+1aq2 + OH-1aq2 [16.33]

The equilibrium-constant expression for this reaction can be written as

Kb =3BH+43OH-4

3B4 [16.34]

Water is the solvent, so it is omitted from the equilibrium-constant expression. One of the most commonly encountered weak bases is ammonia, NH3:

NH31aq2 + H2O1l2 ∆ NH4+1aq2 + OH-1aq2 Kb =

3NH4+43OH-4

3NH34 [16.35]

As with Kw and Ka, the subscript b in Kb denotes that the equilibrium constant refers to a particular type of reaction, namely the ionization of a weak base in water. The constant Kb, the base-dissociation constant, always refers to the equilibrium in which a base reacts with H2O to form the corresponding conjugate acid and OH-.

▶ Table 16.4 lists the Lewis structures, conjugate acids, and Kb values for a number of weak bases in water. Appendix D includes a more extensive list. These bases contain one or more lone pairs of electrons because a lone pair is necessary to form the bond with H+. Notice that in the neutral molecules in Table 16.4, the lone pairs are on nitro-gen atoms. The other bases listed are anions derived from weak acids.

Assuming that y is small relative to 4.0 * 10-5, we have Ka2 =

3H+43CO32-4

3HCO3-4 =

14.0 * 10-521y24.0 * 10-5 = 5.6 * 10-11

y = 5.6 * 10-11 M = 3CO32-4

We see that the value for y is indeed very small compared with 4.0 * 10-5, showing that our assumption was justified. It also shows that the ionization of HCO3

- is negligible relative to that of H2CO3, as far as production of H+ is concerned. However, it is the only source of CO3

2-, which has a very low concentration in the solution.

Our calculations thus tell us that in a solution of carbon dioxide in water, most of the CO2 is in the form of CO2 or H2CO3, only a small fraction ionizes to form H+ and HCO3

-, and an even smaller frac-tion ionizes to give CO3

2-. Notice also that 3CO32-4 is numerically

equal to Ka2.

practice Exercise 1What is the pH of a 0.28 M solution of ascorbic acid (Vitamin C)? (See Table 16.3 for Ka1 and Ka2.)(a) 2.04 (b) 2.32 (c) 2.82 (d) 4.65 (e) 6.17

practice Exercise 2(a) Calculate the pH of a 0.020 M solution of oxalic acid 1H2C2O42. (See Table 16.3 for Ka1 and Ka2.) (b) Calculate the concentration of oxalate ion, 3C2O4

2-4, in this solution.

Comment If we were asked for 3CO32-4

we would need to use Ka2. Let’s illustrate that calculation. Using our calculated values of 3HCO3

-4 and 3H+4 and setting 3CO3

2-4 = y, we have

HCO3-1aq2 ∆ H +1aq2 + CO3

2-1aq2


4.0 * 10-5 4.0 * 10-5 0


-y +y +y


14.0 * 10-5 - y2 14.0 * 10-5 + y2 y

secTiON 16.7 Weak Bases 697

Table 16.4 some Weak Bases in Water at 25 °c

Base structural Formula* conjugate Acid Kb

Ammonia 1NH32 NH4+ 1.8 * 10-5

Pyridine 1C5H5N2 C5H5NH+ 1.7 * 10-9

Hydroxylamine 1HONH22 HONH3+ 1.1 * 10-8

Methylamine 1CH3NH22 CH3NH3+ 4.4 * 10-4

Hydrosulfide ion 1HS-2 H2S 1.8 * 10-7

Carbonate ion 1CO32-2 HCO3

- 1.8 * 10-4

Hypochlorite ion 1ClO-2 HClO 3.3 * 10-7

*The atom that accepts the proton is shown in blue.

2O

O

O O

N

C

S

OH

CH3

H

Cl

H

H

N

H

H

N

−

−

−

H H

H

N

sAmpLE ExERcisE 16.15 using Kb to calculate oh-

Calculate the concentration of OH- in a 0.15 M solution of NH3.

soLuTionAnalyze We are given the concentration of a weak base and asked to determine the concentration of OH-.

Plan We will use essentially the same procedure here as used in solving problems involving the ionization of weak acids—that is, write the chemical equation and tabulate initial and equilibrium concentrations.

NH31aq2 + H2O1l2 ∆ NH4+1aq2 + OH -


0.15 — 0 0


-x — +x +x


10.15 - x2 — x x

Solve The ionization reaction and equilibrium-constant expression are

NH31aq2 + H2O1l2 ∆ NH4+1aq2 + OH -1aq2

Kb =3NH4

+43OH-43NH34

= 1.8 * 10-5

Ignoring the concentration of H2O because it is not involved in the equilibrium-constant expression, the equilibrium concentrations are

Inserting these quantities into the equilibrium-constant expression gives Kb =

3NH4 +43OH-43NH34 =

1x21x20.15 - x

= 1.8 * 10-5

Because Kb is small, the amount of NH3 that reacts with water is much smaller than the NH3 concentration, and so we can neglect x relative to 0.15 M. Then we have

x2

0.15= 1.8 * 10-5

x2 = 10.15211.8 * 10-52 = 2.7 * 10-6

x = 3NH4 +4 = 3OH-4 = 22.7 * 10-6 = 1.6 * 10-3 M


Types of Weak BasesWeak bases fall into two general categories. The first category is neutral substances that have an atom with a nonbonding pair of electrons that can accept a proton. Most of these bases, including all uncharged bases in Table 16.4, contain a nitrogen atom. These sub-stances include ammonia and a related class of compounds called amines (◀ Figure 16.14). In organic amines, at least one N ¬ H bond in NH3 is replaced with an N ¬ C bond. Like NH3, amines can abstract a proton from a water molecule by forming an N ¬ H bond, as shown here for methylamine:

N

H

H

CH3H2O(l) OH (aq)(aq)CH3(aq)NH

H

H

+

−++ [16.36]

Anions of weak acids make up the second general category of weak bases. In an aqueous solution of sodium hypochlorite (NaClO), for example, NaClO dissociates to Na+ and ClO- ions. The Na+ ion is always a spectator ion in acid–base reactions.

(Section 4.3) The ClO- ion, however, is the conjugate base of a weak acid, hypo-chlorous acid. Consequently, the ClO- ion acts as a weak base in water:

ClO-1aq2 + H2O1l2 ∆ HClO1aq2 + OH-1aq2 Kb = 3.3 * 10-7 [16.37]

In Figure 16.6 we saw that bleach is quite basic (pH values of 12–13). Common chlorine bleach is typically a 5% NaOCl solution.

Check The value obtained for x is only about 1% of the NH3 concentration, 0.15 M. Therefore, neglecting x relative to 0.15 was justified.

Comment You may be asked to find the pH of a solution of a weak base. Once you have found 3OH-4, you can proceed as in Sample Exercise 16.9, where we calculated the pH of a strong base. In the present sample exercise, we have seen that the 0.15 M solution of NH3 contains 3OH-4 = 1.6 * 10-3 M. Thus, pOH = - log11.6 * 10-32 = 2.80, and pH = 14.00 - 2.80 = 11.20. The pH of the solution is above 7 because we are dealing with a solution of a base.

practice Exercise 1What is the pH of a 0.65 M solution of pyridine, C5H5N? (See Table 16.4 for Kb.)(a) 4.48 (b) 8.96 (c) 9.52 (d) 9.62 (e) 9.71

practice Exercise 2Which of the following compounds should produce the highest pH as a 0.05 M solution: pyridine, methylamine, or nitrous acid?

▲ Figure 16.14 structures of ammonia and two simple amines.

Go FiGuREWhen hydroxylamine acts as a base, which atom accepts the proton?

AmmoniaNH3

MethylamineCH3NH2

HydroxylamineNH2OH

sAmpLE ExERcisE 16.16 using ph to determine the concentration of a salt

A solution made by adding solid sodium hypochlorite (NaClO) to enough water to make 2.00 L of solution has a pH of 10.50. Using the information in Equation 16.37, calculate the number of moles of NaClO added to the water.

soLuTionAnalyze NaClO is an ionic compound consisting of Na+ and ClO- ion. As such, it is a strong electrolyte that completely dissociates in solution into Na+, a spectator ion, and ClO- ion, a weak base with Kb = 3.3 * 10-7 (Equation 16.37). Given this information we must calculate the number of moles of NaClO needed to increase the pH of 2.00-L of water to 10.50.

Plan From the pH, we can determine the equilibrium concentration of OH-. We can then construct a table of initial and equilibrium concentrations in which the initial concentration of ClO- is our unknown. We can calculate 3ClO-4 using the expression for Kb.

secTiON 16.8 relationship Between Ka and Kb 699

16.8 | Relationship Between Ka and KbWe have seen in a qualitative way that the stronger an acid, the weaker its conjugate base. To see if we can find a corresponding quantitative relationship, let’s consider the NH4

+ and NH3 conjugate acid–base pair. Each species reacts with water. For the acid, NH4

+, the equilibrium is

NH4+1aq2 + H2O1l2 ∆ NH31aq2 + H3O+1aq2

or written in its simpler form:

NH4+1aq2 ∆ NH31aq2 + H+1aq2 [16.38]

For the base, NH3, the equilibrium is

NH31aq2 + H2O1l2 ∆ NH4+1aq2 + OH-1aq2 [16.39]

Each equilibrium is expressed by a dissociation constant:

Ka =3NH343H+43NH4

+4 Kb =3NH4

+43OH-43NH34

When we add Equations 16.38 and 16.39, the NH4+ and NH3 species cancel and we

are left with the autoionization of water:

NH4+1aq2 ∆ NH31aq2 + H+1aq2

NH31aq2 + H2O1l2 ∆ NH4+1aq2 + OH-1aq2

H2O1l2 ∆ H+1aq2 + OH-1aq2Recall that when two equations are added to give a third, the equilibrium constant asso-ciated with the third equation equals the product of the equilibrium constants of the first two equations. (Section 15.3)

ClO-1aq2 + H2O1l2 ∆ HClO1aq2 + OH -1aq2Initial concentration (M)

x — 0 0


-3.2 * 10-4 — +3.2 * 10-4 +3.2 * 10-4


1x - 3.2 * 10-42 — 3.2 * 10-4 3.2 * 10-4

Solve We can calculate 3OH-4 by using either Equation 16.16 or Equation 16.20; we will use the latter method here:

pOH = 14.00 - pH = 14.00 - 10.50 = 3.50 3OH-4 = 10-3.50 = 3.2 * 10-4 M

This concentration is high enough that we can assume that Equation 16.37 is the only source of OH-; that is, we can neglect any OH- produced by the autoionization of H2O. We now assume a value of x for the initial concentration of ClO- and solve the equilib-rium problem in the usual way.

We now use the expression for the base- dissociation constant to solve for x: Kb =

3HClO43OH-43ClO-4 =

13.2 * 10-422

x - 3.2 * 10-4 = 3.3 * 10-7

x =13.2 * 10-422

3.3 * 10-7 + 13.2 * 10-42 = 0.31 M

We say that the solution is 0.31 M in NaClO even though some of the ClO- ions have reacted with water. Because the solution is 0.31 M in NaClO and the total volume of solution is 2.00 L, 0.62 mol of NaClO is the amount of the salt that was added to the water.

practice Exercise 1The benzoate ion, C6H5COO-, is a weak base with Kb = 1.6 * 10-10. How many moles of sodium benzoate are present in 0.50 L of a solution of NaC6H5COO if the pH is 9.04?(a) 0.38 (b) 0.66 (c) 0.76 (d) 1.5 (e) 2.9

practice Exercise 2What is the molarity of an aqueous NH3 solution that has a pH of 11.17?


Applying this rule to our present example, we see that when we multiply Ka and Kb, we obtain

Ka * Kb = a 3NH343H+43NH4 +4 b a 3NH4

+43OH-43NH34 b

= 3H+43OH-4 = Kw

Thus, the product of Ka and Kb is the ion-product constant for water, Kw (Equa-tion 16.16). We expect this result because adding Equations 16.38 and 16.39 gave us the autoionization equilibrium for water, for which the equilibrium constant is Kw.

The above result holds for any conjugate acid–base pair. In general, the product of the acid-dissociation constant for an acid and the base-dissociation constant for its con-jugate base equals the ion-product constant for water:

Ka * Kb = Kw 1for a conjugate acid9base pair2 [16.40]

As the strength of an acid increases (Ka gets larger), the strength of its conjugate base must decrease (Kb gets smaller) so that the product Ka * Kb remains 1.0 * 10-14 at 25 °C. ▼ Table 16.5 demonstrates this relationship. Remember, this important rela-tionship applies only to conjugate acid–base pairs.

By using Equation 16.40, we can calculate Kb for any weak base if we know Ka for its conjugate acid. Similarly, we can calculate Ka for a weak acid if we know Kb for its conjugate base. As a practical consequence, ionization constants are often listed for only one member of a conjugate acid–base pair. For example, Appendix D does not contain Kb values for the anions of weak acids because they can be readily calculated from the tabulated Ka values for their conjugate acids.

Recall that we often express 3H+4 as pH: pH = - log 3H+4. (Section 16.4) This “p” nomenclature is often used for other very small numbers. For example, if you look up the values for acid- or base-dissociation constants in a chemistry handbook, you may find them expressed as pKa or pKb:

pKa = - log Ka and pKb = - log Kb [16.41]

Using this nomenclature, Equation 16.40 can be written in terms of pKa and pKb by taking the negative logarithm of both sides:

pKa + pKb = pKw = 14.00 at 25 °C 1conjugate acid9base pair2 [16.42]

give it some thoughtKa for acetic acid is 1.8 * 10-5. What is the first digit of the pKa value for acetic acid?

Table 16.5 some conjugate Acid–Base pairs

Acid Ka Base Kb

HNO3 (Strong acid) NO3- (Negligible basicity)

HF 6.8 * 10-4 F - 1.5 * 10-11

CH3COOH 1.8 * 10-5 CH3COO- 5.6 * 10-10

H2CO3 4.3 * 10-7 HCO3- 2.3 * 10-8

NH4+ 5.6 * 10-10 NH3 1.8 * 10-5

HCO3- 5.6 * 10-11 CO3

2- 1.8 * 10-4

OH- (Negligible acidity) O2 - (Strong base)

secTiON 16.8 relationship Between Ka and Kb 701

chemistry put to Work

amines and amine Hydrochlorides

Many low-molecular-weight amines have a fishy odor. Amines and NH3 are produced by the anaerobic (absence of O2) decomposition of dead animal or plant matter. Two such amines with very disagreeable aromas are H2N1CH224NH2, putrescine, and H2N1CH225NH2, cadav-erine. The names of these substances are testaments to their repugnant odors!

Many drugs, including quinine, codeine, caffeine, and amphet-amine, are amines. Like other amines, these substances are weak bases; the amine nitrogen is readily protonated upon treatment with an acid. The resulting products are called acid salts. If we use A as the abbreviation for an amine, the acid salt formed by reaction with hydrochloric acid can be written AH+Cl-. It can also be written as A # HCl and referred to as a hydrochloride. Amphetamine hydro-chloride, for example, is the acid salt formed by treating amphet-amine with HCl:

CH2 CH NH2(aq) HCl(aq)

CH3

CH2 CH

CH3

NH3 Cl (aq)

Amphetamine

Amphetamine hydrochloride

+

+ −

Acid salts are much less volatile, more stable, and generally more water soluble than the corresponding amines. For this reason, many drugs that are amines are sold and administered as acid salts. Some

examples of over-the-counter medications that contain amine hydro-chlorides as active ingredients are shown in ▼ Figure 16.15.

Related Exercises: 16.9, 16.73, 16.74, 16.101, 16.114, 16.124

▲ Figure 16.15 some over-the-counter medications in which an amine hydrochloride is a major active ingredient.

sAmpLE ExERcisE 16.17 calculating Ka or Kb for a conjugate

Acid–Base pairCalculate (a) Kb for the fluoride ion, (b) Ka for the ammonium ion.

soLuTionAnalyze We are asked to determine dissociation constants for F-, the conjugate base of HF, and NH4

+, the conjugate acid of NH3.Plan We can use the tabulated K values for HF and NH3 and the rela-tionship between Ka and Kb to calculate the dissociation constants for their conjugates, F- and NH4

+.Solve

(a) For the weak acid HF, Table 16.2 and Appendix D give Ka = 6.8 * 10-4. We can use Equation 16.40 to calculate Kb for the conjugate base, F-:

Kb =Kw

Ka=

1.0 * 10-14

6.8 * 10-4 = 1.5 * 10-11

(b) For NH3, Table 16.4 and in Appendix D give Kb = 1.8 * 10-5, and this value in Equation 16.40 gives us Ka for the conjugate acid, NH4

+:

Ka =Kw

Kb=

1.0 * 10-14

1.8 * 10-5 = 5.6 * 10-10

Check The respective K values for F- and NH4+ are listed in Table 16.5,

where we see that the values calculated here agree with those in Table 16.5.

practice Exercise 1By using information from Appendix D, put the following three substances in order of weakest to strongest base: (i) 1CH323N, (ii) HCOO-, (iii) BrO-.(a) i 6 ii 6 iii (b) ii 6 i 6 iii (c) iii 6 i 6 ii (d) ii 6 iii 6 i (e) iii 6 ii 6 i.

practice Exercise 2(a) Based on information in Appendix D, which of these anions has the largest base-dissociation constant: NO2

-, PO43 - , or N3

-? (b) The base quinoline has the structure

N

Its conjugate acid is listed in handbooks as having a pKa of 4.90. What is the base-dissociation constant for quinoline?


16.9 | Acid–Base properties of salt solutions

Even before you began this chapter, you were undoubtedly aware of many substances that are acidic, such as HNO3, HCl, and H2SO4, and others that are basic, such as NaOH and NH3. However, our discussion up to this point in the chapter has indicated that ions can also exhibit acidic or basic properties. For example, we calculated Ka for NH4

+ and Kb for F- in Sample Exercise 16.17. Such behavior implies that salt solutions can be acidic or basic. Before proceeding with further discussions of acids and bases, let’s examine the way dissolved salts can affect pH.

Because nearly all salts are strong electrolytes, we can assume that any salt dis-solved in water is completely dissociated. Consequently, the acid–base properties of salt solutions are due to the behavior of the cations and anions. Many ions react with water to generate H+1aq2 or OH-1aq2 ions. This type of reaction is often called hydrolysis. The pH of an aqueous salt solution can be predicted qualitatively by considering the salt’s cations and anions.

An Anion’s Ability to React with WaterIn general, an anion A- in solution can be considered the conjugate base of an acid. For example, Cl- is the conjugate base of HCl, and CH3COO- is the conjugate base of CH3COOH. Whether an anion reacts with water to produce hydroxide ions depends on the strength of the anion’s conjugate acid. To identify the acid and assess its strength, we add a proton to the anion’s formula. If the acid HA determined in this way is one of the seven strong acids listed at the beginning of Section 16.5, the anion has a negligible tendency to produce OH- ions from water and does not affect the pH of the solution. The presence of Cl- in an aqueous solution, for example, does not result in the pro-duction of any OH- and does not affect the pH. Thus, Cl- is always a spectator ion in acid–base chemistry.

If HA is not one of the seven common strong acids, it is a weak acid. In this case, the conjugate base A- is a weak base and it reacts to a small extent with water to pro-duce the weak acid and hydroxide ions: A-1aq2 + H2O1l2 ∆ HA1aq2 + OH-1aq2 [16.43]

The OH- ion generated in this way increases the pH of the solution, making it basic. Acetate ion, for example, being the conjugate base of a weak acid, reacts with water to produce acetic acid and hydroxide ions, thereby increasing the pH of the solution: CH3COO-1aq2 + H2O1l2 ∆ CH3COOH1aq2 + OH-1aq2 [16.44]

give it some thoughtWill NO3

- ions affect the pH of a solution? What about CO32- ions?

The situation is more complicated for salts containing anions that have ioniz-able protons, such as HSO3

-. These salts are amphiprotic (Section 16.2), and how they behave in water is determined by the relative magnitudes of Ka and Kb for the ion, as shown in Sample Exercise 16.19. If Ka 7 Kb, the ion causes the solution to be acidic. If Kb 7 Ka, the solution is made basic by the ion.

A Cation’s Ability to React with WaterPolyatomic cations containing one or more protons can be considered the conjugate acids of weak bases. The NH4

+ ion, for example, is the conjugate acid of the weak base NH3. Thus, NH4

+ is a weak acid and will donate a proton to water, producing hydro-nium ions and thereby lowering the pH:

NH4+1aq2 + H2O1l2 ∆ NH31aq2 + H3O+1aq2 [16.45]

Many metal ions react with water to decrease the pH of an aqueous solution. This effect is most pronounced for small, highly charged cations like Fe3+ and Al3+ , as illustrated

secTiON 16.9 Acid–Base properties of salt solutions 703

by the Ka values for metal cations in ▶ Table 16.6. A comparison of Fe2+ and Fe3+ values in the table illustrates how acidity increases as ionic charge increases.

Notice that Ka values for the 3+ ions in Table 16.6 are comparable to the values for familiar weak acids, such as acetic acid 1Ka = 1.8 * 10-52. In contrast, the ions of alkali and alkaline earth metals, being relatively large and not highly charged, do not react with water and therefore do not affect pH. Note that these are the same cations found in the strong bases (Section 16.5). The different tendencies of four cations to lower the pH of a solution are illustrated in ▼ Figure 16.16.

The mechanism by which metal ions produce acidic solutions is shown in ▼ Figure 16.17. Because metal ions are positively charged, they attract the unshared

Table 16.6 Acid-dissociation constants for metal cations in Aqueous solution at 25 °c

cation Ka

Fe2+ 3.2 * 10-10

Zn2+ 2.5 * 10-10

Ni2+ 2.5 * 10-11

Fe3+ 6.3 * 10-3

Cr3+ 1.6 * 10-4

Al3+ 1.4 * 10-5

Interaction between Fe3+ and oxygen of bound H2O molecule weakens O—H bonds

H+ lost, charge of complex ion changes from 3+ to 2+

H3O+ created,

solution becomes acidic

[Fe(H2O)6]3+(aq) + +

+ +

H2O(l) [Fe(H2O)5(OH)]2+(aq) H3O+(aq)

3+ 2++

▲ Figure 16.17 a hydrated Fe3+ ion acts as an acid by donating an H+ to a free H2O molecule, forming H3O+.

▲ Figure 16.16 effect of cations on solution pH. The pH values of 1.0 M solutions of four nitrate salts are estimated using acid–base indicators.

NaNO3Bromothymol bluepH = 7.0

Ca(NO3)2Bromothymol bluepH = 6.9

Zn(NO3)2Methyl redpH = 5.5

Al(NO3)3Methyl orangepH = 3.5

Go FiGuREWhy do we need to use two different acid-base indicators in this figure?


electron pairs of water molecules and become hydrated. (Section 13.1) The larger the charge on the metal ion, the stronger the interaction between the ion and the oxy-gen of its hydrating water molecules. As the strength of this interaction increases, the O ¬ H bonds in the hydrating water molecules become weaker. This facilitates transfer of protons from the hydration water molecules to solvent water molecules.

Combined Effect of Cation and Anion in SolutionTo determine whether a salt forms an acidic, a basic, or a neutral solution when dis-solved in water, we must consider the action of both cation and anion. There are four possible combinations.

1. If the salt contains an anion that does not react with water and a cation that does not react with water, we expect the pH to be neutral. Such is the case when the anion is a conjugate base of a strong acid and the cation is either from group 1A or one of the heavier members of group 2A 1Ca2+, Sr2+, and Ba2+2. Examples: NaCl, Ba1NO322, RbClO4.

2. If the salt contains an anion that reacts with water to produce hydroxide ions and a cation that does not react with water, we expect the pH to be basic. Such is the case when the anion is the conjugate base of a weak acid and the cation is either from group 1A or one of the heavier members of group 2A 1Ca2+, Sr2+, and Ba2+2. Examples: NaClO, RbF, BaSO3.

3. If the salt contains a cation that reacts with water to produce hydronium ions and an anion that does not react with water, we expect the pH to be acidic. Such is the case when the cation is a conjugate acid of a weak base or a small cation with a charge of 2+ or greater. Examples: NH4NO3, AlCl3, Fe1NO323.

4. If the salt contains an anion and a cation both capable of reacting with water, both hydroxide ions and hydronium ions are produced. Whether the solution is basic, neutral, or acidic depends on the relative abilities of the ions to react with water. Examples: NH4ClO, Al1CH3COO23, CrF3.

soLuTionAnalyze We are given the chemical formulas of five ionic compounds (salts) and asked whether their aqueous solutions will be acidic, basic, or neutral.Plan We can determine whether a solution of a salt is acidic, basic, or neutral by identifying the ions in solution and by assessing how each ion will affect the pH.Solve

(a) This solution contains barium ions and acetate ions. The cation is an ion of a heavy alkaline earth metal and will therefore not affect the pH. The anion, CH3COO-, is the conjugate base of the weak acid CH3COOH and will hydrolyze to produce OH - ions, thereby making the solution basic (combination 2).

(b) In this solution, NH4+ is the conjugate acid of a weak base 1NH32

and is therefore acidic. Cl- is the conjugate base of a strong acid (HCl) and therefore has no influence on the pH of the solution. Because the solution contains an ion that is acidic 1NH4

+2 and one that has no influence on pH 1Cl-2, the solution of NH4Cl will be acidic (combination 3).

(c) Here CH3NH3+ is the conjugate acid of a weak base 1CH3NH2,

an amine2 and is therefore acidic, and Br- is the conjugate base of a strong acid (HBr) and therefore pH neutral. Because the solution

sAmpLE ExERcisE 16.18 determining Whether salt solutions Are Acidic, Basic, or neutral

Determine whether aqueous solutions of each of these salts are acidic, basic, or neutral: (a) Ba1CH3COO22, (b) NH4Cl, (c) CH3NH3Br, (d) KNO3, (e) Al1ClO423.

contains one ion that is acidic and one that has no influence on pH, the solution of CH3NH3Br will be acidic (combination 3).

(d) This solution contains the K+ ion, which is a cation of group 1A, and the NO3

- ion, which is the conjugate base of the strong acid HNO3. Neither of the ions will react with water to any appre-ciable extent, making the solution neutral (combination 1).

(e) This solution contains Al3+ and ClO4- ions. Cations, such as Al3+,

that have a charge of 3+ or higher are acidic. The ClO4- ion is the

conjugate base of a strong acid 1HClO42 and therefore does not affect pH. Thus, the solution of Al1ClO423 will be acidic (combination 3).

practice Exercise 1Order the following solutions from lowest to highest pH: (i) 0.10 M NaClO, (ii) 0.10 M KBr, (iii) 0.10 M NH4ClO4.(a) i 6 ii 6 iii (b) ii 6 i 6 iii (c) iii 6 i 6 ii(d) ii 6 iii 6 i (e) iii 6 ii 6 i

practice Exercise 2Indicate which salt in each of the following pairs forms the more acidic (or less basic) 0.010 M solution: (a) NaNO3 or Fe1NO323, (b) KBr or KBrO, (c) CH3NH3Cl or BaCl2, (d) NH4NO2 or NH4NO3.

secTiON 16.10 Acid–Base Behavior and chemical structure 705

16.10 | Acid–Base Behavior and chemical structure

When a substance is dissolved in water, it may behave as an acid, behave as a base, or exhibit no acid–base properties. How does the chemical structure of a substance determine which of these behaviors is exhibited by the substance? For example, why do some substances that contain OH groups behave as bases, releasing OH- ions into solution, whereas others behave as acids, ionizing to release H+ ions? In this section we discuss briefly the effects of chemical structure on acid–base behavior.

Factors That Affect Acid StrengthA molecule containing H will act as a proton donor (an acid) only if the H—A bond is polarized such that the H atom has a partial positive charge. (Section 8.4) Recall that we indicate such polarization in this way:

AH

In ionic hydrides, such as NaH, the bond is polarized in the opposite way: the H atom pos-sesses a negative charge and behaves as a proton acceptor (a base). Nonpolar H ¬ A bonds, such as the H ¬ C bond in CH4, produce neither acidic nor basic aqueous solutions.

Predict whether the salt Na2HPO4 forms an acidic solution or a basic solution when dissolved in water.

soLuTionAnalyze We are asked to predict whether a solution of Na2HPO4 is acidic or basic. This substance is an ionic compound composed of Na+ and HPO4

2- ions.Plan We need to evaluate each ion, predicting whether it is acidic or basic. Because Na+ is a cat-ion of group 1A, it has no influence on pH. Thus, our analysis of whether the solution is acidic or basic must focus on the behavior of the HPO4

2- ion. We need to consider that HPO42- can act as

either an acid or a base: As acid HPO4

2-1aq2 ∆ H+1aq2 + PO43-1aq2 [16.46]

As base HPO42-1aq2 + H2O ∆ H2PO4

-1aq2 + OH-1aq2 [16.47]Of these two reactions, the one with the larger equilibrium constant determines whether the

solution is acidic or basic.Solve The value of Ka for Equation 16.46 is Ka3 for H3PO4: 4.2 * 10-13 (Table 16.3). For Equa-tion 16.47, we must calculate Kb for the base HPO4

2- from the value of Ka for its conjugate acid, H2PO4

-, and the relationship Ka * Kb = Kw 1Equation 16.402. The relevant value of Ka for H2PO4

- is Ka2 for H3PO4: 6.2 * 10-8 (from Table 16.3). We therefore have

Kb1HPO42-2 * Ka1H2PO4

-2 = Kw = 1.0 * 10-14

Kb1HPO42-2 =

1.0 * 10-14

6.2 * 10-8 = 1.6 * 10-7.

This Kb value is more than 105 times larger than Ka for HPO42-; thus, the reaction in Equation

16.47 predominates over that in Equation 16.46, and the solution is basic.

practice Exercise 1How many of the following salts are expected to produce acidic solutions (see Table 16.3 for data): NaHSO4, NaHC2O4, NaH2PO4, and NaHCO3?(a) 0 (b) 1 (c) 2 (d) 3 (e) 4

practice Exercise 2Predict whether the dipotassium salt of citric acid 1K2HC6H5O72 forms an acidic or basic solution in water (see Table 16.3 for data).

sAmpLE ExERcisE 16.19 predicting Whether the solution of

an Amphiprotic Anion is Acidic or Basic


A second factor that helps determine whether a molecule containing an H ¬ A bond will donate a proton is the strength of the bond. (Section 8.8) Very strong bonds are less easily broken than weaker ones. This factor is important, for example, in the hydrogen halides. The H ¬ F bond is the most polar H ¬ A bond. You there-fore might expect HF to be a very strong acid if bond polarity were all that mattered. However, the H ¬ A bond strength increases as you move up the group: 299 kJ>mol in HI, 366 kJ>mol in HBr, 431 kJ>mol in HCl, and 567 kJ>mol in HF. Because HF has the highest bond strength among the hydrogen halides, it is a weak acid, whereas all the other hydrogen halides are strong acids in water.

A third factor that affects the ease with which a hydrogen atom ionizes from HA is the stability of the conjugate base, A-. In general, the greater the stability of the conju-gate base, the stronger the acid.

The strength of an acid is often a combination of all three factors.

Binary AcidsFor a series of binary acids HA in which A represents members of the same group in the periodic table, the strength of the H ¬ A bond is generally the most important factor determining acid strength. The strength of an H ¬ A bond tends to decrease as the ele-ment A increases in size. As a result, the bond strength decreases and acidity increases down a group. Thus, HCl is a stronger acid than HF, and H2S is a stronger acid than H2O.

Bond polarity is the major factor determining acidity for binary acids HA when A represents members of the same period. Thus, acidity increases as the electronegativity of the element A increases, as it generally does moving from left to right across a period. (Section 8.4) For example, the difference in acidity of the period 2 elements is CH4 6 NH3 V H2O 6 HF. Because the C ¬ H bond is essentially nonpolar, CH4 shows no tendency to form H+ and CH3 - ions. Although the N ¬ H bond is polar, NH3 has a nonbonding pair of electrons on the nitrogen atom that dominates its chemistry, so NH3 acts as a base rather than an acid.

The periodic trends in the acid strengths of binary compounds of hydrogen and the nonmetals of periods 2 and 3 are summarized in ▼ Figure 16.18.

Neither acidnor base

SiH4

Neither acidnor base

CH4

Very weak baseKb = 4 × 10−28

PH3

Weak baseKb = 1.8 × 10−5

NH3

Weak acidKa = 9.5 × 10−8

H2S

H2O

Strong acid

HCl


H2SeStrong acid

HBr


HF

4A 5A 6A 7A

Increasing acid strength


▲ Figure 16.18 trends in acid strength for the binary hydrides of periods 2–4.

Go FiGuREAre the acid properties of HI what you would expect from this figure?


OxyacidsMany common acids, such as sulfuric acid, contain one or more O ¬ H bonds:

O

OO

O

H HS

Acids in which OH groups and possibly additional oxygen atoms are bound to a central atom are called oxyacids. At first it may seem confusing that the OH group, which we know behaves as a base, is also present in some acids. Let’s take a closer look at what factors determine whether a given OH group behaves as a base or as an acid.

Consider an OH group bound to some atom Y, which might in turn have other groups attached to it:

O HY

At one extreme, Y might be a metal, such as Na or Mg. Because of the low electronega-tivity of metals, the pair of electrons shared between Y and O is completely transferred to oxygen, and an ionic compound containing OH- is formed. Such compounds are therefore sources of OH- ions and behave as bases, as in NaOH and Mg1OH22.

When Y is a nonmetal, the bond to O is covalent and the substance does not read-ily lose OH-. Instead, these compounds are either acidic or neutral. Generally, as the electronegativity of Y increases, so does the acidity of the substance. This happens for two reasons: First, as electron density is drawn toward Y, the O ¬ H bond becomes weaker and more polar, thereby favoring loss of H+. Second, because the conjugate base of any acid YOH is usually an anion, its stability generally increases as the electronegativity of Y increases. This trend is illustrated by the Ka values of the hypohalous acids (YOH acids where Y is a halide ion), which decrease as the electronegativity of the halogen atom decreases (▼ Figure 16.19).

As the electronegativity of Y increases, electron density shifts toward Y

Protons are more readily transferred to H2O, leading to increased acid strength

The O—H bond becomes more polar

Ka = 3.0 × 10−8

Ka = 2.5 × 10−9

Ka = 2.3 × 10−11

Kw = 1.0 × 10−14

Hypochlorous acidHypobromous acidHypoiodous acidWater

SubstanceElectronegativity

of YDissociation

constant

3.02.82.52.1

Cl—OHBr—OHI—OHH—OH

Y—OH

+ +

−+

1

32

▲ Figure 16.19 acidity of the hypohalous oxyacids (YoH) as a function of electronegativity of Y.

Go FiGuREAt equilibrium, which of the two species with a halogen atom (green) is present in greater concentration?


Many oxyacids contain additional oxygen atoms bonded to the central atom Y. These atoms pull electron density from the O ¬ H bond, further increasing its polarity. Increasing the number of oxygen atoms also helps stabilize the conjugate base by increasing its ability to “spread out” its negative charge. Thus, the strength of an acid increases as additional electronegative atoms bond to the central atom Y. For example, the strength of the chlorine oxyacids 1Y = Cl2 steadily increases as O atoms are added:

O

O O


H Cl OH Cl O

O

OH Cl O

O

OH Cl

Ka 3.0 10−8 Ka 1.1 10−2 Strong acid Strong acid

Hypochlorous Chlorous Chloric Perchloric

×= = ×

Because the oxidation number of Y increases as the number of attached O atoms increases, this correlation can be stated in an equivalent way: In a series of oxyacids, the acidity increases as the oxidation number of the central atom increases.

give it some thoughtWhich acid has the larger acid-dissociation constant, HIO2 or HBrO3?

soLuTionAnalyze We are asked to arrange two sets of compounds in order from weakest acid to strongest acid. In (a), the substances are binary com-pounds containing H, and in (b) the substances are oxyacids.Plan For the binary compounds, we will consider the electronegativi-ties of As, Br, K, and Se relative to the electronegativity of H. The higher the electronegativity of these atoms, the higher the partial positive charge on H and so the more acidic the compound.

For the oxyacids, we will consider both the electronegativities of the central atom and the number of oxygen atoms bonded to the central atom.Solve

(a) Because K is on the left side of the periodic table, it has a very low electronegativity (0.8, from Figure 8.7, p. 310). As a result, the hydrogen in KH carries a negative charge. Thus, KH should be the least acidic (most basic) compound in the series.

Arsenic and hydrogen have similar electronegativities, 2.0 and 2.1, respectively. This means that the As—H bond is nonpo-lar, and so AsH3 has little tendency to donate a proton in aqueous solution.

The electronegativity of Se is 2.4, and that of Br is 2.8. Con-sequently, the H ¬ Br bond is more polar than the H—Se bond, giving HBr the greater tendency to donate a proton. (This expec-tation is confirmed by Figure 16.18, where we see that H2Se is a weak acid and HBr a strong acid.) Thus, the order of increasing acidity is KH 6 AsH3 6 H2Se 6 HBr.

sAmpLE ExERcisE 16.20 predicting Relative Acidities from composition and structure

Arrange the compounds in each series in order of increasing acid strength: (a) AsH3, HBr, KH, H2Se; (b) H2SO4, H2SeO3, H2SeO4.

(b) The acids H2SO4 and H2SeO4 have the same number of O atoms and the same number of OH groups. In such cases, the acid strength increases with increasing electronegativity of the central atom. Because S is slightly more electronegative than Se (2.5 vs 2.4), we predict that H2SO4 is more acidic than H2SeO4.

For acids with the same central atom, the acidity in-creases as the number of oxygen atoms bonded to the cen-tral atom increases. Thus, H2SeO4 should be a stronger acid than H2SeO3. We predict the order of increasing acidity to be H2SeO3 6 H2SeO4 6 H2SO4.

practice Exercise 1Arrange the following substances in order from weakest to stron-gest acid: HClO3, HOI, HBrO2, HClO2, HIO2

(a) HIO2 6 HOI 6 HClO3 6 HBrO2 6 HClO2(b) HOI 6 HIO2 6 HBrO2 6 HClO2 6 HClO3(c) HBrO2 6 HIO2 6 HClO2 6 HOI 6 HClO3(d) HClO3 6 HClO2 6 HBrO2 6 HIO2 6 HOI(e) HOI 6 HClO2 6 HBrO2 6 HIO2 6 HClO3

practice ExerciseIn each pair, choose the compound that gives the more acidic (or less basic) solution: (a) HBr, HF; (b) PH3, H2S; (c) HNO2, HNO3; (d) H2SO3, H2SeO3.


Carboxylic AcidsAnother large group of acids is illustrated by acetic acid, a weak acid 1Ka = 1.8 * 10-52:

H O

H

C C O HH

The portion of the structure shown in red is called the carboxyl group, which is often written COOH. Thus, the chemical formula of acetic acid is written as CH3COOH, where only the hydrogen atom in the carboxyl group can be ionized. Acids that con-tain a carboxyl group are called carboxylic acids, and they form the largest category of organic acids. Formic acid and benzoic acid are further examples of this large and important category of acids:

O

HC OO

HC OH

Formic acid Benzoic acid

Two factors contribute to the acidic behavior of carboxylic acids. First, the addi-tional oxygen atom attached to the carbon of the carboxyl group draws electron den-sity from the O ¬ H bond, increasing its polarity and helping to stabilize the conjugate base. Second, the conjugate base of a carboxylic acid (a carboxylate anion) can exhibit resonance (Section 8.6), which contributes to the stability of the anion by spread-ing the negative charge over several atoms:

resonance

OO

OH

H

H

C H

H

H

CC OC−

−

give it some thoughtWhat group of atoms is present in all carboxylic acids?

chemistry and Life

the amphiprotic behavior of amino acids

As we will discuss in greater detail in Chapter 24, amino acids are the building blocks of proteins. The general structure of amino acids is

O

O H

H

R

C C

Amine group(basic)

Carboxyl group(acidic)

H

H N

where different amino acids have different R groups attached to the central carbon atom. For example, in glycine, the simplest amino acid, R is a hydrogen atom, and in alanine R is a CH3 group:

C

Glycine Alanine

H2N COOH

CH3H

H

CH2N COOH

H

Amino acids contain a carboxyl group and can therefore serve as acids. They also contain an NH2 group, characteristic of amines

(Section 16.7), and thus they can also act as bases. Amino acids, there-fore, are amphiprotic. For glycine, we might expect the acid and base reactions with water to be

Acid: H2N ¬ CH2 ¬ COOH1aq2 + H2O1l2 ∆H2N ¬ CH2 ¬ COO-1aq2 + H3O+1aq2 [16.48]

Base: H2N ¬ CH2 ¬ COOH1aq2 + H2O1l2 ∆+H3N ¬ CH2 ¬ COOH1aq2 + OH -1aq2 [16.49]

The pH of a solution of glycine in water is about 6.0, indicating that it is a slightly stronger acid than base.

The acid–base chemistry of amino acids is more complicated than shown in Equations 16.48 and 16.49, however. Because the COOH group can act as an acid and the NH2 group can act as a base, amino acids un-dergo a “self-contained” Brønsted–Lowry acid–base reaction in which the proton of the carboxyl group is transferred to the basic nitrogen atom:

C C

H HH

H

H

O O

N C CH N+

H

OH

HH

O

Neutral molecule Zwitterion

proton transfer

−


16.11 | Lewis Acids and BasesFor a substance to be a proton acceptor (a Brønsted–Lowry base), it must have an un-shared pair of electrons for binding the proton, as, for example, in NH3. Using Lewis structures, we can write the reaction between H+ and NH3 as

H H

H

H

H

H

NHH+ N

+

+

G. N. Lewis was the first to notice this aspect of acid–base reactions. He proposed a more general definition of acids and bases that emphasizes the shared electron pair: A Lewis acid is an electron-pair acceptor, and a Lewis base is an electron-pair donor.

Every base that we have discussed thus far—whether OH-, H2O, an amine, or an anion—is an electron-pair donor. Everything that is a base in the Brønsted–Lowry sense (a proton acceptor) is also a base in the Lewis sense (an electron-pair donor). In the Lewis theory, however, a base can donate its electron pair to something other than H+. The Lewis definition therefore greatly increases the number of species that can be considered acids; in other words, H+ is a Lewis acid but not the only one. For example, the reaction between NH3 and BF3 occurs because BF3 has a vacant orbital in its valence shell. (Section 8.7) It therefore acts as an electron-pair acceptor (a Lewis acid) toward NH3, which donates the electron pair:

F

F

B F

F

F

B F

H

H

NH

H

H

NH

Lewisacid

Lewisbase

+

give it some thoughtWhat feature must any molecule or ion have in order to act as a Lewis acid?

Our emphasis throughout this chapter has been on water as the solvent and on the proton as the source of acidic properties. In such cases we find the Brønsted–Lowry definition of acids and bases to be the most useful. In fact, when we speak of a substance as being acidic or basic, we are usually thinking of aqueous solutions and using these terms in the Arrhenius or Brønsted–Lowry sense. The advantage of the Lewis defini-tions of acid and base is that they allow us to treat a wider variety of reactions, including

Although the form of the amino acid on the right in this equation is electrically neutral overall, it has a positively charged end and a negatively charged end. A molecule of this type is called a zwitterion (German for “hybrid ion”).

Do amino acids exhibit any properties indicating that they behave as zwitterions? If so, their behavior should be similar to that of ionic substances. (Section 8.2) Crystalline amino acids have relatively

high melting points, usually above 200 °C, which is characteristic of ionic solids. Amino acids are far more soluble in water than in nonpo-lar solvents. In addition, the dipole moments of amino acids are large, consistent with a large separation of charge in the molecule. Thus, the ability of amino acids to act simultaneously as acids and bases has im-portant effects on their properties.

Related Exercise: 16.105, 16.114

secTiON 16.11 Lewis Acids and Bases 711

those that do not involve proton transfer, as acid–base reactions. To avoid confusion, a substance such as BF3 is rarely called an acid unless it is clear from the context that we are using the term in the sense of the Lewis definition. Instead, substances that function as electron-pair acceptors are referred to explicitly as “Lewis acids.”

Lewis acids include molecules that, like BF3, have an incomplete octet of electrons. In addition, many simple cations can function as Lewis acids. For example, Fe3+ inter-acts strongly with cyanide ions to form the ferricyanide ion:

Fe3+ + 63:C ‚ N:4- ¡ 3Fe1C ‚ N:2643-

The Fe3 + ion has vacant orbitals that accept the electron pairs donated by the cya-nide ions. (We will learn more in Chapter 23 about just which orbitals are used by the Fe3+ ion.) The metal ion is highly charged, too, which contributes to the interaction with CN- ions.

Some compounds containing multiple bonds can behave as Lewis acids. For exam-ple, the reaction of carbon dioxide with water to form carbonic acid 1H2CO32 can be pictured as an attack by a water molecule on CO2, in which the water acts as an elec-tron-pair donor and the CO2 as an electron-pair acceptor:

O

O

O C

H

H

H

HH

C

O

C

O

O

O

H O

O

One electron pair of one of the carbon–oxygen double bonds is moved onto the oxygen, leaving a vacant orbital on the carbon, which means the carbon can accept an electron pair donated by H2O. The initial acid–base product rearranges by transferring a proton from the water oxygen to a carbon dioxide oxygen, forming carbonic acid.

The hydrated cations we encountered in Section 16.9, such as 3Fe1H2O2643+ in Figure 16.17, form through the reaction between the cation acting as a Lewis acid and the water molecules acting as Lewis bases. When a water molecule interacts with the positively charged metal ion, electron density is drawn from the oxygen (▼ Figure 16.20). This flow of electron density causes the O ¬ H bond to become more polarized; as a result, water molecules bound to the metal ion are more acidic than

1+

3+

Weak electrostatic interaction means small electron density shift to cation

Strong electrostatic interaction means signi�cant electron density shift to cation

Cation weakens H—O bond strength, solvating H2O can readily donate H+, solution becomes acidic

Cation has little effect on H—O bond strength, solution remains neutral

▲ Figure 16.20 the acidity of a hydrated cation depends on cation charge.


those in the bulk solvent. This effect becomes more pronounced as the charge of the cation increases, which explains why 3+ cations are much more acidic than cations with smaller charges.

The Lewis acid–base concept allows many ideas developed in this chapter to be used more broadly in chemistry, including rections in solvents other than water. If you take a course in organic chemistry, you will see a number of important rections that require the presence of a Lewis acid in order to proceed. The interaction of lone pairs on one molecule or ion with vacant orbitals on another molecule or ion is one of the most important concepts in chemistry, as you will see throughout your studies.

sAmpLE inTEGRATivE ExERcisE putting concepts Together

Phosphorous acid 1H3PO32 has the Lewis structure

P

H

H

H

O

OO

(a) Explain why H3PO3 is diprotic and not triprotic. (b) A 25.0-mL sample of an H3PO3 solution titrated with 0.102 M NaOH requires 23.3 mL of NaOH to neutralize both acidic protons. What is the molarity of the H3PO3 solution? (c) The original solution from part (b) has a pH of 1.59. Calculate the percent ionization and Ka1 for H3PO3, assuming that Ka1 W Ka2. (d) How does the osmotic pressure of a 0.050 M solution of HCl compare qualitatively with that of a 0.050 M solution of H3PO3? Explain.

soLuTionWe will use what we have learned about molecular structure and its impact on acidic behavior to answer part (a). We will then use stoichiometry and the relationship between pH and 3H+4 to answer parts (b) and (c). Finally, we will consider percent ionization in order to compare the osmotic pressure of the two solutions in part (d).(a) Acids have polar H ¬ X bonds. From Figure 8.7 (p. 310) we see that the electronegativity

of H is 2.1 and that of P is also 2.1. Because the two elements have the same electronegativ-ity, the H ¬ P bond is nonpolar. (Section 8.4) Thus, this H cannot be acidic. The other two H atoms, however, are bonded to O, which has an electronegativity of 3.5. The H ¬ O bonds are, therefore, polar with H having a partial positive charge. These two H atoms are consequently acidic.

(b) The chemical equation for the neutralization reaction is

H3PO31aq2 + 2NaOH1aq2 ¡ Na2HPO31aq2 + 2H2O1l2 From the definition of molarity, M = mol>L, we see that moles = M * L. (Section 4.5)

Thus, the number of moles of NaOH added to the solution is

10.0233 L210.102 mol>L2 = 2.38 * 10-3 mol NaOH

The balanced equation indicates that 2 mol of NaOH is consumed for each mole of H3PO3. Thus, the number of moles of H3PO3 in the sample is

12.38 * 10-3 mol NaOH2a 1 mol H3PO3

2 mol NaOHb = 1.19 * 10-3 mol H3PO3

The concentration of the H3PO3 solution, therefore, equals 11.19 * 10-3 mol2>10.0250 L2 = 0.0476 M.

(c) From the pH of the solution, 1.59, we can calculate 3H+4 at equilibrium:

3H+4 = antilog1-1.592 = 10-1.59 = 0.026 M 1two significant figures2 Because Ka1 W Ka2, the vast majority of the ions in solution are from the first ionization

step of the acid.

H3PO31aq2 ∆ H +1aq2 + H2PO3-1aq2

Because one H2PO3- ion forms for each H+ ion formed, the equilibrium concentrations

of H+ and H2PO3- are equal: 3H+4 = 3H2PO3 -4 = 0.026 M. The equilibrium concentra-

tion of H3PO3 equals the initial concentration minus the amount that ionizes to form H+

chapter summary and Key Terms 713

chapter summary and Key TermsintroduCtion to aCids and bases (seCtion 16.1) Acids and bases were first recognized by the properties of their aqueous solutions. For example, acids turn litmus red, whereas bases turn litmus blue. Arrhenius recognized that the properties of acidic solu-tions are due to H+1aq2 ions and those of basic solutions are due to OH-1aq2 ions.

brønsted–lowrY aCids and bases (seCtion 16.2) The Brønsted–Lowry concept of acids and bases is more general than the Arrhenius concept and emphasizes the transfer of a proton 1H+2 from an acid to a base. The H+ ion, which is merely a proton with no surrounding valence electrons, is strongly bound to water. For this reason, the hydronium ion, H3O+1aq2, is often used to represent the predominant form of H+ in water instead of the simpler H+1aq2.

A brønsted–lowry acid is a substance that donates a proton to another substance; a brønsted–lowry base is a substance that accepts a proton from another substance. Water is an amphiprotic substance, one that can function as either a Brønsted–Lowry acid or base, depending on the substance with which it reacts.

The conjugate base of a Brønsted–Lowry acid is the species that remains when a proton is removed from the acid. The conjugate acid of a Brønsted–Lowry base is the species formed by adding a proton to the base. Together, an acid and its conjugate base (or a base and its conjugate acid) are called a conjugate acid–base pair.

The acid–base strengths of conjugate acid–base pairs are related: The stronger an acid, the weaker is its conjugate base; the weaker an acid, the stronger is its conjugate base. In every acid–base reaction, the position of the equilibrium favors the transfer of the proton from the stronger acid to the stronger base.

autoionization of water (seCtion 16.3) Water ionizes to a slight degree, forming H+1aq2 and OH-1aq2. The extent of this autoionization is expressed by the ion-product constant for water: Kw = 3H+43OH-4 = 1.0 * 10-14 125 °C2. This relationship holds

for both pure water and aqueous solutions. The Kw expression indi-cates that the product of 3H+4 and 3OH-4 is a constant. Thus, as 3H+4 increases, 3OH-4 decreases. Acidic solutions are those that contain more H+1aq2 than OH-1aq2, whereas basic solutions contain more OH-1aq2 than H+1aq2. When 3H+4 = 3OH-4, the solution is neutral.

tHe pH sCale (seCtion 16.4) The concentration of H+1aq2 can be expressed in terms of pH: pH = - log3H+4. At 25 °C the pH of a neutral solution is 7.00, whereas the pH of an acidic solution is below 7.00, and the pH of a basic solution is above 7.00. This p notation is also used to represent the negative logarithm of other small quantities, as in pOH and pKw. The pH of a solution can be measured using a pH meter, or it can be estimated using acid–base indicators.

strong aCids and bases (seCtion 16.5) Strong acids are strong electrolytes, ionizing completely in aqueous solution. The common strong acids are HCl, HBr, HI, HNO3, HClO3, HClO4, and H2SO4. The conjugate bases of strong acids have negligible basicity.

Common strong bases are the ionic hydroxides of the alkali met-als and the heavy alkaline earth metals.

weak aCids and bases (seCtions 16.6 and 16.7) Weak acids are weak electrolytes; only a small fraction of the molecules exist in solution in ionized form. The extent of ionization is expressed by the acid-dissociation constant, Ka, which is the equilibrium constant for the reaction HA1aq2 ∆ H+1aq2 + A-1aq2, which can also be writ-ten as HA1aq2 + H2O1l2 ∆ H3O+1aq2 + A-1aq2. The larger the value of Ka, the stronger is the acid. For solutions of the same concen-tration, a stronger acid also has a larger percent ionization. The concen-tration of a weak acid and its Ka value can be used to calculate the pH of a solution.

polyprotic acids, such as H3PO4, have more than one ionizable proton. These acids have acid-dissociation constants that decrease in magnitude in the order Ka1 7 Ka2 7 Ka3. Because nearly all the

and H2PO3-: 3H3PO34 = 0.0476 M - 0.026 M = 0.022 M (two significant figures). These

results can be tabulated as follows:H3PO31aq2 ∆ H+1aq2 + H2PO3

-1aq2Initial concentration (M) 0.0476 0 0

Change in concentration (M) -0.026 +0.026 +0.026

Equilibrium concentration (M) 0.022 0.026 0.026

The percent ionization is

percent ionization =3H+4equilibrium

3H3PO34initial* 100% =

0.026 M0.0476 M

* 100% = 55%

The first acid-dissociation constant is

Ka1 =3H+43H2PO3

-43H3PO34 =

10.026210.02620.022

= 0.031

(d) Osmotic pressure is a colligative property and depends on the total concentration of particles in solution. (Section 13.5) Because HCl is a strong acid, a 0.050 M solution will contain 0.050 M H+1aq2 and 0.050 M Cl-1aq2, or a total of 0.100 mol>L of particles. Because H3PO3 is a weak acid, it ionizes to a lesser extent than HCl and, hence, there are fewer particles in the H3PO3 solution. As a result, the H3PO3 solution will have the lower osmotic pressure.


H+1aq2 in a polyprotic acid solution comes from the first dissociation step, the pH can usually be estimated satisfactorily by considering only Ka1. Weak bases include NH3, amines, and the anions of weak acids. The extent to which a weak base reacts with water to gener-ate the corresponding conjugate acid and OH- is measured by the base-dissociation constant, Kb. Kb is the equilibrium constant for the reaction B1aq2 + H2O1l2 ∆ HB+1aq2 + OH-1aq2, where B is the base.

relationsHip between Ka and Kb (seCtion 16.8) The relationship between the strength of an acid and the strength of its conjugate base is expressed quantitatively by the equa-tion Ka * Kb = Kw, where Ka and Kb are dissociation constants for conjugate acid–base pairs. This equation explains the inverse relationship between the strength of an acid and the strength of its conjugate base.

aCid–base properties of salts (seCtion 16.9) The acid–base properties of salts can be ascribed to the behavior of their respective cations and anions. The reaction of ions with water, with a resultant change in pH, is called hydrolysis. The cations of the alkali metals and the alkaline earth metals as well as the anions of strong acids, such as Cl-, Br-, I-, and NO3

-, do not undergo hydrolysis. They are always spectator ions in acid–base chemistry. A cation that is the conjugate acid of a weak base produces H+ upon hydrolysis.

An anion that is the conjugate base of a weak acid produces OH- upon hydrolysis.

aCid–base beHavior and CHemiCal struCture (seCtion 16.10) The tendency of a substance to show acidic or basic characteristics in water can be correlated with its chemical struc-ture. Acid character requires the presence of a highly polar H ¬ X bond. Acidity is also favored when the H ¬ X bond is weak and when the X- ion is very stable.

For oxyacids with the same number of OH groups and the same number of O atoms, acid strength increases with increasing electro-negativity of the central atom. For oxyacids with the same central atom, acid strength increases as the number of oxygen atoms attached to the central atom increases. Carboxylic acids, which are organic acids containing the COOH group, are the most important class of organic acids. The presence of delocalized p bonding in the conjugate base is a major factor responsible for the acidity of these compounds.

lewis aCids and bases (seCtion 16.11) The Lewis concept of acids and bases emphasizes the shared electron pair rather than the proton. A lewis acid is an electron-pair acceptor, and a lewis base is an electron-pair donor. The Lewis concept is more general than the Brønsted–Lowry concept because it can apply to cases in which the acid is some substance other than H+, and to solvents other than water.

Learning outcomes After studying this chapter, you should be able to:

• Define and identify Arrhenius acids and bases. (Section 16.1)

• Describe the nature of the hydrated proton, represented as either H+(aq) or H3O

+(aq). (Section 16.2)

• Define and identify Brønsted–Lowry acids and bases and identify conjugate acid–base pairs. (Section 16.2)

• Correlate the strength of an acid to the strength of its conjugate base. (Section 16.2)

• Explain how the equilibrium position of a proton-transfer reac-tion relates to the strengths of the acids and bases involved. (Section 16.3)

• Describe the autoionization of water and explain how [H3O+] and [OH−] are related via Kw. (Section 16.3)

• Calculate the pH of a solution given [H3O+] or [OH−]. (Section 16.4)

• Calculate the pH of a strong acid or strong base given its concen-tration. (Section 16.5)

• Calculate Ka or Kb for a weak acid or weak base given its concen-tration and the pH of the solution, and vice versa. (Sections 16.6 and 16.7)

• Calculate the pH of a weak acid or weak base or its percent ioniza-tion given its concentration and Ka or Kb. (Sections 16.6 and 16.7)

• Calculate Kb for a weak base given Ka of its conjugate acid, and similarly calculate Ka from Kb. (Section 16.8)

• Predict whether an aqueous solution of a salt will be acidic, basic, or neutral. (Section 16.9)

• Predict the relative strength of a series of acids from their molecu-lar structures. (Section 16.10)

• Define and identify Lewis acids and bases. (Section 16.11)

Key Equations • Kw = 3H3O+43OH-4 = 3H+43OH-4 = 1.0 * 10-14 [16.16] Ion product of water at 25 °C • pH = - log3H+4 [16.17] Definition of pH

• pOH = - log[OH-] [16.18] Definition of pOH

• pH + pOH = 14.00 [16.20] Relationship between pH and pOH

• Ka =3H3O+43A-4

3HA4 or Ka =3H+43A-43HA4 [16.25] Acid-dissociation constant for a weak acid, HA

• Percent ionization =3H+4equilibrium

3HA4initial* 100% [16.27] Percent ionization of a weak acid

exercises 715

• Kb =3BH+43OH-4

3B4 [16.34] Base-dissociation constant for a weak base, B

• Ka * Kb = Kw [16.40] Relationship between acid- and base-dissociation constants of a conjugate acid–base pair

• pKa = - log Ka and pKb = - log Kb [16.41] Definitions of pKa and pKb

visualizing concepts

16.1 (a) Identify the Brønsted–Lowry acid and base in the reaction

+ +

= H = N = Cl

−

+

(b) Identify the Lewis acid and base in the reaction. [Sections 16.2 and 16.11]

16.2 The following diagrams represent aqueous solutions of two monoprotic acids, HA 1A = X or Y2. The water molecules have been omitted for clarity. (a) Which is the stronger acid, HX or HY? (b) Which is the stronger base, X- or Y-? (c) If you mix equal concentrations of HX and NaY, will the equilibrium

HX1aq2 + Y-1aq2 ∆ HY1aq2 + X-1aq2 lie mostly to the right 1Kc 7 12 or to the left 1Kc 6 12?

[Section 16.2]

HX HY

= HA = A−= H3O+

+

+

+

++

+

+

−−

−

−

−−

−

16.3 The indicator methyl orange has been added to both of the following solutions. Based on the colors, classify each state-ment as true or false:(a) The pH of solution A is definitely less than 7.00.

(b) The pH of solution B is definitely greater than 7.00.(c) The pH of solution B is greater than that of solution A.

[Section 16.4]

Solution A Solution B

16.4 The probe of the pH meter shown here is sitting in a bea-ker that contains a clear liquid. (a) You are told the liquid is pure water, a solution of HCl(aq), or a solution of KOH(aq). Which one is it? (b) If the liquid is one of the solutions, what is its molarity? (c) Why is the temperature given on the pH meter? [Sections 16.4 and 16.5]

16.5 The following diagrams represent aqueous solutions of three acids, HX, HY, and HZ. The water molecules have been omitted for clarity, and the hydrated proton is represented as H+ rather than H3O+. (a) Which of the acids is a strong acid? Explain. (b) Which acid would have the smallest acid-dissociation con-stant, Ka? (c) Which solution would have the highest pH? [Sections 16.5 and 16.6]

Exercises


HX HY HZ

+

+

+

−

++

−+

+

+

+

++

+− −

−−

−−

− −

+−

−

−

16.6 The graph given below shows 3H+4 vs. concentration for an aqueous solution of an unknown substance. (a) Is the sub-stance a strong acid, a weak acid, a strong base, or a weak base? (b) Based on your answer to (a), can you determine the value of the pH of the solution when the concentration is 0.18 M? (c) Would the line go exactly through the origin of the plot? [Sections 16.5 and 16.6]

00

0.060 0.12Concentration (M)

0.18

[H+

]

16.7 (a) Which of these three lines represents the effect of concen-tration on the percent ionization of a weak acid? (b) Explain in qualitative terms why the curve you chose has the shape it does. [Section 16.6]

Acid concentration

00

Perc

ent i

oniz

ed

C

B

A

16.8 Each of the three molecules shown here contains an OH group, but one molecule acts as a base, one as an acid, and the third is neither acid nor base. (a) Which one acts as a base? Why does only this molecule act as a base? (b) Which one acts as an acid? (c) Why is the remaining molecule neither acidic nor basic? [Sections 16.6 and 16.7]

Molecule A Molecule B Molecule C

Molecule A Molecule B Molecule C

16.9 Phenylephrine, an organic substance with molecular formula C9H13NO2, is used as a nasal decongenstant in over-the- counter medications. The molecular structure of phenyleph-rine is shown below using the usual shortcut organic nomen-clature. (a) Would you expect a solution of phenylephrine to be acidic, neutral, or basic? (b) One of the active ingredients in Alka-Seltzer PLUS® cold medication is “phenylephrine hydrochloride.” How does this ingredient differ from the structure shown below? (c) Would you expect a solution of phenylephrine hydrochloride to be acidic, neutral, or basic? [Sections 16.8 and 16.9]

CH3

H

N

OH

HO

16.10 Which of the following diagrams best represents an aque-ous solution of NaF? (For clarity, the water molecules are not shown.) Will this solution be acidic, neutral, or basic? [Section 16.9]

Solution A

Na+ F− OH− HF

Solution B Solution C

− −

+

+

+

+

++

++

++

+ +

++

+

+

+

−

−

−

− −

− −

−−

−−

− −

−

− −

16.11 Consider the molecular models shown here, where X repre-sents a halogen atom. (a) If X is the same atom in both mol-ecules, which molecule will be more acidic? (b) Does the acidity of each molecule increase or decrease as the electro-negativity of the atom X increases? [Section 16.10]

exercises 717

(a) (b)

O

OHX

CO

H

X

16.12 (a) The following diagram represents the reaction of PCl4 + with Cl-. Draw the Lewis structures for the reactants and products, and identify the Lewis acid and the Lewis base in the reaction.

+−

+

(b) The following reaction represents a hydrated cation losing a proton. How does the equilibrium constant for the reaction change as the charge of the cation increases? [Sections 16.9 and 16.11]

++

(n−1)++

n+

Arrhenius and Brønsted–Lowry Acids and Bases (sections 16.1 and 16.2)

16.13 (a) What is the difference between the Arrhenius and the Brønsted–Lowry definitions of an acid? (b) NH31g2 and HCl(g) react to form the ionic solid NH4Cl1s2. Which sub-stance is the Brønsted–Lowry acid in this reaction? Which is the Brønsted–Lowry base?

16.14 (a) What is the difference between the Arrhenius and the Brønsted–Lowry definitions of a base? (b) Can a substance behave as an Arrhenius base if it does not contain an OH group? Explain.

16.15 (a) Give the conjugate base of the following Brønsted–Lowry acids: (i) HIO3, (ii) NH4

+. (b) Give the conjugate acid of the following Brønsted–Lowry bases: (i) O2-, (ii) H2PO4

-. 16.16 (a) Give the conjugate base of the following Brønsted–Lowry

acids: (i) HCOOH, (ii) HPO42-. (b) Give the conjugate acid of

the following Brønsted–Lowry bases: (i) SO42-, (ii) CH3NH2.

16.17 Designate the Brønsted–Lowry acid and the Brønsted–Lowry base on the left side of each of the following equations, and also designate the conjugate acid and conjugate base of each on the right side:(a) NH4

+1aq2 + CN -1aq2 ∆ HCN1aq2 + NH31aq2(b) 1CH323N1aq2 + H2O1l2 ∆

1CH323NH +1aq2 + OH -1aq2(c) HCOOH1aq2 + PO4

3-1aq2 ∆HCOO-1aq2 + HPO4

2-1aq2

16.18 Designate the Brønsted–Lowry acid and the Brønsted–Lowry base on the left side of each equation, and also designate the conjugate acid and conjugate base of each on the right side.(a) HBrO1aq2 + H2O1l2 ∆ H3O+1aq2 + BrO-1aq2(b) HSO4

-1aq2 + HCO3-1aq2 ∆

SO42 - 1aq2 + H2CO31aq2

(c) HSO3-1aq2 + H3O+1aq2 ∆ H2SO31aq2 + H2O1l2

16.19 (a) The hydrogen sulfite ion 1HSO3-2 is amphiprotic. Write

a balanced chemical equation showing how it acts as an acid toward water and another equation showing how it acts as a base toward water. (b) What is the conjugate acid of HSO3

-? What is its conjugate base?

16.20 (a) Write an equat ion for the react ion in which H2C6H7O5

-1aq2 acts as a base in H2O1l2. (b) Write an equa-tion for the reaction in which H2C6H7O5

-1aq2 acts as an acid in H2O1l2. (c) What is the conjugate acid of H2C6H7O5

-1aq2? What is its conjugate base?

16.21 Label each of the following as being a strong base, a weak base, or a species with negligible basicity. In each case write the for-mula of its conjugate acid, and indicate whether the conjugate acid is a strong acid, a weak acid, or a species with negligible acidity: (a) CH3COO-, (b) HCO3

-, (c) O2-, (d) Cl-, (e) NH3. 16.22 Label each of the following as being a strong acid, a weak acid,

or a species with negligible acidity. In each case write the for-mula of its conjugate base, and indicate whether the conjugate base is a strong base, a weak base, or a species with negligible basicity: (a) HCOOH, (b) H2, (c) CH4, (d) HF, (e) NH4

+. 16.23 (a) Which of the following is the stronger Brønsted–Lowry

acid, HBrO or HBr? (b) Which is the stronger Brønsted–Lowry base, F- or Cl-?

16.24 (a) Which of the following is the stronger Brønsted–Lowry acid, HClO3 or HClO2? (b) Which is the stronger Brønsted–Lowry base, HS- or HSO4

-? 16.25 Predict the products of the following acid–base reactions, and

predict whether the equilibrium lies to the left or to the right of the equation:(a) O2-1aq2 + H2O1l2 ∆ (b) CH3COOH1aq2 + HS-1aq2 ∆ (c) NO2

-1aq2 + H2O1l2 ∆ 16.26 Predict the products of the following acid–base reactions, and

predict whether the equilibrium lies to the left or to the right of the equation:(a) NH4

+1aq2 + OH-1aq2 ∆ (b) CH3COO-1aq2 + H3O+1aq2 ∆ (c) HCO3

-1aq2 + F-1aq2 ∆

Autoionization of Water (section 16.3)

16.27 If a neutral solution of water, with pH = 7.00, is cooled to 10 °C, the pH rises to 7.27. Which of the following three statements is correct for the cooled water: (i) 3H+4 7 3OH-4, (ii) 3H+4 = 3OH-4, or (iii) 3H+4 6 3OH-4?

16.28 (a) Write a chemical equation that illustrates the autoioniza-tion of water. (b) Write the expression for the ion-product constant for water, Kw. (c) If a solution is described as ba-sic, which of the following is true: (i) 3H+4 7 3OH-4, (ii) 3H+4 = 3OH-4, or (iii) 3H+4 6 3OH-4?


the solution acidic, neutral, or basic? (b) Which of the follow-ing can you establish about the solution: (i) A minimum pH, (ii) A maximum pH, or (iii) A specific range of pH values? (c) What other indicator or indicators would you want to use to determine the pH of the solution more precisely?

strong Acids and Bases (section 16.5)

16.41 Is each of the following statements true or false? (a) All strong acids contain one or more H atoms. (b) A strong acid is a strong electrolyte. (c) A 1.0-M solution of a strong acid will have pH = 1.0.

16.42 Determine whether each of the following is true or false: (a) All strong bases are salts of the hydroxide ion. (b) The addition of a strong base to water produces a solution of pH 7 7.0. (c) Because Mg1OH22 is not very soluble, it can-not be a strong base.

16.43 Calculate the pH of each of the following strong acid solu-tions: (a) 8.5 * 10-3M HBr, (b) 1.52 g of HNO3 in 575 mL of solution, (c) 5.00 mL of 0.250 M HClO4 diluted to 50.0 mL, (d) a solution formed by mixing 10.0 mL of 0.100 M HBr with 20.0 mL of 0.200 M HCl.

16.44 Calculate the pH of each of the following strong acid solu-tions: (a) 0.0167 M HNO3, (b) 0.225 g of HClO3 in 2.00 L of solution, (c) 15.00 mL of 1.00 M HCl diluted to 0.500 L, (d) a mixture formed by adding 50.0 mL of 0.020 M HCl to 125 mL of 0.010 M HI.

16.45 Calculate 3OH-4 and pH for (a) 1.5 * 10-3M Sr1OH22, (b) 2.250 g of LiOH in 250.0 mL of solution, (c) 1.00 mL of 0.175 M NaOH diluted to 2.00 L, (d) a solution formed by adding 5.00 mL of 0.105 M KOH to 15.0 mL of 9.5 * 10-2 M Ca1OH22.

16.46 Calculate 3OH-4 and pH for each of the following strong base solutions: (a) 0.182 M KOH, (b) 3.165 g of KOH in 500.0 mL of solution, (c) 10.0 mL of 0.0105 M Ca1OH22 diluted to 500.0 mL, (d) a solution formed by mixing 20.0 mL of 0.015 M Ba1OH22 with 40.0 mL of 8.2 * 10-3 M NaOH.

16.47 Calculate the concentration of an aqueous solution of NaOH that has a pH of 11.50.

16.48 Calculate the concentration of an aqueous solution of Ca1OH22 that has a pH of 10.05.

Weak Acids (section 16.6)

16.49 Write the chemical equation and the Ka expression for the ionization of each of the following acids in aqueous solution. First show the reaction with H+1aq2 as a product and then with the hydronium ion: (a) HBrO2, (b) C2H5COOH.

16.50 Write the chemical equation and the Ka expression for the acid dissociation of each of the following acids in aqueous solution. First show the reaction with H+1aq2 as a prod-uct and then with the hydronium ion: (a) C6H5COOH, (b) HCO3

-. 16.51 Lactic acid 1CH3CH(OH2COOH) has one acidic hydrogen. A

0.10 M solution of lactic acid has a pH of 2.44. Calculate Ka. 16.52 Phenylacetic acid 1C6H5CH2COOH2 is one of the substances

that accumulates in the blood of people with phenylketonuria, an inherited disorder that can cause mental retardation or even death. A 0.085 M solution of C6H5CH2COOH has a pH of 2.68. Calculate the Ka value for this acid.

16.29 Calculate 3H+4 for each of the following solutions, and in-dicate whether the solution is acidic, basic, or neutral: (a) 3OH-4 = 0.00045 M; (b) 3OH-4 = 8.8 * 10-9 M; (c) a solution in which 3OH-4 is 100 times greater than 3H+4.

16.30 Calculate 3OH-4 for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) 3H+4 = 0.0505 M; (b) 3H+4 = 2.5 * 10-10 M; (c) a solu-tion in which 3H+4 is 1000 times greater than 3OH-4.

16.31 At the freezing point of water 10 °C2, Kw = 1.2 * 10-15. Calculate 3H+4 and 3OH-4 for a neutral solution at this temperature.

16.32 Deuterium oxide 1D2O, where D is deuterium, the hydrogen-2 isotope) has an ion-product constant, Kw, of 8.9 * 10-16 at 20 °C. Calculate 3D+4 and 3OD-4 for pure (neutral) D2O at this temperature.

The ph scale (section 16.4)

16.33 By what factor does 3H+4 change for a pH change of (a) 2.00 units, (b) 0.50 units?

16.34 Consider two solutions, solution A and solution B. 3H+4 in solution A is 250 times greater than that in solution B. What is the difference in the pH values of the two solutions?

16.35 Complete the following table by calculating the missing en-tries and indicating whether the solution is acidic or basic.

3h+ 4 oh− 1aq 2 ph poh Acidic or Basic?

7.5 * 10-3 M

3.6 * 10-10 M8.25

5.70

16.36 Complete the following table by calculating the missing entries. In each case indicate whether the solution is acidic or basic.

ph poh 3h+ 4 3oh− 4 Acidic or Basic?

5.252.02

4.4 * 10-10 M

8.5 * 10-2 M

16.37 The average pH of normal arterial blood is 7.40. At normal body temperature 137 °C2, Kw = 2.4 * 10-14. Calculate 3H+4, 3OH-4, and pOH for blood at this temperature.

16.38 Carbon dioxide in the atmosphere dissolves in raindrops to produce carbonic acid 1H2CO32, causing the pH of clean, unpolluted rain to range from about 5.2 to 5.6. What are the ranges of 3H+4 and 3OH-4 in the raindrops?

16.39 Addition of the indicator methyl orange to an unknown so-lution leads to a yellow color. The addition of bromthymol blue to the same solution also leads to a yellow color. (a) Is the solution acidic, neutral, or basic? (b) What is the range (in whole numbers) of possible pH values for the solution? (c) Is there another indicator you could use to narrow the range of possible pH values for the solution?

16.40 Addition of phenolphthalein to an unknown colorless solu-tion does not cause a color change. The addition of bromthy-mol blue to the same solution leads to a yellow color. (a) Is

exercises 719

Weak Bases (section 16.7)

16.67 Consider the base hydroxylamine, NH2OH. (a) What is the conjugate acid of hydroxylamine? (b) When it acts as a base, which atom in hydroxylamine accepts a proton? (c) There are two atoms in hydroxylamine that have nonbonding electron pairs that could act as proton acceptors. Use Lewis structures and formal charges (Section 8.5) to rationalize why one of these two atoms is a much better proton acceptor than the other.

16.68 The hypochlorite ion, ClO-, acts as a weak base. (a) Is ClO- a stronger or weaker base than hydroxylamine? (b) When ClO- acts as a base, which atom, Cl or O, acts as the proton accep-tor? (c) Can you use formal charges to rationalize your an-swer to part (b)?

16.69 Write the chemical equation and the Kb expression for the re-action of each of the following bases with water: (a) dimethyl-amine, 1CH322NH; (b) carbonate ion, CO3

2-; (c) formate ion, CHO2

-. 16.70 Write the chemical equation and the Kb expression for the

reaction of each of the following bases with water: (a) propyl-amine, C3H7NH2; (b) monohydrogen phosphate ion, HPO4

2-; (c) benzoate ion, C6H5CO2

-. 16.71 Calculate the molar concentration of OH- in a 0.075 M solu-

tion of ethylamine 1C2H5NH2; Kb = 6.4 * 10-42. Calculate the pH of this solution.

16.72 Calculate the molar concentration of OH- in a 0.724 M solu-tion of hypobromite ion 1BrO-; Kb = 4.0 * 10-62. What is the pH of this solution?

16.73 Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base:

C10H15ON1aq2 + H2O1l2 ∆ C10H15ONH+1aq2 + OH-1aq2 A 0.035 M solution of ephedrine has a pH of 11.33. (a) What are

the equilibrium concentrations of C10H15ON, C10H15ONH+, and OH-? (b) Calculate Kb for ephedrine.

16.74 Codeine 1C18H21NO32 is a weak organic base. A 5.0 * 10-3M solution of codeine has a pH of 9.95. Calculate the value of Kb for this substance. What is the pKb for this base?

The Ka − Kb Relationship; Acid–Base properties of salts (sections 16.8 and 16.9)

16.75 Although the acid-dissociation constant for phenol 1C6H5OH2 is listed in Appendix D, the base-dissociation con-stant for the phenolate ion 1C6H5O-2 is not. (a) Explain why it is not necessary to list both Ka for phenol and Kb for the phenolate ion. (b) Calculate Kb for the phenolate ion. (c) Is the phenolate ion a weaker or stronger base than ammonia?

16.76 Use the acid-dissociation constants in Table 16.3 to ar-range these oxyanions from strongest base to weakest: SO4

2-, CO32-, SO3

2-, and PO43-.

16.77 (a) Given that Ka for acetic acid is 1.8 * 10-5 and that for hypochlorous acid is 3.0 * 10-8, which is the stronger acid? (b) Which is the stronger base, the acetate ion or the hypo-chlorite ion? (c) Calculate Kb values for CH3COO- and ClO-.

16.78 (a) Given that Kb for ammonia is 1.8 * 10-5 and that for hydroxylamine is 1.1 * 10-8, which is the stronger base?

16.53 A 0.100 M solution of chloroacetic acid 1ClCH2COOH2 is 11.0% ionized. Using this information, calculate 3ClCH2COO-4, 3H+4, 3ClCH2COOH4, and Ka for chloroace-tic acid.

16.54 A 0.100 M solution of bromoacetic acid 1BrCH2COOH2 is 13.2% ionized. Calculate 3H+4, 3BrCH2COO-4, 3BrCH2COOH4 and Ka for bromoacetic acid.

16.55 A particular sample of vinegar has a pH of 2.90. If acetic acid is the only acid that vinegar contains 1Ka = 1.8 * 10-52, calculate the concentration of acetic acid in the vinegar.

16.56 If a solution of HF 1Ka = 6.8 * 10-42 has a pH of 3.65, calculate the concentration of hydrofluoric acid.

16.57 The acid-dissociation constant for benzoic acid 1C6H5COOH2 is 6.3 * 10-5. Calculate the equilibrium concentrations of H3O+, C6H5COO-, and C6H5COOH in the solution if the initial concentration of C6H5COOH is 0.050 M.

16.58 The acid-dissociation constant for chlorous acid 1HClO22 is 1.1 * 10-2. Calculate the concentrations of H3O+, ClO2

-, and HClO2 at equilibrium if the initial concentration of HClO2 is 0.0125 M.

16.59 Calculate the pH of each of the following solutions (Ka and Kb values are given in Appendix D): (a) 0.095 M propionic acid 1C2H5COOH2, (b) 0.100 M hydrogen chromate ion 1HCrO4

-2, (c) 0.120 M pyridine 1C5H5N2. 16.60 Determine the pH of each of the following solutions

(Ka and Kb values are given in Appendix D): (a) 0.095 M hypochlorous acid, (b) 0.0085 M hydrazine, (c) 0.165 M hydroxylamine.

16.61 Saccharin, a sugar substitute, is a weak acid with pKa = 2.32 at 25 °C. It ionizes in aqueous solution as follows:

HNC7H4SO31aq2 ∆ H+1aq2 + NC7H4SO3-1aq2

What is the pH of a 0.10 M solution of this substance? 16.62 The active ingredient in aspirin is acetylsalicylic acid

1HC9H7O42, a monoprotic acid with Ka = 3.3 * 10-4 at 25 °C. What is the pH of a solution obtained by dissolv-ing two extra-strength aspirin tablets, containing 500 mg of acetylsalicylic acid each, in 250 mL of water?

16.63 Calculate the percent ionization of hydrazoic acid 1HN32 in solutions of each of the following concentrations (Ka is given in Appendix D): (a) 0.400 M, (b) 0.100 M, (c) 0.0400 M.

16.64 Calculate the percent ionization of propionic acid 1C2H5COOH2 in solutions of each of the following con-centrations 1Ka is given in Appendix D): (a) 0.250 M, (b) 0.0800 M, (c) 0.0200 M.

16.65 Citric acid, which is present in citrus fruits, is a triprotic acid (Table 16.3). (a) Calculate the pH of a 0.040 M solution of citric acid. (b) Did you have to make any approximations or assumptions in completing your calculations? (c) Is the con-centration of citrate ion 1C6H5O7

3-2 equal to, less than, or greater than the H+ ion concentration?

16.66 Tartaric acid is found in many fruits, including grapes, and is partially responsible for the dry texture of certain wines. Calculate the pH and the tartrate ion 1C4H4O6

2-2 concen-tration for a 0.250 M solution of tartaric acid, for which the acid-dissociation constants are listed in Table 16.3. Did you have to make any approximations or assumptions in your calculation?


16.90 Based on their compositions and structures and on conjugate acid–base relationships, select the stronger base in each of the following pairs: (a) NO3

- or NO2-, (b) PO4

3- or AsO43-,

(c) HCO3- or CO3

2-. 16.91 Indicate whether each of the following statements is true or

false. For each statement that is false, correct the statement to make it true. (a) In general, the acidity of binary acids in-creases from left to right in a given row of the periodic table. (b) In a series of acids that have the same central atom, acid strength increases with the number of hydrogen atoms bonded to the central atom. (c) Hydrotelluric acid 1H2Te2 is a stronger acid than H2S because Te is more electronegative than S.

16.92 Indicate whether each of the following statements is true or false. For each statement that is false, correct the statement to make it true. (a) Acid strength in a series of H ¬ A mol-ecules increases with increasing size of A. (b) For acids of the same general structure but differing electronegativities of the central atoms, acid strength decreases with increasing electro-negativity of the central atom. (c) The strongest acid known is HF because fluorine is the most electronegative element.

Lewis Acids and Bases (section 16.11)

16.93 If a substance is an Arrhenius base, is it necessarily a Brønsted–Lowry base? Is it necessarily a Lewis base?

16.94 If a substance is a Lewis acid, is it necessarily a Brønsted–Lowry acid? Is it necessarily an Arrhenius acid?

16.95 Identify the Lewis acid and Lewis base among the reactants in each of the following reactions:(a) Fe1ClO4231s2 + 6 H2O1l2 ∆

Fe1H2O263+1aq2 + 3 ClO4

-1aq2(b) CN-1aq2 + H2O1l2 ∆ HCN1aq2 + OH-1aq2(c) 1CH323N1g2 + BF31g2 ∆ 1CH323NBF31s2(d) HIO1lq2 + NH2

-1lq2 ∆ NH31lq2 + IO-1lq2 (lq denotes liquid ammonia as solvent)

16.96 Identify the Lewis acid and Lewis base in each of the following reactions:(a) HNO21aq2 + OH-1aq2 ∆ NO2

-1aq2 + H2O1l2(b) FeBr31s2 + Br-1aq2 ∆ FeBr4

-1aq2(c) Zn2+1aq2 + 4 NH31aq2 ∆ Zn1NH324

2+1aq2(d) SO21g2 + H2O1l2 ∆ H2SO31aq2

16.97 Predict which member of each pair produces the more acidic aqueous solution: (a) K+ or Cu2+, (b) Fe2+ or Fe3+, (c) Al3+ or Ga3+.

16.98 Which member of each pair produces the more acidic aque-ous solution: (a) ZnBr2 or CdCl2, (b) CuCl or Cu1NO322, (c) Ca1NO322 or NiBr2?

(b) Which is the stronger acid, the ammonium ion or the hy-droxylammonium ion? (c) Calculate Ka values for NH4

+ and H3NOH+.

16.79 Using data from Appendix D, calculate 3OH-4 and pH for each of the following solutions: (a) 0.10 M NaBrO, (b) 0.080 M NaHS, (c) a mixture that is 0.10 M in NaNO2 and 0.20 M in Ca1NO222.

16.80 Using data from Appendix D, calculate 3OH-4 and pH for each of the following solutions: (a) 0.105 M NaF, (b) 0.035 M Na2S, (c) a mixture that is 0.045 M in NaCH3COO and 0.055 M in Ba1CH3COO22.

16.81 A solution of sodium acetate 1NaCH3COO2 has a pH of 9.70. What is the molarity of the solution?

16.82 Pyridinium bromide 1C5H5NHBr2 is a strong electrolyte that dissociates completely into C5H5NH+ and Br-. A solution of pyridinium bromide has a pH of 2.95. What is the molarity of the solution?

16.83 Predict whether aqueous solutions of the following com-pounds are acidic, basic, or neutral: (a) NH4Br, (b) FeCl3, (c) Na2CO3, (d) KClO4, (e) NaHC2O4.

16.84 Predict whether aqueous solutions of the following sub-stances are acidic, basic, or neutral: (a) AlCl3, (b) NaBr, (c) NaClO, (d) 3CH3NH34NO3, (e) Na2SO3.

16.85 An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the pH of the solution is 8.08. What is the identity of the salt?

16.86 An unknown salt is either KBr, NH4Cl, KCN, or K2CO3. If a 0.100 M solution of the salt is neutral, what is the identity of the salt?

Acid–Base character and chemical structure (section 16.10)

16.87 Explain the following observations: (a) HNO3 is a stron-ger acid than HNO2; (b) H2S is a stronger acid than H2O; (c) H2SO4 is a stronger acid than HSO4

-; (d) H2SO4 is a stron-ger acid than H2SeO4; (e) CCl3COOH is a stronger acid than CCl3COOH.

16.88 Explain the following observations: (a) HCl is a stronger acid than H2S; (b) H3PO4 is a stronger acid than H3AsO4; (c) HBrO3 is a stronger acid than HBrO2; (d) H2C2O4 is a stronger acid than HC2O4 -; (e) benzoic acid 1C6H5COOH2 is a stronger acid than phenol 1C6H5OH2.

16.89 Based on their compositions and structures and on conju-gate acid–base relationships, select the stronger base in each of the following pairs: (a) BrO- or ClO-, (b) BrO- or BrO2

-, (c) HPO4

2- or H2PO4-.

Additional Exercises 16.99 Indicate whether each of the following statements is correct or

incorrect.(a) Every Brønsted–Lowry acid is also a Lewis acid.(b) Every Lewis acid is also a Brønsted–Lowry acid.(c) Conjugate acids of weak bases produce more acidic

solutions than conjugate acids of strong bases.

(d) K+ ion is acidic in water because it causes hydrating water molecules to become more acidic.

(e) The percent ionization of a weak acid in water increases as the concentration of acid decreases.

16.100 A solution is made by adding 0.300 g Ca1OH221s2, 50.0 mL of 1.40 M HNO3, and enough water to make a final volume

Additional exercises 721

16.107 Oxalic acid 1H2C2O42 is a diprotic acid. By using data in Ap-pendix D as needed, determine whether each of the following statements is true: (a) H2C2O4 can serve as both a Brønsted–Lowry acid and a Brønsted–Lowry base. (b) C2O4

2- is the conjugate base of HC2O4

-. (c) An aqueous solution of the strong electrolyte KHC2O4 will have pH 6 7.

16.108 Succinic acid 1H2C4H6O42, which we will denote H2Suc, is a biologically relevant diprotic acid with the structure shown below. It is closely related to tartaric acid and malic acid (Figure 16.1). At 25 °C, the acid-dissociation constants for succinic acid are Ka1 = 6.9 * 10-5 and Ka2 = 2.5 * 10-6. (a) Determine the pH of a 0.32 M solution of H2Suc at 25 °C, assuming that only the first dissociation is relevant. (b) Deter-mine the molar concentration of Suc2 - in the solution in part (a). (c) Is the assumption you made in part (a) justified by the re-sult from part (b)? (d) Will a solution of the salt NaHSuc be acidic, neutral, or basic?

OH

O

CC

CC

O

OH

H H

HH

16.109 Butyric acid is responsible for the foul smell of rancid but-ter. The pKa of butyric acid is 4.84. (a) Calculate the pKb for the butyrate ion. (b) Calculate the pH of a 0.050 M solution of butyric acid. (c) Calculate the pH of a 0.050 M solution of sodium butyrate.

16.110 Arrange the following 0.10 M solutions in order of increas-ing acidity (decreasing pH): (i) NH4NO3, (ii) NaNO3, (iii) CH3COONH4, (iv) NaF, (v) CH3COONa.

16.111 A 0.25 M solution of a salt NaA has pH = 9.29. What is the value of Ka for the parent acid HA?

[16.112] The following observations are made about a diprotic acid H2A: (i) A 0.10 M solution of H2A has pH = 3.30. (ii) A 0.10 M solution of the salt NaHA is acidic. Which of the following could be the value of pKa2 for H2A: (i) 3.22, (ii) 5.30, (iii) 7.47, or (iv) 9.82?

16.113 Many moderately large organic molecules containing basic nitrogen atoms are not very soluble in water as neutral mol-ecules, but they are frequently much more soluble as their acid salts. Assuming that pH in the stomach is 2.5, indicate whether each of the following compounds would be present in the stomach as the neutral base or in the protonated form: nicotine, Kb = 7 * 10-7; caffeine, Kb = 4 * 10-14; strych-nine, Kb = 1 * 10-6; quinine, Kb = 1.1 * 10-6.

16.114 The amino acid glycine 1H2N ¬ CH2 ¬ COOH2 can partici-pate in the following equilibria in water:

H2N ¬ CH2 ¬ COOH + H2O ∆H2N ¬ CH2 ¬ COO- + H3O+ Ka = 4.3 * 10-3

H2N ¬ CH2 ¬ COOH + H2O ∆+H3N ¬ CH2 ¬ COOH + OH- Kb = 6.0 * 10-5

(a) Use the values of Ka and Kb to estimate the equilibrium constant for the intramolecular proton transfer to form a zwitterion:

H2N ¬ CH2 ¬ COOH ∆ +H3N ¬ CH2 ¬ COO-

of 75.0 mL Assuming that all of the solid dissolves, what is the pH of the final solution?

16.101 The odor of fish is due primarily to amines, especially methylamine 1CH3NH22. Fish is often served with a wedge of lemon, which contains citric acid. The amine and the acid react forming a product with no odor, thereby mak-ing the less-than-fresh fish more appetizing. Using data from Appendix D, calculate the equilibrium constant for the reaction of citric acid with methylamine, if only the first proton of the citric acid 1Ka12 is important in the neutraliza-tion reaction.

[16.102] For solutions of a weak acid, a graph of pH versus the loga-rithm of the initial acid concentration should be a straight line. What is the magnitude of the slope of that line?

16.103 Hemoglobin plays a part in a series of equilibria involving protonation-deprotonation and oxygenation-deoxygenation. The overall reaction is approximately as follows:

HbH+1aq2 + O21aq2 ∆ HbO21aq2 + H+1aq2

where Hb stands for hemoglobin and HbO2 for oxyhemoglobin. (a) The concentration of O2 is higher in the lungs and lower in the tissues. What effect does high 3O24 have on the posi-tion of this equilibrium? (b) The normal pH of blood is 7.4. Is the blood acidic, basic, or neutral? (c) If the blood pH is low-ered by the presence of large amounts of acidic metabolism products, a condition known as acidosis results. What effect does lowering blood pH have on the ability of hemoglobin to transport O2?

[16.104] Calculate the pH of a solution made by adding 2.50 g of lithium oxide 1Li2O2 to enough water to make 1.500 L of solution.

16.105 Benzoic acid 1C6H5COOH2 and aniline 1C6H5NH22 are both derivatives of benzene. Benzoic acid is an acid with Ka = 6.3 * 10-5 and aniline is a base with Ka = 4.3 * 10-10.

HHN

OHOC

Benzoic acid Aniline

(a) What are the conjugate base of benzoic acid and the con-jugate acid of aniline? (b) Anilinium chloride 1C6H5NH3Cl2 is a strong electrolyte that dissociates into anilinium ions 1C6H5NH3

+2 and chloride ions. Which will be more acidic, a 0.10 M solution of benzoic acid or a 0.10 M solution of ani-linium chloride? (c) What is the value of the equilibrium con-stant for the following equilibrium?

C6H5COOH1aq2 + C6H5NH21aq2 ∆C6H5COO-1aq2 + C6H5NH3

+1aq2

16.106 What is the pH of a solution that is 2.5 * 10-9 M in NaOH? Does your answer make sense? What assumption do we nor-mally make that is not valid in this case?


Using Lewis structures as the basis of your discussion, explain the observed trend in acidities in the series. Calculate the pH of a 0.010 M solution of each acid.

(b) What is the pH of a 0.050 M aqueous solution of glycine? (c) What would be the predominant form of glycine in a solu-tion with pH 13? With pH 1?

16.115 The structural formula for acetic acid is shown in Table 16.2. Replacing hydrogen atoms on the carbon with chlorine atoms causes an increase in acidity, as follows:

Acid Formula Ka 125 °C 2Acetic CH3COOH 1.8 * 10-5

Chloroacetic CH2ClCOOH 1.4 * 10-3

Dichloroacetic CHCl2COOH 3.3 * 10-2

Trichloroacetic CCl3COOH 2 * 10-1

integrative Exercises 16.116 Calculate the number of H+1aq2 ions in 1.0 mL of pure water

at 25 °C. 16.117 How many milliliters of concentrated hydrochloric acid solu-

tion (36.0% HCl by mass, density = 1.18 g>mL) are required to produce 10.0 L of a solution that has a pH of 2.05?

16.118 The volume of an adult’s stomach ranges from about 50 mL when empty to 1 L when full. If the stomach volume is 400 mL and its contents have a pH of 2, how many moles of H+ does the stomach contain? Assuming that all the H+ comes from HCl, how many grams of sodium hydrogen carbonate will totally neutralize the stomach acid?

16.119 Atmospheric CO2 levels have risen by nearly 20% over the past 40 years from 315 ppm to 380 ppm. (a) Given that the average pH of clean, unpolluted rain today is 5.4, determine the pH of unpolluted rain 40 years ago. Assume that carbonic acid 1H2CO32 formed by the reaction of CO2 and water is the only factor influencing pH.

CO21g2 + H2O1l2 ∆ H2CO31aq2 (b) What volume of CO2 at 25 °C and 1.0 atm is dissolved in a

20.0-L bucket of today’s rainwater? 16.120 At 50 °C, the ion-product constant for H2O has the value

Kw = 5.48 * 10-14. (a) What is the pH of pure water at 50 °C? (b) Based on the change in Kw with temperature, pre-dict whether ∆H is positive, negative, or zero for the autoion-ization reaction of water:

2 H2O1l2 ∆ H3O+1aq2 + OH-1aq2 16.121 In many reactions the addition of AlCl3 produces the same

effect as the addition of H+. (a) Draw a Lewis structure for AlCl3 in which no atoms carry formal charges, and determine its structure using the VSEPR method. (b) What characteris-tic is notable about the structure in part (a) that helps us un-derstand the acidic character of AlCl3? (c) Predict the result of the reaction between AlCl3 and NH3 in a solvent that does not participate as a reactant. (d) Which acid–base theory is most suitable for discussing the similarities between AlCl3 and H+?

16.122 What is the boiling point of a 0.10 M solution of NaHSO4 if the solution has a density of 1.002 g>mL?

16.123 Use average bond enthalpies from Table 8.4 to estimate the enthalpies of the following gas-phase reactions:

Reaction 1: HF1g2 + H2O1g2 ∆ F-1g2 + H3O+1g2Reaction 2: HCl1g2 + H2O1g2 ∆ Cl-1g2 + H3O+1g2

Are both reactions exothermic? How do these values relate to the different strengths of hydrofluoric and hydrochloric acid?

16.124 Cocaine is a weak organic base whose molecular formula is C17H21NO4. An aqueous solution of cocaine was found to have a pH of 8.53 and an osmotic pressure of 52.7 torr at 15 °C. Calculate Kb for cocaine.

[16.125] The iodate ion is reduced by sulfite according to the following reaction:

IO3-1aq2 + 3SO3

2-1aq2 ¡ I-1aq2 + 3 SO42-1aq2

The rate of this reaction is found to be first order in IO3-, first

order in SO32-, and first order in H+. (a) Write the rate law for

the reaction. (b) By what factor will the rate of the reaction change if the pH is lowered from 5.00 to 3.50? Does the reac-tion proceed more quickly or more slowly at the lower pH? (c) By using the concepts discussed in Section 14.6, explain how the reaction can be pH-dependent even though H+ does not appear in the overall reaction.

16.126 (a) Using dissociation constants from Appendix D, determine the value for the equilibrium constant for each of the follow-ing reactions.(i) HCO3

-1aq2 + OH-1aq2 ∆ CO32-1aq2 + H2O1l2

(ii) NH4+1aq2 + CO3

2-1aq2 ∆ NH31aq2 + HCO3-1aq2

(b) We usually use single arrows for reactions when the for-ward reaction is appreciable (K much greater than 1) or when products escape from the system, so that equilibrium is never established. If we follow this convention, which of these equi-libria might be written with a single arrow?

Design an experiment 723

design an ExperimentYour professor gives you a bottle that contains a clear liquid. You are told that the liquid is a pure substance that is volatile, soluble in water, and might be an acid or a base. Design experi-ments to elucidate the following about this unknown sample. (a) Determine whether the sub-stance in the sample is an acid or a base. (b) Suppose the substance is an acid. How would you determine whether it is a strong acid or a weak acid? (c) If the substance is a weak acid, how would you determine the value of Ka for the substance? (d) Suppose the substance were a weak acid and that you are also given a solution of NaOH(aq) of known molarity. What procedure would you use to isolate a pure sample of the sodium salt of the substance? (e) Now suppose that the substance were a base rather than an acid. How would you adjust the procedures in parts (b) and (c) to determine if the substance were a strong or weak base, and, if weak, the value of Kb?

17Additional Aspects of Aqueous EquilibriaWater, the most common and most important solvent on Earth, occupies its position of importance because of its abundance and its exceptional ability to dissolve a wide variety of substances. Coral reefs are a striking example of aqueous chemistry at work in nature. Coral reefs are built by tiny animals called stony corals, which secrete a hard calcium carbonate exoskeleton. Over time, the stony corals build up large networks of calcium carbonate upon which a reef is built. The size of such structures can be immense, as illustrated by the Great Barrier Reef.

17.3 Acid–BAse TiTrATions We examine acid–base titrations and explore how to determine pH at any point in an acid–base titration.

17.4 soluBiliTy equiliBriA We learn how to use solubility-product constants to determine to what extent a sparingly soluble salt dissolves in water.

17.5 FAcTors ThAT AFFecT soluBiliTy We investigate some of the factors that affect solubility, including the common-ion effect and the effect of acids.

17.1 The common-ion eFFecT We begin by considering a specific example of Le Châtelier’s principle known as the common-ion effect.

17.2 BuFFers We consider the composition of buffered solutions and learn how they resist pH change when small amounts of a strong acid or strong base are added to them.

WhAt’s AhEAd

▶ A microscopic view of corAl And sAnd. This image shows coral and sand particles, magnified by 100x.

Stony corals make their exoskeletons from dissolved Ca2 + and CO32 - ions. This pro-

cess is aided by the fact that the CO32 - concentration is supersaturated in most parts

of the ocean. However, well documented increases in the amount of CO2 in the atmo-sphere threaten to upset the aqueous chemistry that stony corals depend on. As atmo-spheric CO2 levels increase, the amount of CO2 dissolved in the ocean also increases. This lowers the pH of the ocean and leads to a decrease in the CO3

2 - concentration. As a result it becomes more difficult for stony corals and other important ocean creatures to maintain their exoskeletons. We will take a closer look at the consequences of ocean acidification later in the chapter.

To understand the chemistry that underlies coral reef formation and other pro-cesses in the ocean and in aqueous systems such as living cells, we must understand the concepts of aqueous equilibria. In this chapter, we take a step toward understand-ing such complex solutions by looking first at further applications of acid–base equi-libria. The idea is to consider not only solutions in which there is a single solute but

17.7 quAliTATive AnAlysis For meTAllic elemenTs We explain how the principles of solubility and complexation equilibria can be used to identify ions in solution.

17.6 PreciPiTATion And sePArATion oF ions We learn how differences in solubility can be used to separate ions through selective precipitation.

726 CHapter 17 additional aspects of aqueous equilibria

also those containing a mixture of solutes. We then broaden our discussion to include two additional types of aqueous equilibria: those involving slightly soluble salts and those involving the formation of metal complexes in solution. For the most part, the discussions and calculations in this chapter are extensions of those in Chapters 15 and 16.

17.1 | the Common-Ion EffectIn Chapter 16, we examined the equilibrium concentrations of ions in solutions con-taining a weak acid or a weak base. We now consider solutions that contain a weak acid, such as acetic acid 1CH3COOH2, and a soluble salt of that acid, such as sodium acetate 1CH3COONa2. Notice that these solutions contain two substances that share a common ion, CH3COO-. It is instructive to view these solutions from the perspective of Le Châtelier’s principle. (Section 15.7)

Sodium acetate is a soluble ionic compound and therefore a strong electrolyte. (Section 4.1) Consequently, it dissociates completely in aqueous solution to form

Na+ and CH3COO- ions:

CH3COONa1aq2 ¡ Na+1aq2 + CH3COO-1aq2In contrast, CH3COOH is a weak electrolyte that ionizes only partially, represented by the dynamic equilibrium

CH3COOH1aq2 ∆ H+1aq2 + CH3COO-1aq2 [17.1]

The equilibrium constant for Equation 17.1 is Ka = 1.8 * 10-5 at 25 °C (Table 16.2). If we add sodium acetate to a solution of acetic acid in water, the CH3COO- from CH3COONa causes the equilibrium concentrations of the substances in Equation 17.1 to shift to the left, thereby decreasing the equilibrium concentration of H+1aq2:

CH3COOH1aq2 ∆ H+1aq2 + CH3COO-1aq2

In other words, the presence of the added acetate ion causes the acetic acid to ionize less than it normally would. The equilibrium constant itself does not change; it is the relative concentrations of products and reactants in the equilibrium expression that change.

Whenever a weak electrolyte and a strong electrolyte containing a common ion are together in solution, the weak electrolyte ionizes less than it would if it were alone in solu-tion. We call this observation the common-ion effect.

solutIonAnalyze We are asked to determine the pH of a solution of a weak electrolyte 1CH3COOH2 and a strong electrolyte 1CH3COONa2 that share a common ion, CH3COO-.

sAmplE ExErCIsE 17.1 Calculating the ph When a Common Ion Is Involved

What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution?

Plan In any problem in which we must determine the pH of a solution containing a mixture of solutes, it is helpful to proceed by a series of logical steps:(1) Consider which solutes are strong electrolytes and which are

weak electrolytes, and identify the major species in solution.

(2) Identify the important equilibrium reaction that is the source of H+ and therefore determines pH.

(3) Tabulate the concentrations of ions involved in the equilibrium.(4) Use the equilibrium-constant expression to calculate 3H+4 and

then pH.

Addition of CH3COO− shifts equilibrium concentrations, lowering [H+]

seCtion 17.1 the Common-ion effect 727

Solve First, because CH3COOH is a weak electrolyte and CH3COONa is a strong electrolyte, the major species in the solution are CH3COOH (a weak acid), Na+ (which is neither acidic nor basic and is therefore a spectator in the acid–base chemistry), and CH3COO- (which is the con-jugate base of CH3COOH).

Second, 3H+4 and, therefore, the pH are controlled by the dissociation equilibrium of CH3COOH: CH3COOH1aq2 ∆ H+1aq2 + CH3COO-1aq2(We have written the equilibrium using H+1aq2 rather than H3O+1aq2, but both rep-resentations of the hydrated hydrogen ion are equally valid.)Third, we tabulate the initial and equilibrium concentrations as we did in solving other equi-librium problems in Chapters 15 and 16:

CH3COOH1aq2 ∆ H+1aq2 + CH3COO-1aq2

Initial (M) 0.30 0 0.30Change (M) -x +x +xEquilibrium (M) (0.30 - x) x 10.30 + x2

The equilibrium concentration of CH3COO- (the common ion) is the initial concentration that is due to CH3COONa (0.30 M) plus the change in concentration (x) that is due to the ionization of CH3COOH.Now we can use the equilibrium-constant expression: Ka = 1.8 * 10-5 =

3H+43CH3COO-43CH3COOH4

The dissociation constant for CH3COOH at 25 °C is from Table 16.2, or Appendix D; addi-tion of CH3COONa does not change the value of this constant. Substituting the equilibrium-constant concentrations from our table into the equilibrium expression gives: Ka = 1.8 * 10-5 =

x10.30 + x20.30 - x

Because Ka is small, we assume that x is small compared to the original concentrations of CH3COOH and CH3COO- (0.30 M each). Thus, we can ignore the very small x relative to 0.30 M, giving Ka = 1.8 * 10-5 =

x10.3020.30

x = 1.8 * 10-5 M = 3H+4The resulting value of x is indeed small relative to 0.30, justifying the approximation made in simplifying the problem.Finally, we calculate the pH from the equilibrium concentration of H+1aq2: pH = - log11.8 * 10-52 = 4.74

Comment In Section 16.6, we calculated that a 0.30 M solution of CH3COOH has a pH of 2.64, corresponding to 3H+4 = 2.3 * 10-3 M. Thus, the addition of CH3COONa has substantially decreased 3H+4, as we expect from Le Châtelier’s principle.

practice Exercise 1For the generic equilibrium HA 1aq2 ∆ H+ 1aq2 + A-1aq2, which of these statements is true?(a) The equilibrium constant for this reaction changes as the pH changes.(b) If you add the soluble salt KA to a solution of HA that is at equilibrium, the concentra-

tion of HA would decrease.(c) If you add the soluble salt KA to a solution of HA that is at equilibrium, the concentration

of A- would decrease.(d) If you add the soluble salt KA to a solution of HA that is at equilibrium, the pH would

increase.

practice Exercise 2Calculate the pH of a solution containing 0.085 M nitrous acid 1HNO2, Ka = 4.5 * 10-42 and 0.10 M potassium nitrite 1KNO22.


Comment Notice that for all practical purposes, the hydrogen ion concentration is due entirely to the HCl; the HF makes a negligible contribution by comparison.

practice Exercise 1Calculate the concentration of the lactate ion in a solution that is 0.100 M in lactic acid 1CH3CH(OH2COOH, pKa = 3.86) and 0.080 M in HCl.(a) 4.83 M, (b) 0.0800 M, (c) 7.3 * 10-3 M, (d) 3.65 * 10-3 M, (e) 1.73 * 10-4 M.

practice Exercise 2Calculate the formate ion concentration and pH of a solution that is 0.050 M in formic acid 1HCOOH, Ka = 1.8 * 10-42 and 0.10 M in HNO3.

sAmplE ExErCIsE 17.2 Calculating Ion Concentrations When a Common Ion Is Involved

Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.

solutIonAnalyze We are asked to determine the concentration of F- and the pH in a solution containing the weak acid HF and the strong acid HCl. In this case the common ion is H+.Plan We can again use the four steps outlined in Sample Exercise 17.1.

Solve Because HF is a weak acid and HCl is a strong acid, the major species in solution are HF, H+, and Cl-. The Cl-, which is the conju-gate base of a strong acid, is merely a spectator ion in any acid–base chemistry. The problem asks for 3F-4, which is formed by ionization of HF. Thus, the important equilibrium is HF1aq2 ∆ H+1aq2 + F-1aq2The common ion in this problem is the hydro-gen (or hydronium) ion. Now we can tabulate the initial and equilibrium concentrations of each species involved in this equilibrium:

HF1aq2 ∆ H+1aq2 + F-1aq2Initial (M) 0.20 0.10 0Change (M) -x +x +xEquilibrium (M) 10.20 - x2 10.10 + x2 x

The equilibrium constant for the ionization of HF, from Appendix D, is 6.8 * 10-4. Substitut-ing the equilibrium-constant concentrations into the equilibrium expression gives Ka = 6.8 * 10-4 =

3H+43F-43HF4 =

10.10 + x21x20.20 - x

If we assume that x is small relative to 0.10 or 0.20 M, this expression simplifies to

10.1021x20.20

= 6.8 * 10-4

x =0.200.10

16.8 * 10-42 = 1.4 * 10-3 M = 3F-4This F- concentration is substantially smaller than it would be in a 0.20 M solution of HF with no added HCl. The common ion, H+, suppresses the ionization of HF. The concen-tration of H+1aq2 is 3H+4 = 10.10 + x2 M ≃ 0.10 MThus, pH = 1.00

Sample Exercises 17.1 and 17.2 both involve weak acids. The ionization of a weak base is also decreased by the addition of a common ion. For example, the addition of NH4

+ (as from the strong electrolyte NH4Cl) causes the equilibrium concentrations of the reagents in Equation 17.2 to shift to the left, decreasing the equilibrium concentration of OH- and lowering the pH: NH31aq2 + H2O1l2 ∆ NH4

+1aq2 + OH-1aq2 [17.2]

Addition of NH4+ shifts equilibrium

concentrations, lowering [OH−]

seCtion 17.2 Buffers 729

Give it some ThoughtIf solutions of NH4Cl1aq2 and NH31aq2 are mixed, which ions in the resulting solution are spectator ions in any acid–base chemistry occurring in the solution? What equilibrium reaction determines 3OH-4 and, therefore, the pH of the solution?

17.2 | BuffersSolutions that contain high concentrations (10-3 M or more) of a weak conjugate acid–base pair and that resist drastic changes in pH when small amounts of strong acid or strong base are added to them are called buffered solutions (or merely buffers). Human blood, for example, is a complex buffered solution that maintains the blood pH at about 7.4. (Section 17.2, “Blood as a Buffered Solution”) Much of the chemical be-havior of seawater is determined by its pH, buffered at about 8.1 to 8.3 near the surface. (Section 17.5, “Ocean Acidification”) Buffers find many important applications in the laboratory and in medicine (▶ Figure 17.1). Many biological reactions occur at the op-timal rates only when properly buffered. If you ever work in a biochemistry lab, you will very likely to have prepare specific buffers in which to run your biochemical reactions.

Composition and Action of BuffersA buffer resists changes in pH because it contains both an acid to neutralize added OH- ions and a base to neutralize added H+ ions. The acid and base that make up the buffer, however, must not consume each other through a neutralization reaction.

(Section 4.3) These requirements are fulfilled by a weak acid–base conjugate pair, such as CH3COOH>CH3COO- or NH4

+>NH3. The key is to have roughly equal con-centrations of both the weak acid and its conjugate base.

There are two ways to make a buffer:

1. Mix a weak acid or a weak base with a salt of that acid or base. For example, the CH3COOH>CH3COO- buffer can be prepared by adding CH3COONa to a solu-tion of CH3COOH. Similarly, the NH4 +>NH3 buffer can be prepared by adding NH4Cl to a solution of NH3.

2. Make the conjugate acid or base from a solution of weak base or acid by the ad-dition of strong acid or base. For example, to make the CH3COOH>CH3COO- buffer, you could start with a solution of CH3COOH and add some NaOH to the solution—enough to neutralize about half of CH3COOH according to the reaction.

CH3COOH + OH- ¡ CH3COO- + H2O

(Section 4.3) Neutralization reactions have very large equilibrium constants, and so the amount of acetate formed will only be limited by the relative amounts of the acid and strong base that are mixed. The resulting solution is the same as if you added sodium acetate to the acetic acid solution: You will have comparable quantities of both acetic acid and its conjugate base in solution.

By choosing appropriate components and adjusting their relative concentrations, we can buffer a solution at virtually any pH.

Give it some ThoughtWhich of these conjugate acid–base pairs will not function as a buffer: C2H5COOH and C2H5COO-, HCO3

- and CO32 -, or HNO3 and NO3 -? Explain.

To understand how a buffer works, let’s consider one composed of a weak acid HA and one of its salts MA, where M+ could be Na+, K+, or any other cation that does not react with water. The acid-dissociation equilibrium in this buffered solution involves both the acid and its conjugate base:

HA1aq2 ∆ H+1aq2 + A-1aq2 [17.3]

▲ Figure 17.1 standard buffers. For laboratory work, prepackaged buffers at specific pH values can be purchased.

2 4 6 9 12


The corresponding acid-dissociation-constant expression is

Ka =3H+43A-43HA4 [17.4]

Solving this expression for 3H+4, we have

3H+4 = Ka3HA43A-4 [17.5]

We see from this expression that 3H+4 and, thus, the pH are determined by two factors: the value of Ka for the weak-acid component of the buffer and the ratio of the concen-trations of the conjugate acid–base pair, 3HA4>3A-4.

If OH- ions are added to this buffered solution, they react with the buffer acid component to produce water and A-:

OH-1aq2 + HA1aq2 ¡ H2O1l2 + A-1aq2 [17.6]added base

This neutralization reaction causes [HA] to decrease and 3A-4 to increase. As long as the amounts of HA and A- in the buffer are large relative to the amount of OH- added, the ratio 3HA4>3A-4 does not change much and, thus, the change in pH is small.

If H+ ions are added, they react with the base component of the buffer:

H +1aq2 + A-1aq2 ¡ HA1aq2 [17.7]added acid

This reaction can also be represented using H3O+:

H3O+1aq2 + A-1aq2 ¡ HA1aq2 + H2O1l2

Using either equation, we see that this reaction causes 3A-4 to decrease and [HA] to increase. As long as the change in the ratio 3HA4>3A-4 is small, the change in pH will be small.

▼ Figure 17.2 shows an HA/A- buffer consisting of equal concentrations of hydro-fluoric acid and fluoride ion (center). The addition of OH- reduces [HF] and increases 3F-4, whereas the addition of 3H+4 reduces 3F-4 and increases 3HF4.

▲ Figure 17.2 Buffer action. The pH of an HF>F- buffered solution changes by only a small amount in response to addition of an acid or base.

After additionof OH−

Add OH− Add H+

After additionof H+

Initial buffered solution

HF F− HF

[HF] = [F−]

F−+ H+

F− HF F−

HFHF +H2O F−+ OH−

Presence of HF counteracts addition of base; pH increase is small

Presence of F– counteracts addition of acid; pH decrease is small


sAmplE ExErCIsE 17.3 Calculating the ph of a Buffer

What is the pH of a buffer that is 0.12 M in lactic acid 3CH3CH1OH2COOH, or HC3H5O34 and 0.10 M in sodium lactate 3CH3CH1OH2COONa or NaC3H5O34? For lactic acid, Ka = 1.4 * 10-4.

solutIonAnalyze We are asked to calculate the pH of a buffer containing lactic acid 1HC3H5O32 and its conjugate base, the lactate ion 1C3H5O3

-2.

Plan We will first determine the pH using the method described in Section 17.1. Because HC3H5O3 is a weak electrolyte and NaC3H5O3 is a strong electrolyte, the major species in solution are HC3H5O3, Na+,

and C3H5O3-. The Na+ ion is a spectator ion. The HC3H5O3>C3H5O3

- conjugate acid–base pair determines 3H+4 and, thus, pH; 3H+4 can be determined using the acid-dissociation equilibrium of lactic acid.

It is possible to overwhelm a buffer by adding too much strong acid or strong base. We will examine this in more detail a little later in this chapter.

Give it some Thought(a) What happens when NaOH is added to a buffer composed of

CH3COOH and CH3COO-?(b) What happens when HCl is added to this buffer?

Calculating the pH of a BufferBecause conjugate acid–base pairs share a common ion, we can use the same proce-dures to calculate the pH of a buffer that we used to treat the common-ion effect in Sample Exercise 17.1. Alternatively, we can take an approach based on an equation de-rived from Equation 17.5. Taking the negative logarithm of both sides of Equation 17.5, we have

- log[H+4 = - logaKa3HA43A-4 b = - logKa - log

3HA43A-4

Because - log3H+4 = pH and - logKa = pKa, we have

pH = pKa - log 3HA43A-4 = pKa + log

3A-43HA4 [17.8]

(Remember the logarithm rules in Appendix A.2, if you are not sure how this calcula-tion works.)

In general,

pH = pKa + log 3base43acid4 [17.9]

where [acid] and [base] refer to the equilibrium concentrations of the conjugate acid–base pair. Note that when 3base4 = 3acid4, we have pH = pKa.

Equation 17.9 is known as the Henderson–Hasselbalch equation. Biologists, biochemists, and others who work frequently with buffers often use this equation to calculate the pH of buffers. In doing equilibrium calculations, we have seen that we can normally neglect the amounts of the acid and base of the buffer that ionize. Therefore, we can usually use the initial concentrations of the acid and base components of the buffer directly in Equation 17.9, as seen in Sample Exercise 17.3. However, the assump-tion that the initial concentrations of the acid and base components in the buffer are equal to the equilibrium concentrations is just that: an assumption. There may be times when you will need to be more careful, as seen in Sample Exercise 17.4.


Solve The initial and equilibrium concentra-tions of the species involved in this equilib-rium are

Because Ka is small and a common ion is pres-ent, we expect x to be small relative to either 0.12 or 0.10 M. Thus, our equation can be simplified to give Ka = 1.4 * 10-4 =

x10.1020.12

Solving for x gives a value that justifies our approximation: 3H+4 = x = a 0.12

0.10b11.4 * 10-42 = 1.7 * 10-4 M

Alternatively, we can use the Henderson– Hasselbalch equation with the initial concentra-tions of acid and base to calculate pH directly:

pH = pKa + log3base43acid4 = 3.85 + loga 0.10

0.12b

= 3.85 + 1-0.082 = 3.77

practice Exercise 1If the pH of a buffer solution is equal to the pKa of the acid in the buffer, what does this tell you about the relative concentrations of the acid and conjugate base forms of the buffer components?(a) The acid concentration must be zero. (b) The base concentration must be zero.(c) The acid and base concentrations must be equal. (d) The acid and base concentrations must be equal to the Ka. (e) The base concentration must be 2.3 times as large as the acid concentration.

practice Exercise 2Calculate the pH of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. (Refer to Appendix D.)

sAmplE ExErCIsE 17.4 Calculating ph When the henderson–hasselbalch Equation

may not Be Accurate

Calculate the pH of a buffer that initially contains 1.00 * 10-3 M CH3COOH and 1.00 * 10-4 M CH3COONa in the following two ways: (i) using the Henderson–Hasselbalch equation and (ii) making no assumptions about quantities (which means you will need to use the quadratic equation). The Ka of CH3COOH is 1.80 * 10-5.

solutIonAnalyze We are asked to calculate the pH of a buffer two different ways. We know the initial concen-trations of the weak acid and its conjugate base, and the Ka of the weak acid.Plan We will first use the Henderson–Hasselbalch equation, which relates pKa and ratio of acid–base concentrations to the pH. This will be straightforward. Then, we will redo the calculation making no assumptions about any quantities, which means we will need to write out the initial/change/equilibrium concentrations, as we have done before. In addition, we will need to solve for quantities using the quadratic equation (since we cannot make assumptions about unknowns being small).

Solve

(i) The Henderson–Hasselbalch equation is pH = pKa + log3base43acid4

We know the Ka of the acid 11.8 * 10-52, so we know pKa 1pKa = - log Ka = 4.742. We know the initial concentrations of the base, sodium acetate, and the acid, acetic acid, which we will assume are the same as the equilibrium concentrations.

Initial (M) 0.12 0 0.10Change (M) -x +x +xEquilibrium (M) (0.12 - x) x 10.10 + x2

CH3CH1OH2COOH 1aq2 ∆ H+ 1aq2 + CH3CH1OH2COO-1aq2

The equilibrium concentrations are governed by the equilibrium expression: Ka = 1.4 * 10-4 =

3H+43C3H5O3-4

3HC3H5O34 =x10.10 + x2

0.12 - x

Then, we can solve for pH: pH = - log11.7 * 10-42 = 3.77


Therefore, we havepH = 4.74 + log

11.00 * 10 - 4211.00 * 10 - 32

Therefore, pH = 4.74 - 1.00 = 3.74

(ii) Now we will redo the calculation, with-out making any assumptions at all. We will solve for x, which represents the H+ concentration at equilibrium, in order to calculate pH.

CH3COOH 1aq2 ∆ CH3COO-1aq2 + H+ 1aq2

Initial (M) 1.00 * 10-3 1.00 * 10-4 0

Change (M) -x +x +xEquilibrium (M) 11.00 * 10-3 - x2 11.00 * 10-4 + x2 x

3CH3COO-4[H+ 43CH3COOH]

= Ka

11.00 * 10-4 + x21x211.00 * 10-3 - x2 = 1.8 * 10-5

1.00 * 10-4x + x2 = 1.8 * 10-511.00 * 10-3 - x2 x2 + 1.00 * 10-4x = 1.8 * 10-8 - 1.8 * 10-5x

x2 + 1.18 * 10-4x - 1.8 * 10-8 = 0

x =-1.18 * 10-4 { 211.18 * 10-422 - 41121-1.8 * 10-82

2112

=-1.18 * 10-4 { 28.5924 * 10-8

2

= 8.76 * 10-5 = 3H+4 pH = 4.06

Comment In Sample Exercise 17.3, the calculated pH is the same whether we solve exactly using the quadratic equation or make the simplifying as-sumption that the equilibrium concentrations of acid and base are equal to their initial concentrations. The simplifying assumption works because the concentrations of the acid–base conjugate pair are both a thousand times larger than Ka. In this Sample Exercise, the acid–base conjugate pair concentrations are only 10–100 as large as Ka. Therefore, we cannot

assume that x is small compared to the initial concentrations (that is, that the initial concentrations are essentially equal to the equilibrium concentrations). The best answer to this Sample Exercise is pH = 4.06, obtained without assuming x is small. Thus we see that the assumptions behind the Henderson–Hasselbalch equation become less accurate as the acid/base becomes stronger and/or its concentration becomes smaller.

practice Exercise 1A buffer is made with sodium acetate 1CH3COONa2 and acetic acid 1CH3COOH2; the Ka for acetic acid is 1.80 * 10-5. The pH of the buffer is 3.98. What is the ratio of the equilib-rium concentration of sodium acetate to that of acetic acid? (a) -0.760, (b) 0.174, (c) 0.840, (d) 5.75, (e) Not enough information is given to answer this question.

practice Exercise 2Calculate the final, equilibrium pH of a buffer that initially contains 6.50 * 10-4 M HOCl and 7.50 * 10-4 M NaOCl. The Ka of HOCl is 3.0 * 10-5.

In Sample Exercise 17.3 we calculated the pH of a buffered solution. Often we will need to work in the opposite direction by calculating the amounts of the acid and its conjugate base needed to achieve a specific pH. This calculation is illustrated in Sample Exercise 17.5.

solutIonAnalyze We are asked to determine the amount of NH4

+ ion required to prepare a buffer of a specific pH.

sAmplE ExErCIsE 17.5 preparing a Buffer

How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH is 9.00? (Assume that the addition of NH4Cl does not change the volume of the solution.)


Plan The major species in the solution will be NH4+, Cl-, and NH3. Of these, the Cl- ion is a

spectator (it is the conjugate base of a strong acid). Thus, the NH4+>NH3 conjugate acid–base

pair will determine the pH of the buffer. The equilibrium relationship between NH4+ and NH3 is

given by the base-dissociation reaction for NH3:

NH31aq2 + H2O1l2 ∆ NH4+1aq2 + OH-1aq2 Kb =

3NH4+43OH-4

3NH34 = 1.8 * 10-5

The key to this exercise is to use this Kb expression to calculate 3NH4+4.

Solve We obtain 3OH-4 from the given pH:pOH = 14.00 - pH = 14.00 - 9.00 = 5.00

and so3OH-4 = 1.0 * 10-5 M

Because Kb is small and the common ion 3NH4+4 is present, the equilibrium concentration of

NH3 essentially equals its initial concentration:3NH34 = 0.10 M

We now use the expression for Kb to calculate 3NH4+4 :

3NH4+4 = Kb

3NH343OH-4 = 11.8 * 10-52 10.102

11.0 * 10-52 = 0.18 M

Thus, for the solution to have pH = 9.00, 3NH4+4 must equal 0.18 M. The number of moles of

NH4Cl needed to produce this concentration is given by the product of the volume of the solu-tion and its molarity:

12.0 L210.18 mol NH4Cl>L2 = 0.36 mol NH4Cl

Comment Because NH4+ and NH3 are a conjugate acid–base pair, we could use the Henderson–

Hasselbalch equation (Equation 17.9) to solve this problem. To do so requires first using Equa-tion 16.41 to calculate pKa for NH4

+ from the value of pKb for NH3. We suggest you try this ap-proach to convince yourself that you can use the Henderson–Hasselbalch equation for buffers for which you are given Kb for the conjugate base rather than Ka for the conjugate acid.

practice Exercise 1Calculate the number of grams of ammonium chloride that must be added to 2.00 L of a 0.500 M ammonia solution to obtain a buffer of pH = 9.20. Assume the volume of the solution does not change as the solid is added. Kb for ammonia is 1.80 * 10-5. (a) 60.7 g, (b) 30.4 g, (c) 1.52 g, (d) 0.568 g, (e) 1.59 * 10-5 g.

practice Exercise 2Calculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid 1C6H5COOH2 to produce a pH of 4.00. Refer to Appendix D.

Buffer Capacity and pH RangeTwo important characteristics of a buffer are its capacity and its effective pH range. Buffer capacity is the amount of acid or base the buffer can neutralize before the pH begins to change to an appreciable degree. The buffer capacity depends on the amount of acid and base used to prepare the buffer. According to Equation 17.5, for example, the pH of a 1-L solution that is 1 M in CH3COOH and 1 M in CH3COONa is the same as the pH of a 1-L solution that is 0.1 M in CH3COOH and 0.1 M in CH3COONa. The first solution has a greater buffering capacity, however, because it contains more CH3COOH and CH3COO-.

The pH range of any buffer is the pH range over which the buffer acts effectively. Buffers most effectively resist a change in pH in either direction when the concentra-tions of weak acid and conjugate base are about the same. From Equation 17.9 we see that when the concentrations of weak acid and conjugate base are equal, pH = pKa. This relationship gives the optimal pH of any buffer. Thus, we usually try to select a buffer whose acid form has a pKa close to the desired pH. In practice, we find that if the concentration of one component of the buffer is more than 10 times the concentration of the other component, the buffering action is poor. Because log 10 = 1, buffers usually have a usable range within {1 pH unit of pKa (that is, a range of pH = pKa {1).


Give it some ThoughtThe Ka values for nitrous acid 1HNO22 and hypochlorous (HClO) acid are 4.5 * 10-4 and 3.0 * 10-8, respectively. Which one would be more suitable for use in a solution buffered at pH = 7.0? What other substances would be needed to make the buffer?

1 2

Use Ka, [HA],and [A−] to

calculate [H+]

A− + H+ HA

Buffer containingHA and A−

Calculate newvalues of

[HA] and [A−]

HA + OH− A− + H2O

pH

Add strong base

Add strong acid

Strong acid reacts withconjugate base component of buffer

Strong base reacts withweak acid component of buffer

Stoichiometrycalculation

Equilibriumcalculation

▲ Figure 17.3 calculating the pH of a buffer after addition of a strong acid or strong base.

Addition of Strong Acids or Bases to BuffersLet’s now consider in a more quantitative way how a buffered solution responds to ad-dition of a strong acid or base. In this discussion, it is important to understand that neutralization reactions between strong acids and weak bases proceed essentially to com-pletion, as do those between strong bases and weak acids. This is because water is a pro-duce of the reaction, and you have an equilibrium constant of 1>Kw = 1014 in your favor when making water. (Section 16.3) Thus, as long as we do not exceed the buffering capacity of the buffer, we can assume that the strong acid or strong base is completely consumed by reaction with the buffer.

Consider a buffer that contains a weak acid HA and its conjugate base A-. When a strong acid is added to this buffer, the added H+ is consumed by A- to produce HA; thus, [HA] increases and 3A-4 decreases. (See Equation 17.7.) Upon addition of a strong base, the added OH- is consumed by HA to produce A-; in this case [HA] decreases and 3A-4 increases. (See Equation 17.6.) These two situations are summa-rized in Figure 17.2.

To calculate how the pH of the buffer responds to the addition of a strong acid or a strong base, we follow the strategy outlined in ▼ Figure 17.3:

1. Consider the acid–base neutralization reaction and determine its effect on [HA] and 3A-4. This step is a limiting reactant stoichiometry calculation. (Section 3.6 and 3.7)

2. Use the calculated values of [HA] and 3A-4 along with Ka to calculate 3H+4. This step is an equilibrium calculation and is most easily done using the Henderson–Hasselbalch equation (if the concentrations of the weak acid–base pair are very large compared to Ka for the acid).


A buffer is made by adding 0.300 mol CH3COOH and 0.300 mol CH3COONa to enough water to make 1.000 L of solution. The pH of the buffer is 4.74 (Sample Exercise 17.1). (a) Calculate the pH of this solution after 5.0 mL of 4.0 M NaOH1aq2 solution is added. (b) For comparison, calculate the pH of a solution made by adding 5.0 mL of 4.0 M NaOH1aq2 solution to 1.000 L of pure water.

solutIonAnalyze We are asked to determine the pH of a buffer after addition of a small amount of strong base and to compare the pH change with the pH that would result if we were to add the same amount of strong base to pure water.Plan Solving this problem involves the two steps outlined in Figure 17.3. First we do a stoichi-ometry calculation to determine how the added OH- affects the buffer composition. Then we use the resultant buffer composition and either the Henderson–Hasselbalch equation or the equilibrium-constant expression for the buffer to determine the pH.Solve

(a) Stoichiometry Calculation: The OH- provided by NaOH reacts with CH3COOH, the weak acid component of the buffer. Since volumes are changing, it is prudent to fig-ure out how many moles of reactants and products would be produced, then divide by the final volume later to obtain concentrations. Prior to this neutralization reaction, there are 0.300 mol each of CH3COOH and CH3COO-. The amount of base added is 0.0050 L * 4.0 mol>L = 0.020 mol. Neutralizing the 0.020 mol OH- requires 0.020 mol of CH3COOH. Consequently, the amount of CH3COOH decreases by 0.020 mol, and the amount of the product of the neutralization, CH3COO-, increases by 0.020 mol. We can create a table to see how the composition of the buffer changes as a result of its reaction with OH-:

sAmplE ExErCIsE 17.6 Calculating ph Changes in Buffers

Before reaction (mol) 0.300 0.020 — 0.300Change (limiting reactant) (mol)

-0.020 -0.020 — +0.020

After reaction (mol) 0.280 0 — 0.320

CH3COOH1aq2 + OH-1aq2 ¡ H2O1l2 + CH3COO-1aq2

Equilibrium Calculation: We now turn our attention to the equilibrium for the ionization of acetic acid, the relationship that determines the buffer pH:

CH3COOH1aq2 ∆ H+1aq2 + CH3COO-1aq2Using the quantities of CH3COOH and CH3COO- remaining in the buffer after the reaction with strong base, we determine the pH using the Henderson–Hasselbalch equation. The volume of the solution is now 1.000 L + 0.0050 L = 1.005 L due to addition of the NaOH solution:

pH = 4.74 + log 0.320 mol>1.005 L0.280 mol>1.005 L

= 4.80

(b) To determine the pH of a solution made by adding 0.020 mol of NaOH to 1.000 L of pure water, we first determine the concentration of OH- ions in solution,

3OH-4 = 0.020 mol>1.005 L = 0.020 M

We use this value in Equation 16.18 to calculate pOH and then use our calculated pOH value in Equation 16.20 to obtain pH:

pOH = - log3OH-4 = - log10.0202 = +1.70 pH = 14 - 1+1.702 = 12.30

Comment Note that the small amount of added NaOH changes the pH of water significantly. In contrast, the pH of the buffer changes very little when the NaOH is added, as summarized in ◀ Figure 17.4.

practice Exercise 1Which of these statements is true? (a) If you add strong acid or base to a buffer, the pH will never change. (b) In order to do calculations in which strong acid or base is added to a buffer, you only need to use the Henderson–Hasselbalch equation. (c) Strong bases react with strong acids, but not weak acids. (d) If you add a strong acid or base to a buffer, the buffer’s pKa or pKb will change. (e) In order to do calculations in which strong acid or base is added to a buffer, you need to cal-culate the amounts of substances from the neutralization reaction and then equilibrate.▲ Figure 17.4 effect of adding a strong

base to a buffered solution and to water.

1.000 L buffer0.300 M CH3COOH0.300 M CH3COO−

Add 5.0 mL of4.0 M NaOH(aq)

pH = 4.74

pH = 4.80

pH increases by0.06 pH units

1.000 L H2O

Add 5.0 mL of4.0 M NaOH(aq)

pH = 7.00

pH = 12.30

pH increases by5.30 pH units


practice Exercise 2Determine (a) the pH of the original buffer described in Sample Exercise 17.6 after the addi-tion of 0.020 mol HCl and (b) the pH of the solution that would result from the addition of 0.020 mol HCl to 1.000 L of pure water.

Chemistry and life

Blood as a Buffered solution

Chemical reactions that occur in living systems are often extremely sensitive to pH. Many of the enzymes that catalyze important bio-chemical reactions are effective only within a narrow pH range. For this reason, the human body maintains a remarkably intricate system of buffers, both within cells and in the fluids that transport cells. Blood, the fluid that transports oxygen to all parts of the body, is one of the most prominent examples of the importance of buffers in living beings.

Human blood has a normal pH of 7.35 to 7.45. Any deviation from this range can have extremely disruptive effects on the stability of cell membranes, the structures of proteins, and the activities of enzymes. Death may result if the blood pH falls below 6.8 or rises above 7.8. When the pH falls below 7.35, the condition is called acidosis; when it rises above 7.45, the condition is called alkalosis. Acidosis is the more com-mon tendency because metabolism generates several acids in the body.

The major buffer system used to control blood pH is the carbonic acid–bicarbonate buffer system. Carbonic acid 1H2CO32 and bicarbon-ate ion 1HCO3

-2 are a conjugate acid–base pair. In addition, carbonic acid decomposes into carbon dioxide gas and water. The important equilibria in this buffer system are

H+1aq2 + HCO3-1aq2 ∆ H2CO31aq2 ∆ H2O1l2 + CO21g2

[17.10]

Several aspects of these equilibria are notable. First, although carbonic acid is diprotic, the carbonate ion 1CO3

2 - 2 is unimportant in this system. Second, one component of this equilibrium, CO2, is a gas, which provides a mechanism for the body to adjust the equilib-ria. Removal of CO2 via exhalation shifts the equilibria to the right, consuming H+ ions. Third, the buffer system in blood operates at pH 7.4, which is fairly far removed from the pKa1 value of H2CO3 (6.1 at physiological temperatures). For the buffer to have a pH of 7.4, the ratio [base]/[acid] must be about 20. In normal blood plasma, the con-centrations of HCO3

- and H2CO3 are about 0.024 M and 0.0012 M, respectively. Consequently, the buffer has a high capacity to neutralize additional acid but only a low capacity to neutralize additional base.

The principal organs that regulate the pH of the carbonic acid–bi-carbonate buffer system are the lungs and kidneys. When the concen-tration of CO2 rises, the equilibrium concentrations in Equation 17.10 shift to the left, which leads to the formation of more H+ and a drop in pH. This change is detected by receptors in the brain that trigger a reflex to breathe faster and deeper, increasing the rate at which CO2 is expelled from the lungs and thereby shifting the equilibrium concen-trations back to the right. When the blood pH becomes too high, the kidneys remove HCO3

- from the blood. This shifts the equilibrium concentrations to the left, increasing the concentration of H+. As a re-sult, the pH decreases.

Regulation of blood pH relates directly to the effective trans-port of O2 throughout the body. The protein hemoglobin, found in red blood cells (▶ Figure 17.5), carries oxygen. Hemoglobin (Hb)

reversibly binds both O2 and H+. These two substances compete for the Hb, which can be represented approximately by the equilibrium

HbH+ + O2 ∆ HbO2 + H+ [17.11]

Oxygen enters the blood through the lungs, where it passes into the red blood cells and binds to Hb. When the blood reaches tissue in which the concentration of O2 is low, the equilibrium concentrations in Equation 17.11 shift to the left and O2 is released.

During periods of strenuous exertion, three factors work together to ensure delivery of O2 to active tissues. The role of each factor can be understood by applying Le Châtelier’s principle to Equation 17.11: 1. O2 is consumed, causing the equilibrium concentrations to shift

to the left, releasing more O2. 2. Large amounts of CO2 are produced by metabolism, which in-

creases 3H+4 and causes the equilibrium concentrations to shift to the left, releasing O2.

3. Body temperature rises. Because Equation 17.11 is exothermic, the increase in temperature shifts the equilibrium concentrations to the left, releasing O2.

In addition to the factors causing release of O2 to tissues, the decrease in pH stimulates an increase in breathing rate, which furnishes more O2 and eliminates CO2. Without this elaborate series of equilibrium shifts and pH changes, the O2 in tissues would be rapidly depleted, making further activity impossible. Under such conditions, the buffer-ing capacity of the blood and the exhalation of CO2 through the lungs are essential to keep the pH from dropping too low, thereby triggering acidosis.

Related Exercises: 17.29, 17.97

▲ Figure 17.5 red blood cells. A scanning electron micrograph of red blood cells traveling through a small branch of an artery. The red blood cells are approximately 0.010 mm in diameter.


17.3 | Acid–Base titrationsTitrations are procedures in which one reactant is slowly added into a solution of another reactant, while equilibrium concentrations along the way are monitored. (Section 4.6) There are two main reasons to do titrations: (1) you want to know the concentration of one of the reactants or (2) you want to know the equilibrium constant for the reaction.

In an acid–base titration, a solution containing a known concentration of base is slowly added to an acid (or the acid is added to the base). (Section 4.6) Acid–base indicators can be used to signal the equivalence point of a titration (the point at which stoichiometrically equivalent quantities of acid and base have been brought together). Alternatively, a pH meter can be used to monitor the progress of the reaction (◀ Figure 17.6), producing a pH titration curve, a graph of the pH as a function of the volume of titrant added. The shape of the titration curve makes it possible to determine the equivalence point. The curve can also be used to select suitable indicators and to determine the Ka of the weak acid or the Kb of the weak base being titrated.

To understand why titration curves have certain characteristic shapes, we will examine the curves for three kinds of titrations: (1) strong acid–strong base, (2) weak acid–strong base, and (3) polyprotic acid–strong base. We will also briefly consider how these curves relate to those involving weak bases.

Strong Acid–Strong Base TitrationsThe titration curve produced when a strong base is added to a strong acid has the gen-eral shape shown in ▼ Figure 17.7, which depicts the pH change that occurs as 0.100 M

Burette containingNaOH(aq) of knownconcentration

pH meter

Beaker containingHCl(aq) of unknownconcentration

▲ Figure 17.6 measuring pH during a titration.

Go FIGurEIn which direction do you expect the pH to change as NaOH is added to the HCl solution?

1 2 3 4

0

2

4

6

8pH

10

12

14

10 20 30 40

mL NaOH

H+ consumed as OH− added, forming H2O (pH < 7.0)

H+ completelyneutralized by OH−(pH = 7.0)

No H+ left to react with excess OH− (pH > 7.0)

50 60 70 80

Equivalence point

OH−

Only HCl(aq)present before titration

H+

Na+Cl−

pH = 7.0 at equivalence point, NaCl(aq) salt solution

Equivalence point occurs when moles base = moles acid

▲ Figure 17.7 Titration of a strong acid with a strong base. The pH curve for titration of 50.0 mL of a 0.100 M solution of hydrochloric acid with a 0.100 M solution of NaOH1aq2. For clarity, water molecules have been omitted from the molecular art.

Go FIGurEWhat volume of NaOH(aq) would be needed to reach the equivalence point if the concentration of the added base were 0.200 M?

seCtion 17.3 acid–Base titrations 739

NaOH is added to 50.0 mL of 0.100 M HCl. The pH can be calculated at various stages of the titration. To help understand these calculations, we can divide the curve into four regions:

1. Initial pH: The pH of the solution before the addition of any base is de-termined by the initial concentration of the strong acid. For a solution of 0.100 M HCl, 3H+4 = 0.100 M and pH = - log10.1002 = 1.000. Thus, the initial pH is low.

2. Between initial pH and equivalence point: As NaOH is added, the pH increases slowly at first and then rapidly in the vicinity of the equivalence point. The pH be-fore the equivalence point is determined by the concentration of acid not yet neu-tralized. This calculation is illustrated in Sample Exercise 17.7(a).

3. Equivalence point: At the equivalence point an equal number of moles of NaOH and HCl have reacted, leaving only a solution of their salt, NaCl. The pH of the solu-tion is 7.00 because the cation of a strong base (in this case Na+) and the anion of a strong acid (in this case Cl-) are neither acids nor bases and, therefore, have no ap-preciable effect on pH. (Section 16.9)

4. After equivalence point: The pH of the solution after the equivalence point is de-termined by the concentration of excess NaOH in the solution. This calculation is illustrated in Sample Exercise 17.7(b).

solutIonAnalyze We are asked to calculate the pH at two points in the titration of a strong acid with a strong base. The first point is just before the equivalence point, so we expect the pH to be determined by the small amount of strong acid that has not yet been neutralized. The second point is just after the equivalence point, so we expect this pH to be determined by the small amount of excess strong base.

sAmplE ExErCIsE 17.7 Calculations for a strong Acid–strong Base titration

Calculate the pH when (a) 49.0 mL and (b) 51.0 mL of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.

Plan (a) As the NaOH solution is added to the HCl solution, H+1aq2 reacts with OH-1aq2 to form H2O. Both Na+ and Cl- are spectator ions, having negligible effect on the pH. To determine the pH of the solution, we must first determine how many moles of H+ were origi-nally present and how many moles of OH- were added. We can then calculate how many moles of each ion remain after the neutralization reaction. To calculate 3H+4, and hence pH, we must also remember that the volume of the solution increases as we add titrant, thus dilut-ing the concentration of all solutes present. Therefore, it is best to deal with moles first, and then convert to molarities using total solution volumes (volume of acid plus volume of base).

Solve The number of moles of H+ in the origi-nal HCl solution is given by the product of the volume of the solution and its molarity:

10.0500 L soln2a 0.100 mol H+

1 L solnb = 5.00 * 10-3 mol H+

Likewise, the number of moles of OH-, in 49.0 mL of 0.100 M NaOH is 10.0490 L soln2a 0.100 mol OH-

1 L solnb = 4.90 * 10-3 mol OH-

Because we have not reached the equivalence point, there are more moles of H+ present than OH-. Therefore, OH- is the limiting reactant. Each mole of OH- reacts with 1 mol of H+.Using the convention introduced in Sample Exercise 17.6, we have

H+1aq2 + OH-1aq2 ¡ H2O1l2Before reaction (mol) 5.00 * 10-3 4.90 * 10-3 —

Change (limiting reactant) (mol) -4.90 * 10-3 -4.90 * 10-3 —

After reaction (mol) 0.10 * 10-3 0 —

The volume of the reaction mixture increases as the NaOH solution is added to the HCl so-lution. Thus, at this point in the titration, the volume in the titration flask is 50.0 mL + 49.0 mL = 99.0 mL = 0.0990 L

Thus, the concentration of H+1aq2 in the flask is 3H+4 =

moles H+1aq2liters soln

=0.10 * 10-3 mol

0.09900 L= 1.0 * 10-3 M

The corresponding pH is - log11.0 * 10-32 = 3.00


Weak Acid–Strong Base TitrationsThe curve for titration of a weak acid by a strong base is similar to the curve in Figure 17.7. Consider, for example, the curve for titration of 50.0 mL of 0.100 M acetic acid with 0.100 M NaOH shown in ▶ Figure 17.9. We can calculate the pH at points along this curve, using principles we discussed earlier, which means again dividing the curve into four regions:

1. Initial pH: We use Ka to calculate this pH, as shown in Section 16.6. The calculated pH of 0.100 M CH3COOH is 2.89.

Titration of a solution of a strong base with a solution of a strong acid yields an analogous curve of pH versus added acid. In this case, how-ever, the pH is high at the outset of the titration and low at its completion (◀ Figure 17.8). The pH at the equivalence point is still 7.0 (at 25 °C), just like the strong acid–strong base titration.

Plan (b) We proceed in the same way as we did in part (a) except we are now past the equivalence point and have more OH- in the solu-tion than H+. As before, the initial number of moles of each reactant

is determined from their volumes and concentrations. The reactant present in smaller stoichiometric amount (the limiting reactant) is consumed completely, leaving an excess of hydroxide ion.

Solve H+1aq2 + OH-1aq2 ¡ H2O1l2

Before reaction (mol) 5.00 * 10-3 5.10 * 10-3 —

Change (limiting reactant) (mol)

-5.00 * 10-3 -5.00 * 10-3 —

After reaction (mol) 0 0.10 * 10-3 —

In this case, the volume in the titration flask is 50.0 mL + 51.0 mL = 101.0 mL = 0.1010 L

Hence, the concentration of OH-1aq2 in the flask is 3OH-4 =

moles OH-1aq2liters soln

=0.10 * 10-3 mol

0.1010 L= 1.0 * 10-3 M

and we have pOH = - log11.0 * 10-32 = 3.00 pH = 14.00 - pOH = 14.00 - 3.00 = 11.00

Comment Note that the pH increased by only two pH units, from 1.00 (Figure 17.7) to 3.00, after the first 49.0 mL of NaOH solution was added, but jumped by eight pH units, from 3.00 to 11.00, as 2.0 mL of base solution was added near the equivalence point. Such a rapid rise in pH near the equivalence point is a characteristic of titrations involving strong acids and strong bases.

practice Exercise 1An acid–base titration is performed: 250.0 mL of an unknown concentration of HCl (aq) is titrated to the equivalence point with 36.7 mL of a 0.1000 M aqueous solution of NaOH. Which of the following statements is not true of this titration?(a) The HCl solution is less concentrated than the NaOH solu-tion. (b) The pH is less than 7 after adding 25 mL of NaOH solution. (c) The pH at the equivalence point is 7.00. (d) If an

▲ Figure 17.8 Titration of a strong base with a strong acid. The pH curve for titration of 50.0 mL of a 0.100 M solution of a strong base with a 0.100 M solution of a strong acid.

0

2

4

6

8pH

10

12

14

10 20 30 40mL acid

50 60 70 80

Equivalence point

Give it some ThoughtWhat is the pH at the equivalence point when 0.10 M HNO3 is used to titrate a volume of solution containing 0.30 g of KOH?

1 2 3 4

pH change near the equivalence point is smaller than in strong acid–strong base titration

0

2

4

6

8pH

10

12

14

10 20 30 40mL NaOH

50 60 70 80

Equivalence point

CH3COOHCH3COO−

OH−

CH3COOH(aq)solution before titration

Added OH− convertsCH3COOH(aq) intoCH3COO−(aq),forming buffer solution

Acid completelyneutralized by addedbase, CH3COONa(aq)salt solution results

No acid left to reactwith excess OH−

Na+

pH > 7.0 at equivalence point because CH3COO− is a weak base that reacts with water to make OH−

Equivalence point occurs when moles base = moles acidadditional 1.00 mL of NaOH solution is added beyond the

equivalence point, the pH of the solution is more than 7.00. (e) At the equivalence point, the OH- concentration in the solution is 3.67 * 10-3 M.

practice Exercise 2Calculate the pH when (a) 24.9 mL and (b) 25.1 mL of 0.100 M HNO3 have been added to 25.0 mL of 0.100 M KOH solution.


2. Between initial pH and equivalence point: Prior to reaching the equivalence point, the acid is being neutralized, and its conjugate base is being formed:

CH3COOH1aq2 + OH-1aq2 ¡ CH3COO-1aq2 + H2O1l2 [17.12]

Thus, the solution contains a mixture of CH3COOH and CH3COO-. Calculating the pH in this region involves two steps. First, we consider the neutralization reac-tion between CH3COOH and OH- to determine 3CH3COOH] and 3CH3COO-4. Next, we calculate the pH of this buffer pair using procedures developed in Sec-tions 17.1 and 17.2. The general procedure is diagrammed in Figure 17.10 and illustrated in Sample Exercise 17.7.

3. Equivalence point: The equivalence point is reached when 50.0 mL of 0.100 M NaOH has been added to the 50.0 mL of 0.100 M CH3COOH. At this point, the 5.00 * 10-3 mol of NaOH completely reacts with the 5.00 * 10-3 mol

1 2 3 4

pH change near the equivalence point is smaller than in strong acid–strong base titration

0

2

4

6

8pH

10

12

14

10 20 30 40mL NaOH

50 60 70 80

Equivalence point

CH3COOHCH3COO−

OH−

CH3COOH(aq)solution before titration

Added OH− convertsCH3COOH(aq) intoCH3COO−(aq),forming buffer solution

Acid completelyneutralized by addedbase, CH3COONa(aq)salt solution results

No acid left to reactwith excess OH−

Na+

pH > 7.0 at equivalence point because CH3COO− is a weak base that reacts with water to make OH−

Equivalence point occurs when moles base = moles acid

▲ Figure 17.9 Titration of a weak acid with a strong base. The pH curve for titration of 50.0 mL of a 0.100 M solution of acetic acid with a 0.100 M solution of NaOH1aq2. For clarity, water molecules have been omitted from the molecular art.

Go FIGurEIf the acetic acid being titrated here were replaced by hydrochloric acid, would the amount of base needed to reach the equivalence point change? Would the pH at the equivalence point change?


of CH3COOH to form 5.00 * 10-3 mol of CH3COONa. The Na+ ion of this salt has no significant effect on the pH. The CH3COO- ion, however, is a weak base whose reaction with water cannot be neglected, and the pH at the equiva-lence point is therefore greater than 7. In general, the pH at the equivalence point is always above 7 in a weak acid–strong base titration because the anion of the salt formed is a weak base. The procedure for calculating the pH of the solution of a weak base is described in Section 16.7 and is shown in Sample Exercise 17.8.

4. After equivalence point (excess base): In this region, 3OH-4 from the reaction of CH3COO- with water is negligible relative to 3OH-4 from the excess NaOH. Thus, the pH is determined by the concentration of OH- from the excess NaOH. The method for calculating pH in this region is therefore like that illustrated in Sample Exercise 17.7(b). Thus, the addition of 51.0 mL of 0.100 M NaOH to 50.0 mL of either 0.100 M HCl or 0.100 M CH3COOH yields the same pH, 11.00. Notice by comparing Figures 17.7 and 17.9 that the titration curves for a strong acid and a weak acid are the same after the equivalence point.

▲ Figure 17.10 procedure for calculating pH when a weak acid is partially neutralized by a strong base.

1 2

HA + OH−

Addstrongbase

Neutralization reaction:[HA] decreases[A−] increases

A− + H2OSolution

containingweak acid

Calculate newvalues of

[HA] and [A−]

Use Ka, [HA],and [A−] to

calculate [H+]pH

Stoichiometrycalculation

Equilibriumcalculation

sAmplE ExErCIsE 17.8 Calculations for a Weak Acid–strong Base titration

Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH 1Ka = 1.8 * 10-52.

solutIonAnalyze We are asked to calculate the pH before the equivalence point of the titration of a weak acid with a strong base.Plan We first must determine the number of moles of CH3COOH and CH3COO- present after the neutralization reaction (the stoichiometry calculation). We then calculate pH using Ka, 3CH3COOH], and 3CH3COO-4 (the equilibrium calculation).

Solve Stoichiometry Calculation: The product of the volume and concentration of each solu-tion gives the number of moles of each reactant present before the neutralization:

10.0500 L soln2a 0.100 mol CH3COOH1 L soln

b = 5.00 * 10-3 mol CH3COOH

10.0450 L soln2a 0.100 mol NaOH1 L soln

b = 4.50 * 10-3 mol NaOH

The 4.50 * 10-3 of NaOH consumes 4.50 * 10-3 of CH3COOH:

CH3COOH1aq2 + OH-1aq2 ¡ CH3COO-1aq2 + H2O1l2Before reaction (mol) 5.00 * 10-3 4.50 * 10-3 0 —

Change (limiting reactant) (mol)

-4.50 * 10-3 -4.50 * 10-3 +4.50 * 10-3

After reaction (mol) 0.50 * 10-3 0 4.50 * 10-3 —

The total volume of the solution is 45.0 mL + 50.0 mL = 95.0 mL = 0.0950 L


In order to further monitor the evolution of pH as a function of added base, we can calculate the pH at the equivalence point.

The resulting molarities of CH3COOH and CH3COO- after the reaction are therefore 3CH3COOH4 =

0.50 * 10-3 mol0.0950 L

= 0.0053 M

3CH3COO-4 =4.50 * 10-3 mol

0.0950 L= 0.0474 M

Equilibrium Calculation: The equilibrium between CH3COOH and CH3COO- must obey the equilibrium-constant expression for CH3COOH:

Ka =3H+43CH3COO-43CH3COOH4 = 1.8 * 10-5

Solving for 3H+4 gives 3H+4 = Ka *

3CH3COOH43CH3COO-4 = 11.8 * 10-52 * a 0.0053

0.0474b = 2.0 * 10-6 M

pH = - log12.0 * 10-62 = 5.70

Comment We could have solved for pH equally well using the Henderson–Hasselbalch equation in the last step.

practice Exercise 1If you think carefully about what happens during the course of a weak acid–strong base titra-tion, you can learn some very interesting things. For example, let’s look back at Figure 17.9 and pretend you did not know that acetic acid was the acid being titrated. You can figure out the pKa of a weak acid just by thinking about the definition of Ka and looking at the right place on the titration curve! Which of the following choices is the best way to do this? (a) At the equivalence point, pH = pKa. (b) Halfway to the equivalence point, pH = pKa. (c) Before any base is added, pH = pKa. (d) At the top of the graph with excess base added, pH = pKa.

practice Exercise 2(a) Calculate the pH in the solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic acid 1C6H5COOH, Ka = 6.3 * 10-52. (b) Calculate the pH in the solu-tion formed by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NH3.

sAmplE ExErCIsE 17.9 Calculating the ph at the Equivalence point

Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.

solutIonAnalyze We are asked to determine the pH at the equivalence point of the titration of a weak acid with a strong base. Because the neutralization of a weak acid produces its anion, a conjugate base that can react with water, we expect the pH at the equivalence point to be greater than 7.Plan The initial number of moles of acetic acid equals the number of moles of acetate ion at the equivalence point. We use the volume of the solution at the equivalence point to calculate the concentration of acetate ion. Because the acetate ion is a weak base, we can calculate the pH using Kb and 3CH3COO-4.Solve The number of moles of acetic acid in the initial solution is obtained from the volume and molarity of the solution:

Moles = M * L = 10.100 mol>L210.0500 L2 = 5.00 * 10-3 mol CH3COOH

Hence, 5.00 * 10-3 mol of CH3COO- is formed. It will take 50.0 mL of NaOH to reach the equivalence point (Figure 17.9). The volume of this salt solution at the equivalence point is the sum of the volumes of the acid and base, 50.0 mL + 50.0 mL = 100.0 mL = 0.1000 L. Thus, the concentration of CH3COO- is

3CH3COO-4 =5.00 * 10-3 mol

0.1000 L= 0.0500 M

The CH3COO- ion is a weak base:

CH3COO-1aq2 + H2O1l2 ∆ CH3COOH1aq2 + OH-1aq2


The Kb for CH3COO- can be calculated from the Ka value of its conjugate acid, Kb = Kw>Ka = 11.0 * 10-142>11.8 * 10-52 = 5.6 * 10-10. Using the Kb expression, we have

Kb =3CH3COOH43OH-4

3CH3COO-4 =1x21x2

0.0500 - x= 5.6 * 10-10

Making the approximation that 0.0500 - x ≃ 0.0500, and then solving for x, we have x = 3OH-4 = 5.3 * 10-6 M, which gives pOH = 5.28 and pH = 8.72.

Check The pH is above 7, as expected for the salt of a weak acid and strong base.

practice Exercise 1Why is pH at the equivalence point larger than 7 when you titrate a weak acid with a strong base? (a) There is excess strong base at the equivalence point. (b) There is excess weak acid at the equivalence point. (c) The conjugate base that is formed at the equivalence point is a strong base. (d) The conjugate base that is formed at the equivalence point reacts with water. (e) This statement is false: the pH is always 7 at an equivalence point in a pH titration.

practice Exercise 2Calculate the pH at the equivalence point when (a) 40.0 mL of 0.025 M benzoic acid 1C6H5COOH, Ka = 6.3 * 10-52 is titrated with 0.050 M NaOH and (b) 40.0 mL of 0.100 M NH3 is titrated with 0.100 M HCl.

The titration curve for a weak acid–strong base titration (Figure 17.9) differs from the curve for a strong acid–strong base titration (Figure 17.7) in three noteworthy ways:

1. The solution of the weak acid has a higher initial pH than a solution of a strong acid of the same concentration.

2. The pH change in the rapid-rise portion of the curve near the equivalence point is smaller for the weak acid than for the strong acid.

3. The pH at the equivalence point is above 7.00 for the weak acid titration.

0 10 20 30 40 50 60

14

12

10

8

6

4

2

0

mL NaOH

Equivalence point

Ka = 10−8

Ka = 10−6

Ka = 10−4

Ka = 10−2

pH

Strong acid

Initial pH increases as Ka decreases

pH at equivalence point (marked with circles) increases as Ka decreases

▲ Figure 17.11 A set of curves showing the effect of acid strength on the characteristics of the titration curve when a weak acid is titrated by a strong base. Each curve represents titration of 50.0 mL of 0.10 M acid with 0.10 M NaOH.

Go FIGurEHow does the pH at the equivalence point change as the acid being titrated becomes weaker? How does the volume of NaOH(aq) needed to reach the equivalence point change?

Give it some ThoughtDescribe the reasons why statement 3 above is true.

The weaker the acid, the more pronounced these differences become. To illustrate this consider the family of titration curves shown in ◀ Figure 17.11. Notice that as the acid becomes weaker (that is, as Ka becomes smaller), the initial pH increases and the pH change near the equivalence point becomes less marked. Fur-thermore, the pH at the equivalence point steadily increases as Ka decreases, because the strength of the conjugate base of the weak acid increases. It is virtually impossible to determine the equivalence point when pKa is 10 or higher because the pH change is too small and gradual.

Titrating with an Acid–Base IndicatorOftentimes in an acid–base titration, an indicator is used rather than a pH meter. An indicator is a compound that changes color in solution over a specific pH range. Optimally, an indicator should change color at the equivalence point in a titration. In practice, however, an indica-tor need not precisely mark the equivalence point. The pH changes very rapidly near the equivalence point, and in this region one drop of titrant can change the pH by several units. Thus, an indicator begin-ning and ending its color change anywhere on the rapid-rise portion of the titration curve gives a sufficiently accurate measure of the titrant volume needed to reach the equivalence point. The point in a titration where the indicator changes color is called the end point to distinguish it from the equivalence point that it closely approximates.


▶ Figure 17.12 shows the curve for titration of a strong base (NaOH) with a strong acid (HCl). We see from the vertical part of the curve that the pH changes rapidly from roughly 11 to 3 near the equivalence point. Consequently, an indicator for this titration can change color anywhere in this range. Most strong acid–strong base titrations are carried out using phe-nolphthalein as an indicator because it changes color in this range (see Figure 16.8, page 684). Several other indicators would also be satisfactory, including methyl red, which, as the lower color band in Figure 17.12 shows, changes color in the pH range from about 4.2 to 6.0 (see Figure 16.8, page 684).

As noted in our discussion of Figure 17.11, because the pH change near the equivalence point becomes smaller as Ka decreases, the choice of indica-tor for a weak acid–strong base titration is more criti-cal than it is for titrations where both acid and base are strong. When 0.100 M CH3COOH 1Ka = 1.8 * 10-52 is titrated with 0.100 M NaOH, for example, the pH increases rapidly only over the pH range from about 7 to 11 (▼ Figure 17.13). Phenolphthalein is therefore an ideal indicator because it changes color from pH 8.3 to 10.0, close to the pH at the equivalence point. Methyl red is a poor choice, however, because its color change, from 4.2 to 6.0, begins well before the equiva-lence point is reached.

Titration of a weak base (such as 0.100 M NH3) with a strong acid solution (such as 0.100 M HCl) leads to the titration curve shown in Figure 17.14. In this example, the equivalence point occurs at pH 5.28. Thus, methyl red is an ideal indicator but phenol-phthalein would be a poor choice.

0

21

43

65

87pH

1011121314

9

10 20 30 40mL HCl

50 60 70 80

Methyl redcolor-changeinterval

Phenolphthaleincolor-changeinterval

Equivalence point

▲ Figure 17.12 Using color indicators for titration of a strong base with a strong acid. Both phenolphthalein and methyl red change color in the rapid-rise portion of the titration curve.

Go FIGurEIs methyl red a suitable indicator when you are titrating a strong acid with a strong base? Explain your answer.

2

0

4

6

8pH

10

12

14

10 20 30 40mL NaOH

Phenolphthalein indicatorColor-change interval 8.3 < pH < 10.0

Methyl red indicatorColor-change interval 4.2 < pH < 6.0

50 60 70 80

2

0

4

6

8pH

10

12

14

10 20 30 40mL NaOH

50 60 70 80

Equivalence point Equivalence point

Suitable indicator for titration of a weak acid with a strong base because equivalence point falls within the color-change interval

Unsatisfactory indicator for titration of a weak acid with a strong base because color changes before reaching equivalence point

▲ Figure 17.13 Good and poor indicators for titration of a weak acid with a strong base.


Give it some ThoughtWhy is the choice of indicator more crucial for a weak acid–strong base titration than for a strong acid–strong base titration?

Titrations of Polyprotic AcidsWhen weak acids contain more than one ionizable H atom, the reaction with OH- oc-curs in a series of steps. Neutralization of H3PO3, for example, proceeds in two steps (the third H is bonded to the P and does not ionize):

H3PO31aq2 + OH-1aq2 ¡ H2PO3-1aq2 + H2O1l2 [17.13]

H2PO3-1aq2 + OH-1aq2 ¡ HPO3

2 - 1aq2 + H2O1l2 [17.14]

When the neutralization steps of a polyprotic acid or polybasic base are sufficiently separated, the titration has multiple equivalence points. ▶ Figure 17.15 shows the two equivalence points corresponding to Equations 17.13 and 17.14.

Give it some ThoughtSketch an approximate titration curve for the titration of Na2CO3 with HCl.

You can use titration data such as that shown in Figure 17.15 to figure out the pKas of the weak polyprotic acid. For example, let’s write the Ka1 and Ka2 reactions for phos-phorous acid:

H3PO31aq2 ∆ H2PO3-1aq2 + H+1aq2 Ka1 =

3H2PO3-43H+4

H3PO3

H2PO3-1aq2 ∆ HPO3

2 - 1aq2 + H+1aq2 Ka2 =3HPO3

2 - 43H+4H2PO3

-

0

2

4

6

8pH

10

12

14

10 20 30 40mL HCl

50 60 70 80 0

2

4

6

8pH

10

12

14

10 20 30 40mL HCl

50 60 70 80

Phenolphthalein indicatorColor-change interval 8.3 < pH < 10.0

Methyl red indicatorColor-change interval 4.2 < pH < 6.0

Equivalence point Equivalence point

Unsatisfactory indicator for titration of a weak base with a strong acid because colorchanges before reaching equivalence point

Suitable indicator for titration of a weak base with a strong acid because equivalence point falls within the color-change interval

▲ Figure 17.14 Good and poor indicators for titration of a weak base with a strong acid.


If we rearrange these equilibrium expressions, you see that we obtain Henderson– Hasselbalch equations:

pH = pKa1 + log 3H2PO3

- 43H3PO34

pH = pKa2 + log 3HPO3

2 - 43H2PO3

-4Therefore, if the concentrations of the each acid and base conjugate pairs were identical for each equilibrium, log 112 = 0 and so pH = pKa. When does this happen during the titration? At the beginning of the titration, the acid is H3PO3 initially; at the first equivalence point, it is all converted to H2PO3

-. Therefore, halfway to the first equiva-lence point, half of the H3PO3 is converted to H2PO3

-. Thus, halfway to the equiva-lence point, the concentration of H3PO3 is equal to that of H2PO3

-, and at that point, pH = pKa1. Similar logic is true for the second equilibrium reaction: halfway toward its equivalence point, pH = pKa2.

We can then just look at titration data and estimate the pKas for the polyprotic acid directly from the titration curve. This procedure is especially useful if you are trying to identify an unknown polyprotic acid. In Figure 17.15, for instance, the first equivalence point occurs for 50 mL NaOH added. Halfway to the equivalence point corresponds to 25 mL NaOH. Because the pH at 25 mL NaOH is about 1.5 we may estimate pKa1 = 1.5 for phosphorous acid. The second equivalence point occurs at 100 mL NaOH added; halfway there (from the first equivalence point) is at 75 mL NaOH added. The graph indicates the pH at 75 mL NaOH added is about 6.5, and we therefore estimate that pKa2 for phosphorous acid is 6.5. The actual values for the two pKas are pKa1 = 1.3 and pKa2 = 6.7 (close to our estimates).

▲ Figure 17.15 Titration curve for a diprotic acid. The curve shows the pH change when 50.0 mL of 0.10 M H3PO3 is titrated with 0.10 M NaOH.

0 10 20 30 40 60 70 80 90 110100500

2

4

6pH

8

10

12

14

− 2−

mL NaOH

H3PO3 is the dominant species

H2PO3– is the

dominant speciesHPO3

2– is the dominant species


17.4 | solubility EquilibriaThe equilibria we have considered thus far in this chapter have involved acids and bases. Furthermore, they have been homogeneous; that is, all the species have been in the same phase. Through the rest of the chapter, we will consider the equilibria involved in the dissolution or precipitation of ionic compounds. These reactions are heterogeneous.

Dissolution and precipitation occur both within us and around us. Tooth enamel dissolves in acidic solutions, for example, causing tooth decay, and the precipitation of certain salts in our kidneys produces kidney stones. The waters of Earth contain salts dis-solved as water passes over and through the ground. Coral reefs are principally made of CaCO3, as we saw in the beginning of this chapter. Precipitation of CaCO3 from ground-water is responsible for the formation of stalactites and stalagmites within limestone caves.

In our earlier discussion of precipitation reactions, we considered general rules for predicting the solubility of common salts in water. (Section 4.2) These rules give us a qualitative sense of whether a compound has a low or high solubility in water. By consid-ering solubility equilibria, however, we can make quantitative predictions about solubility.

The Solubility-Product Constant, Ksp

Recall that a saturated solution is one in which the solution is in contact with undis-solved solute. (Section 13.2) Consider, for example, a saturated aqueous solution of BaSO4 in contact with solid BaSO4. Because the solid is an ionic compound, it is a strong electrolyte and yields Ba2 + 1aq2 and SO4

2 - 1aq2 ions when dissolved in water, readily establishing the equilibrium

BaSO41s2 ∆ Ba2 + 1aq2 + SO42 - 1aq2 [17.15]

As with any other equilibrium, the extent to which this dissolution reaction occurs is expressed by the magnitude of the equilibrium constant. Because this equilibrium equation describes the dissolution of a solid, the equilibrium constant indicates how soluble the solid is in water and is referred to as the solubility-product constant (or simply the solubility product). It is denoted Ksp, where sp stands for solubility product.

The equilibrium-constant expression for the equilibrium between a solid and an aqueous solution of its component ions 1Ksp2 is written according to the rules that apply to any other equilibrium-constant expression. Remember, however, that solids do not appear in the equilibrium-constant expressions for heterogeneous equilibrium. (Section 15.4)

Thus, the solubility-product expression for BaSO4, which is based on Equation 17.15, is

Ksp = 3Ba2 + 43SO42 - 4 [17.16]

In general, the solubility product Ksp of a compound equals the product of the concentra-tion of the ions involved in the equilibrium, each raised to the power of its coefficient in the equilibrium equation. The coefficient for each ion in the equilibrium equation also equals its subscript in the compound’s chemical formula.

The values of Ksp at 25 °C for many ionic solids are tabulated in Appendix D. The value of Ksp for BaSO4 is 1.1 * 10-10, a very small number indicating that only a very small amount of the solid dissolves in 25 °C water.

Write the expression for the solubility-product constant for CaF2, and look up the correspond-ing Ksp value in Appendix D.

solutIonAnalyze We are asked to write an equilibrium-constant expression for the process by which CaF2 dissolves in water.Plan We apply the general rules for writing an equilibrium-constant expression, excluding the solid reactant from the expression. We assume that the compound dissociates com-pletely into its component ions:

CaF21s2 ∆ Ca2 + 1aq2 + 2 F -1aq2

sAmplE ExErCIsE 17.10 Writing solubility-product 1Ksp 2 Expressions

seCtion 17.4 solubility equilibria 749

Solubility and Ksp

It is important to distinguish carefully between solubility and the sol-ubility-product constant. The solubility of a substance is the quan-tity that dissolves to form a saturated solution. (Section 13.2) Solubility is often expressed as grams of solute per liter of solution 1g>L2. Molar solubility is the number of moles of solute that dis-solve in forming 1 L of saturated solution of the solute 1mol>L2. The solubility-product constant 1Ksp2 is the equilibrium constant for the equilibrium between an ionic solid and its saturated solution and is a unitless number. Thus, the magnitude of Ksp is a measure of how much of the solid dissolves to form a saturated solution.

Give it some ThoughtWithout doing a calculation, predict which of these compounds has the greatest molar solubility in water: AgCl 1Ksp = 1.8 * 10-102, AgBr 1Ksp = 5.0 * 10-132, or AgI 1Ksp = 8.3 * 10-172.

The solubility of a substance can change considerably in response to a number of fac-tors. For example, the solubilities of hydroxide salts, like Mg1OH22, are dependent upon the pH of the solution. The solubility is also affected by concentrations of other ions in solution, especially common ions. In other words, the numeric value of the solubility of a given solute does change as the other species in solution change. In contrast, the solubility-product con-stant, Ksp, has only one value for a given solute at any specific temperature.* ▲ Figure 17.16 summarizes the relationships among various expressions of solubility and Ksp.

*This is strictly true only for very dilute solutions, for Ksp values change somewhat when the concentration of ionic substances in water is increased. However, we will ignore these effects, which are taken into consider-ation only for work that requires exceptional accuracy.

▲ Figure 17.16 procedure for converting between solubility and Ksp. Starting from the mass solubility, follow the green arrows to determine Ksp. Starting from Ksp, follow the red arrows to determine either molar solubility or mass solubility.

Mass solubilityof compound

(g/L)

Molar solubilityof compound

(mol/L)

Molarconcentration

of ions

Ksp

Formulaweight

Solubilityequilibrium

Empiricalformula

Plan The equilibrium equation and the expression for Ksp areAg2CrO41s2 ∆ 2 Ag+1aq2 + CrO4

2 - 1aq2Ksp = 3Ag+423CrO4

2 - 4

sAmplE ExErCIsE 17.11 Calculating Ksp from solubility

Solid silver chromate is added to pure water at 25 °C, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag2CrO41s2 and the solution. Analysis of the equilibrated solution shows that its silver ion concen-tration is 1.3 * 10-4 M. Assuming that the Ag2CrO4 solution is saturated and that there are no other important equilibria involving Ag+ or CrO4

2 - ions in the solution, calculate Ksp for this compound.

solutIonAnalyze We are given the equilibrium concentration of Ag+ in a saturated solution of Ag2CrO4 and asked to determine the value of Ksp for Ag2CrO4.

To calculate Ksp, we need the equilibrium concentrations of Ag+ and CrO4

2 - . We know that at equilibrium 3Ag+4 = 1.3 * 10-4 M. All the Ag+ and CrO4

2 - ions in the solution come from the Ag2CrO4 that dissolves. Thus, we can use 3Ag+4 to calculate 3CrO4

2 - 4.

Solve The expression for Ksp isKsp = 3Ca2 + 43F-42

Appendix D gives 3.9 * 10-11 for this Ksp.

practice Exercise 1Which of these expressions correctly expresses the solubility-product constant for Ag3PO4 in water? (a) 3Ag43PO44, (b) 3Ag+43PO4

3 - 4, (c) 3Ag+433PO43 - 4, (d) 3Ag+43PO4

3 - 43, (e) 3Ag+433PO4

3 - 43.

practice Exercise 2Give the solubility-product-constant expressions and Ksp values (from Appendix D) for (a) barium carbonate and (b) silver sulfate.


In principle, it is possible to use the Ksp value of a salt to calculate solubility under a variety of conditions. In practice, great care must be taken in doing so for the reasons indicated in “A Closer Look: Limitations of Solubility Products” at the end of this sec-tion. Agreement between the measured solubility and that calculated from Ksp is usu-ally best for salts whose ions have low charges (1+ and 1- ) and do not react with water.

Solve From the chemical formula of silver chromate, we know that there must be two Ag+ ions in solution for each CrO4

2 - ion in solution. Con-sequently, the concentration of CrO4

2 - is half the concentration of Ag+:

3CrO42 - 4 = a1.3 * 10-4 mol Ag+

Lb a1 mol CrO4

2 -

2 mol Ag+ b = 6.5 * 10-5 M

and Ksp isKsp = 3Ag+423CrO4

2 - 4 = 11.3 * 10-42216.5 * 10-52 = 1.1 * 10-12

Check We obtain a small value, as expected for a slightly soluble salt. Furthermore, the calculated value agrees well with the one given in Appendix D, 1.2 * 10-12.

practice Exercise 1You add 10.0 grams of solid copper(II) phosphate, Cu31PO422, to a beaker and then add 100.0 mL of water to the beaker at

T = 298 K. The solid does not appear to dissolve. You wait a long time, with occasional stirring and eventually measure the equilibrium concentration of Cu2 + 1aq2 in the water to be 5.01 * 10-8 M. What is the Ksp of copper(II) phosphate?(a) 5.01 * 10-8 (b) 2.50 * 10-15 (c) 4.20 * 10-15

(d) 3.16 * 10-37 (e) 1.40 * 10-37

practice Exercise 2A saturated solution of Mg1OH22 in contact with undissolved Mg1OH221s2 is prepared at 25 °C. The pH of the solution is found to be 10.17. Assuming that there are no other simultaneous equilibria involving the Mg2 + or OH- ions, calculate Ksp for this compound.

Plan To go from Ksp to solubility, we follow the steps indicated by the red arrows in Figure 17.16. We first write the chemical equation for the dissolution and set up a table of initial and equilibrium concen-trations. We then use the equilibrium-constant expression. In this

sAmplE ExErCIsE 17.12 Calculating solubility from Ksp

The Ksp for CaF2 is 3.9 * 10-11 at 25 °C. Assuming equilibrium is established between solid and dissolved CaF2, and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF2 in grams per liter.

solutIonAnalyze We are given Ksp for CaF2 and asked to determine solubility. Recall that the solubility of a substance is the quantity that can dissolve in solvent, whereas the solubility-product constant, Ksp, is an equilibrium constant.

case we know Ksp, and so we solve for the concentrations of the ions in solution. Once we know these concentrations, we use the formula weight to determine solubility in g>L.

Solve Assume that initially no salt has dissolved, and then allow x mol/L of CaF2 to dissociate com-pletely when equilibrium is achieved:

The stoichiometry of the equilibrium dictates that 2x mol>L of F- are produced for each x mol>L of CaF2 that dissolve. We now use the expression for Ksp and substitute the equilibrium concentrations to solve for the value of x:

CaF21s2 ∆ Ca2 + 1aq2 + 2 F-1aq2Initial concentration (M) — 0 0Change (M) — +x +2xEquilibrium concentration (M) — x 2x

Ksp = 3Ca2+ 43F-42 = 1x212x22 = 4x3 = 3.9 * 10-11

x =3A3.9 * 10-11

4= 2.1 * 10-4(Remember that 32y = y1>3.) Thus, the molar

solubility of CaF2 is 2.1 * 10-4 mol>L.

The mass of CaF2 that dissolves in water to form 1 L of solution is

Check We expect a small number for the solubility of a slightly soluble salt. If we reverse the calculation, we should be able to recalculate the solubility product: Ksp = 12.1 * 10-4214.2 * 10-422 = 3.7 * 10-11, close to the value given in the problem statement, 3.9 * 10-11.Comment Because F- is the anion of a weak acid, you might expect hydrolysis of the ion to affect the solubility of CaF2. The basicity of F- is so small 1Kb = 1.5 * 10-112, however, that the hydrolysis occurs to only a slight extent and does not sig-nificantly influence the solubility. The reported solubility is 0.017 g/L at 25 °C, in good agreement with our calculation.

a 2.1 * 10-4 mol CaF2

1 L solnb a 78.1 g CaF2

1 mol CaF2b = 1.6 * 10-2 g CaF2>L soln

seCtion 17.5 Factors that affect solubility 751

17.5 | Factors that Affect solubilitySolubility is affected by temperature and by the presence of other solutes. The presence of an acid, for example, can have a major influence on the solubility of a substance. In Section 17.4, we considered the dissolving of ionic compounds in pure water. In this section, we examine three factors that affect the solubility of ionic compounds: (1) pres-ence of common ions, (2) solution pH, and (3) presence of complexing agents. We will also examine the phenomenon of amphoterism, which is related to the effects of both pH and complexing agents.

Common-Ion EffectThe presence of either Ca2 + 1aq2 or F-1aq2 in a solution reduces the solubility of CaF2, shifting the equilibrium concentrations to the left:

CaF21s2 ∆ Ca2 + 1aq2 + 2 F-1aq2

Addition of Ca2+ or F− shifts equilibriumconcentrations, reducing solubility

This reduction in solubility is another manifestation of the common-ion effect we looked at in Section 17.1. In general, the solubility of a slightly soluble salt is decreased by the presence of a second solute that furnishes a common ion, as ▶ Figure 17.17 shows for CaF2.

practice Exercise 1Of the five salts listed below, which has the highest concentration of its cation in water? Assume that all salt solutions are saturated and that the ions do not undergo any additional reactions in water.(a) lead (II) chromate, Ksp = 2.8 * 10-13 (b) cobalt(II) hydroxide, Ksp = 1.3 * 10-15 (c) cobalt(II) sulfide, Ksp = 5 * 10-22 (d) chromium(III) hydroxide, Ksp = 1.6 * 10-30

(e) silver sulfide, Ksp = 6 * 10-51

practice Exercise 2The Ksp for LaF3 is 2 * 10-19. What is the solubility of LaF3 in water in moles per liter?

Another common source of error in calculating ion concentra-tions from Ksp is ignoring other equilibria that occur simultaneously in the solution. It is possible, for example, that acid–base equilibria take place simultaneously with solubility equilibria. In particular, both basic anions and cations with high charge-to-size ratios undergo hydrolysis reactions that can measurably increase the solubilities of their salts. For example, CaCO3 contains the basic carbonate ion 1Kb = 1.8 * 10-42, which reacts with water:

CO32 - 1aq2 + H2O1l2 ∆ HCO3

-1aq2 + OH-1aq2If we consider the effect of ion–ion interactions as well as simul-taneous solubility and Kb equilibria, we calculate a solubility of 1.4 * 10-4 mol>L, in agreement with the measured value for calcite.

Finally, we generally assume that ionic compounds dissoci-ate completely when they dissolve, but this assumption is not always valid. When MgF2 dissolves, for example, it yields not only Mg2 + and F- ions but also MgF+ ions.

A Closer look

limitations of solubility products

Ion concentrations calculated from Ksp values sometimes deviate ap-preciably from those found experimentally. In part, these deviations are due to electrostatic interactions between ions in solution, which can lead to ion pairs. (Section 13.5, “The van’t Hoff Factor”) These interactions increase in magnitude both as the concentrations of the ions increase and as their charges increase. The solubility cal-culated from Ksp tends to be low unless corrected to account for these interactions.

As an example of the effect of these interactions, consider CaCO3 (calcite), whose solubility product, 4.5 * 10-9, gives a cal-culated solubility of 6.7 * 10-5 mol>L; correcting for ionic interac-tions in the solution yields 7.3 * 10-5 mol>L. The reported solubility, however, is 1.4 * 10-4 mol>L, indicating that there must be additional factors involved.

0 0.05 0.10 0.15 0.20

10−9

10−8

10−7

10−6

10−5

10−4

10−3

Concentration ofNaF (mol/L)

Mol

ar s

olub

ility

of C

aF2

(mol

/L

)

Pure water

Solubility of CaF2 decreases sharply as a common ion (F–) is added to the solution

▲ Figure 17.17 common-ion effect. Notice that the CaF2 solubility is on a logarithmic scale.


Analyze We are asked to determine the solubility of CaF2 in the pres-ence of two strong electrolytes, each containing an ion common to CaF2. In (a) the common ion is Ca2 + , and NO3

- is a spectator ion. In (b) the common ion is F-, and Na+ is a spectator ion.

sAmplE ExErCIsE 17.13 Calculating the Effect of a Common Ion on solubility

Calculate the molar solubility of CaF2 at 25 °C in a solution that is (a) 0.010 M in Ca1NO322 and (b) 0.010 M in NaF.

solutIonPlan Because the slightly soluble compound is CaF2, we need to use Ksp for this compound, which Appendix D gives as 3.9 * 10-11. The value of Ksp is unchanged by the presence of additional solutes. Because of the common-ion effect, however, the solubility of the salt decreases in the presence of common ions. We use our standard equilibrium techniques of starting with the equation for CaF2 dissolution, setting up a table of initial and equilibrium concentrations, and using the Ksp expression to determine the concentration of the ion that comes only from CaF2.

Solve

(a) The initial concentration of Ca2 + is 0.010 M because of the dissolved Ca1NO322:

Substituting into the solubility-product expression gives

CaF21s2 ∆ Ca2 + 1aq2 + 2 F-1aq2Initial Concentration (M) — 0.010 0Change (M) — +x +2xEquilibrium Concentration (M) — 10.010 + x2 2x

Ksp = 3.9 * 10-11 = 3Ca2 + 43F-42 = 10.010 + x212x22

If we assume that x is small compared to 0.010, we have 3.9 * 10-11 = 10.010212x22

x2 =3.9 * 10-11

410.0102 = 9.8 * 10-10

x = 29.8 * 10-10 = 3.1 * 10-5 M

This very small value for x validates the simplifying assumption we made. Our calculation indicates that 3.1 * 10-5 mol of solid CaF2 dissolves per liter of 0.010 M Ca1NO322 solution.

(b) The common ion is F-, and at equilibrium we have 3Ca2 + 4 = x and 3F-4 = 0.010 + 2x

Assuming that 2x is much smaller than 0.010 M (that is, 0.010 + 2x ≃ 0.010), we have

3.9 * 10-11 = 1x210.010 + 2x22 ≃ x10.01022

x =3.9 * 10-11

10.01022 = 3.9 * 10-7 M

Thus, 3.9 * 10-7 mol of solid CaF2 should dissolve per liter of 0.010 M NaF solution.

Comment The molar solubility of CaF2 in water is 2.1 * 10-4 M (Sample Exercise 17.12). By comparison, our cal-culations here give a CaF2 solubility of 3.1 * 10-5 M in the presence of 0.010 M Ca2 + and 3.9 * 10-7 M in the presence of 0.010 M F- ion. Thus, the addition of either Ca2 + or F- to a solution of CaF2 decreases the solubility. However, the effect of F- on the solubility is more pronounced than that of Ca2 + because 3F-4 appears to the second power in the Ksp expression for CaF2, whereas 3Ca2 + 4 appears to the first power.

practice Exercise 1Consider a saturated solution of the salt MA3, in which M is a metal cation with a 3+ charge and A is an anion with a 1- charge, in water at 298 K. Which of the following will affect the Ksp of MA3 in water? (a) The addition of more M3 + to the solution. (b) The addition of more A- to the solution. (c) Diluting the solution. (d) Raising the temperature of the solution. (e) More than one of the above factors.

practice Exercise 2For manganese(II) hydroxide, Mn1OH22, Ksp = 1.6 * 10-13. Calculate the molar solubility of Mn1OH22 in a solution that contains 0.020 M NaOH.


Solubility and pHThe pH of a solution affects the solubility of any substance whose anion is basic. Con-sider Mg1OH22, for which the solubility equilibrium is

Mg1OH221s2 ∆ Mg2 + 1aq2 + 2 OH-1aq2 Ksp = 1.8 * 10-11 [17.17]

A saturated solution of Mg1OH22 has a calculated pH of 10.52 and its Mg2 + concentra-tion is 1.7 * 10-4 M. Now suppose that solid Mg1OH22 is equilibrated with a solution buffered at pH 9.0. The pOH, therefore, is 5.0, so 3OH-4 = 1.0 * 10-5. Inserting this value for 3OH-4 into the solubility-product expression, we have

Ksp = 3Mg2 + 43OH-42 = 1.8 * 10-11

3Mg2 + 411.0 * 10-522 = 1.8 * 10-11

3Mg2 + 4 =1.8 * 10-11

11.0 * 10-522 = 0.18 M

Thus, the Mg1OH22 dissolves until 3Mg2 + 4 = 0.18 M. It is apparent that Mg1OH22 is much more soluble in this solution.

Chemistry and life

ocean Acidification

Seawater is a weakly basic solution, with pH values typically between 8.0 and 8.3. This pH range is maintained through a carbonic acid buf-fer system similar to the one in blood (see Equation 17.10). Because the pH of seawater is higher than that of blood (7.35–7.45), the second dissociation of carbonic acid cannot be neglected and CO3

2 - becomes an important aqueous species.

The availability of carbonate ions plays an important role in shell formation for a number of marine organisms, including stony corals (▶ Figure 17.18). These organisms, which are referred to as ma-rine calcifiers and play an important role in the food chains of nearly all oceanic ecosystems, depend on dissolved Ca2 + and CO3

2 - ions to form their shells and exoskeletons. The relatively low solubility- product constant of CaCO3,

CaCO31s2 ∆ Ca2 + 1aq2 + CO32 - 1aq2 Ksp = 4.5 * 10-9

and the fact that the ocean contains saturated concentrations of Ca2 + and CO3

2 - mean that CaCO3 is usually quite stable once formed. In fact, calcium carbonate skeletons of creatures that died millions of years ago are not uncommon in the fossil record.

Just as in our bodies, the carbonic acid buffer system can be per-turbed by removing or adding CO21g2. The concentration of dissolved CO2 in the ocean is sensitive to changes in atmospheric CO2 levels. As discussed in Chapter 18, the atmospheric CO2 concentration has risen by approximately 30% over the past three centuries to the pres-ent level of 400 ppm. Human activity has played a prominent role in this increase. Scientists estimate that one-third to one-half of the CO2 emissions resulting from human activity have been absorbed by Earth’s oceans. While this absorption helps mitigate the greenhouse gas effects of CO2, the extra CO2 in the ocean produces carbonic acid, which lowers the pH. Because CO3

2 - is the conjugate base of the weak

▲ Figure 17.18 marine calcifiers. Many sea-dwelling organisms use CaCO3 for their shells and exoskeletons. Examples include stony coral, crustaceans, some phytoplankton, and echinoderms, such as sea urchins and starfish.

acid HCO3-, the carbonate ion readily combines with the hydrogen

ion:CO3

2 - 1aq2 + H+1aq2 ¡ HCO3-1aq2

This consumption of carbonate ion shifts the CaCO3 dissolution equi-librium to the right, increasing the solubility of CaCO3, which can lead to partial dissolution of calcium carbonate shells and exoskeletons. If the amount of atmospheric CO2 continues to increase at the present rate, scientists estimate that seawater pH will fall to 7.9 sometime over the next 50 years. While this change might sound small, it has dra-matic ramifications for oceanic ecosystems.

Related Exercises: 17.99


If 3OH-4 were reduced further by making the solution even more acidic, the Mg2 + concentration would have to increase to maintain the equilibrium condition. Thus, a sample of Mg1OH221s2 dissolves completely if sufficient acid is added, as we saw in Figure 4.9 (page 137).

The solubility of almost any ionic compound is affected if the solution is made suf-ficiently acidic or basic. The effects are noticeable, however, only when one (or both) ions in the compound are at least moderately acidic or basic. The metal hydroxides, such as Mg1OH22, are examples of compounds containing a strongly basic ion, the hydroxide ion.

In general, the solubility of a compound containing a basic anion (that is, the anion of a weak acid) increases as the solution becomes more acidic. As we have seen, the sol-ubility of Mg1OH22 greatly increases as the acidity of the solution increases. The solu-bility of PbF2 increases as the solution becomes more acidic, too, because F- is a base (it is the conjugate base of the weak acid HF). As a result, the solubility equilibrium of PbF2 is shifted to the right as the concentration of F- is reduced by protonation to form HF. Thus, the solution process can be understood in terms of two consecutive reactions:

PbF21s2 ∆ Pb2 + 1aq2 + 2 F-1aq2 [17.18]

F-1aq2 + H+1aq2 ∆ HF1aq2 [17.19]

The equation for the overall process is

PbF21s2 + 2 H+1aq2 ∆ Pb2 + 1aq2 + 2 HF1aq2 [17.20]

The processes responsible for the increase in solubility of PbF2 in acidic solution are illustrated in ▼ Figure 17.19(a).

Other salts that contain basic anions, such as CO32 - , PO4

3 - , CN-, or S2 - , behave similarly. These examples illustrate a general rule: The solubility of slightly soluble salts containing basic anions increases as 3H+4 increases (as pH is lowered). The more basic the anion, the more the solubility is influenced by pH. The solubility of salts with anions of negligible basicity (the anions of strong acids), such as Cl-, Br-, I-, and NO3

-, is unaffected by pH changes, as shown in Figure 17.19(b).

▲ Figure 17.19 response of two ionic compounds to addition of a strong acid. (a) The solubility of PbF2 increases upon addition of acid. (b) The solubility of Pbl2 is not affected by addition of acid. The water molecules and the anion of the strong acid have been omitted for clarity.

Add H+ Add H+

(b)

[F−] decreases[Pb2+] increases

(a)

Pb2+

PbF2 PbF2 PbI2 PbI2

Pb2+ Pb2+ Pb2+

F−F− I−

I−

H+H+ + F− HF No reaction

HF

Salt whose anion is conjugatebase of weak acid:Solubility increases as pH decreases

Salt whose anion is conjugatebase of strong acid:Solubility unaffected by changes in pH


Which of these substances are more soluble in acidic solution than in basic solution: (a) Ni1OH221s2, (b) CaCO31s2, (c) BaF21s2, (d) AgCl(s)?

solutIonAnalyze The problem lists four sparingly soluble salts, and we are asked to determine which are more soluble at low pH than at high pH.Plan We will identify ionic compounds that dissociate to produce a basic anion, as these are especially soluble in acid solution.Solve

(a) Ni1OH221s2 is more soluble in acidic solution because of the basicity of OH-; the H+ reacts with the OH- ion, forming water:

Overall:

Ni1OH221s2 ∆ Ni2 + 1aq2 + 2 OH-1aq2 2 OH-1aq2 + 2 H+1aq2 ¡ 2 H2O1l2Ni1OH221s2 + 2 H+1aq2 ∆ Ni2 + 1aq2 + 2 H2O1l2

(b) Similarly, CaCO31s2 dissolves in acid solutions because CO32 - is a basic anion:

Overall:

CaCO31s2 ∆ Ca2 + 1aq2 + CO32 - 1aq2

CO32 - 1aq2 + 2 H+1aq2 ∆ H2CO31aq2

H2CO31aq2 ∆ CO21g2 + H2O1l2 CaCO31s2 + 2 H+1aq2 ∆ Ca2 + 1aq2 + CO21g2 + H2O1l2

The reaction between CO32 - and H+ occurs in steps, with HCO3

- forming first and H2CO3 forming in appreciable amounts only when 3H+4 is sufficiently high.(c) The solubility of BaF2 is enhanced by lowering the pH because F- is a basic anion:

Overall:

BaF21s2 ∆ Ba2 + 1aq2 + 2 F-1aq22 F-1aq2 + 2 H+1aq2 ∆ 2 HF1aq2 BaF21s2 + 2 H+1aq2 ∆ Ba2 + 1aq2 + 2 HF1aq2

(d) The solubility of AgCl is unaffected by changes in pH because Cl- is the anion of a strong acid and therefore has negligible basicity.

practice Exercise 1Which of the following actions will increase the solubility of AgBr in water? (a) increasing the pH, (b) decreasing the pH, (c) adding NaBr, (d) adding NaNO3, (e) none of the above.

practice Exercise 2Write the net ionic equation for the reaction between a strong acid and (a) CuS, (b) Cu1N322.

sAmplE ExErCIsE 17.14 predicting the Effect of Acid on solubility

Chemistry and life

Tooth decay and fluoridation

Tooth enamel consists mainly of the mineral hydroxyapatite, Ca101PO4261OH22, the hardest substance in the body. Tooth cavities form when acids dissolve tooth enamel:

Ca101PO4261OH221s2 + 8 H+1aq2 ¡10 Ca2 + 1aq2 + 6 HPO4

2 - 1aq2 + 2 H2O1l2The Ca2 + and HPO4

2 - ions diffuse out of the enamel and are washed away by saliva. The acids that attack the hydroxyapatite are formed by the action of bacteria on sugars and other carbohydrates present in the plaque adhering to the teeth.

Fluoride ion, which is added to municipal water systems and toothpastes, can react with hydroxyapatite to form fluoroapatite,

Ca101PO426F2. This mineral, in which F- has replaced OH-, is much more resistant to attack by acids because the fluoride ion is a much weaker Brønsted–Lowry base than the hydroxide ion.

The usual concentration of F- in municipal water systems is 1 mg>L 11 ppm2. The compound added may be NaF or Na2SiF6. The silicon-fluorine anion reacts with water to release fluoride ions:

SiF62 - 1aq2 + 2 H2O1l2 ¡ 6 F-1aq2 + 4 H+1aq2 + SiO21s2

About 80% of all toothpastes now sold in the United States con-tain fluoride compounds, usually at the level of 0.1% fluoride by mass. The most common compounds in toothpastes are sodium fluoride (NaF), sodium monofluorophosphate 1Na2PO3F2, and stannous fluo-ride 1SnF22.



Formation of Complex IonsA characteristic property of metal ions is their ability to act as Lewis acids toward water molecules, which act as Lewis bases. (Section 16.11) Lewis bases other than water can also interact with metal ions, particularly transition-metal ions. Such interactions can dramatically affect the solubility of a metal salt. For example, AgCl 1Ksp = 1.8 * 10-102 dissolves in the presence of aqueous ammonia because Ag+ interacts with the Lewis base NH3, as shown in ▼ Figure 17.20. This process can be viewed as the sum of two reactions:

AgCl1s2 ∆ Ag+1aq2 + Cl-1aq2 [17.21]

Ag+1aq2 + 2 NH31aq2 ∆ Ag1NH322+1aq2 [17.22]

Overall: AgCl1s2 + 2 NH31aq2 ∆ Ag1NH322+1aq2 + Cl-1aq2 [17.23]

The presence of NH3 drives the reaction, the dissolution of AgCl, to the right as Ag+1aq2 is consumed to form Ag1NH322

+, which is a very soluble species.

▲ Figure 17.20 concentrated NH3 1aq 2 dissolves Agcl(s), which otherwise has very low solubility in water.

NH3 reacts with Ag+,forming Ag(NH3)2

+

Addition of suf�cientNH3 leads to complete

dissolution of AgCl

Reaction with NH3 reducesconcentration of free Ag+ and

increases solubility of AgCl

AgCl(s) + 2 NH3(aq) Ag(NH3)2+(aq) + Cl−(aq)

Ag+

AgCl

Cl−

NH3Ag(NH3)2+

Cl−

For a Lewis base such as NH3 to increase the solubility of a metal salt, the base must be able to interact more strongly with the metal ion than water does. In other words, the NH3 must displace solvating H2O molecules (Sections 13.1 and 16.11) in order to form 3Ag1NH3224+:

Ag+1aq2 + 2 NH31aq2 ∆ Ag1NH322+1aq2 [17.24]


An assembly of a metal ion and the Lewis bases bonded to it, such as Ag1NH322+, is

called a complex ion. Complex ions are very soluble in water. The stability of a complex ion in aqueous solution can be judged by the size of the equilibrium constant for its formation from the hydrated metal ion. For example, the equilibrium constant for Equation 17.24 is

Kf =3Ag1NH322

+43Ag+43NH342 = 1.7 * 107 [17.25]

Note that the equilibrium constant for this kind of reaction is called a formation con-stant, Kf. The formation constants for several complex ions are listed in ▲ table 17.1.

table 17.1 Formation Constants for some metal Complex Ions in Water at 25 °C

Complex Ion Kf Equilibrium Equation

Ag1NH322+ 1.7 * 107 Ag+1aq2 + 2 NH31aq2 ∆ Ag1NH322

+ 1aq2Ag1CN22

- 1 * 1021 Ag+1aq2 + 2 CN-1aq2 ∆ Ag1CN22- 1aq2

Ag1S2O3223 - 2.9 * 1013 Ag+1aq2 + 2 S2O3

2 - 1aq2 ∆ Ag1S2O3223 - 1aq2

CdBr42 - 5 * 103 Cd2 + 1aq2 + 4 Br-1aq2 ∆ CdBr4

2 - 1aq2Cr1OH24

- 8 * 1029 Cr3 + 1aq2 + 4 OH-1aq2 ∆ Cr1OH24- 1aq2

Co1SCN242 - 1 * 103 Co2 + 1aq2 + 4 SCN-1aq2 ∆ Co1SCN24

2 - 1aq2Cu1NH324

2 + 5 * 1012 Cu2 + 1aq2 + 4 NH3 1aq2 ∆ Cu1NH3242+ 1aq2

Cu1CN242 - 1 * 1025 Cu2 + 1aq2 + 4 CN-1aq2 ∆ Cu1CN24

2+ 1aq2Ni1NH326

2 + 1.2 * 109 Ni2 + 1aq2 + 6 NH3 1aq2 ∆ Ni1NH3262 + 1aq2

Fe1CN264 - 1 * 1035 Fe2 + 1aq2 + 6 CN- 1aq2 ∆ Fe1CN326

4 - 1aq2Fe1CN26

3 - 1 * 1042 Fe3 + 1aq2 + 6 CN- 1aq2 ∆ Fe1CN263 - 1aq2

Calculate the concentration of Ag+ present in solution at equilibrium when concentrated ammonia is added to a 0.010 M solution of AgNO3 to give an equilibrium concentration of 3NH34 = 0.20 M. Neglect the small volume change that occurs when NH3 is added.

solutIonAnalyze Addition of NH31aq2 to Ag+1aq2 forms Ag1NH322

+, as shown in Equation 17.22. We are asked to determine what concentration of Ag+1aq2 remains uncombined when the NH3 con-centration is brought to 0.20 M in a solution originally 0.010 M in AgNO3.Plan We assume that the AgNO3 is completely dissociated, giving 0.010 M Ag+. Because Kf for the formation of Ag1NH322

+ is quite large, we assume that essentially all the Ag+ is converted to Ag1NH322

+ and approach the problem as though we are concerned with the dissociation of Ag1NH322

+ rather than its formation. To facilitate this approach, we need to reverse Equation 17.22 and make the corresponding change to the equilibrium constant:

Ag1NH322+ 1aq2 ∆ Ag+1aq2 + 2 NH31aq2

1Kf

=1

1.7 * 107 = 5.9 * 10-8

Solve If 3Ag+4 is 0.010 M initially, 3Ag1NH322+4 will be 0.010 M following addition of the NH3.

We construct a table to solve this equilibrium problem. Note that the NH3 concentration given in the problem is an equilibrium concentration rather than an initial concentration.

Ag1NH322+1aq2 ∆ Ag+1aq2 + 2 NH31aq2

Initial (M) 0.010 0 –Change (M) -x +x –Equilibrium (M) (0.010 – x) x 0.20

sAmplE ExErCIsE 17.15 Evaluating an Equilibrium Involving a

Complex Ion


The general rule is that the solubility of metal salts increases in the presence of suit-able Lewis bases, such as NH3, CN-, or OH-, provided the metal forms a complex with the base. The ability of metal ions to form complexes is an extremely important aspect of their chemistry.

AmphoterismSome metal oxides and hydroxides that are relatively insoluble in water dissolve in strongly acidic and strongly basic solutions. These substances, called amphoteric oxides and amphoteric hydroxides,* are soluble in strong acids and bases because they themselves are capable of behaving as either an acid or base. Examples of amphoteric substances include the oxides and hydroxides of Al3 + , Cr3 + , Zn2 + , and Sn2 + .

Like other metal oxides and hydroxides, amphoteric species dissolve in acidic solutions because their anions, O2 - or OH-, react with acids. What makes amphoteric oxides and hydroxides special, though, is that they also dissolve in strongly basic solu-tions. This behavior results from the formation of complex anions containing several (typically four) hydroxides bound to the metal ion (▼ Figure 17.21): Al1OH231s2 + OH-1aq2 ∆ Al1OH24

-1aq2 [17.26]

Because 3Ag+4 is very small, we can assume x is small compared to 0.010. Substituting these values into the equilibrium-constant expression for the dissociation of Ag1NH322

+, we obtain3Ag+43NH342

3Ag1NH322+4 =

1x210.2022

0.010= 5.9 * 10-8

x = 1.5 * 10-8 M = 3Ag+4Formation of the Ag1NH322

+ complex drastically reduces the concentration of free Ag+ ion in solution.

practice Exercise 1You have an aqueous solution of chromium(III) nitrate that you titrate with an aqueous solu-tion of sodium hydroxide. After a certain amount of titrant has been added, you observe a precipitate forming. You add more sodium hydroxide solution and the precipitate dissolves, leaving a solution again. What has happened? (a) The precipitate was sodium hydroxide, which redissolved in the larger volume. (b) The precipitate was chromium hydroxide, which dissolved once more solution was added, forming Cr3 + 1aq2. (c) The precipitate was chro-mium hydroxide, which then reacted with more hydroxide to produce a soluble complex ion, Cr1OH24

-. (d) The precipitate was sodium nitrate, which reacted with more nitrate to pro-duce the soluble complex ion Na1NO322 - .

practice Exercise 2Calculate 3Cr3 + 4 in equilibrium with Cr1OH24

- when 0.010 mol of Cr1NO323 is dissolved in 1 L of solution buffered at pH 10.0.

Al(OH)3(s) + OH−(aq) Al(OH)4−(aq)

Al(OH)3 dissolves instrongly acidic solutionsdue to an acid–basereaction

Al(OH)3 dissolves instrongly basic solutionsdue to complex ionformation

Add H+

Add OH−

Al3+(aq) + 3 H2O(l) 3 H+(aq) + Al(OH)3(s)

▲ Figure 17.21 Amphoterism. Some metal oxides and hydroxides, such as Al1OH23, are amphoteric, which means they dissolve in both strongly acidic and strongly basic solutions.

*Notice that the term amphoteric is applied to the behavior of insoluble oxides and hydroxides that dissolve in acidic or basic solutions. The similar term amphiprotic (Section 16.2) relates more generally to any molecule or ion that can either gain or lose a proton.

seCtion 17.6 precipitation and separation of ions 759

The extent to which an insoluble metal hydroxide reacts with either acid or base varies with the particular metal ion involved. Many metal hydroxides—such as Ca1OH22, Fe1OH22, and Fe1OH23:are capable of dissolving in acidic solution but do not react with excess base. These hydroxides are not amphoteric.

The purification of aluminum ore in the manufacture of aluminum metal provides an interesting application of amphoterism. As we have seen, Al1OH23 is amphoteric, whereas Fe1OH23 is not. Aluminum occurs in large quantities as the ore bauxite, which is essentially hydrated Al2O3 contaminated with Fe2O3. When bauxite is added to a strongly basic solution, the Al2O3 dissolves because the aluminum forms complex ions, such as Al1OH24

-. The Fe2O3 impurity, however, is not amphoteric and remains as a solid. The solution is filtered, getting rid of the iron impurity. Aluminum hydroxide is then precipitated by addition of acid. The purified hydroxide receives further treatment and eventually yields aluminum metal.

Give it some ThoughtWhat is the difference between an amphoteric substance and an amphiprotic substance?

17.6 | precipitation and separation of Ions

Equilibrium can be achieved starting with the substances on either side of a chemical equation. For example, the equilibrium that exists between BaSO41s2, Ba2 + 1aq2, and SO4

2 - 1aq2 (Equation 17.15), can be achieved either by starting with BaSO41s2 or by starting with solutions containing Ba2 + and SO4

2 - . If we mix, say, a BaCl2 aqueous solution with a Na2SO4 aqueous solution, BaSO4 might precipitate out. How can we predict whether a precipitate will form under various conditions?

Recall that we used the reaction quotient Q in Section 15.6 to determine the direc-tion in which a reaction must proceed to reach equilibrium. The form of Q is the same as the equilibrium-constant expression for a reaction, but instead of only equilibrium concentrations, you can use whatever concentrations are being considered. The direc-tion in which a reaction proceeds to reach equilibrium depends on the relationship between Q and K for the reaction. If Q 6 K, the product concentrations are too low and reactant concentrations are too high relative to the equilibrium concentrations, and so the reaction will proceed to the right (toward products) to achieve equilibrium. If Q 7 K, product concentrations are too high and reactant concentrations are too low, and so the reaction will proceed to the left to achieve equilibrium. If Q = K, the reac-tion is at equilibrium.

For solubility-product equilibria, the relationship between Q and Ksp is exactly like that for other equilibria. For Ksp reactions, products are always the soluble ions, and the reactant is always the solid.

Therefore, for solubility equilibria, • If Q = Ksp, the system is at equilibrium, which means the solution is saturated;

this is the highest concentration the solution can have without precipitating. • If Q 6 Ksp, the reaction will proceed to the right, towards the soluble ions; no

precipitate will form. • If Q 7 Ksp, the reaction will proceed to the left, towards the solid; precipitate will form.For the case of the barium sulfate solution, then we would calculate Q = 3Ba2 + 43SO4

2 - 4, and compare this quantity to the Ksp for barium sulfate.


Plan We should determine the concentrations of all ions just after the solutions are mixed and compare the value of Q with Ksp for any potentially insoluble product. The possible metathesis products are

sAmplE ExErCIsE 17.16 predicting Whether a precipitate Forms

Does a precipitate form when 0.10 L of 8.0 * 10-3 M Pb1NO322 is added to 0.40 L of 5.0 * 10-3 M Na2SO4?

solutIonAnalyze The problem asks us to determine whether a precipitate forms when two salt solutions are combined.

PbSO4 and NaNO3. Like all sodium salts NaNO3 is soluble, but PbSO4 has a Ksp of 6.3 * 10-7 (Appendix D) and will precipitate if the Pb2+ and SO4

2 - concentrations are high enough for Q to exceed Ksp.

Solve When the two solutions are mixed, the volume is 0.10 L + 0.40 L = 0.50 L. The number of moles of Pb2 + in 0.10 L of 8.0 * 10-3 M Pb1NO322 is

10.10 L2a 8.0 * 10-3 molL

b = 8.0 * 10-4 mol

The concentration of Pb2 + in the 0.50-L mixture is therefore 3Pb2 + 4 =

8.0 * 10-4 mol0.50 L

= 1.6 * 10-3 M

The number of moles of SO42 - in 0.40 L of

5.0 * 10-3 M Na2SO4 is 10.40 L2a 5.0 * 10-3 molL

b = 2.0 * 10-3 mol

Therefore 3SO4 2 - 4 =2.0 * 10-3 mol

0.50 L= 4.0 * 10-3 M

and Q = 3Pb2 + 43SO42 - 4 = 11.6 * 10-3214.0 * 10-32 = 6.4 * 10-6

Because Q 7 Ksp, PbSO4 precipitates.

practice Exercise 1An insoluble salt MA has a Ksp of 1.0 * 10-16. Two solutions, MNO3 and NaA are mixed, to yield a final solution that is 1.0 * 10-8 M in M+1aq2 and 1.00 * 10-7 M in A- (aq). Will a precipitate form?(a) Yes. (b) No.

practice Exercise 2Does a precipitate form when 0.050 L of 2.0 * 10-2 M NaF is mixed with 0.010 L of 1.0 * 10-2 M Ca1NO322?

Selective Precipitation of IonsIons can be separated from each other based on the solubilities of their salts. Con-sider a solution containing both Ag+ and Cu2 + . If HCl is added to the solution, AgCl 1Ksp = 1.8 * 10-102 precipitates, while Cu2 + remains in solution because CuCl2 is soluble. Separation of ions in an aqueous solution by using a reagent that forms a pre-cipitate with one or more (but not all) of the ions is called selective precipitation.

solutIonAnalyze We are asked to determine the concentration of Cl- neces-sary to begin the precipitation from a solution containing Ag+ and Pb2 + ions, and to predict which metal chloride will begin to precipi-tate first.

sAmplE ExErCIsE 17.17 selective precipitation

A solution contains 1.0 * 10-2 M Ag+ and 2.0 * 10-2 M Pb2 + . When Cl- is added, both AgCl 1Ksp = 1.8 * 10-102 and PbCl21Ksp = 1.7 * 10-52 can precipitate. What concentration of Cl- is necessary to begin the precipitation of each salt? Which salt precipitates first?

Plan We are given Ksp values for the two precipitates. Using these and the metal ion concentrations, we can calculate what Cl- concentration is necessary to precipitate each salt. The salt requiring the lower Cl- ion concentration precipitates first.

seCtion 17.6 precipitation and separation of ions 761

Sulfide ion is often used to separate metal ions because the solubilities of sulfide salts span a wide range and depend greatly on solution pH. For example, Cu2 + and Zn2 + can be separated by bubbling H2S gas through an acidified solution containing these two cat-ions. Because CuS 1Ksp = 6 * 10-372 is less soluble than ZnS 1Ksp = 2 * 10-252, CuS precipitates from an acidified solution pH ≈ 1 while ZnS does not (▼ Figure 17.22):

Cu2 + 1aq2 + H2S1aq2 ∆ CuS1s2 + 2 H+1aq2 [17.27]

Solve For AgCl we have Ksp = 3Ag+43Cl-4 = 1.8 * 10-10.Because 3Ag+4 = 1.0 * 10-2 M, the greatest concentration of Cl- that can be present without causing precipitation of AgCl can be calculated from the Ksp expression:

Ksp = 11.0 * 10-223Cl-4 = 1.8 * 10-10

3Cl-4 =1.8 * 10-10

1.0 * 10-2 = 1.8 * 10-8 M

Any Cl- in excess of this very small concentration will cause AgCl to precipitate from solution. Proceeding similarly for PbCl2, we have

Ksp = 3Pb2 + 43Cl-42 = 1.7 * 10-5

12.0 * 10-223Cl-42 = 1.7 * 10-5

3Cl-42 =1.7 * 10-5

2.0 * 10-2 = 8.5 * 10-4

3Cl-4 = 28.5 * 10-4 = 2.9 * 10-2 M

Thus, a concentration of Cl- in excess of 2.9 * 10-2 M causes PbCl2 to precipitate.

Comparing the Cl- concentration required to precipitate each salt, we see that as Cl- is added, AgCl precipitates first because it requires a much smaller concentration of Cl-. Thus, Ag+ can be separated from Pb2 + by slowly adding Cl- so that the chloride ion concentration re-mains between 1.8 * 10-8 M and 2.9 * 10-2 M.Comment Precipitation of AgCl will keep the Cl- concentration low until the number of moles of Cl- added exceeds the number of moles of Ag+ in the solution. Once past this point, 3Cl-4 rises sharply and PbCl2 will soon begin to precipitate.

practice Exercise 1Under what conditions does an ionic compound precipitate from a solution of the constituent ions? (a) always, (b) when Q = Ksp, (c) when Q exceeds Ksp, (d) when Q is less than Ksp, (e) never, if it is very soluble.

practice Exercise 2A solution consists of 0.050 M Mg2 + and Cu2 + . Which ion precip-itates first as OH- is added? What concentration of OH- is neces-sary to begin the precipitation of each cation? 3Ksp = 1.8 * 10-11 for Mg1OH22, and Ksp = 4.8 * 10-20 for Cu1OH22.4

▲ Figure 17.22 selective precipitation. In this example, Cu2 + ions are separated from Zn2 + ions.

Go FIGurEWhat would happen if the pH were raised to 8 first and then H2S were added?

Zn2+HS−

Cu2+pH ≈ 1 pH ≈ 1 pH ≈ 8

H+

H2S

CuS(s) ZnS(s)

When H2S is added to a solution whose pH exceeds 0.6,

CuS precipitates

Solution containing Zn2+(aq) and Cu2+(aq)

After CuS is removed, the pH is increased, allowing ZnS to

precipitate

Add H2S

Remove CuSand

increase pH


The CuS can be separated from the Zn2 + solution by filtration. The separated CuS can then be dissolved by raising the concentration of H+ even further, shifting the equilib-rium concentrations of the compounds in Equation 17.27 to the left.

17.7 | Qualitative Analysis for metallic Elements

In this final section, we look at how solubility equilibria and complex-ion formation can be used to detect the presence of particular metal ions in solution. Before the development of modern analytical instrumentation, it was necessary to analyze mix-tures of metals in a sample by what were called wet chemical methods. For example, an ore sample that might contain several metallic elements was dissolved in a con-centrated acid solution that was then tested in a systematic way for the presence of various metal ions.

Qualitative analysis determines only the presence or absence of a particular metal ion relative to some threshold, whereas quantitative analysis determines how much of a given substance is present. Even though wet methods of qualitative analy-sis have become less important in the chemical industry, they are frequently used in general chemistry laboratory programs to illustrate equilibria, to teach the proper-ties of common metal ions in solution, and to develop laboratory skills. Typically, such analyses proceed in three stages: (1) The ions are separated into broad groups on the basis of solubility properties. (2) The ions in each group are separated by selectively dissolving members in the group. (3) The ions are identified by means of specific tests.

A scheme in general use divides the common cations into five groups (▶ Figure 17.23). The order in which reagents are added is important in this scheme. The most selective separations—those that involve the smallest number of ions—are carried out first. The reactions used must proceed so far toward completion that any concentration of cations remaining in the solution is too small to interfere with subsequent tests.

Let’s look at each of these five groups of cations, briefly examining the logic used in this qualitative analysis scheme.

Group 1. Insoluble chlorides: Of the common metal ions, only Ag+, Hg22+, and

Pb2 + form insoluble chlorides. When HCl is added to a mixture of cations, there-fore, only AgCl, Hg2Cl2, and PbCl2 precipitate, leaving the other cations in solu-tion. The absence of a precipitate indicates that the starting solution contains no Ag+, Hg2

2 + , or Pb2 + .Group 2. Acid-insoluble sulfides: After any insoluble chlorides have been re-moved, the remaining solution, now acidic from HCl treatment, is treated with H2S. Since H2S is a weak acid compared to HCl, its role here is to act as a source for small amounts of sulfide. Only the most insoluble metal sulfides—CuS, Bi2S3, CdS, PbS, HgS, As2S3, Sb2S3, and SnS2—precipitate. (Note the very small values of Ksp for some of these sulfides in Appendix D.) Those metal ions whose sulfides are somewhat more soluble—for example, ZnS or NiS—remain in solution.Group 3. Base-insoluble sulfides and hydroxides: After the solution is filtered to remove any acid-insoluble sulfides, it is made slightly basic, and 1NH422S is added. In basic solutions the concentration of S2 - is higher than in acidic solu-tions. Under these conditions, the ion products for many of the more soluble sulfides exceed their Ksp values and thus precipitation occurs. The metal ions precipitated at this stage are Al3 + , Cr3 + , Fe3 + , Zn2 + , Ni2 + , Co2 + , and Mn2 + . (The Al3 + , Fe3 + , and Cr3 + ions do not form insoluble sulfides; instead they precipitate as insoluble hydroxides, as Figure 17.23 shows.)

seCtion 17.7 Qualitative analysis for Metallic elements 763

Group 4. Insoluble phosphates: At this point, the solution contains only metal ions from groups 1A and 2A of the periodic table. Adding 1NH422HPO4 to a basic solution precipitates the group 2A elements Mg2 + , Ca2 + , Sr2 + , and Ba2 + because these metals form insoluble phosphates.Group 5. The alkali metal ions and NH4

+: The ions that remain after removing the insoluble phosphates are tested for individually. A flame test can be used to determine the presence of K+, for example, because the flame turns a characteris-tic violet color if K+ is present (Figure 7.22).

Give it some ThoughtIf a precipitate forms when HCl is added to an aqueous solution, what conclusions can you draw about the contents of the solution?

▲ Figure 17.23 Qualitative analysis. A flowchart showing a common scheme for identifying cations.

Solution containingunknown metal cations

Add 6 M HCl

Precipitate Decantate

Remainingcations

Remainingcations

Remainingcations

Group 1Insoluble chlorides:AgCl, PbCl2, Hg2Cl2

Group 2Acid-insoluble sul�des:

CuS, CdS, Bi2S3, PbS,HgS, As2S3, Sb2S3, SnS2

Group 3Base-insoluble sul�des and hydroxides:Al(OH)3, Fe(OH)3, Cr(OH)3, ZnS, NiS,

MnS, CoS

Group 4Insoluble phosphates:

Ca3(PO4)2, Sr3(PO4)2, Ba3(PO4)2,MgNH4PO4

Group 5Alkali metal ions

and NH4+

Add H2S and 0.2 M HCl


Add (NH4)2S at pH = 8


Add (NH4)2HPO4 and NH3


Go FIGurEIf a solution contained a mixture of Cu2 + and Zn2 + ions, would this separation scheme work? After which step would the first precipitate be observed?


putting Concepts together

A sample of 1.25 L of HCl gas at 21 °C and 0.950 atm is bubbled through 0.500 L of 0.150 M NH3 solution. Calculate the pH of the resulting solution assuming that all the HCl dissolves and that the volume of the solution remains 0.500 L.

solutIonThe number of moles of HCl gas is calculated from the ideal-gas law,

n =PVRT

=10.950 atm211.25 L2

10.0821 L@atm>mol@K21294 K2 = 0.0492 mol HCl

The number of moles of NH3 in the solution is given by the product of the volume of the solu-tion and its concentration,

Moles NH3 = 10.500 L210.150 mol NH3>L2 = 0.0750 mol NH3

The acid HCl and base NH3 react, transferring a proton from HCl to NH3, producing NH4+

and Cl- ions,HCl1g2 + NH31aq2 ¡ NH4

+1aq2 + Cl-1aq2To determine the pH of the solution, we first calculate the amount of each reactant and each product present at the completion of the reaction. Because you can assume this neu-tralization reaction proceeds as far toward the product side as possible, this is a limiting reactant problem.

HCl1g2 + NH31aq2 ¡ NH4+1aq2 + Cl-1aq2

Before reaction (mol) 0.0492 0.0750 0 0Change (limiting reactant) (mol)

-0.0492 -0.0492 +0.0492 +0.0492

After reaction (mol) 0 0.0258 0.0492 0.0492

Thus, the reaction produces a solution containing a mixture of NH3, NH4+, and Cl-. The

NH3 is a weak base 1Kb = 1.8 * 10-52, NH4+ is its conjugate acid, and Cl- is neither acidic nor

basic. Consequently, the pH depends on 3NH34 and 3NH4+4,

3NH34 =0.0258 mol NH3

0.500 L soln= 0.0516 M

3NH4+4 =

0.0492 mol NH4+

0.500 L soln= 0.0984 M

We can calculate the pH using either Kb for NH3 or Ka for NH4+. Using the Kb expression, we

have

NH31aq2 + H2O1l2 ∆ NH4+ 1aq2 + OH -1aq2

Initial (M) 0.0516 — 0.0984 0Change (M) -x — +x +xEquilibrium (M) 10.0516 - x2 — 10.0984 + x2 x

Kb =3NH4

+43OH-43NH34

=10.0984 + x21x210.0516 - x2 _

10.09842x0.0516

= 1.8 * 10-5

x = 3OH-4 =10.0516211.8 * 10-52

0.0984= 9.4 * 10-6 M

Hence, pOH = - log19.4 * 10-62 = 5.03 and pH = 14.00 - pOH = 14.00 - 5.03 = 8.97.

sAmplE IntEGrAtIvE ExErCIsE

Learning outcomes 765

Chapter summary and Key termsTHe common ion effecT (secTion 17.1) In this chapter, we have considered several types of important equilibria that occur in aqueous solution. Our primary emphasis has been on acid–base equi-libria in solutions containing two or more solutes and on solubility equilibria. The dissociation of a weak acid or weak base is repressed by the presence of a strong electrolyte that provides an ion common to the equilibrium (the common-ion effect).

BUffers (secTion 17.2) A particularly important type of acid–base mixture is that of a weak conjugate acid–base pair that functions as a buffered solution (buffer). Addition of small amounts of a strong acid or a strong base to a buffered solution causes only small changes in pH because the buffer reacts with the added acid or base. (Strong acid–strong base, strong acid–weak base, and weak acid–strong base reactions proceed essentially to completion.) Buffered solutions are usually prepared from a weak acid and a salt of that acid or from a weak base and a salt of that base. Two important characteristics of a buffered solution are its buffer capacity and its pH range. The optimal pH of a buffer is equal to pKa (or pKb) of the acid (or base) used to prepare the buffer. The relationship between pH, pKa, and the con-centrations of an acid and its conjugate base can be expressed by the Henderson–Hasselbalch equation. It is important to realize that the Henderson–Hasselbalch equation is an approximation, and more detailed calculations may need to be performed to obtain equilibrium concentrations.

Acid–BAse TiTrATions (secTion 17.3) The plot of the pH of an acid (or base) as a function of the volume of added base (or acid) is called a pH titration curve. The titration curve of a strong acid–strong base titration exhibits a large change in pH in the immediate vicinity of the equivalence point; at the equivalence point for such a titration pH = 7. For strong acid–weak base or weak acid–strong base titra-tions, the pH change in the vicinity of the equivalence point is not as large as the strong acid–strong base titration, nor will the pH equal 7 at the equivalence point in these cases. Instead, what determines the pH at the equivalence point is the conjugate base or acid salt solu-tion that results from the neutralization reaction. For this reason, it is important to choose an indicator whose color change is near the pH at the equivalence point for titrations involving either weak acids or weak bases. It is possible to calculate the pH at any point of the titra-tion curve by first considering the effects of the acid–base reaction on solution concentrations and then examining equilibria involving the remaining solute species.

solUBiliTy eQUiliBriA (secTion 17.4) The equilibrium between a solid compound and its ions in solution provides an example of heterogeneous equilibrium. The solubility-product constant (or simply the solubility product), Ksp, is an equilibrium constant that expresses quantitatively the extent to which the compound dissolves. The Ksp can be used to calculate the solubility of an ionic compound, and the solubility can be used to calculate Ksp.

fAcTors THAT AffecT solUBiliTy (secTion 17.5) Several experimental factors, including temperature, affect the solubili-ties of ionic compounds in water. The solubility of a slightly soluble ionic compound is decreased by the presence of a second solute that furnishes a common ion (the common-ion effect). The solubility of compounds containing basic anions increases as the solution is made more acidic (as pH decreases). Salts with anions of negligible basicity (the anions of strong acids) are unaffected by pH changes.

The solubility of metal salts is also affected by the presence of certain Lewis bases that react with metal ions to form stable complex ions. Complex-ion formation in aqueous solution involves the dis-placement by Lewis bases (such as NH3 and CN-) of water molecules attached to the metal ion. The extent to which such complex forma-tion occurs is expressed quantitatively by the formation constant for the complex ion. Amphoteric oxides and hydroxides are those that are only slightly soluble in water but dissolve on addition of either acid or base.

precipiTATion And sepArATion of ions (secTion 17.6) Comparison of the reaction quotient, Q, with the value of Ksp can be used to judge whether a precipitate will form when solutions are mixed or whether a slightly soluble salt will dissolve under vari-ous conditions. Precipitates form when Q 7 Ksp. If two salts have sufficiently different solubilities, selective precipitation can be used to precipitate one ion while leaving the other in solution, effectively separating the two ions.

QUAliTATive AnAlysis for meTAllic elemenTs (secTion 17.7) Metallic elements vary a great deal in the solubilities of their salts, in their acid–base behavior, and in their tendencies to form com-plex ions. These differences can be used to separate and detect the presence of metal ions in mixtures. Qualitative analysis determines the presence or absence of species in a sample, whereas quantitative analy-sis determines how much of each species is present. The qualitative analysis of metal ions in solution can be carried out by separating the ions into groups on the basis of precipitation reactions and then ana-lyzing each group for individual metal ions.

learning outcomes After studying this chapter, you should be able to:

• Describe the common-ion effect. (Section 17.1)

• Explain how a buffer functions. (Section 17.2)

• Calculate the pH of a buffered solution. (Section 17.2)

• Calculate the pH of a buffer after the addition of small amounts of a strong acid or a strong base. (Section 17.2)

• Calculate the appropriate quantities of compounds to make a buf-fer at a given pH. (Section 17.2)

• Calculate the pH at any point in a strong acid–strong base titra-tion. (Section 17.3)

• Calculate the pH at any point in a weak acid–strong base or weak base–strong acid titration. (Section 17.3)

• Describe the differences between the titration curves for a strong acid–strong base titration and those when either the acid or base is weak. (Section 17.3)

• Estimate the pKa for monoprotic or polyprotic acids from titration curves. (Section 17.3)

• Given either Ksp, molar solubility or mass solubility for a substance, calculate the other two quantities. (Section 17.4)


• Calculate molar solubility in the presence of a common ion. (Section 17.5)

• Predict the effect of pH on solubility. (Section 17.5)

• Predict whether a precipitate will form when solutions are mixed, by comparing Q and Ksp. (Section 17.6)

• Calculate the ion concentrations required to begin precipitation. (Section 17.6)

• Explain the effect of complex-ion formation on solubility. (Section 17.6)

• Explain the logic of identification of metal ions in aqueous solution by a series of reactions. (Section 17.7)

Key Equations • pH = pKa + log

3base43acid4 [17.9] The Henderson–Hasselbalch equation, used to estimate the pH of a buffer from the concentrations of

a conjugate acid–base pair

Exercisesvisualizing Concepts 17.1 The following boxes represent aqueous solutions containing

a weak acid, HA and its conjugate base, A-. Water molecules, hydronium ions, and cations are not shown. Which solution has the highest pH? Explain. [Section 17.1]

= HA = A−

17.2 The beaker on the right contains 0.1 M acetic acid solution with methyl orange as an indicator. The beaker on the left contains a mixture of 0.1 M acetic acid and 0.1 M sodium ace-tate with methyl orange. (a) Using Figure 16.7, which solution has a higher pH? (b) Which solution is better able to main-tain its pH when small amounts of NaOH are added? Explain. [Sections 17.1 and 17.2]

17.3 A buffer contains a weak acid, HA, and its conjugate base. The weak acid has a pKa of 4.5, and the buffer has a pH of 4.3. Without doing a calculation, state which of these pos-sibilities are correct. (a) 3HA4 = 3A-4, (b) 3HA4 7 3A-4, or (c) 3HA4 6 3A-4. [Section 17.2]

17.4 The following diagram represents a buffer composed of equal concentrations of a weak acid, HA, and its conjugate base, A-. The heights of the columns are proportional to the concentra-tions of the components of the buffer.

(a) Which of the three drawings, (1), (2), or (3), represents the buffer after the addition of a strong acid? (b) Which of the

three represents the buffer after the addition of a strong base? (c) Which of the three represents a situation that cannot arise from the addition of either an acid or a base? [Section 17.2]

HA HA

(1)

HA

(2)

HAA− A− A− A−

(3)

17.5 The following figure represents solutions at various stages of the titration of a weak acid, HA, with NaOH. (The Na+ ions and water molecules have been omitted for clarity.) To which of the following regions of the titration curve does each draw-ing correspond: (a) before addition of NaOH, (b) after addition of NaOH but before the equivalence point, (c) at the equiva-lence point, (d) after the equivalence point? [Section 17.3]

(i) (ii) (iii) (iv)

= HA = A− = OH−

17.6 Match the following descriptions of titration curves with the diagrams: (a) strong acid added to strong base, (b) strong base added to weak acid, (c) strong base added to strong acid, (d) strong base added to polyprotic acid. [Section 17.3]

mL titrant

(i)

pH

mL titrant

(ii)

pH

mL titrant

(iii)

pH

mL titrant

(iv)

pH

exercises 767

(i) 50 mL1.0 M HCl(aq)

(ii) 50 mL1.0 M NaCl(aq)

(iii) 50 mL1.0 M CaCl2(aq)

(iv) 50 mL0.10 M CaCl2(aq)

(i) 50 mL1.0 M HCl(aq)

(ii) 50 mL1.0 M NaCl(aq)

(iii) 50 mL1.0 M CaCl2(aq)

(iv) 50 mL0.10 M CaCl2(aq)

17.11 The graph below shows the solubility of a salt as a function of pH. Which of the following choices explain the shape of this graph? (a) None; this behavior is not possible. (b) A soluble salt reacts with acid to form a precipitate, and additional acid reacts with this product to dissolve it. (c) A soluble salt forms an insoluble hydroxide, then additional base reacts with this product to dissolve it. (d) The solubility of the salt increases with pH then decreases because of the heat generated from the neutralization reactions. [Section 17.5]

pH

Solubility

17.12 Three cations, Ni2 + , Cu2 + , and Ag+, are separated using two different precipitating agents. Based on Figure 17.23, what two precipitating agents could be used? Using these agents, indicate which of the cations is A, which is B, and which is C. [Section 17.7]

Mixture ofcations A,B,C

CationsB,C

CationC

Cation Aremoved

Cation Bremoved

Add 1stprecipitating

agent

Add 2ndprecipitating

agentDecantliquid

Decantliquid

Cation A Cation B Cation C

the Common-Ion Effect (section 17.1)

17.13 Which of these statements about the common-ion effect is most correct? (a) The solubility of a salt MA is decreased in a solution that already contains either M+ or A-. (b) Common ions alter the equilibrium constant for the reaction of an ionic

17.7 Equal volumes of two acids are titrated with 0.10 M NaOH resulting in the two titration curves shown in the following figure. (a) Which curve corresponds to the more concen-trated acid solution? (b) Which corresponds to the acid with the larger Ka? [Section 17.3]

pH

0

2

4

6

8

10

12

0 10 20 30 40 50mL NaOH

17.8 A saturated solution of Cd1OH22 is shown in the middle bea-ker. If hydrochloric acid solution is added, the solubility of

AddHCl(aq)

AddHCl(aq)

Cd(OH)2(s)

Cd2+(aq) OH−(aq)

Beaker A Beaker B

Saturated solution

Cd(OH)2(s) Cd(OH)2(s)

Cd1OH22 will increase, causing additional solid to dissolve. Which of the two choices, Beaker A or Beaker B, accurately represents the solution after equilibrium is reestablished? (The water molecules and CI- ions are omitted for clarity.) [Sections 17.4 and 17.5]

17.9 The following graphs represent the behavior of BaCO3 under different circumstances. In each case, the vertical axis indi-cates the solubility of the BaCO3 and the horizontal axis rep-resents the concentration of some other reagent. (a) Which graph represents what happens to the solubility of BaCO3 as HNO3 is added? (b) Which graph represents what happens to the BaCO3 solubility as Na2CO3 is added? (c) Which rep-resents what happens to the BaCO3 solubility as NaNO3 is added? [Section 17.5]

Conc

Solubility

Conc

Solubility

Conc

Solubility

17.10 Ca1OH22 has a Ksp of 6.5 * 10-6. (a) If 0.370 g of Ca1OH22 is added to 500 mL of water and the mixture is allowed to come to equilibrium, will the solution be saturated? (b) If 50 mL of the so-lution from part (a) is added to each of the beakers shown here, in which beakers, if any, will a precipitate form? In those cases where a precipitate forms, what is its identity? [Section 17.6]


buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of sodium hydroxide solution are added to the buffer.

17.24 A buffer is prepared by adding 10.0 g of ammonium chloride 1NH4Cl2 to 250 mL of 1.00 M NH3 solution. (a) What is the pH of this buffer? (b) Write the complete ionic equation for the reaction that occurs when a few drops of nitric acid are added to the buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of potassium hy-droxide solution are added to the buffer.

17.25 You are asked to prepare a pH = 3.00 buffer solution start-ing from 1.25 L of a 1.00 M solution of hydrofluoric acid (HF) and any amount you need of sodium fluoride (NaF). (a) What is the pH of the hydrofluoric acid solution prior to adding sodium fluoride? (b) How many grams of sodium fluoride should be added to prepare the buffer solution? Neglect the small volume change that occurs when the sodium fluoride is added.

17.26 You are asked to prepare a pH = 4.00 buffer starting from 1.50 L of 0.0200 M solution of benzoic acid 1C6H5COOH2 and any amount you need of sodium benzoate 1C6H5COONa2. (a) What is the pH of the benzoic acid solution prior to add-ing sodium benzoate? (b) How many grams of sodium benzo-ate should be added to prepare the buffer? Neglect the small volume change that occurs when the sodium benzoate is added.

17.27 A buffer contains 0.10 mol of acetic acid and 0.13 mol of so-dium acetate in 1.00 L. (a) What is the pH of this buffer? (b) What is the pH of the buffer after the addition of 0.02 mol of KOH? (c) What is the pH of the buffer after the addition of 0.02 mol of HNO3?

17.28 A buffer contains 0.15 mol of propionic acid 1C2H5COOH2 and 0.10 mol of sodium propionate 1C2H5COONa2 in 1.20 L. (a) What is the pH of this buffer? (b) What is the pH of the buffer after the addition of 0.01 mol of NaOH? (c) What is the pH of the buffer after the addition of 0.01 mol of HI?

17.29 (a) What is the ratio of HCO3 - to H2CO3 in blood of pH 7.4? (b) What is the ratio of HCO3 - to H2CO3 in an exhausted marathon runner whose blood pH is 7.1?

17.30 A buffer, consisting of H2PO4 - and HPO4 2 -, helps control the pH of physiological fluids. Many carbonated soft drinks also use this buffer system. What is the pH of a soft drink in which the major buffer ingredients are 6.5 g of NaH2PO4 and 8.0 g of Na2HPO4 per 355 mL of solution?

17.31 You have to prepare a pH 3.50 buffer, and you have the follow-ing 0.10 M solutions available: HCOOH, CH3COOH, H3PO4,HCOONa, CH3COONa, and NaH2PO4. Which solutions would you use? How many milliliters of each solution would you use to make approximately 1 L of the buffer?

17.32 You have to prepare a pH 5.00 buffer, and you have the fol-lowing 0.10 M solutions available: HCOOH, HCOONa, CH3COOH, CH3COONa, HCN, and NaCN. Which solutions would you use? How many milliliters of each solution would you use to make approximately 1 L of the buffer?

Acid–Base titrations (section 17.3)

17.33 The accompanying graph shows the titration curves for two monoprotic acids. (a) Which curve is that of a strong acid? (b) What is the approximate pH at the equivalence point of

solid with water. (c) The common-ion effect does not apply to unusual ions like SO3

2 - . (d) The solubility of a salt MA is af-fected equally by addition of either A- or a non-common ion.

17.14 Consider the equilibrium B1aq2 + H2O1l2 ∆ HB+1aq2 + OH-1aq2. Suppose that a salt of HB+ is added to a solution of B at equi-

librium. (a) Will the equilibrium constant for the reaction in-crease, decrease, or stay the same? (b) Will the concentration of B(aq) increase, decrease, or stay the same? (c) Will the pH of the solution increase, decrease, or stay the same?

17.15 Use information from Appendix D to calculate the pH of (a) a solution that is 0.060 M in potassium propionate 1C2H5COOK or KC3H5O22 and 0.085 M in propionic acid 1C2H5COO K or HC3H5O22; (b) a solution that is 0.075 M in trimethylamine, 1CH323N, and 0.10 M in trimethylammo-nium chloride, 1CH323NHCl; (c) a solution that is made by mixing 50.0 mL of 0.15 M acetic acid and 50.0 mL of 0.20 M sodium acetate.

17.16 Use information from Appendix D to calculate the pH of (a) a solution that is 0.250 M in sodium formate (HCOONa) and 0.100 M in formic acid (HCOOH), (b) a solution that is 0.510 M in pyridine 1C5H5N2 and 0.450 M in pyridinium chloride 1C5H5NHCl2, (c) a solution that is made by combining 55 mL of 0.050 M hydrofluoric acid with 125 mL of 0.10 M sodium fluoride.

17.17 (a) Calculate the percent ionization of 0.0075 M butanoic acid 1Ka = 1.5 * 10-52. (b) Calculate the percent ionization of 0.0075 M butanoic acid in a solution containing 0.085 M sodium butanoate.

17.18 (a) Calculate the percent ionization of 0.125 M lactic acid 1Ka = 1.4 * 10-42. (b) Calculate the percent ionization of 0.125 M lactic acid in a solution containing 0.0075 M sodium lactate.

Buffers (section 17.2)

17.19 Which of the following solutions is a buffer? (a) 0.10 MCH3COOH and 0.10 M CH3COONa, (b) 0.10 M CH3COOH, (c) 0.10 M HCl and 0.10 M NaCl, (d) both a and c, (e) all of a, b, and c.

17.20 Which of the following solutions is a buffer? (a) A solution made by mixing 100 mL of 0.100 M CH3COOH and 50 mL of 0.100 M NaOH, (b) a solution made by mixing 100 mL of 0.100 M CH3COOH and 500 mL of 0.100 M NaOH, (c) A so-lution made by mixing 100 mL of 0.100 M CH3COOH and 50 mL of 0.100 M HCl, (d) A solution made by mixing 100 mL of 0.100 M CH3COOK and 50 mL of 0.100 M KCl.

17.21 (a) Calculate the pH of a buffer that is 0.12 M in lactic acid and 0.11 M in sodium lactate. (b) Calculate the pH of a buffer formed by mixing 85 mL of 0.13 M lactic acid with 95 mL of 0.15 M sodium lactate.

17.22 (a) Calculate the pH of a buffer that is 0.105 M in NaHCO3 and 0.125 M in Na2CO3. (b) Calculate the pH of a solution formed by mixing 65 mL of 0.20 M NaHCO3 with 75 mL of 0.15 M Na2CO3.

17.23 A buffer is prepared by adding 20.0 g of sodium ac-etate 1CH3COONa2 to 500 mL of a 0.150 M acetic acid 1CH3COOH2 solution. (a) Determine the pH of the buffer. (b) Write the complete ionic equation for the reaction that occurs when a few drops of hydrochloric acid are added to the

exercises 769

with NaOH, (b) NH3 titrated with HCl, (c) KOH titrated with HBr.

17.38 Predict whether the equivalence point of each of the following titrations is below, above, or at pH 7: (a) formic acid titrated with NaOH, (b) calcium hydroxide titrated with perchloric acid, (c) pyridine titrated with nitric acid.

17.39 As shown in Figure 16.8, the indicator thymol blue has two color changes. Which color change will generally be more suitable for the titration of a weak acid with a strong base?

17.40 Assume that 30.0 mL of a 0.10 M solution of a weak base B that accepts one proton is titrated with a 0.10 M solution of the monoprotic strong acid HA. (a) How many moles of HA have been added at the equivalence point? (b) What is the predomi-nant form of B at the equivalence point? (c) Is the pH 7, less than 7, or more than 7 at the equivalence point? (d) Which indicator, phenolphthalein or methyl red, is likely to be the better choice for this titration?

17.41 How many milliliters of 0.0850 M NaOH are required to ti-trate each of the following solutions to the equivalence point: (a) 40.0 mL of 0.0900 M HNO3, (b) 35.0 mL of 0.0850 M CH3COOH, (c) 50.0 mL of a solution that contains 1.85 g of HCl per liter?

17.42 How many milliliters of 0.105 M HCl are needed to titrate each of the following solutions to the equivalence point: (a) 45.0 mL of 0.0950 M NaOH, (b) 22.5 mL of 0.118 M NH3, (c) 125.0 mL of a solution that contains 1.35 g of NaOH per liter?

17.43 A 20.0-mL sample of 0.200 M HBr solution is titrated with 0.200 M NaOH solution. Calculate the pH of the solution after the following volumes of base have been added: (a) 15.0 mL, (b) 19.9 mL, (c) 20.0 mL, (d) 20.1 mL, (e) 35.0 mL.

17.44 A 20.0-mL sample of 0.150 M KOH is titrated with 0.125 M HClO4 solution. Calculate the pH after the following volumes of acid have been added: (a) 20.0 mL, (b) 23.0 mL, (c) 24.0 mL, (d) 25.0 mL, (e) 30.0 mL.

17.45 A 35.0-mL sample of 0.150 M acetic acid 1CH3COOH2 is ti-trated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added: (a) 0 mL, (b) 17.5 mL, (c) 34.5 mL, (d) 35.0 mL, (e) 35.5 mL, (f) 50.0 mL.

17.46 Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of ti-trant have been added: (a) 0 mL, (b) 20.0 mL, (c) 59.0 mL, (d) 60.0 mL, (e) 61.0 mL, (f) 65.0 mL.

17.47 Calculate the pH at the equivalence point for titrating 0.200 M solutions of each of the following bases with 0.200 M HBr: (a) sodium hydroxide (NaOH), (b) hydroxylamine 1NH2OH2, (c) aniline 1C6H5NH22.

17.48 Calculate the pH at the equivalence point in titrating 0.100 M solutions of each of the following with 0.080 M NaOH: (a) hydrobromic acid (HBr), (b) chlorous acid 1HClO22, (c) benzoic acid 1C6H5COOH2.

solubility Equilibria and Factors Affecting solubility (sections 17.4 and 17.5)

17.49 For each statement, indicate whether it is true or false.(a) The solubility of a slightly soluble salt can be expressed in

units of moles per liter.(b) The solubility product of a slightly soluble salt is simply

the square of the solubility.

each titration? (c) 40.0 mL of each acid was titrated with 0.100 M base. Which acid is more concentrated? (d) Estimate the pKa of the weak acid.

0 10

A

B

20 30 40 50 60

2

4

pH6

8

10

12

14

0

mL NaOH

17.34 Compare the titration of a strong, monoprotic acid with a strong base to the titration of a weak, monoprotic acid with a strong base. Assume the strong and weak acid solutions initially have the same concentrations. Indicate whether the following statements are true or false. (a) More base is re-quired to reach the equivalence point for the strong acid than the weak acid. (b) The pH at the beginning of the titration is lower for the weak acid than the strong acid. (c) The pH at the equivalence point is 7 no matter which acid is titrated.

17.35 The samples of nitric and acetic acids shown here are both titrated with a 0.100 M solution of NaOH(aq).

25.0 mL of 1.0 M HNO3(aq) 25.0 mL of 1.0 M CH3COOH(aq)

Determine whether each of the following statements concern-ing these titrations is true or false.(a) A larger volume of NaOH1aq2 is needed to reach the

equivalence point in the titration of HNO3.(b) The pH at the equivalence point in the HNO3 titration

will be lower than the pH at the equivalence point in the CH3COOH titration.

(c) Phenolphthalein would be a suitable indicator for both titrations.

17.36 Determine whether each of the following statements concern-ing the titrations in Problem 17.35 is true or false.(a) The pH at the beginning of the two titrations will be the

same.(b) The titration curves will both be essentially the same af-

ter passing the equivalence point.(c) Methyl red would be a suitable indicator for both

titrations. 17.37 Predict whether the equivalence point of each of the follow-

ing titrations is below, above, or at pH 7: (a) NaHCO3 titrated


17.62 Calculate the molar solubility of Ni1OH22 when buffered at pH (a) 8.0, (b) 10.0, (c) 12.0.

17.63 Which of the following salts will be substantially more soluble in acidic solution than in pure water: (a) ZnCO3, (b) ZnS, (c) BiI3, (d) AgCN, (e) Ba31PO422?

17.64 For each of the following slightly soluble salts, write the net ionic equation, if any, for reaction with a strong acid: (a) MnS, (b) PbF2, (c) AuCl3, (d) Hg2C2O4, (e) CuBr.

17.65 From the value of Kf listed in Table 17.1, calculate the concentration of Ni2 + in 1.0 L of a solution that contains a total of 1 * 10-3 mol of nickel(II) ion and that is 0.20 M in NH3.

17.66 To what final concentration of NH3 must a solution be ad-justed to just dissolve 0.020 mol of NiC2O4 1Ksp = 4 * 10-102 in 1.0 L of solution? (Hint: You can neglect the hydrolysis of C2O4 2 - because the solution will be quite basic.)

17.67 Use values of Ksp for AgI and Kf for Ag1CN22 - to (a) calculate the molar solubility of AgI in pure water, (b) calculate the equilibrium constant for the reaction AgI1s2 + 2 CN-1aq2 ∆ Ag1CN22 -1aq2 + I-1aq2 , ( c ) determine the molar solubility of AgI in a 0.100 M NaCN solution.

17.68 Using the value of Ksp for Ag2S, Ka1 and Ka2 for H2S, and Kf = 1.1 * 105 for AgCl2 -, calculate the equilibrium con-stant for the following reaction:

Ag2S1s2 + 4 Cl-1aq2 + 2 H+1aq2 ∆ 2 AgCl2 -1aq2 + H2S1aq2

precipitation and separation of Ions (section 17.6)

17.69 (a) Will Ca1OH22 precipitate from solution if the pH of a 0.050 M solution of CaCl2 is adjusted to 8.0? (b) Will Ag2SO4 precipitate when 100 mL of 0.050 M AgNO3 is mixed with 10 mL of 5.0 * 10-2 M Na2SO4 solution?

17.70 (a) Will Co1OH22 precipitate from solution if the pH of a 0.020 M solution of Co1NO322 is adjusted to 8.5? (b) Will AgIO3 precipitate when 20 mL of 0.010 M AgIO3 is mixed with 10 mL of 0.015 M NaIO3? 1Ksp of AgIO3 is 3.1 * 10-82.

17.71 Calculate the minimum pH needed to precipitate Mn1OH22 so completely that the concentration of Mn2 + is less than 1 mg per liter [1 part per billion (ppb)].

17.72 Suppose that a 10-mL sample of a solution is to be tested for I- ion by addition of 1 drop (0.2 mL) of 0.10 M Pb1NO322. What is the minimum number of grams of I- that must be present for Pbl21s2 to form?

17.73 A solution contains 2.0 * 10-4 M Ag+ and 1.5 * 10-3 M Pb2 + . If NaI is added, will AgI 1Ksp = 8.3 * 10-172 or PbI2 1Ksp = 7.9 * 10-92 precipitate first? Specify the concentration of I- needed to begin precipitation.

17.74 A solution of Na2SO4 is added dropwise to a solution that is 0.010 M in Ba2 + and 0.010 M in Sr2 + . (a) What concen-tration of SO4

2 - is necessary to begin precipitation? (Ne-glect volume changes. BaSO4: Ksp = 1.1 * 10-10; SrSO4: Ksp = 3.2 * 10-7.) (b) Which cation precipitates first? (c) What is the concentration of SO4

2 - when the second cat-ion begins to precipitate?

17.75 A solution contains three anions with the following concentrations: 0.20 M CrO4

2 - , 0.10 M CO32 - , and 0.010 M Cl-. If a dilute

(c) The solubility of a slightly soluble salt is independent of the presence of a common ion.

(d) The solubility product of a slightly soluble salt is inde-pendent of the presence of a common ion.

17.50 The solubility of two slightly soluble salts of M2 + , MA and MZ2, are the same, 4 * 10-4 mol/L. (a) Which has the larger numeri-cal value for the solubility product constant? (b) In a saturated solution of each salt in water, which has the higher concentra-tion of M2 + ? (c) If you added an equal volume of a solution saturated in MA to one saturated in MZ2, what would be the equilibrium concentration of the cation, M2 + ?

17.51 (a) Why is the concentration of undissolved solid not ex-plicitly included in the expression for the solubility-product constant? (b) Write the expression for the solubility-prod-uct constant for each of the following strong electrolytes: AgI, SrSO4, Fe1OH22, and Hg2Br2.

17.52 (a) True or false: “solubility” and “solubility-product con-stant” are the same number for a given compound. (b) Write the expression for the solubility-product constant for each of the following ionic compounds: MnCO3, Hg1OH22, and Cu31PO422.

17.53 ( a ) I f t h e m o l a r s o l u b i l i t y o f CaF2 a t 35 °C i s 1.24 * 10-3 mol>L, what is Ksp at this temperature? (b) It is found that 1.1 * 10-2 g SrF2 dissolves per 100 mL of aque-ous solution at 25 °C. Calculate the solubility product for SrF2. (c) The Ksp of Ba1IO322 at 25 °C is 6.0 * 10-10. What is the molar solubility of Ba1IO322?

17.54 (a) The molar solubility of PbBr2 at 25 °C is 1.0 * 10-2 mol>L. Calculate Ksp. (b) If 0.0490 g of AgIO3 dissolves per liter of so-lution, calculate the solubility-product constant. (c) Using the appropriate Ksp value from Appendix D, calculate the pH of a saturated solution of Ca1OH22.

17.55 A 1.00-L solution saturated at 25 °C with calcium oxa-late 1CaC2O42 contains 0.0061 g of CaC2O4. Calculate the solubility-product constant for this salt at 25 °C.

17.56 A 1.00-L solution saturated at 25 °C with lead(II) iodide con-tains 0.54 g of PbI2. Calculate the solubility-product constant for this salt at 25 °C.

17.57 Using Appendix D, calculate the molar solubility of AgBr in (a) pure water, (b) 3.0 * 10-2 M AgNO3 solution, (c) 0.10 M NaBr solution.

17.58 Calculate the solubility of LaF3 in grams per liter in (a) pure water, (b) 0.010 M KF solution, (c) 0.050 M LaCl3 solution.

17.59 Consider a beaker containing a saturated solution of CaF2 in equilibrium with undissolved CaF21s2. Solid CaCl2 is then added to the solution. (a) Will the amount of solid CaF2 at the bottom of the beaker increase, decrease, or remain the same? (b) Will the concentration of Ca2 + ions in solution increase or decrease? (c) Will the concentration of F- ions in solution increase or decrease?

17.60 Consider a beaker containing a saturated solution of Pbl2 in equilibrium with undissolved Pbl21s2. Now solid KI is added to this solution. (a) Will the amount of solid Pbl2 at the bot-tom of the beaker increase, decrease, or remain the same? (b) Will the concentration of Pb2 + ions in solution increase or decrease? (c) Will the concentration of I- ions in solution in-crease or decrease?

17.61 Calculate the solubility of Mn1OH22 in grams per liter when buffered at pH (a) 7.0, (b) 9.5, (c) 11.8.

additional exercises 771

17.83 Derive an equation similar to the Henderson–Hasselbalch equation relating the pOH of a buffer to the pKb of its base component.

[17.84] Rainwater is acidic because CO21g2 dissolves in the water, cre-ating carbonic acid, H2CO3. If the rainwater is too acidic, it will react with limestone and seashells (which are principally made of calcium carbonate, CaCO3). Calculate the concentra-tions of carbonic acid, bicarbonate ion 1HCO3

-2 and carbon-ate ion 1CO3

2 - 2 that are in a raindrop that has a pH of 5.60, assuming that the sum of all three species in the raindrop is 1.0 * 10-5 M.

17.85 Furoic acid 1HC5H3O32 has a Ka value of 6.76 * 10-4 at 25 °C. Calculate the pH at 25 °C of (a) a solution formed by adding 25.0 g of furoic acid and 30.0 g of sodium furoate 1NaC5H3O32 to enough water to form 0.250 L of solution, (b) a solution formed by mixing 30.0 mL of 0.250 M HC5H3O3 and 20.0 mL of 0.22 M NaC5H3O3 and diluting the total vol-ume to 125 mL, (c) a solution prepared by adding 50.0 mL of 1.65 M NaOH solution to 0.500 L of 0.0850 M HC5H3O3.

17.86 The acid–base indicator bromcresol green is a weak acid. The yellow acid and blue base forms of the indicator are present in equal concentrations in a solution when the pH is 4.68. What is the pKa for bromcresol green?

17.87 Equal quantities of 0.010 M solutions of an acid HA and a base B are mixed. The pH of the resulting solution is 9.2. (a) Write the chemical equation and equilibrium-constant expression for the reaction between HA and B. (b) If Ka for HA is 8.0 * 10-5, what is the value of the equilibrium con-stant for the reaction between HA and B? (c) What is the value of Kb for B?

17.88 Two buffers are prepared by adding an equal number of moles of formic acid (HCOOH) and sodium formate (HCOONa) to enough water to make 1.00 L of solution. Buffer A is prepared using 1.00 mol each of formic acid and sodium formate. Buf-fer B is prepared by using 0.010 mol of each. (a) Calculate the pH of each buffer. (b) Which buffer will have the greater buf-fer capacity? (c) Calculate the change in pH for each buffer upon the addition of 1.0 mL of 1.00 M HCl. (d) Calculate the change in pH for each buffer upon the addition of 10 mL of 1.00 M HCl.

17.89 A biochemist needs 750 mL of an acetic acid–sodium acetate buffer with pH 4.50. Solid sodium acetate 1CH3COONa2 and glacial acetic acid 1CH3COOH2 are available. Glacial acetic acid is 99% CH3COOH by mass and has a density of 1.05 g>mL. If the buffer is to be 0.15 M in CH3COOH, how many grams of CH3COONa and how many milliliters of gla-cial acetic acid must be used?

17.90 A sample of 0.2140 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0950 M NaOH. The acid required 27.4 mL of base to reach the equiv-alence point. (a) What is the molar mass of the acid? (b) After 15.0 mL of base had been added in the titration, the pH was found to be 6.50. What is the Ka for the unknown acid?

17.91 A sample of 0.1687 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.1150 M NaOH. The acid required 15.5 mL of base to reach the equivalence point. (a) What is the molecular weight of the acid? (b) After 7.25 mL of base had been added in the titra-tion, the pH was found to be 2.85. What is the Ka for the unknown acid?

No precipitate forms upon addition of 1NH422HPO4. (See Figure 7.23.) The remaining solution shows a yellow color in a flame test (see Figure 7.22). Based on these observations, which of the following compounds might be present, which are definitely present, and which are definitely absent: CdS, Pb1NO322, HgO, ZnSO4, Cd1NO322, and Na2SO4?

17.79 In the course of various qualitative analysis procedures, the following mixtures are encountered: (a) Zn2 + and Cd2 + , (b) Cr1OH23 and Fe1OH23, (c) Mg2 + and K+, (d) Ag+ and Mn2 + . Suggest how each mixture might be separated.

17.80 Suggest how the cations in each of the following solution mixtures can be separated: (a) Na+ and Cd2 + , (b) Cu2 + and Mg2 + , (c) Pb2 + and Al3 + , (d) Ag+ and Hg2 + .

17.81 (a) Precipitation of the group 4 cations of Figure 17.23 re-quires a basic medium. Why is this so? (b) What is the most significant difference between the sulfides precipitated in group 2 and those precipitated in group 3? (c) Suggest a pro-cedure that would serve to redissolve the group 3 cations fol-lowing their precipitation.

17.82 A student who is in a great hurry to finish his laboratory work decides that his qualitative analysis unknown contains a metal ion from group 4 of Figure 17.23. He therefore tests his sam-ple directly with 1NH422HPO4, skipping earlier tests for the metal ions in groups 1, 2, and 3. He observes a precipitate and concludes that a metal ion from group 4 is indeed present. Why is this possibly an erroneous conclusion?

AgNO3 solution is slowly added to the solution, what is the first compound to precipitate: Ag2CrO4 1Ksp = 1.2 * 10-122, Ag2CO3 1Ksp = 8.1 * 10-122, or AgCl 1Ksp = 1.8 * 10-102?

17.76 A 1.0 M Na2SO4 solution is slowly added to 10.0 mL of a so-lution that is 0.20 M in Ca2 + and 0.30 M in Ag+. (a) Which compound will precipitate first: CaSO41Ksp = 2.4 * 10-52or Ag2SO4 1Ksp = 1.5 * 10-52? (b) How much Na2SO4 solu-tion must be added to initiate the precipitation?

Qualitative Analysis for metallic Elements (section 17.7)

17.77 A solution containing an unknown number of metal ions is treated with dilute HCl; no precipitate forms. The pH is adjusted to about 1, and H2S is bubbled through. Again, no precipitate forms. The pH of the solution is then adjusted to about 8. Again, H2S is bubbled through. This time a pre-cipitate forms. The filtrate from this solution is treated with 1NH422HPO4. No precipitate forms. Which metal ions dis-cussed in Section 17.7 are possibly present? Which are defi-nitely absent within the limits of these tests?

17.78 An unknown solid is entirely soluble in water. On addition of dilute HCl, a precipitate forms. After the precipitate is filtered off, the pH is adjusted to about 1 and H2S is bubbled in; a pre-cipitate again forms. After filtering off this precipitate, the pH is adjusted to 8 and H2S is again added; no precipitate forms.

Additional Exercises


point. What is the molecular weight of the unknown? (b) As the acid is titrated, the pH of the solution after the addition of 11.05 mL of the base is 4.89. What is the Ka for the acid? (c) Using Appendix D, suggest the identity of the acid.

17.112 A sample of 7.5 L of NH3 gas at 22 °C and 735 torr is bubbled into a 0.50-L solution of 0.40 M HCl. Assuming that all the NH3 dissolves and that the volume of the solution remains 0.50 L, calculate the pH of the resulting solution.

17.110 (a) Write the net ionic equation for the reaction that occurs when a solution of hydrochloric acid (HCl) is mixed with a solution of sodium formate 1NaCHO22. (b) Calculate the equilibrium constant for this reaction. (c) Calculate the equi-librium concentrations of Na+, Cl-, H+, CHO2

-, and HCHO2 when 50.0 mL of 0.15 M HCl is mixed with 50.0 mL of 0.15 M NaCHO2.

17.111 (a) A 0.1044-g sample of an unknown monoprotic acid requires 22.10 mL of 0.0500 M NaOH to reach the end

17.100 Tooth enamel is composed of hydroxyapatite, whose sim-plest formula is Ca51PO423OH, and whose corresponding Ksp = 6.8 * 10-27. As discussed in the “Chemistry and Life” box on page 755, fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, Ca51PO423F, whose Ksp = 1.0 * 10-60. (a) Write the expression for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds.

17.101 Use the solubi l i ty-product constant for Cr1OH23 1Ksp = 6.7 * 10-312 and the formation constant for Cr1OH24

- from Table 17.1 to determine the concentration of Cr1OH24

- in a solution that is buffered at pH = 10.0 and is in equilibrium with solid Cr1OH23.

17.102 Calculate the solubility of Mg1OH22 in 0.50 M NH4Cl. 17.103 The solubility-product constant for barium permanganate,

Ba1MnO422, is 2.5 * 10-10. Assume that solid Ba1MnO422 is in equilibrium with a solution of KMnO4. What concen-tration of KMnO4 is required to establish a concentration of 2.0 * 10-8 M for the Ba2 + ion in solution?

17.104 Calculate the ratio of 3Ca2 + 4 to 3Fe2 + 4 in a lake in which the water is in equilibrium with deposits of both CaCO3 and FeCO3. Assume that the water is slightly basic and that the hydrolysis of the carbonate ion can therefore be ignored.

17.105 The solubility product constants of PbSO4 and SrSO4 are 6.3 * 10-7 and 3.2 * 10-7, respectively. What are the values of 3SO4

2 - 4, 3Pb2 + 4, and 3Sr2 + 4 in a solution at equilibrium with both substances?

17.106 A buffer of what pH is needed to give a Mg2 + concentration of 3.0 * 10-2 M in equilibrium with solid magnesium oxalate?

17.107 The value of Ksp for Mg31AsO422 is 2.1 * 10-20. The AsO4

3 - ion is derived from the weak acid H3AsO4 1pKa1 =2.22; pKa2 = 6.98; pKa3 = 11.502. When asked to calculate the molar solubility of Mg31AsO422 in water, a student used the Ksp expression and assumed that 3Mg2 + 4 = 1.53AsO4

3 - 4. Why was this a mistake?

17.108 The solubility product for Zn1OH22 is 3.0 * 10-16. The for-mation constant for the hydroxo complex, Zn1OH24

2 - , is 4.6 * 1017. What concentration of OH- is required to dis-solve 0.015 mol of Zn1OH22 in a liter of solution?

17.109 The value of Ksp for Cd1OH22 is 2.5 * 10-14. (a) What is the molar solubility of Cd1OH22? (b) The solubility of Cd1OH22 can be increased through formation of the com-plex ion CdBr4

2 - 1Kf = 5 * 1032. If solid Cd1OH22 is added to a NaBr solution, what is the initial concentration of NaBr needed to increase the molar solubility of Cd1OH22 to 1.0 * 10-3 mol/L?

17.92 Mathematically prove that the pH at the halfway point of a titration of a weak acid with a strong base (where the volume of added base is half of that needed to reach the equivalence point) is equal to pKa for the acid.

17.93 A weak monoprotic acid is titrated with 0.100 M NaOH. It requires 50.0 mL of the NaOH solution to reach the equiv-alence point. After 25.0 mL of base is added, the pH of the solution is 3.62. Estimate the pKa of the weak acid.

17.94 What is the pH of a solution made by mixing 0.30 mol NaOH, 0.25 mol Na2HPO4, and 0.20 mol H3PO4 with water and diluting to 1.00 L?

17.95 Suppose you want to do a physiological experiment that calls for a pH 6.50 buffer. You find that the organism with which you are working is not sensitive to the weak acid H2A 1Ka1 = 2 * 10-2; Ka2 = 5.0 * 10-72 or its sodium salts. You have available a 1.0 M solution of this acid and a 1.0 M solution of NaOH. How much of the NaOH solution should be added to 1.0 L of the acid to give a buffer at pH 6.50? (Ig-nore any volume change.)

17.96 How many microliters of 1.000 M NaOH solution must be added to 25.00 mL of a 0.1000 M solution of lactic acid 3CH3CH1OH2COOH or HC3H5O34 to produce a buffer with pH = 3.75?

17.97 A person suffering from anxiety begins breathing rapidly and as a result suffers alkalosis, an increase in blood pH. (a) Using Equation 17.10, explain how rapid breathing can cause the pH of blood to increase. (b) One cure for this problem is breath-ing in a paper bag. Why does this procedure lower blood pH?

17.98 For each pair of compounds, use Ksp values to determine which has the greater molar solubility: (a) CdS or CuS, (b) PbCO3 or BaCrO4, (c) Ni1OH22 or NiCO3, (d) AgI or Ag2SO4.

17.99 The solubility of CaCO3 is pH dependent. (a) Calculate the molar solubility of CaCO3 1Ksp = 4.5 * 10-92 neglecting the acid–base character of the carbonate ion. (b) Use the Kb expression for the CO3

2 - ion to determine the equilibrium constant for the reaction

CaCO31s2 + H2O1l2 ∆Ca2 + 1aq2 + HCO3

-1aq2 + OH-1aq2 (c) If we assume that the only sources of Ca2 + , HCO3

-, and OH- ions are from the dissolution of CaCO3, what is the mo-lar solubility of CaCO3 using the equilibrium expression from part (b)? (d) What is the molar solubility of CaCO3 at the pH of the ocean (8.3)? (e) If the pH is buffered at 7.5, what is the molar solubility of CaCO3?

Integrative Exercises

Design an experiment 773

concentration of Ag+ in parts per billion that would exist in equilibrium with (a) AgCl, (b) AgBr, (c) AgI.

17.118 Fluoridation of drinking water is employed in many places to aid in the prevention of tooth decay. Typically the F- ion concentration is adjusted to about 1 ppb. Some water supplies are also “hard”; that is, they contain certain cations such as Ca2 + that interfere with the action of soap. Consider a case where the concentration of Ca2 + is 8 ppb. Could a precipitate of CaF2 form under these conditions? (Make any necessary approximations.)

17.119 Baking soda (sodium bicarbonate, NaHCO3) reacts with ac-ids in foods to form carbonic acid 1H2CO32, which in turn decomposes to water and carbon dioxide gas. In a cake bat-ter, the CO21g2 forms bubbles and causes the cake to rise. (a) A rule of thumb in baking is that 1/2 teaspoon of bak-ing soda is neutralized by one cup of sour milk. The acid component in sour milk is lactic acid, CH3CH1OH2COOH. Write the chemical equation for this neutralization reac-tion. (b) The density of baking soda is 2.16 g>cm3. Calcu-late the concentration of lactic acid in one cup of sour milk (assuming the rule of thumb applies), in units of mol/L. 1One cup = 236.6 mL = 48 teaspoons2. (c) If 1/2 teaspoon of baking soda is indeed completely neutralized by the lac-tic acid in sour milk, calculate the volume of carbon dioxide gas that would be produced at 1 atm pressure, in an oven set to 350 °F.

17.120 In nonaqueous solvents, it is possible to react HF to create H2F+. Which of these statements follows from this observa-tion? (a) HF can act like a strong acid in nonaqueous solvents, (b) HF can act like a base in nonaqueous solvents, (c) HF is thermodynamically unstable, (d) There is an acid in the non-aqueous medium that is a stronger acid than HF.

17.113 Aspirin has the structural formula

O

C CH3

C OH

O

O

At body temperature 137 °C2, Ka for aspirin equals 3 * 10-5. If two aspirin tablets, each having a mass of 325 mg, are dissolved in a full stomach whose volume is 1 L and whose pH is 2, what percent of the aspirin is in the form of neutral molecules?

17.114 What is the pH at 25 °C of water saturated with CO2 at a par-tial pressure of 1.10 atm? The Henry’s law constant for CO2 at 25 °C is 3.1 * 10-2 mol>L@atm.

17.115 Excess Ca1OH22 is shaken with water to produce a saturated solution. The solution is filtered, and a 50.00-mL sample titrated with HCl requires 11.23 mL of 0.0983 M HCl to reach the end point. Calculate Ksp for Ca1OH22. Compare your result with that in Appendix D. 25 °C. Suggest a reason for any differences you find between your value and the one in Appendix D.

17.116 The osmotic pressure of a saturated solution of strontium sul-fate at 25 °C is 21 torr. What is the solubility product of this salt at 25 °C?

17.117 A concentration of 10–100 parts per billion (by mass) of Ag+ is an effective disinfectant in swimming pools. However, if the concentration exceeds this range, the Ag+ can cause adverse health effects. One way to maintain an appropriate concentration of Ag+ is to add a slightly soluble salt to the pool. Using Ksp values from Appendix D, calculate the equilibrium

design an Experiment The pKa of acetic acid is 4.74. The pKa of chloroacetic acid, CH2ClCOOH, is 2.86. The pKa of trichloroacetic acid, CCl3COOH, is 0.66. (a) Why might this be the case? For example, one hypothesis is that the O ¬ H bond of trichloroacetic acid is significantly more polar than the O ¬ H bond in acetic acid due to chlorine being more electron-withdrawing than hydrogen, making the O ¬ H bond in trichloroacetic acid weak. Another hypothesis is that the chlorines thermodynamically stabilize the conjugate base forms of these acids, and the more chlorines there are, the more stable the conjugate bases. Design a set of experiments or calculations to test these hypotheses. (b) Would you predict that the differences in pKas of these acids would lead to differences in aqueous solubilities of their sodium salts? Design an experiment to test this hypothesis.

18 Chemistry of the EnvironmentThe richness of life on Earth, represented in the chapter-opening photograph, is made possible by our planet’s supportive atmosphere, the energy received from the Sun, and an abundance of water. These are the signature environmental features believed to be necessary for life.

involving human-made gases and how acid rain and smog are the result of atmospheric reactions involving compounds produced by human activity.

18.3 Earth’s WatEr We examine the global water cycle, which describes how water moves from the ground to surface to the atmosphere and back into the ground. We compare the chemical compositions of seawater, freshwater, and groundwater.

18.1 Earth’s atmosphErE We begin with a look at the temperature profile, pressure profile, and chemical composition of Earth’s atmosphere. We then examine photoionization and photodissociation, reactions that result from atmospheric absorption of solar radiation.

18.2 human activitiEs and Earth’s atmosphErE We next examine the effect human activities have on the atmosphere. We discuss how atmospheric ozone is depleted by reactions

WhaT’s ahEad

▶ Tahquomenon falls in The upper peninsula of michigan, one of the largest waterfalls east of the Mississippi in the United States.

As technology has advanced and the world human population has increased, humans have put new and greater stresses on the environment. Paradoxically, the very technol-ogy that can cause pollution also provides the tools to help understand and manage the environment in a beneficial way. Chemistry is often at the heart of environmental issues. The economic growth of both developed and developing nations depends criti-cally on chemical processes that range from treatment of water supplies to industrial processes. Some of these processes produce products or by-products that are harmful to the environment.

We are now in a position to apply the principles we have learned in preceding chapters to an understanding of how our environment operates and how human activi-ties affect it. To understand and protect the environment in which we live, we must understand how human-made and natural chemical compounds interact on land and in the sea and sky. Our daily actions as consumers turn on the same choices made by leading experts and governmental leaders: Each decision should reflect the costs versus the benefits of our choices. Unfortunately, the environmental impacts of our decisions are often subtle and not immediately evident.

18.5 GrEEn chEmistry We conclude by examining green chemistry, an international initiative to make industrial products, processes, and chemical reactions compatible with a sustainable society and environment.

18.4 human activitiEs and WatEr QuaLity We consider how Earth’s water is connected to the global climate and examine one measure of water quality: dissolved oxygen concentration. Water for drinking and for irrigation must be free of salts and pollutants.

776 chaptEr 18 chemistry of the Environment

18.1 | Earth’s atmosphereBecause most of us have never been very far from Earth’s surface, we often take for granted the many ways in which the atmosphere determines the environment in which we live. In this section we examine some of the important characteristics of our planet’s atmosphere.

The temperature of the atmosphere varies with altitude (▼ Figure 18.1), and the atmosphere is divided into four regions based on this temperature profile. Just above the surface, in the troposphere, the temperature normally decreases with increasing altitude, reaching a minimum of about 215 K at about 10 km. Nearly all of us live our

entire lives in the troposphere. Howling winds and soft breezes, rain, and sunny skies—all that we normally think of as “weather”—occur in this region. Commercial jet aircraft typically fly about 10 km (33,000 ft) above Earth, an altitude that defines the upper limit of the troposphere, which we call the tropopause.

Above the tropopause, air temperature increases with altitude, reaching a maximum of about 275 K at about 50 km. The region from 10 km to 50 km is the stratosphere, and above it are the mesosphere and thermosphere. Notice in Figure 18.1 that the temperature extremes that form the boundaries between adjacent regions are denoted by the suffix -pause. The boundar-ies are important because gases mix across them relatively slowly. For example, pollutant gases generated in the troposphere pass through the tropopause and find their way into the strato-sphere only very slowly.

Atmospheric pressure decreases with increas-ing elevation (Figure 18.1), declining much more rapidly at lower elevations than at higher ones because of the atmosphere’s compressibility. Thus, the pressure decreases from an average value of 760 torr (101 kPa) at sea level to 2.3 * 10-3 torr 13.1 * 10-4 kPa2 at 100 km, to only 1.0 * 10-6 torr 11.3 * 10-7 kPa2 at 200 km.

The troposphere and stratosphere together account for 99.9% of the mass of the atmo-sphere, 75% of which is the mass in the tropo-sphere. Nevertheless the thin upper atmosphere plays many important roles in determining the conditions of life at the surface.

Composition of the AtmosphereEarth’s atmosphere is constantly bombarded by radiation and energetic particles from the Sun. This barrage of energy has profound chemical and physical effects, especially in the upper regions of the atmosphere, above about 80 km (▶ Figure 18.2). In addi-tion, because of Earth’s gravitational field, heavier atoms and molecules tend to sink in the atmosphere, leaving lighter atoms and molecules at the top of the atmosphere. (This is why, as just noted, 75% of the atmosphere’s mass is in the troposphere.) Because of all these factors, the composition of the atmosphere is not uniform.

▶ Table 18.1 shows the composition of dry air near sea level. Note that although traces of many substances are present, N2 and O2 make up about 99% of sea-level air. The noble gases and CO2 make up most of the remainder.

170 190 210 230 250 270 2900

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Troposphere

Stratosphere

Mesosphere

Thermosphere

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m)

Mesopause

Stratopause

Tropopause

200 400 600 8000

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Pressure (torr)

Alt

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▲ Figure 18.1 Temperature and pressure in the atmosphere vary as a function of altitude above sea level.

Go FiGurEAt what altitude is the atmospheric temperature lowest?

sEction 18.1 Earth’s atmosphere 777

give it some ThoughtHow would you expect the ratio of atmospheric helium to argon to differ at 100 km elevation as compared with sea level?

When applied to substances in aqueous solution, the concentration unit parts per million (ppm) refers to grams of substance per million grams of solution. (Section 13.4) When dealing with gases, however, 1 ppm means one part by volume in 1 million volumes of the whole. Because volume is proportional to number of moles of gas via the ideal-gas equation 1PV = nRT2, volume fraction and mole fraction are the same. Thus, 1 ppm of a trace constituent of the atmosphere amounts to 1 mol of that constituent in 1 million moles of air; that is, the concentration in parts per million is equal to the mole fraction times 106. For example, Table 18.1 lists the mole fraction of CO2 in the atmosphere as 0.000400, which means its concentration in parts per million is 0.000400 * 106 = 400 ppm.

Other minor constituents of the troposphere, in addition to CO2, are listed in ▼ Table 18.2.

Before we consider the chemical processes that occur in the atmosphere, let’s review some of the properties of the two major components, N2 and O2. Recall that the N2 mole-cule possesses a triple bond between the nitrogen atoms. (Section 8.3) This very strong bond (bond energy 941 kJ>mol) is largely responsible for the very low reactivity of N2. The bond energy in O2 is only 495 kJ>mol, making O2 much more reactive than N2. For example, oxygen reacts with many substances to form oxides. The oxides of nonmetals,

Table 18.1 The Major Components of dry air near sea Level

Component* Content (mole fraction) Molar Mass (g/mol)

Nitrogen 0.78084 28.013Oxygen 0.20948 31.998Argon 0.00934 39.948Carbon dioxide 0.000400 44.0099Neon 0.00001818 20.183Helium 0.00000524 4.003Methane 0.000002 16.043Krypton 0.00000114 83.80Hydrogen 0.0000005 2.0159Nitrous oxide 0.0000005 44.0128Xenon 0.000000087 131.30*Ozone, sulfur dioxide, nitrogen dioxide, ammonia, and carbon monoxide are present as trace gases in variable amounts.

▲ Figure 18.2 The aurora borealis (northern lights).

High-energy solar particles create excited N and O atoms; visible light results as electrons in these atoms fall from upper-level states to lower-level states

Table 18.2 sources and Typical Concentrations of some Minor atmospheric Constituents

Constituent sources Typical Concentration

Carbon dioxide, CO2 Decomposition of organic matter, release from oceans, fossil-fuel combustion

400 ppm throughout troposphere

Carbon monoxide, CO Decomposition of organic matter, industrial processes, fossil-fuel combustion

0.05 ppm in unpolluted air; 1–50 ppm in urban areas

Methane, CH4 Decomposition of organic matter, natural-gas seepage, livestock emissions

1.82 ppm throughout troposphere

Nitric oxide, NO Atmospheric electrical discharges, internal combustion engines, combustion of organic matter

0.01 ppm in unpolluted air; 0.2 ppm in smog

Ozone, O3 Atmospheric electrical discharges, diffusion from the stratosphere, photochemical smog

0–0.01 ppm in unpolluted air; 0.5 ppm in photochemical smog

Sulfur dioxide, SO2 Volcanic gases, forest fires, bacterial action, fossil-fuel combustion, industrial processes

0–0.01 ppm in unpolluted air; 0.1–2 ppm in polluted urban areas


such as SO2, usually form acidic solutions when dissolved in water. The oxides of active metals, such as CaO, form basic solutions when dissolved in water. (Section 7.7)

Photochemical Reactions in the AtmosphereAlthough the atmosphere beyond the stratosphere contains only a small fraction of the atmospheric mass, it forms the outer defense against the hail of radiation and high-energy particles that continuously bombard Earth. As the bombarding radiation passes through the upper atmosphere, it causes two kinds of chemical changes: photodissociation and photoionization. These processes protect us from high-energy radiation by absorbing most of the radiation before it reaches the troposphere. If it were not for these photo-chemical processes, plant and animal life as we know it could not exist on Earth.

The Sun emits radiant energy over a wide range of wavelengths (▼ Figure 18.3). To understand the connection between the wavelength of radiation and its effect on

soLuTionAnalyze We are given the partial pressure of water vapor and the total pressure of an air sample and asked to determine the water vapor concentration.Plan Recall that the partial pressure of a component in a mixture of gases is given by the product of its mole fraction and the total pressure of the mixture (Section 10.6):

PH2O = XH2OPt

Solve Solving for the mole fraction of water vapor in the mixture, XH2O, gives

XH2O =PH2O

Pt=

0.80 torr735 torr

= 0.0011

saMpLE ExErCisE 18.1 Calculating Concentration from partial pressure

What is the concentration, in parts per million, of water vapor in a sample of air if the partial pres-sure of the water is 0.80 torr and the total pressure of the air is 735 torr?

The concentration in ppm is the mole fraction times 106:0.0011 * 106 = 1100 ppm

practice Exercise 1From the data in Table 18.1, the partial pressure of argon in dry air at an atmospheric pressure of 668 mm Hg is (a) 3.12 mm Hg, (b) 7.09 mm Hg, (c) 6.24 mm Hg, (d) 9.34 mm Hg, (e) 39.9 mm Hg.

practice Exercise 2The concentration of CO in a sample of air is 4.3 ppm. What is the partial pressure of the CO if the total air pressure is 695 torr?

▲ Figure 18.3 The solar spectrum above earth’s atmosphere compared to that at sea level. The more structured curve at sea level is due to gases in the atmosphere absorbing specific wavelengths of light. “Flux,” the unit on the vertical axis, is light energy per area per unit of time.

Go FiGurEWhy does not the solar spectrum at sea level perfectly match the solar spectrum outside the atmosphere?

Incr

easi

ng �

ux

200 600 1000Wavelength (nm)

1400 1800 2200 2600 3000

InfraredVisible

Ultraviolet

Solar spectrum at sea level

Solar spectrum outside atmosphere


atoms and molecules, recall that electromagnetic radiation can be pictured as a stream of photons. (Section 6.2) The energy of each photon is given by E = hn, where h is Planck constant and n is the radiation frequency. For a chemical change to occur when radiation strikes atoms or molecules, two conditions must be met. First, the incoming photons must have sufficient energy to break a chemical bond or remove an electron from the atom or molecule. Second, the atoms or molecules being bombarded must absorb these photons. When these requirements are met, the energy of the photons is used to do the work associated with some chemical change.

The rupture of a chemical bond resulting from absorption of a photon by a mol-ecule is called photodissociation. No ions are formed when the bond between two atoms is cleaved by photodissociation. Instead, half the bonding electrons stay with one atom and half stay with the other atom. The result is two electrically neutral particles.

One of the most important processes occurring above an altitude of about 120 km is photodissociation of the oxygen molecule:

+ +O O hν O O [18.1]

The minimum energy required to cause this change is determined by the bond energy (or dissociation energy) of O2, 495 kJ>mol.

soLuTionAnalyze We are asked to determine the wavelength of a photon that has just enough energy to break the O “ O double bond in O2.Plan We first need to calculate the energy required to break the O “ O double bond in one molecule and then find the wavelength of a photon of this energy.Solve The dissociation energy of O2 is 495 kJ>mol. Using this value and Avogadro’s number, we can calculate the amount of energy needed to break the bond in a single O2 molecule:

a495 * 103 Jmol

b a 1 mol6.022 * 1023 molecules

b = 8.22 * 10-19 Jmolecule

We next use the Planck relationship, E = hn, (Equation 6.2) to calculate the frequency n of a photon that has this amount of energy:

n =Eh=

8.22 * 10-19 J6.626 * 10-34 J@s

= 1.24 * 1015 s-1

Finally, we use the relationship between frequency and wavelength (Section 6.1) to calculate the wavelength of the light:

l =cn= a 3.00 * 108 m>s

1.24 * 1015>sb a 109 nm

1 mb = 242 nm

saMpLE ExErCisE 18.2 Calculating the Wavelength required to Break a Bond

What is the maximum wavelength of light, in nanometers, that has enough energy per photon to dissociate the O2 molecule?

Thus, light of wavelength 242 nm, which is in the ultraviolet region of the electromagnetic spectrum, has sufficient energy per photon to photodissociate an O2 molecule. Because photon energy increases as wavelength decreases, any photon of wave-length shorter than 242 nm will have sufficient energy to dissociate O2.

practice Exercise 1The bond dissociation energy of the Br ¬ Br bond is 193 kJ>mol. What wavelength of light has just sufficient energy to cause Br ¬ Br bond dissociation?(a) 620 nm (b) 310 nm (c) 148 nm (d) 6200 nm (e) 563 nm

practice Exercise 2The bond energy in N2 is 941 kJ>mol. What is the longest wave-length a photon can have and still have sufficient energy to dissociate N2?

Fortunately for us, O2 absorbs much of the high-energy, short-wavelength radia-tion from the solar spectrum before that radiation reaches the lower atmosphere. As it does, atomic oxygen, O, is formed. The dissociation of O2 is very extensive at higher elevations. At 400 km, for example, only 1% of the oxygen is in the form of O2; 99% is atomic oxygen. At 130 km, O2 and atomic oxygen are just about equally abundant. Below 130 km, O2 is more abundant than atomic oxygen because most of the solar energy has been absorbed in the upper atmosphere.

The dissociation energy of N2 is very high, 941 kJ>mol. As you should have seen in working out Practice Exercise 2 of Sample Exercise 18.2, only photons having a


wavelength shorter than 127 nm possess sufficient energy to dissociate N2. Further-more, N2 does not readily absorb photons, even when they possess sufficient energy. As a result, very little atomic nitrogen is formed in the upper atmosphere by photodis-sociation of N2.

Other photochemical processes besides photodissociation occur in the upper atmo-sphere, although their discovery has taken many twists and turns. In 1901 Guglielmo Marconi received a radio signal in St. John’s, Newfoundland, that had been transmit-ted from Land’s End, England, 2900 km away. Because people at the time thought radio waves traveled in straight lines, they assumed that the curvature of Earth’s sur-face would make radio communication over large distances impossible. Marconi’s suc-cessful experiment suggested that Earth’s atmosphere in some way substantially affects radio-wave propagation. His discovery led to intensive study of the upper atmosphere. In about 1924, the existence of electrons in the upper atmosphere was established by experimental studies.

The electrons in the upper atmosphere result mainly from photoionization, which occurs when a molecule in the upper atmosphere absorbs solar radiation and the absorbed energy causes an electron to be ejected from the molecule. The mol-ecule then becomes a positively charged ion. For photoionization to occur, there-fore, a molecule must absorb a photon, and the photon must have enough energy to remove an electron. (Section 7.4) Notice that this is a very different process from photodissociation.

Four important photoionization processes occurring in the atmosphere above about 90 km are shown in ▼ Table 18.3. Photons of any wavelength shorter than the maximum lengths given in the table have enough energy to cause photoionization. A look back at Figure 18.3 shows you that virtually all of these high-energy photons are filtered out of the radiation reaching Earth because they are absorbed by the upper atmosphere.

give it some ThoughtExplain the difference between photoionization and photodissociation.

Ozone in the StratosphereAlthough N2, O2, and atomic oxygen absorb photons having wavelengths shorter than 240 nm, ozone, O3, is the key absorber of photons having wavelengths ranging from 240 to 310 nm, in the ultraviolet region of the electromagnetic spectrum. Ozone in the upper atmosphere protects us from these harmful high-energy photons, which would otherwise penetrate to Earth’s surface. Let’s consider how ozone forms in the upper atmosphere and how it absorbs photons.

By the time radiation from the Sun reaches an altitude of 90 km above Earth’s surface, most of the short-wavelength radiation capable of photoionization has been absorbed. At this altitude, however, radiation capable of dissociating the O2 molecule is sufficiently intense for photodissociation of O2 (Equation 18.1) to remain important down to an altitude of 30 km. In the region between 30 and 90 km, however, the con-centration of O2 is much greater than the concentration of atomic oxygen. From this

Table 18.3 photoionization reactions for Four Components of the atmosphere

process ionization Energy 1kJ ,mol 2 Lmax 1nm 2N2 + hn ¡ N2

+ + e- 1495 80.1

O2 + hn ¡ O2+ + e- 1205 99.3

O + hn ¡ O+ + e- 1313 91.2

NO + hn ¡ NO+ + e- 890 134.5


finding, we conclude that the oxygen atoms formed by photodissociation of O2 in this region frequently collide with O2 molecules and form ozone:

O2 O3*O + [18.2]

The asterisk on O3 denotes that the product contains an excess of energy, because the reac-tion is exothermic. The 105 kJ/mol that is released must be transferred away from the O3* molecule quickly or else the molecule will fly apart into O2 and atomic O—a decomposition that is the reverse of the reaction by which O3* is formed.

An energy-rich O3* molecule can release its excess energy by colliding with another atom or molecule and transferring some of the excess energy to it. Let’s use M to repre-sent the atom or molecule with which O3* collides. (Usually M is N2 or O2 because these are the most abundant molecules in the atmosphere.) The formation of O3* and the transfer of excess energy to M are summarized by the equations

O1g2 + O21g2 ∆ O3*1g2 [18.3] O3*1g2 + M1g2 ¡ O31g2 + M*1g2 [18.4] O1g2 + O21g2 + M1g2 ¡ O31g2 + M*1g2 [18.5]

The rate at which the reactions of Equations 18.3 and 18.4 proceed depends on two factors that vary in opposite directions with increasing altitude. First, the Equa-tion 18.3 reaction depends on the presence of O atoms. At low altitudes, most of the radiation energetic enough to dissociate O2 into O atoms has been absorbed; thus, O atoms are more plentiful at higher altitudes. Second, Equations 18.3 and 18.4 both depend on molecular collisions. (Section 14.5) The concentration of molecules is greater at low altitudes, and so the rates of both reactions are greater at lower altitudes. Because these two reactions vary with altitude in opposite directions, the highest rate of O3 formation occurs in a band at an altitude of about 50 km, near the stratopause (Figure 18.1). Overall, roughly 90% of Earth’s ozone is found in the stratosphere.

give it some ThoughtWhy do O2 and N2 molecules fail to filter out ultraviolet light with wavelengths between 240 and 310 nm?

The photodissociation of ozone reverses the reaction that forms it. We thus have a cycle of ozone formation and decomposition, summarized as follows:

O21g2 + hn ¡ O1g2 + O1g2 O1g2 + O21g2 + M1g2 ¡ O31g2 + M*1g2 1heat released2

O31g2 + hn ¡ O21g2 + O1g2 O1g2 + O1g2 + M1g2 ¡ O21g2 + M*1g2 1heat released2

The first and third processes are photochemical; they use a solar photon to ini-tiate a chemical reaction. The second and fourth are exothermic chemical reactions. The net result of the four reactions is a cycle in which solar radiant energy is con-verted into thermal energy. The ozone cycle in the stratosphere is responsible for the rise in temperature that reaches its maximum at the stratopause (Figure 18.1).

The reactions of the ozone cycle account for some, but not all, of the facts about the ozone layer. Many chemical reactions occur that involve substances other than oxygen. We must also consider the effects of turbulence and winds that mix up the stratosphere. A complicated picture results. The overall result of ozone formation and removal reactions, coupled with atmospheric turbulence and other factors, is to produce the upper-atmosphere ozone profile shown in ▶ Figure 18.4, with a maximum ozone concentration occurring at an altitude of about 25 km. This band of relatively high ozone concentration is referred to as the “ozone layer” or the “ozone shield.”

1010 1011 1012 10130

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Ozone concentration (molecules/cm3)

Alt

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Stra

tosp

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▲ Figure 18.4 Variation in ozone concentration in the atmosphere as a function of altitude.

Go FiGurEEstimate the ozone concentration in moles per liter for the peak value in this graph.


Photons with wavelengths shorter than about 300 nm are energetic enough to break many kinds of single chemical bonds. Thus, the “ozone shield” is essential for our continued well-being. The ozone molecules that form this essential shield against high-energy radiation represent only a tiny fraction of the oxygen atoms present in the stratosphere, however, because these molecules are continually destroyed even as they are formed.

18.2 | human activities and Earth’s atmosphere

Both natural and anthropogenic (human-caused) events can modify Earth’s atmosphere. One impressive natural event was the eruption of Mount Pinatubo in June 1991 (◀ Figure 18.5). The volcano ejected approximately 10 km3 of material into the stratosphere, causing a 10% drop in the amount of sunlight reaching Earth’s surface during the next 2 years. That drop in sunlight led to a temporary 0.5 °C drop in Earth’s surface temperature. The volcanic particles that made it to the stratosphere remained there for approximately 3 years, raising the temperature of the stratosphere by several degrees due to light ab-sorption. Measurements of the stratospheric ozone concentration showed significantly increased ozone decomposition in this 3-year period.

The Ozone Layer and Its DepletionThe ozone layer protects Earth’s surface from the damaging ultraviolet (UV) radiation. Therefore, if the concentration of ozone in the stratosphere decreases substantially, more UV radiation will reach Earth’s surface, causing unwanted photochemical reac-tions, including reactions correlated with skin cancer. Satellite monitoring of ozone, which began in 1978, has revealed a depletion of ozone in the stratosphere that is partic-ularly severe over Antarctica, a phenomenon known as the ozone hole (◀ Figure 18.6). The first scientific paper on this phenomenon appeared in 1985, and the National Aeronautics and Space Administration (NASA) maintains an “Ozone Hole Watch” website with daily updates and data from 1999 to the present.

In 1995 the Nobel Prize in Chemistry was awarded to F. Sherwood Rowland, Mario Molina, and Paul Crutzen for their studies of ozone depletion. In 1970 Crutzen showed that naturally occurring nitrogen oxides catalytically destroy ozone. Rowland and Molina recognized in 1974 that chlorine from chlorofluorocarbons (CFCs) may deplete the ozone layer. These substances, principally CFCl3 and CF2Cl2, do not occur in nature and have been widely used as propellants in spray cans, as refrigerant and air-conditioner gases, and as foaming agents for plastics. They are virtually unreactive in the lower atmosphere. Furthermore, they are relatively insoluble in water and are therefore not removed from the atmosphere by rainfall or by dissolution in the oceans. Unfortunately, the lack of reactivity that makes them commercially useful also allows them to survive in the atmosphere and to diffuse into the stratosphere. It is estimated that several million tons of chlorofluorocarbons are now present in the atmosphere.

As CFCs diffuse into the stratosphere, they are exposed to high-energy radiation, which can cause photodissociation. Because C¬Cl bonds are considerably weaker than C¬F bonds, free chlorine atoms are formed readily in the presence of light with wavelengths in the range from 190 to 225 nm, as shown in this typical reaction:

CF2Cl21g2 + hn ¡ CF2Cl1g2 + Cl1g2 [18.6]

Calculations suggest that chlorine atom formation occurs at the greatest rate at an alti-tude of about 30 km, the altitude at which ozone is at its highest concentration.

Atomic chlorine reacts rapidly with ozone to form chlorine monoxide and molec-ular oxygen:

Cl1g2 + O31g2 ¡ ClO1g2 + O21g2 [18.7]

▲ Figure 18.6 ozone present in the southern hemisphere, sept. 24, 2006. The data were taken from an orbiting satellite. This day had the lowest stratospheric ozone concentration yet recorded. One “Dobson unit” corresponds to 2.69 * 1016 ozone molecules in a 1 cm2 column of atmosphere.

110 550440330220

Total ozone (Dobson units)

▲ Figure 18.5 mount pinatubo erupts, June 1991.

sEction 18.2 human activities and Earth’s atmosphere 783

This reaction follows a second-order rate law with a very large rate constant:

Rate = k3Cl43O34 k = 7.2 * 109 M-1 s-1 at 298 K [18.8]

Under certain conditions, the ClO generated in Equation 18.7 can react to regener-ate free Cl atoms. One way that this can happen is by photodissociation of ClO:

ClO1g2 + hn ¡ Cl1g2 + O1g2 [18.9]

The Cl atoms generated in Equations 18.6 and 18.9 can react with more O3, according to Equation 18.7. The result is a sequence of reactions that accomplishes the Cl-catalyzed decomposition of O3 to O2:

2 Cl1g2 + 2 O31g2 ¡ 2 ClO1g2 + 2 O21g2 2 ClO1g2 + hn ¡ 2 Cl1g2 + 2 O1g2 O1g2 + O1g2 ¡ O21g22 Cl1g2 + 2 O31g2 + 2 ClO1g2 + 2 O1g2 ¡ 2 Cl1g2 + 2 ClO1g2 + 3 O21g2 + 2 O1g2

The equation can be simplified by eliminating like species from each side to give

2 O31g2 ¡Cl 3 O21g2 [18.10]

Because the rate of Equation 18.7 increases linearly with [Cl], the rate at which ozone is destroyed increases as the quantity of Cl atoms increases. Thus, the greater the amount of CFCs that diffuse into the stratosphere, the faster the destruction of the ozone layer. Even though troposphere-to-stratosphere diffusion rates are slow, a substantial thin-ning of the ozone layer over the South Pole has been observed, particularly during September and October (Figure 18.6).

give it some ThoughtSince the rate of ozone destruction depends on [Cl], can Cl be considered a catalyst for the reaction of Equation 18.10?

Because of the environmental problems associated with CFCs, steps have been taken to limit their manufacture and use. A major step was the signing in 1987 of the Montreal Protocol on Substances That Deplete the Ozone Layer, in which participating nations agreed to reduce CFC production. More stringent limits were set in 1992, when representatives of approximately 100 nations agreed to ban the production and use of CFCs by 1996, with some exceptions for “essential uses.” Since then, the production of CFCs has indeed dropped precipitously. Images such as that shown in Figure 18.6 taken annually reveal that the depth and size of the ozone hole has begun to decline. Nevertheless, because CFCs are unreactive and because they diffuse so slowly into the stratosphere, scientists estimate that ozone depletion will continue for many years to come. What substances have replaced CFCs? At this time, the main alternatives are hydrofluorocarbons (HFCs), compounds in which C¬H bonds replace the C¬Cl bonds of CFCs. One such compound in current use is CH2FCF3, known as HFC-134a. While the HFCs are a big improvement over the CFCs because they contain no C¬Cl bonds, it turns out that they are potent greenhouse warming gases, with which we will deal shortly.

There are no naturally occurring CFCs, but some natural sources contribute chlorine and bromine to the atmosphere, and, just like halogens from CFC, these nat-urally occurring Cl and Br atoms can participate in ozone-depleting reactions. The principal natural sources are methyl bromide and methyl chloride, which are emitted from the oceans. It is estimated that these molecules contribute less than a third of the total Cl and Br in the atmosphere; the remaining two-thirds is a result of human activities.

Volcanoes are a source of HCl, but generally the HCl they release reacts with water in the troposphere and does not make it to the upper atmosphere.


Sulfur Compounds and Acid RainSulfur-containing compounds are present to some extent in the natural, unpolluted at-mosphere. They originate in the bacterial decay of organic matter, in volcanic gases, and from other sources listed in Table 18.2. The amount of these compounds released into the atmosphere worldwide from natural sources is about 24 * 1012 g per year, less than the amount from human activities, about 80 * 1012 g per year (principally related to combustion of fuels).

Sulfur compounds, chiefly sulfur dioxide, SO2, are among the most unpleasant and harmful of the common pollutant gases. ◀ Table 18.4 shows the concentrations of sev-eral pollutant gases in a typical urban environment (where by typical we mean one that is not particularly affected by smog). According to these data, the level of sulfur dioxide is 0.08 ppm or higher about half the time. This concentration is considerably lower than that of other pollutants, notably carbon monoxide. Nevertheless, SO2 is regarded as the most serious health hazard among the pollutants shown, especially for people with respiratory difficulties.

Combustion of coal accounts for about 65% of the SO2 released annually in the United States, and combustion of oil accounts for another 20%. The majority of this amount is from coal-burning electrical power plants, which generate about 50% of our electricity. The extent to which SO2 emissions are a problem when coal is burned depends on the amount of sulfur in the coal. Because of concern about SO2 pollution, low-sulfur coal is in greater demand and is thus more expensive. Much of the coal from east of the Mississippi is relatively high in sulfur content, up to 6% by mass. Much of the coal from the western states has a lower sulfur content, but also a lower heat content per unit mass, so the difference in sulfur content per unit of heat produced is not as large as is often assumed.

In 2010, the U.S. Environmental Protection Agency set new standards to reduce SO2 emissions, the first change in nearly 40 years. The old standard of 140 parts per billion, measured over 24 h, has been replaced by a standard of 75 parts per billion, measured over 1 h. The impact of SO2 emissions is not restricted to the United States, however. China, which generates about 70% of its energy from coal, is the world’s largest generator of SO2, producing about 35 * 1012 g annually from coal and other sources. As a result, that nation has a major problem with SO2 pollution and has set targets to reduce emissions with some success. India, which is projected to surpass China as the largest importer of coal by 2014, is also concerned about increased SO2 emissions. Nations will need to work together to address what has truly become a global issue.

Sulfur dioxide is harmful to both human health and property; furthermore, atmo-spheric SO2 can be oxidized to SO3 by several pathways (such as reaction with O2 or O3). When SO3 dissolves in water, it produces sulfuric acid:

SO31g2 + H2O1l2 ¡ H2SO41aq2Many of the environmental effects ascribed to SO2 are actually due to H2SO4.

The presence of SO2 in the atmosphere and the sulfuric acid it produces result in the phenomenon of acid rain. (Nitrogen oxides, which form nitric acid, are also major contributors to acid rain.) Uncontaminated rainwater generally has a pH value of about 5.6. The primary source of this natural acidity is CO2, which reacts with water to form carbonic acid, H2CO3. Acid rain typically has a pH value of about 4. This shift toward greater acidity has affected many lakes in northern Europe, the northern United States, and Canada, reducing fish populations and affecting other parts of the ecological net-work in the lakes and surrounding forests.

The pH of most natural waters containing living organisms is between 6.5 and 8.5, but as ▶ Figure 18.7 shows, freshwater pH values are far below 6.5 in many parts of the continental United States. At pH levels below 4.0, all vertebrates, most invertebrates, and many microorganisms are destroyed. The lakes most susceptible to damage are those with low concentrations of basic ions, such as HCO3

-, that would act as a buffer to mini-mize changes in pH. Some of these lakes are recovering as sulfur emissions from fossil

Table 18.4 Median Concentrations of atmospheric pollutants in a Typical urban atmosphere

pollutantConcentration (ppm)

Carbon monoxide 10Hydrocarbons 3Sulfur dioxide 0.08Nitrogen oxides 0.05Total oxidants 0.02(ozone and others)


▲ Figure 18.7 Water ph values from freshwater sites across the united states, 2008. The numbered dots indicate the locations of monitoring stations.

Go FiGurEWhy is the pH found in freshwater sources in the Eastern half of the United States dramatically lower than found in the western half?

Lab pH≥ 5.35.2–5.35.1–5.25.0–5.14.9–5.04.8–4.94.7–4.84.6–4.74.5–4.64.4–4.54.3–4.4< 4.3

(a) (b)

▲ Figure 18.8 Damage from acid rain. The right photograph, recently taken, shows how the statue has lost detail in its carvings.

fuel combustion decrease, in part because of the Clean Air Act. In the period 1990–2010 the average ambient air concentration of SO2 nationwide has declined by 75%.

Because acids react with metals and with carbonates, acid rain is corrosive both to metals and to stone building materials. Marble and limestone, for example, whose major constituent is CaCO3, are readily attacked by acid rain (▲ Figure 18.8). Billions of dollars each year are lost because of corrosion due to SO2 pollution.


One way to reduce the quantity of SO2 released into the environment is to remove sulfur from coal and oil before these fuels are burned. Although difficult and expensive, several methods have been developed. Powdered limestone 1CaCO32, for example, can be injected into the furnace of a power plant, where it decomposes into lime (CaO) and carbon dioxide:

CaCO31s2 ¡ CaO1s2 + CO21g2The CaO then reacts with SO2 to form calcium sulfite:

CaO1s2 + SO21g2 ¡ CaSO31s2The solid particles of CaSO3, as well as much of the unreacted SO2, can be removed from the furnace gas by passing it through an aqueous suspension of CaO (▲ Figure 18.9). Not all the SO2 is removed, however, and given the enormous quantities of coal and oil burned worldwide, pollution by SO2 will probably remain a problem for some time.

give it some ThoughtWhat chemical behavior associated with sulfur oxides gives rise to acid rain?

Nitrogen Oxides and Photochemical SmogNitrogen oxides are primary components of smog, a phenomenon with which city dwellers are all too familiar. The term smog refers to the pollution condition that occurs in certain urban environments when weather conditions produce a relatively stagnant air mass. The smog made famous by Los Angeles, but now common in many other urban areas as well, is more accurately described as photochemical smog because pho-tochemical processes play a major role in its formation (◀ Figure 18.10).

The majority of nitrogen oxide emissions (about 50%) comes from cars, buses, and other forms of transportation. Nitric oxide, NO, forms in small quantities in the cylin-ders of internal combustion engines in the reaction

N21g2 + O21g2 ∆ 2 NO1g2 ∆H = 180.8 kJ [18.11]

As noted in the “Chemistry Put to Work” box in Section 15.7, the equilibrium con-stant for this reaction increases from about 10-15 at 300 K to about 0.05 at 2400 K

▲ Figure 18.9 one method for removing so2 from combusted fuel.

Go FiGurEWhat is the major solid product resulting from removal of SO2 from furnace gas?

High-sulfur coalCaCO3(s)

CaO(s) + CO2(g)

Furnace

CaO(s) + SO2(g) CaSO3(s)

Water slurryremoved

CaCO3 decomposes to CaO (lime) and CO2

2

Powdered limestone (CaCO3) and air injected into furnace

1 CaO reacts with SO2 from reaction S + O2 SO2 to form CaSO3

3 CaSO3 and unreacted SO2 removed by passing through aqueous suspension of CaO

4

CaSO3 precipitated into watery slurry

5

Cleaner airexpelledthrough stack

6

▲ Figure 18.10 photochemical smog is produced largely by the action of sunlight on vehicle exhaust gases.


(approximate temperature in the cylinder of an engine during combustion). Thus, the reaction is more favorable at higher temperatures. In fact, some NO is formed in any high-temperature combustion. As a result, electrical power plants are also major con-tributors to nitrogen oxide pollution.

Before the installation of pollution-control devices on automobiles, typical emis-sion levels of NOx were 4 g>mi. (The x is either 1 or 2 because both NO and NO2 are formed, although NO predominates.) Starting in 2004, the auto emission standards for NOx called for a phased-in reduction to 0.07 g>mi by 2009, which was achieved.

In air, nitric oxide is rapidly oxidized to nitrogen dioxide:

2 NO1g2 + O21g2 ∆ 2 NO21g2 ∆H = -113.1 kJ [18.12]

The equilibrium constant for this reaction decreases from about 1012 at 300 K to about 10-5 at 2400 K.

The photodissociation of NO2 initiates the reactions associated with photochemi-cal smog. Dissociation of NO2 requires 304 kJ>mol, which corresponds to a photon wavelength of 393 nm. In sunlight, therefore, NO2 dissociates to NO and O:

NO21g2 + hn ¡ NO1g2 + O1g2 [18.13]

The atomic oxygen formed undergoes several reactions, one of which gives ozone, as described earlier:

O1g2 + O2 + M1g2 ¡ O31g2 + M*1g2 [18.14]

Although it is an essential UV screen in the upper atmosphere, ozone is an undesir-able pollutant in the troposphere. It is extremely reactive and toxic, and breathing air that contains appreciable amounts of ozone can be especially dangerous for asthma suf-ferers, exercisers, and the elderly. We therefore have two ozone problems: excessive amounts in many urban environments, where it is harmful, and depletion in the strato-sphere, where its presence is vital.

In addition to nitrogen oxides and carbon monoxide, an automobile engine also emits unburned hydrocarbons as pollutants. These organic compounds are the prin-cipal components of gasoline and of many compounds we use as fuel (propane, C3H8, and butane, C4H10; for example), but are major ingredients of smog. A typical engine without effective emission controls emits about 10 to 15 g of hydrocarbons per mile. Current standards require that hydrocarbon emissions be less than 0.075 g>mi. Hydro-carbons are also emitted naturally from living organisms (see “A Closer Look” box later in this section).

Reduction or elimination of smog requires that the ingredients essential to its for-mation be removed from automobile exhaust. Catalytic converters reduce the levels of NOx and hydrocarbons, two of the major ingredients of smog. (See the “Chemistry Put to Work: Catalytic Converters” in Section 14.7.)

give it some ThoughtWhat photochemical reaction involving nitrogen oxides initiates the formation of photochemical smog?

Greenhouse Gases: Water Vapor, Carbon Dioxide, and ClimateIn addition to screening out harmful short-wavelength radiation, the atmosphere is essential in maintaining a reasonably uniform and moderate temperature on Earth’s surface. Earth is in overall thermal balance with its surroundings. This means that the planet radiates energy into space at a rate equal to the rate at which it absorbs energy from the Sun. Figure 18.11 shows the distribution of radiation to and from Earth’s surface, and Figure 18.12 shows which portion of the infrared radiation leaving the surface is absorbed by atmospheric water vapor and carbon dioxide. In absorbing this radiation, these two atmospheric gases help maintain a livable uniform temperature at the surface by holding in, as it were, the infrared radiation, which we feel as heat.


▲ Figure 18.11 earth’s thermal balance. The amount of radiation reaching the surface of the planet is approximately equal to the amount radiated back into space.

▲ Figure 18.12 portions of the infrared radiation emitted by earth’s surface that are absorbed by atmospheric co2 and h2o.

Infrared radiation emitted from Earth’s surface

Wavelengths absorbed by atmospheric CO2 and H2O

Wavelength (nm)

Rad

iati

on in

tens

ity

10,000 20,000

CO2

H2O

30,000

Go FiGurEWhat fraction of the incoming solar radiation is absorbed by Earth’s surface?

Backradiation

IR absorbedby surface324 W/m2

IR emittedfrom surface

390 W/m2

IR and longerwavelengths of

radiation out throughatmosphere

235 W/m2

= 100 watts for each square meter

Greenhouse gasesLatent heat

of condensation78 W/m2

Solar radiationre�ected by surface

30 W/m2Solar radiation

re�ected by clouds,asteroids andatmosphere

77 W/m2

Incomingsolar radiation

342 W/m2

Solar radiationabsorbed byatmosphereand clouds

67 W/m2

Solar radiationabsorbed by surface

168 W/m2


The influence of H2O, CO2, and certain other atmospheric gases on Earth’s temper-ature is called the greenhouse effect because in trapping infrared radiation these gases act much like the glass of a greenhouse. The gases themselves are called greenhouse gases.

Water vapor makes the largest contribution to the greenhouse effect. The partial pressure of water vapor in the atmosphere varies greatly from place to place and time to time but is generally highest near Earth’s surface and drops off with increasing elevation. Because water vapor absorbs infrared radiation so strongly, it plays the major role in main-taining the atmospheric temperature at night, when the surface is emitting radiation into space and not receiving energy from the Sun. In very dry desert climates, where the water-vapor concentration is low, it may be extremely hot during the day but very cold at night. In the absence of a layer of water vapor to absorb and then radiate part of the infrared radi-ation back to Earth, the surface loses this radiation into space and cools off very rapidly.

Carbon dioxide plays a secondary but very important role in maintaining the sur-face temperature. The worldwide combustion of fossil fuels, principally coal and oil, on a prodigious scale in the modern era has sharply increased carbon dioxide levels in the atmosphere. To get a sense of the amount of CO2 produced—for example, by the com-bustion of hydrocarbons and other carbon-containing substances, which are the compo-nents of fossil fuels—consider the combustion of butane, C4H10. Combustion of 1.00 g of C4H10 produces 3.03 g of CO2. (Section 3.6) Similarly, a gallon (3.78 L) of gaso-line (density 0.7 g>mL, approximate composition C8H18) produces about 8 kg of CO2. Combustion of fossil fuels releases about 2.2 * 1016 g (24 billion tons) of CO2 into the atmosphere annually, with the largest quantity coming from transportation vehicles.

Much CO2 is absorbed into oceans or used by plants. Nevertheless, we are now generating CO2 much faster than it is being absorbed or used. Analysis of air trapped in ice cores taken from Antarctica and Greenland makes it possible to determine the atmospheric levels of CO2 during the past 160,000 years. These measurements reveal that the level of CO2 remained fairly constant from the last Ice Age, some 10,000 years ago, until roughly the beginning of the Industrial Revolution, about 300 years ago. Since that time, the concentration of CO2 has increased by about 30% to a current high of about 400 ppm (▼ Figure 18.13). Climate scientists believe that the CO2 level has not been this high since 3 to 5 million years ago.

▲ Figure 18.13 rising CO2 levels. The sawtooth shape of the graph is due to regular seasonal variations in CO2 concentration for each year.

1960 1970 1980 1990 2000 2010Year

CO

2 co

ncen

trat

ion

(ppm

) 380

400

360

340

320

Mauna Loa Observatory, Hawaii

Go FiGurEWhat is the source of the slight but steady increase in slope of this curve over time?


The consensus among climate scientists is that the increase in atmospheric CO2 is perturbing Earth’s climate and is very likely playing a role in the observed increase in the average global air temperature of 0.390.6 °C over the past century. Scientists often use the term climate change instead of global warming to refer to this effect because as the Earth’s temperature increases, winds and ocean currents are affected in ways that cool some areas and warm others.

On the basis of present and expected future rates of fossil-fuel use, the atmospheric CO2 level is expected to double from its present level sometime between 2050 and 2100. Computer models predict that this increase will result in an average global temperature increase of 1 °C to 3 °C. Because so many factors go into determining climate, we cannot predict with certainty what changes will occur because of this warming. Clearly, however, humanity has acquired the potential, by changing the concentrations of CO2 and other heat-trapping gases in the atmosphere, to substantially alter the climate of the planet.

The climate change threat posed by atmospheric CO2 has sparked considerable research into ways of capturing the gas at its largest combustion sources and storing it under ground or under the seafloor. There is also much interest in developing new ways to use CO2 as a chemical feedstock. However, the approximately 115 million tons of CO2 used annually by the global chemical industry is but a small fraction of the approx-imately 24 billion tons of annual CO2 emissions. The use of CO2 as a raw material will probably never be great enough to significantly reduce its atmospheric concentration.

give it some ThoughtExplain why nighttime temperatures remain higher in locations where there is higher humidity.

but there are indirect effects to consider. Methane is oxidized in the stratosphere, producing water vapor, a powerful greenhouse gas that is otherwise virtually absent from the stratosphere. In the troposphere, methane is attacked by reactive species such as OH radicals or nitro-gen oxides, eventually producing other greenhouse gases, such as O3. It has been estimated that on a per-molecule level, the global warm-ing potential of CH4 is about 21 times that of CO2. Given this large contribution, important reductions of the greenhouse effect could be achieved by reducing methane emissions or capturing the emissions for use as a fuel.


a Closer Look

other greenhouse gases

Although CO2 receives most of the attention, other gases contribute to the greenhouse effect, including methane, CH4, hydrofluorocarbons (HFCs), and chlorofluorocarbons (CFCs).

HFCs have replaced CFCs in a host of applications, including refrigerants and air-conditioner gases. Although they do not con-tribute to the depletion of the ozone layer, HFCs are nevertheless potent greenhouse gases. For example, one of the byproduct mol-ecules from production of HFCs that are used in commerce is HCF3, which is estimated to have a global warming potential, gram for gram, more than 14,000 times that of CO2. The total concentration of HFCs in the atmosphere has been increasing about 10% per year. Thus, these substances are becoming increasingly important con-tributors to the greenhouse effect. Methane already makes a signifi-cant contribution to the greenhouse effect. Studies of atmospheric gas trapped long ago in the Greenland and Antarctic ice sheets show that the atmospheric methane concentration has increased from pre-industrial values of 0.3 to 0.7 ppm to the present value of about 1.8 ppm. The major sources of methane are associated with agriculture and fossil-fuel use.

Methane is formed in biological processes that occur in low-oxygen environments. Anaerobic bacteria, which flourish in swamps and land-fills, near the roots of rice plants, and in the digestive systems of cows and other ruminant animals, produce methane (▶ Figure 18.14). It also leaks into the atmosphere during natural-gas extraction and transport. It is estimated that about two-thirds of present-day methane emissions, which are increasing by about 1% per year, are related to human activities.

Methane has a half-life in the atmosphere of about 10 years, whereas CO2 is much longer-lived. This might seem a good thing,

▲ Figure 18.14 methane production. Ruminant animals, such as cows and sheep, produce methane in their digestive systems.

sEction 18.3 Earth’s Water 791

18.3 | Earth’s WaterWater covers 72% of Earth’s surface and is essential to life. Our bodies are about 65% water by mass. Because of extensive hydrogen bonding, water has unusually high melting and boiling points and a high heat capacity. (Section 11.2) Water’s highly polar char-acter is responsible for its exceptional ability to dissolve a wide range of ionic and polar-covalent substances. Many reactions occur in water, including reactions in which H2O itself is a reactant. Recall, for example, that H2O can participate in acid–base reactions as either a proton donor or a proton acceptor. (Section 16.3) All these properties play a role in our environment.

The Global Water CycleAll the water on Earth is connected in a global water cycle (▼ Figure 18.15). Most of the processes depicted here rely on the phase changes of water. For instance, warmed by the Sun, liquid water in the oceans evaporates into the atmosphere as water va-por and condenses into liquid water droplets that we see as clouds. Water droplets in the clouds can crystallize to ice, which can precipitate as hail or snow. Once on the ground, the hail or snow melts to liquid water, which soaks into the ground. If condi-tions are right, it is also possible for ice on the ground to sublime to water vapor in the atmosphere.

give it some ThoughtConsider the phase diagram for water shown in Figure 11.28 (page 465). In what pressure range and in what temperature range must H2O exist in order for H2O1s2 to sublime to H2O1g2?

Condensation

Waterstorage in theatmosphere

Water storage inice and snow

Water storagein oceans

Evaporation

Groundwater storage

Groundwaterdischarge

Snowmelt runoffto streams

Precipitation

Evaporation

Freshwaterstorage

Plantuptake

Surface runoff

Fog dripEvaporation,transpiration

Deposition

Sublimation

▲ Figure 18.15 The global water cycle.

Go FiGurEWhich processes shown in this figure involve the phase transition H2O1l 2 ¡ H2O1g2?


Salt Water: Earth’s Oceans and SeasThe vast layer of salty water that covers so much of the planet is in actuality one large connected body and is generally constant in composition. For this reason, oceanog-raphers speak of a world ocean rather than of the separate oceans we learn about in geography books.

The world ocean is huge, having a volume of 1.35 * 109 km3 and containing 97.2% of all the water on Earth. Of the remaining 2.8%, 2.1% is in the form of ice caps and glaciers. All the freshwater—in lakes, in rivers, and in the ground—amounts to only 0.6%. Most of the remaining 0.1% is in brackish (salty) water, such as that in the Great Salt Lake in Utah.

Seawater is often referred to as saline water. The salinity of seawater is the mass in grams of dry salts present in 1 kg of seawater. In the world ocean, salinity averages about 35. To put it another way, seawater contains about 3.5% dissolved salts by mass. The list of elements present in seawater is very long. Most, however, are present only in very low concentrations. ▼ Table 18.5 lists the 11 ionic species most abundant in seawater.

Seawater temperature varies as a function of depth (▶ Figure 18.16), as does salin-ity and density. Sunlight penetrates well only 200 m into the water; the region between 200 m and 1000 m deep is the “twilight zone,” where visible light is faint. Below 1000 m, the ocean is pitch-black and cold, about 4 °C. The transport of heat, salt, and other chemicals throughout the ocean is influenced by these changes in the physical proper-ties of seawater, and in turn the changes in the way heat and substances are transported affects ocean currents and the global climate.

The sea is so vast that if the concentration of a substance in seawater is 1 part per billion (1 * 10-6 g>kg of water), there is 1 * 1012 kg of the substance in the world ocean. Nevertheless, because of high extracting costs, only three substances are obtained from seawater in commercially important amounts: sodium chloride, bromine (from bromide salts), and magnesium (from its salts).

Absorption of CO2 by the ocean plays a large role in global climate. Because carbon dioxide and water form carbonic acid, the H2CO3 concentration in the ocean increases as the water absorbs atmospheric CO2. Most of the carbon in the ocean, however, is in the form of HCO3

- and CO32- ions, which form a buffer system that maintains the

ocean’s pH between 8.0 and 8.3. The pH of the ocean is predicted to decrease as the concentration of CO2 in the atmosphere increases, as discussed in the “Chemistry and Life” box on ocean acidification in Section 17.5.

Freshwater and GroundwaterFreshwater is the term used to denote natural waters that have low concentrations (less than 500 ppm) of dissolved salts and solids. Freshwater includes the waters of

Table 18.5 ionic Constituents of seawater present in Concentrations Greater than 0.001 g ,kg 11 ppm 2ionic Constituent salinity Concentration (M)

Chloride, Cl- 19.35 0.55

Sodium, Na+ 10.76 0.47

Sulfate, SO42- 2.71 0.028

Magnesium, Mg2+ 1.29 0.054

Calcium, Ca2+ 0.412 0.010

Potassium, K+ 0.40 0.010

Carbon dioxide* 0.106 2.3 * 10-3

Bromide, Br- 0.067 8.3 * 10-4

Boric acid, H3BO3 0.027 4.3 * 10-4

Strontium, Sr2+ 0.0079 9.1 * 10-5

Fluoride, F- 0.0013 7.0 * 10-5

*CO2 is present in seawater as HCO3- and CO3

2-.

sEction 18.3 Earth’s Water 793

lakes, rivers, ponds, and streams. The United States is fortunate in its abundance of freshwater:1.7 * 1015 L (660 trillion gallons) is the estimated reserve, which is re-newed by rainfall. An estimated 9 * 1011 L of freshwater is used every day in the United States. Most of this is used for agriculture (41%) and hydroelectric power (39%), with small amounts for industry (6%), household needs (6%), and drinking water (1%). An adult drinks about 2 L of water per day. In the United States, our daily use of water per person far exceeds this subsistence level, amounting to an average of about 300 L/day for personal consumption and hygiene. We use about 8 L/person for cooking and drinking, about 120 L/person for cleaning (bathing, laundering, and housecleaning), 80 L>person for flushing toilets, and 80 L/person for watering lawns.

The total amount of freshwater on Earth is not a very large fraction of the total water present. Indeed, freshwater is one of our most precious resources. It forms by evaporation from the oceans and the land. The water vapor that accumulates in the atmosphere is transported by global atmospheric circulation, eventually returning to Earth as rain, snow, and other forms of precipitation (Figure 18.15).

As water runs off the land on its way to the oceans, it dissolves a variety of cat-ions (mainly Na+, K+, Mg2+, Ca2+, and Fe2+), anions (mainly Cl-, SO4

2-, and HCO3-),

and gases (principally O2, N2, and CO2). As we use water, it becomes laden with addi-tional dissolved material, including the wastes of human society. As our population and output of environmental pollutants increase, ever-increasing amounts of money and resources must be spent to guarantee a supply of freshwater.

The availability and cost of fresh water that is clean enough to sustain daily life varies greatly among nations. To illustrate, daily fresh water use in the United States approaches 600 L/person, whereas in the relatively underdeveloped nations of sub-Sahara Africa it is only about 30 L. To make matters worse, for many people, water is not only scarce, it is so contaminated that it is a continuing source of diseases.

Approximately 20% of the world’s freshwater is under the soil, in the form of groundwater. Groundwater resides in aquifers, which are layers of porous rock that hold water. The water in aquifers can be very pure, and accessible for human consumption if near the surface. Dense underground formations that do not allow water to read-ily penetrate can hold groundwater for years or even millennia. When their water is removed by drilling and pumping, such aquifers are slow to recharge via the diffusion of surface water.

0

5

10

20

15

250 500 750 1000Depth (m)

Tem

pera

ture

(°C

)

1250 1500 17500

▲ Figure 18.16 Typical average temperature of mid-latitude seawater as a function of depth.

Go FiGurEHow would you expect the temperature variation to affect the density of seawater in the range 0 to 100 m depth?


The nature of the rock that contains the groundwater has a large influence on the water’s chemical composition. If minerals in the rock are water soluble to some extent, ions can leach out of the rock and remain dissolved in the groundwater. Arsenic in the form of HAsO4

2-, H2AsO4-, and H3AsO3 is found in many groundwater sources across

the world, most infamously in Bangladesh, at concentrations poisonous to humans.

18.4 | human activities and Water Quality

All life on Earth depends on the availability of suitable water. Many human activities entail waste disposal into natural waters without any treatment. These practices result in contaminated water that is detrimental to both plant and animal aquatic life. Unfor-tunately people in many parts of the world do not have access to water that has been treated to remove harmful contaminants, including disease-bearing bacteria.

Dissolved Oxygen and Water QualityThe amount of O2 dissolved in water is an important indicator of water quality. Water fully saturated with air at 1 atm and 20 °C contains about 9 ppm of O2. Oxygen

a Closer Look

The ogallala aquifer— a shrinking resource

The Ogallala Aquifer, also referred to as the High Plains Aquifer, is an enormous underground water body of wa-ter lying beneath the Great Plains of the United States. One of the world’s largest aquifers, it covers an area of approxi-mately 450,000 km2 1170,000 mi22 encompassing portions of the eight states of South Dakota, Nebraska, Wyoming, Colorado, Kansas, Oklahoma, New Mexico, and Texas. (▶ Figure 18.17). The depth of the Ogallala formation that gives rise to the aquifer ranges from about 120 m, to more than 300 m, particularly in the northern portion. Be-fore large-scale pumping in the modern era, the depth of water in the aquifer ranged up to more than 120 m in the northern portion.

Anyone who has flown over the Great Plains is fa-miliar with the view of huge circles made by the center pivot irrigators nearly covering the land. The center post irrigation system, developed in the post–World War II era, permitted application of water onto large areas. As a re-sult, the Great Plains became one of the most productive agricultural areas in the world. Unfortunately, the premise that the aquifer is an inexhaustible source of fresh water proved false. Recharge of the aquifer from surface water is slow, taking hundreds, perhaps thousands of years. Re-cently, water levels in many regions of the Ogallala have declined to the point where the costs of bringing water to the surface have become prohibitive. As the aquifer levels continue to drop, less water will be available for the needs of cities, residences and businesses.


▲ Figure 18.17 A map showing the extent of the Ogallala (High Plains) aquifer.Note that the elevation of the land varies greatly. The aquifer follows the topography of the formations that underlie the area.

MidlandOdessa

Lubbock

Amarillo

Guymon

WichitaGarden City

North Platte

Scottsbluff

Cheyenne

TEXAS

COLORADO

KANSAS

NEW MEXICO

NEBRASKA

OKLAHOMA

WYOMING SOUTH DAKOTA

Elevation, in feet8,000

1,000

Northern High Plains

Central High Plains

Southern High Plains

0 50 100 miles

0 50 100 kilometers

sEction 18.4 human activities and Water Quality 795

is necessary for fish and most other aquatic life. Cold-water fish require water contain-ing at least 5 ppm of dissolved oxygen for survival. Aerobic bacteria consume dissolved oxygen to oxidize organic materials for energy. The organic material the bacteria are able to oxidize is said to be biodegradable.

Excessive quantities of biodegradable organic materials in water are detrimen-tal because they remove the oxygen necessary to sustain normal animal life. Typical sources of these biodegradable materials, which are called oxygen-demanding wastes, include sewage, industrial wastes from food-processing plants and paper mills, and liq-uid waste from meatpacking plants.

In the presence of oxygen, the carbon, hydrogen, nitrogen, sulfur, and phosphorus in biodegradable material end up mainly as CO2, HCO3

-, H2O, NO3-, SO4

2-, and phos-phates. The formation of these oxidation products sometimes reduces the amount of dissolved oxygen to the point where aerobic bacteria can no longer survive. Anaerobic bacteria then take over the decomposition process, forming CH4, NH3, H2S, PH3, and other products, several of which contribute to the offensive odors of some polluted waters.

Plant nutrients, particularly nitrogen and phosphorus, contribute to water pollution by stimulating excessive growth of aquatic plants. The most visible results of excessive plant growth are floating algae and murky water. What is more significant, however, is that as plant growth becomes excessive, the amount of dead and decaying plant matter increases rapidly, a process called eutrophication (▶ Figure 18.18). The processes by which plants decay consumes O2, and without sufficient oxygen, the water cannot sustain animal life.

The most significant sources of nitrogen and phosphorus compounds in water are domestic sewage (phosphate-containing detergents and nitrogen-containing body wastes), runoff from agricultural land (fertilizers contain both nitrogen and phospho-rus), and runoff from livestock areas (animal wastes contain nitrogen).

give it some ThoughtIf a test on a sample of polluted water shows a considerable decrease in dissolved oxygen over a five-day period, what can we conclude about the nature of the pollutants present?

Water Purification: DesalinationBecause of its high salt content, seawater is unfit for human consumption and for most of the uses to which we put water. In the United States, the salt content of munici-pal water supplies is restricted by health codes to no more than about 0.05% by mass. This amount is much lower than the 3.5% dissolved salts present in seawater and the 0.5% or so present in brackish water found underground in some regions. The removal of salts from seawater or brackish water to make the water usable is called desalination.

Water can be separated from dissolved salts by distillation because water is a volatile sub-stance and the salts are nonvolatile. (Section 1.3, “Separation of Mixtures”) The principle of distillation is simple enough, but carrying out the process on a large scale presents many problems. As water is distilled from seawater, for example, the salts become more and more concentrated and eventually precipitate out. Distillation is also an energy-intensive process.

Seawater can also be desalinated using reverse osmosis. Recall that osmosis is the net movement of solvent molecules, but not solute molecules, through a semiperme-able membrane. (Section 13.5) In osmosis, the solvent passes from the more dilute solution into the more concentrated one. However, if sufficient external pressure is applied, osmosis can be stopped and, at still higher pressures, reversed. When reverse osmosis occurs, solvent passes from the more concentrated into the more dilute solu-tion. In a modern reverse-osmosis facility, hollow fibers are used as the semipermeable membrane (▶ Figure 18.19). Saline water (water containing significant salts) is intro-duced under pressure into the fibers, and desalinated water is recovered.

The world’s largest desalination plant, in Jubail, Saudi Arabia, provides 50% of that country’s drinking water by using reverse osmosis to desalinate seawater from the Persian Gulf. An even larger plant, which will produce 600 million L/day (160 mil-lion gallons) of drinking water, is scheduled for completion in Saudi Arabia in 2018. Such plants are becoming increasingly common in the United States. The largest, near

▲ Figure 18.18 eutrophication. This rapid accumulation of dead and decaying plant matter in a body of water uses up the water’s oxygen supply, making the water unsuitable for aquatic animals.

▲ Figure 18.19 reverse osmosis.

Go FiGurEWhat feature of this process is responsible for its being called reverse osmosis?

Desalinated water to collector

Permeator

Fiber

Hollow �bers ofsemipermeablemembrane

Seawater pumped through at high pressure

Few solute particles enter hollow �bers

Water molecules pushed into hollow �bers


Tampa Bay, Florida, has been operating since 2007 and produces 35 million gallons of drinking water a day by reverse osmosis. Small-scale, manually operated reverse-osmosis desalinators are used in camping, traveling, and at sea.

Water Purification: Municipal TreatmentThe water needed for domestic, agricultural, and industrial use is taken either from lakes, rivers, and underground sources or from reservoirs. Much of the water that finds its way into municipal water systems is “used” water, meaning it has already passed through one or more sewage systems or industrial plants. Consequently, this water must be treated before it is distributed to our faucets.

Municipal water treatment usually involves five steps (▲ Figure 18.20). After coarse filtration through a screen, the water is allowed to stand in large sedimentation tanks where sand and other minute particles settle out. To aid in removing very small particles, the water may first be made slightly basic with CaO. Then Al21SO423 is added and reacts with OH- ions to form a spongy, gelatinous precipitate of Al1OH23 1Ksp = 1.3 * 10-332. This precipitate settles slowly, carrying suspended particles down with it, thereby remov-ing nearly all finely divided matter and most bacteria. The water is then filtered through a sand bed. Following filtration, the water may be sprayed into the air (aeration) to hasten oxidation of dissolved inorganic ions of iron and manganese, reduce concentrations of any H2S or NH3 that may be present and reduce bacterial concentrations.

The final step normally involves treating the water with a chemical agent to ensure the destruction of bacteria. Ozone is more effective, but chlorine is less expensive. Liq-uefied Cl2 is dispensed from tanks through a metering device directly into the water supply. The amount used depends on the presence of other substances with which the chlorine might react and on the concentrations of bacteria and viruses to be removed. The sterilizing action of chlorine is probably due not to Cl2 itself but to hypochlorous acid, which forms when chlorine reacts with water: Cl21aq2 + H2O1l2 ¡ HClO1aq2 + H+1aq2 + Cl-1aq2 [18.15]

It is estimated that about 800 million people worldwide lack access to clean water. According to the United Nations, 95% of the world’s cities still dump raw sewage into their water supplies. Thus, it should come as no surprise that 80% of all the health maladies in developing countries can be traced to waterborne diseases associated with unsanitary water.

One promising development is a device called the LifeStraw (◀ Figure 18.21). When a person sucks water through the straw, the water first encounters a textile

Waterintake

Coarse�ltrationscreen

Sedimentation tanks

Aeration

Chlorinesterilizers

Storagetank

To users

Sand�lter

CaO,Al2(SO4)3added

▲ Figure 18.20 common steps in treating water for a public water system.

Go FiGurEWhat is the primary function of the aeration step in water treatment?

100-μm textile �lterremoves debris

15-μm textile �lterremoves debris

Iodine-impregnatedbeads kill bacteria,viruses, and parasites

Carbon removesiodine smellsand parasites

▲ Figure 18.21 a lifestraw purifies water as it is drunk.

sEction 18.4 human activities and Water Quality 797

filter with a mesh opening of 100 μm followed by a second textile filter with a mesh opening of 15 μm. These filters remove debris and even clusters of bacteria. The water next encounters a chamber of iodine-impregnated beads, where bacteria, viruses, and parasites are killed. Finally, the water passes through granulated active carbon, which removes the smell of iodine as well as the parasites that have not been taken by the fil-ters or killed by the iodine. At present the Lifestraw is too costly to permit widespread use in underdeveloped countries, but there is hope that its cost can be greatly reduced.

Water disinfection is one of the greatest public health innovations in human his-tory. It has dramatically decreased the incidences of waterborne bacterial diseases such as cholera and typhus. However, this great benefit comes at a price.

In 1974 scientists in Europe and the United States discovered that chlorination of water produces a group of by-products previously undetected. These by-products are called trihalomethanes (THMs) because all have a single carbon atom and three halo-gen atoms: CHCl3, CHCl2Br, CHClBr2, and CHBr3. These and many other chlorine- and bromine-containing organic substances are produced by the reaction of dissolved chlorine with the organic materials present in nearly all natural waters, as well as with

a Closer Look

fracking and Water quality

In recent years fracking, short for hydrau-lic fracturing, has become widely used to greatly increase the availability of petro-leum reserves. In fracking, a large volume of water, typically two million gallons or more, mixed with various additives, is injected at high pressure into wellbores extended horizontally into rock formations (▶ Figure 18.22). The water is laden with sand, ceramic materi-als and other additives, including gels, foams, and compressed gases, that serve to increase the yield in the process. The high pressure fluid finds its way into tiny faults in geological formations, releasing petroleum and natural gas. Fracking has greatly increased petroleum reserves, particularly of natural gas, in many parts of the world. The technique has been so productive that more than 20,000 new wells are being drilled annually in the United States alone, in all areas of the country.

Unfortunately, the potential for en-vironmental damage from fracking is sig-nificant. The large volume of fracking fluid required to create a well must be returned to the surface. Without purification the fluid is rendered unfit for other uses, and becomes a large-scale environmental problem. Often the waste water is allowed to sit in open waste water pits. The 2005 Energy Policy act and other federal legislation exempts hydraulic fracturing operations from certain provisions of the Safe Drinking Water act and other regulations. Some areas of the country that are already facing water shortages thus have one more large demand for a limited supply. Because fracturing of rock formations increases the pathways for flows of petroleum and various gases, bodies of underground water that have been serving as municipal water supplies or wells for in-dividual homes in some locales have become contaminated with petroleum, hydrogen sulfide and other toxic substances. The escape of

Waste water ponds

Fracking �uid

Shallow aquifer

Deep aquifer

Impermeable layer

Impermeable layer

Gas-bearing formation

Hydraulic fractures

Methane

Migration of released gases

Contaminants escape to surroundings

▲ Figure 18.22 a schematic of a well site employing fracking. The yellow arrows indicate the avenues through which contaminants enter the environment.

a variety of gases, including methane and other hydrocarbons, from the wellheads contributes to air pollution. In a study published in 2013, methane emissions to the atmosphere during hydraulic frack-ing operations in Utah were estimated to be in the range of 6–12% of the amount of methane produced. As related in the Closer Look box on page 790, methane is a potent greenhouse gas.

The many environmental issues surrounding the practice of frack-ing have generated widespread concern and adverse public reaction. Fracking represents yet one more instance of the conflict between those who advocate the availability of low cost energy and those who are more focused on sustaining long term the quality of the environment.


substances that are by-products of human activity. Recall that chlorine dissolves in water to form the oxidizing agent HClO:

Cl21g2 + H2O1l2 ¡ HClO1aq2 + H+1aq2 + Cl-1aq2 [18.16]

The HClO in turn reacts with organic substances to form THMs. Bromine enters the reaction sequence through the reaction of HClO with dissolved bromide ion:

HClO1aq2 + Br-1aq2 ¡ HBrO1aq2 + Cl-1aq2 [18.17]

Then both HBrO(aq) and HClO(aq) can halogenate organic substances to form the THMs.Some THMs and other halogenated organic substances are suspected carcino-

gens; others interfere with the body’s endocrine system. As a result, the World Health Organization and the U.S. Environmental Protection Agency have placed concentra-tion limits of 80 mg>L (80 ppb) on the total quantity of THMs in drinking water. The goal is to reduce the levels of THMs and other disinfection by-products in the drinking water supply while preserving the antibacterial effectiveness of the water treatment. In some cases, lowering the concentration of chlorine may provide adequate disinfection while reducing the concentrations of THMs formed. Alternative oxidizing agents, such as ozone or chlorine dioxide, produce less of the halogenated substances but have their own disadvantages. For example, each is capable of oxidizing dissolved bromide, as shown here for ozone:

O31aq2 + Br-1aq2 + H2O1l2 ¡ HBrO1aq2 + O21aq2 + OH-1aq2 [18.18] HBrO1aq2 + 2 O31aq2 ¡ BrO3

-1aq2 + 2 O21aq2 + H+1aq2 [18.19]

Bromate ion, BrO3-, has been shown to cause cancer in animal tests.

At present, there seem to be no completely satisfactory alternatives to chlorination or ozonation, and we are faced with a consideration of benefit versus risk. In this case, the risks of cancer from THMs and related substances in municipal water are very low rela-tive to the risks of cholera, typhus, and gastrointestinal disorders from untreated water. When the water supply is cleaner to begin with, less disinfectant is needed and thus the risk of THMs is lowered. Once THMs form, their concentrations in the water supply can be reduced by aeration because the THMs are more volatile than water. Alternatively, they can be removed by adsorption onto activated charcoal or other adsorbents.

18.5 | Green ChemistryThe planet on which we live is, to a large extent, a closed system, one that exchanges en-ergy but not matter with its surroundings. If humankind is to thrive in the future, all the processes we carry out should be in balance with Earth’s natural processes and physical resources. This goal requires that no toxic materials be released to the environment, that our needs be met with renewable resources, and that we consume the least possible amount of energy. Although the chemical industry is but a small part of human activity, chemical processes are involved in nearly all aspects of modern life. Chemistry is there-fore at the heart of efforts to accomplish these goals.

Green chemistry is an initiative that promotes the design and application of chem-ical products and processes that are compatible with human health and that preserve the environment. Green chemistry rests on a set of 12 principles:

1. Prevention It is better to prevent waste than to clean it up after it has been created. 2. Atom Economy Methods to make chemical compounds should be designed to

maximize the incorporation of all starting atoms into the final product. 3. Less Hazardous Chemical Syntheses Wherever practical, synthetic methods

should be designed to use and generate substances that possess little or no toxicity to human health and the environment.

4. Design of Safer Chemicals Chemical products should be designed to minimze toxicity and yet maintain their desired function.

sEction 18.5 Green chemistry 799

5. Safer Solvents and Auxiliaries Auxiliary substances (for example, solvents, sepa-ration agents, etc.) should be used as little as possible. Those that are used should be as nontoxic as possible.

6. Design for Energy Efficiency Energy requirements of chemical processes should be recognized for their environmental and economic impacts and should be mini-mized. If possible, chemical reactions should be conducted at room temperature and pressure.

7. Use of Renewable Feedstocks A raw material or feedstock should be renewable whenever technically and economically practical.

8. Reduction of Derivatives Unnecessary derivatization (intermediate compound formation, temporary modification of physical/chemical processes) should be minimized or avoided if possible, because such steps require additional reagents and can generate waste.

9. Catalysis Catalytic reagents (as selective as possible) improve product yields within a given time and with a lower energy cost compared to noncatalytic processes and are, therefore, preferred to noncatalytic alternatives.

10. Design for Degradation The end products of chemical processing should break down at the end of their useful lives into innocuous degradation products that do not persist in the environment.

11. Real-Time Analysis for Pollution Prevention Analytical methods need to be de-veloped that allow for real-time, in-process monitoring and control prior to the formation of hazardous substances.

12. Inherently Safer Chemistry for Accident Prevention Reagents and solvents used in a chemical process should be chosen to minimize the potential for chemical ac-cidents, including releases, explosions, and fires.*

give it some ThoughtExplain how a chemical reaction that uses a catalyst can be “greener” than the same reaction run without a catalyst.

To illustrate how green chemistry works, consider the manufacture of styrene, an important building block for many polymers, including the expanded polystyrene packages used to pack eggs and restaurant takeout food. The global demand for styrene is more than 25 billion kg per year. For many years, styrene has been produced in a two-step process: Benzene and ethylene react to form ethyl benzene, followed by the ethyl benzene being mixed with high-temperature steam and passed over an iron oxide catalyst to form styrene:

H2C CH2

CH CH2CH2CH3

+−H2

Acidcatalyst

Iron oxidecatalyst

Benzene Ethylene Ethyl benzene Styrene

This process has several shortcomings. One is that both benzene, which is formed from crude oil, and ethylene, formed from natural gas, are high-priced starting mate-rials for a product that should be a low-priced commodity. Another is that benzene is a known carcinogen. In a recently-developed process that bypasses some of these

*Adapted from P. T. Anastas and J. C. Warner, Green Chemistry: Theory and Practice. New York: Oxford University Press 1998, p. 30. See also, Mike Lancaster, Green Chemistry: An Introductory Text. Cambridge, UK: RSC Publishing, 2010, Second Edition, Chapter 1.

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