chapter eleven - unamdepa.fquim.unam.mx/amyd/archivero/unidad_acids-bases... · 2015. 11. 17. ·...

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Chapter ElevenACIDS AND BASES

11.1 Properties of Acids and Bases

11.2 The Arrhenius Definition of Acids and Bases

11.3 The Brønsted–Lowry Definition of Acids and Bases

11.4 Conjugate Acid–Base Pairs

11.5 The Role of Water in the Brønsted Model

11.6 To What Extent Does Water Dissociate to Form Ions?

11.7 pH as a Measure of the Concentration of the H3O� Ion

11.8 Relative Strengths of Acids and Bases

11.9 Relative Strengths of Conjugate Acid–Base Pairs

11.10 Relative Strengths of Different Acids and Bases

11.11 Relationship of Structure to Relative Strengths of Acids and Bases

11.12 Strong Acid pH Calculations

11.13 Weak Acid pH Calculations

11.14 Base pH Calculations

11.15 Mixtures of Acids and Bases: Buffers

11.16 Buffers and Buffer Capacity

11.17 Buffers in the Body

11.18 Acid–Base Reactions

11.19 pH Titration Curves

Special Topics11A.1 Diprotic Acids

11A.2 Diprotic Bases

11A.3 Compounds That Could Be Either Acids or Bases

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11.1 Properties of Acids and BasesFor more than 300 years, chemists have classified substances that behave likevinegar as acids and substances that have properties like wood ash as bases (oralkalies). The word acid comes from the Latin acidus, which means “sour,” andrefers to the sharp odor and sour taste of many acids. Vinegar, for example, tastessour because it is a dilute solution of acetic acid in water. Lemon juice tastes sourbecause it contains citric acid. Milk turns sour when it spoils because lactic acidis formed, and the unpleasant, sour odor of rotten meat or butter can be attrib-uted to compounds such as butyric acid that form when fat spoils.

One of the characteristic properties of an acid is its ability to dissolve mostmetals. Zinc metal, for example, rapidly reacts with hydrochloric acid to form anaqueous solution of ZnCl2 and hydrogen gas.

Another characteristic property of acids is their ability to change the color of veg-etable dyes such as litmus. Litmus is a mixture of blue dyes that turns red in thepresence of acid. Litmus has been used to test for acids for more than 300 years.

Bases also have characteristic properties. They taste bitter and often feelslippery. They change the color of litmus from red to blue, thereby reversingthe change in color that occurs when litmus comes in contact with an acid.Bases become less alkaline when they react with acids, and acids lose theircharacteristic sour taste and ability to dissolve metals when they are mixed withbases or alkalies.

11.2 The Arrhenius Definition of Acids and BasesIn 1887, Svante Arrhenius took a major step toward answering the important ques-tion, “What factors determine whether a compound is an acid or a base?” Arrhe-nius suggested that acids dissociate or ionize when they dissolve in water to giveH� ions and a corresponding negative ion. According to this model, hydrogenchloride is an acid because it dissociates, or ionizes, when it dissolves in waterto give H� and Cl� ions (Figure 11.1). This aqueous solution is known ashydrochloric acid and is often written as HCl(aq). It is important to recognize,however, that HCl is assumed to dissociate almost completely to form the H� andCl� ions when it dissolves in water.

Arrhenius argued that bases are compounds that dissociate in water to giveOH� ions and a positive ion. NaOH is an Arrhenius base because it dissociatesin water to give the hydroxide (OH�) and sodium (Na�) ions.

An Arrhenius acid therefore can be defined as any substance that ionizeswhen it dissolves in water to give the hydrogen ion, H�. An Arrhenius base isany substance that gives the hydroxide ion, OH�, when it dissolves in water. Arrhe-

NaOH(s) ¡

H2ONa+(aq) + OH-(aq)

HCl(g) ¡

H2OH+(aq) + Cl-(aq)

Zn(s) + 2 HCl(aq) ¡ ZnCl2(aq) + H2(g)

11.2 THE ARRHENIUS DEFINITION OF ACIDS AND BASES 469

Fig. 11.1 The Arrhenius modelassumes that HCl dissociates intoH� and Cl� ions when it dissolves inwater.

Cl−

Cl−

Cl−

H+

H+

H+

H2O


nius acids include compounds such as HCl, HCN, and H2SO4 that ionize in waterto give the H� ion. Arrhenius bases include ionic compounds that contain theOH� ion, such as NaOH, KOH, and Ca(OH)2.

11.3 The Brønsted–Lowry Definition of Acids and BasesIn 1923, Johannes Brønsted and Thomas Lowry independently proposed a morepowerful set of definitions of acids and bases. The Brønsted, or Brønsted–Lowry,model is based on the assumption that acids donate H� ions to another ion ormolecule, which acts as a base. According to this model, HCl doesn’t dissociatein water to form H� and Cl� ions. Instead, an H� ion is transferred from HCl toa water molecule to form an H3O� ion and a Cl� ion.

The H3O� ion is known as the hydronium ion. The Brønsted model of the reac-tion between HCl and water is shown in Figure 11.2.

HCl(aq) + H2O(l) ¡ H3O+(aq) + Cl-(aq)

470 CHAPTER 11 / ACIDS AND BASES

Fig. 11.2 The Brønsted model assumes that HCl moleculesdonate an H� ion to water molecules to form H3O� and Cl�

ions when HCl dissolves in water.

H3O+

H3O+

H3O+

Cl−

Cl−

Cl−

Because it is a proton, an H� ion is several orders of magnitude smallerthan the smallest atom. As a result, the charge on an isolated H� ion is distrib-uted over such a small surface area that the H� ion is attracted toward any sourceof negative charge that exists in the solution. Thus the instant that an H� ion iscreated in an aqueous solution, it bonds to the electronegative oxygen atom of awater molecule. The Brønsted model, in which H� ions are transferred from oneion or molecule to another, therefore seems more reasonable than the Arrheniusmodel, which assumes that H� ions exist in aqueous solution.

Even the Brønsted model is naive, however. Each H� ion that an acid donatesto water is actually bound in a complex of four neighboring water molecules, asshown in Figure 11.3. A more realistic formula for the substance produced whenan acid loses an H� ion in water is therefore H(H2O)4

�, or H9O4�. For practical

purposes, however, this substance can be represented as the H3O� ion.The reaction between HCl and water provides the basis for understanding

the definitions of a Brønsted acid and a Brønsted base. According to the Brøn-sted model, when HCl dissociates in water, HCl acts as an H� ion donor and H2Oacts as an H� ion acceptor.

donor acceptorH+ ionH+ ion


Fig. 11.3 Structure of the H(H2O)4�

ion formed when an acid reacts withwater. For practical purposes, the ioncan be thought of as an H3O� ion.

H

H

O

O

O

H

H H

H

H

OH

H

+

➤ CHECKPOINTClassify the following compounds asArrhenius acids or bases: HNO3,Mg(OH)2, CH3CO2H.

H+(aq) + CH3CO2-(aq)

CH3CO2H(aq) uvH2O

Mg2+(aq) + 2 OH-(aq)Mg(OH)2(s) uv

H2O

H+(aq) + NO3-(aq)

HNO3(aq) ¡

H2O


A Brønsted acid is therefore any substance (such as HCl) that can donate an H�

ion to a base. A Brønsted base is any substance (such as H2O) that can acceptan H� ion from an acid.

There are two ways of naming the H� ion. Some chemists call it a hydro-gen ion; others call it a proton. As a result, Brønsted acids are known as eitherhydrogen-ion donors or proton donors. Brønsted bases are hydrogen-ionacceptors or proton acceptors. The main body of this chapter will deal prima-rily with monoprotic acids (and the corresponding monoprotic bases) that havea single H� ion that they can donate. The Special Topics section at the end of thechapter will discuss more complex acids and bases.

From the perspective of the Brønsted model, reactions between acids andbases always involve the transfer of an H� ion from a proton donor to a protonacceptor. Acids can be uncharged molecules.

Acid Base

They can also be positive ions

Acid Base

or negative ions.

Acid Base

Brønsted bases can be identified from their Lewis structures. According tothe Brønsted model, a base is any ion or molecule that can accept a proton. Tounderstand the implications of this definition, look at how the prototypical base,the OH� ion, accepts a proton.

The only way to accept an H� ion is to form a covalent bond to it. In orderto form a covalent bond to an H� ion that has no valence electrons, the base mustprovide both of the electrons needed to form the covalent bond. Thus only com-pounds that have pairs of nonbonding valence electrons can act as H� ion accep-tors, or Brønsted bases. The following compounds, for example, can all act asBrønsted bases because they all contain nonbonding pairs of electrons.

The Brønsted model therefore includes within the category of bases any ionor molecule that contains one or more pairs of nonbonding valence electrons thatcan accept a proton. Many molecules and ions satisfy the definition of a Brønsted

O OOANHNH3 H

H

OQO OHH2O HO

BS

S

SO

SSOS

S S

CO32−

2−

O

CD G

H� � O OH�

OOO HO OHO

OO

H2PO4-(aq) + H2O(l) uv H3O

+(aq) + HPO42 -(aq)

NH4+(aq) + OH-(aq) uv NH3(aq) + H2O(l)

HCl(aq) + NH3(aq) ¡ Cl-(aq) + NH4+(aq)

11.3 THE BRØNSTED–LOWRY DEFINITION OF ACIDS AND BASES 471

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base. However, the following substances are not Brønsted bases because they haveno nonbonding valence electrons.

AO O

ACH HCH4

H

H

H2 OH H

AO O

ANHNH4

� H

H

H �


E x e r c i s e 1 1 . 1Identify the reactant that behaves as a Brønsted acid and the reactant thatbehaves as a Brønsted base in each of the following reactions.

(a)

(b)

(c)

Solution(a) acid: HF base: OH�

(b) acid: CH3CO2H base: H2O

(c) acid: HNO3 base: C6H5NH2

C6H5NH2(aq) + HNO3(aq) uv C6H5NH3+(aq) + NO3

-(aq)

CH3CO2H(aq) + H2O(l) uv CH3CO2-(aq) + H3O

+(aq)

HF(aq) + OH-(aq) uv H2O(l) + F-(aq)

11.4 Conjugate Acid–Base PairsAn important consequence of the Brønsted theory is the recognition that acids andbases are linked or coupled to form conjugate acid–base pairs. The term conju-gate comes from the Latin stem meaning “joined together” and refers to thingsthat are joined, particularly in pairs. It is therefore the perfect term to describethe relationship between Brønsted acids and bases.

We can write the formula for a generic acid as HA. When the acid donatesan H� ion to water, one product of the reaction is A�, which could act as an H�

ion acceptor, or Brønsted base. In other words, every time a Brønsted acid actsas a proton donor, it forms a conjugate base.

Acid Conjugate base

Conversely, if A� accepts an H� ion from water, it forms HA, which could actas a proton donor or Brønsted acid. Thus, every time a base gains an H� ion, itforms the conjugate acid.

Base Conjugate acid

A-(aq) + H2O(l) uv HA(aq) + OH-(aq)

HA(aq) + H2O(l) uv H3O+(aq) + A-(aq)


Acids and bases in the Brønsted model therefore exist as conjugate acid–base pairswhose formulas are related by the gain or loss of a hydrogen ion.

In order to determine whether a substance behaves as a Brønsted acid orbase, the substance must be examined in the context of the chemical reaction inwhich it participates. In the chemical equation shown above in which HA reactswith water, water behaves as a base. In the second equation in which A� reactswith water, water behaves as an acid.

Our use of the symbols HA and A� for a conjugate acid–base pair doesn’tmean that all acids are electrically neutral molecules or that all bases are nega-tive ions. It signifies only that the acid contains an H� ion that isn’t present inthe conjugate base. As noted earlier, Brønsted acids and bases can be electricallyneutral molecules, positive ions, or negative ions. Various Brønsted acids and theirconjugate bases are given in Table 11.1.

It is important to recognize that some compounds can be both a Brønstedacid and a Brønsted base. H2O and HSO4

�, for example, can be found in bothcolumns in Table 11.1. Water is the perfect example of this behavior because itsimultaneously acts as an acid and a base when it reacts with itself to form theH3O� and OH� ions.

The concept of conjugate acid–base pairs plays a vital role in explainingreactions between acids and bases. According to the Brønsted model, an acidalways reacts with a base to form the conjugate base and conjugate acid. Con-sider the following reaction, for example.

Acid Base Conjugate Conjugateacid base

In the course of this reaction, nitric acid donates an H� ion to form its conjugatebase, the nitrate ion (NO3

�). At the same time, ammonia acts as a base, accept-ing an H� ion to form its conjugate acid, the ammonium ion (NH4

�).The products of the reaction between nitric acid and ammonia are often

combined and written as an aqueous solution of an ionic compound, or salt.

Because the products of the reaction are neither as acidic as nitric acid nor asbasic as ammonia, the reaction is often called a neutralization reaction. Thisdoesn’t imply that the products have no acid or base properties. It only suggeststhat the products are less acidic and less basic than the starting materials.

Water is often one of the products of a neutralization reaction. Consider thereaction between formic acid and sodium hydroxide, for example.

Acid Base Conjugate Conjugateacid base

The salt produced in this reaction is sodium formate, NaHCO2. The formate ion,HCO2

�, is the conjugate base of formic acid, and water is the conjugate acid ofthe hydroxide ion in sodium hydroxide. The Na� is a spectator ion.

HCO2H(aq) + NaOH(aq) ¡ H2O(l) + Na+(aq) + HCO2-(aq)

HNO3(aq) + NH3(aq) ¡ NH4NO3(aq)

HNO3(aq) + NH3(aq) ¡ NH4+(aq) + NO3

-(aq)

H2O(l) + H2O(l) uv H3O+(aq) + OH-(aq)

11.4 CONJUGATE ACID–BASE PAIRS 473

Table 11.1 Typical Brønsted Acids andTheir Conjugate Bases

Acid Base

H3O� H2OH2O OH�

HCl Cl�

H2SO4 HSO4�

HSO4� SO4

2�

NH4� NH3

➤ CHECKPOINTPhosphoric acid, H3PO4, is an impor-tant component of many carbonatedbeverages. Write the chemical equationfor the dissociation of phosphoric acidin water and predict the chemical for-mula of its conjugate base. Aniline,C6H5NH2, is a base used in the manu-facture of dyes. Write the chemicalequation for the reaction of aniline withwater and predict the chemical formulaof its conjugate acid.


11.5 The Role of Water in the Brønsted ModelThe Lewis structure of water can help us understand why H3O� and OH� ionsplay such an important role in the chemistry of aqueous solutions. The Lewisstructure suggests that the hydrogen and oxygen atoms in a water molecule arebound together by sharing a pair of electrons. But oxygen (EN � 3.61) is muchmore electronegative than hydrogen (EN � 2.30), so the electrons in the covalentbonds are not shared equally by the hydrogen and oxygen atoms. The electronsare drawn toward the oxygen atom in the center of the molecule and away fromthe hydrogen atoms on either end. As a result, the water molecule is polar. Theoxygen atom carries a partial negative charge (�0.8), and each hydrogen atomcarries a partial positive charge (�0.4).

When neutral water molecules dissociate to form ions, positively chargedH3O� ions and negatively charged OH� ions are formed.

The opposite reaction can also occur: H3O� ions can combine with OH� ions toform neutral water molecules.

The fact that water molecules dissociate to form H3O� and OH� ions, which canthen recombine to form water molecules, is indicated by the following equation.

The pair of arrows that separate the “reactants” and the “products” of this reac-tion indicate that the reaction occurs in both directions.

2 H2O(l) uv H3O+(aq) + OH-(aq)

H3O+(aq) + OH-(aq) ¡ 2 H2O(l)

2 H2O(l) ¡ H3O+(aq) + OH-(aq)

δ−

δ+δ+

HH

O


E x e r c i s e 1 1 . 2Write the chemical equation for the reaction of chlorous acid, HClO2, withpotassium hydroxide. Identify the acid and base among the reactants and theconjugate acid and conjugate base formed in this reaction.

Solution

Reactants Products

acid: HClO2 conjugate base: ClO2�

base: KOH conjugate acid: H2O

HClO2(aq) + KOH(aq) ¡ H2O(l) + K+(aq) + ClO2-(aq)


The Brønsted model explains water’s role in acid–base reactions.

● Water dissociates to form ions by transferring an H� ion from one mole-cule acting as an acid to another molecule acting as a base.

Acid Base

● Acids react with water by donating an H� ion to a neutral water moleculeacting as a base to form the H3O� ion.

Acid Base

● Bases react with water by accepting an H� ion from a water molecule act-ing as an acid to form the OH� ion.

Base Acid

● Water molecules can act as intermediates in acid–base reactions by gainingH� ions from an acid.

Acid Base

and then losing the H� ions to a base.

Base Acid

Adding these reactions so that species that are on both sides of the equation can-cel gives the overall equation for the acid–base reaction.

It is important to recognize that acid–base reactions don’t have to occur inwater; they can also occur in the gas phase. Many students encounter this phe-nomenon in their first chemistry courses. When bottles of concentrated hydrochlo-ric acid and concentrated aqueous ammonia are both open at the same time, awhite cloud of ammonium chloride often forms in the air above these bottles.

11.6 To What Extent Does Water Dissociate to Form Ions?At 25�C, the density of water is 0.9971 g/cm3, or 0.9971 g/mL. The concentra-tion of pure water is therefore 55.35 M.

0.9971 g H2O

1 mL*

1000 mL1 L

*

1 mol H2O

18.015 g H2O= 55.35 mol H2O/L

HCl(g) + NH3(g) Δ NH4Cl(s)

CH3CO2H(aq) + NH3(aq) uv NH4+(aq) + CH3CO2

-(aq)

NH3(aq) + H3O+(aq) uv NH4

+(aq) + H2O(l)


+(aq)

NH3(aq) + H3O+(aq) uv NH4

+(aq) + H2O(l)


+(aq)

NH3(aq) + H2O(l) uv NH4+(aq) + OH-(aq)

HF(aq) + H2O(l) uv H3O+(aq) + F-(aq)

H2O(l) + H2O(l) uv H3O+(aq) + OH-(aq)

11.6 TO WHAT EXTENT DOES WATER DISSOCIATE TO FORM IONS? 475

HCl gas and NH3 gas that collectabove open bottles of concentratedsolutions of these compounds react inthe gas phase to form a cloud of solidNH4Cl.


The concentration of the H3O� and OH� ions formed by the dissociation of pureneutral H2O at that temperature is only 1.0 � 10�7 mol/L. The ratio of the con-centration of the H3O� (or OH�) ions to the concentration of the neutral H2Omolecules is therefore 1.8 � 10�9.

In other words, only about 2 parts per billion (ppb) of the water molecules dis-sociate into ions at room temperature.

It is difficult to imagine what 2 ppb means. One way to visualize this num-ber is to assume that 2500 letters appear on a typical page of this book. If typo-graphical errors occurred with a frequency of 2 ppb, the book would have to be400,000 pages long in order to contain two errors. Figure 11.4 shows a model of20 water molecules, one of which has dissociated to form H3O� and OH� ions.If this illustration were a high-resolution photograph of the structure of water, wewould encounter a pair of H3O� and OH� ions on the average of only once forevery 25 million such photographs.

The rate at which water molecules react to form the H3O� and OH� ionsin pure water is equal to the rate at which these ions combine to form a pair ofneutral water molecules.

This reaction is therefore said to be at equilibrium. The extent to which water dis-sociates to form ions can be described in terms of the following equilibrium con-stant expression.

As we have seen, the concentration of water at 25�C is 55.35 mol/L. Theconcentration of the H3O� and OH� ions, on the other hand, is only about 1.0 �10�7M. The concentration of H2O molecules is therefore so much larger than theconcentration of the H3O� and OH� ions that it remains effectively constant whenwater dissociates to form these ions. Chemists therefore rearrange the equilibriumconstant expression for the dissociation of water to give the following equation.

This effectively places two variables––the H3O� and OH� ion concentrations––onone side of the equation and two constants––Kc and [H2O]2––on the other sideof the equation. We then replace the terms on the left-hand side of the equationwith a constant known as the water dissociation equilibrium constant, Kw.

Because the H3O� and OH� concentrations in pure water at 25�C are each 1.0 � 10�7M, the value of Kw at that temperature is 1.0 � 10�14.

Equilibrium constants are determined experimentally and are subject tosome uncertainty. Therefore, equilibrium constants in this text are generally

Kw = [1.0 * 10- 7][1.0 * 10- 7] = 1.0 * 10- 14 (at 25°C)

Kw = [H3O+][OH-]

Kc * [H2O]2= [H3O

+][OH-]

Kc =

[H3O+][OH-]

[H2O]2


1.0 * 10- 7M H3O+

55.35 M H2O= 1.8 * 10- 9


Fig. 11.4 If this figure is thought of asa high-resolution snapshot of 20 H2Omolecules, one of which hasdissociated to form H3O� and OH�

ions, we would have to look at anaverage of 25 million such snapshotsbefore we encountered another pairof these ions.

H3O+

OH−

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reported to only one or at most two significant figures. The equilibrium constantexpression for the dissociation of water is still useful, however, for obtainingapproximate concentrations of the H3O� and OH� ions in water. Regardless ofthe source of the H3O� and OH� ions in water, the product of the concentrationsof these ions at equilibrium at 25�C is approximately 1.0 � 10�14.

What happens to the H3O� and OH� ion concentrations that come from thedissociation of water when we add a strong acid to water? Suppose, for example,that we add enough acid to a beaker of water to raise the H3O� concentration to0.010 M. Note that there are now two sources of H3O�. One source is the acidand the other is the dissociation of water.

According to Le Châtelier’s principle, the additional H3O� from the acid shoulddrive the equilibrium between water and its ions to the left, reducing the numberof H3O� and OH� ions from the dissociation of water. Adding an acid to watertherefore decreases the extent to which the water dissociates into ions.

This is another example of the “common-ion effect” encountered in the pre-vious chapter during the discussion of solubility product equilibria. The additionof a second source of an ion that is shared in common with an ion formed in anequilibrium that involves the dissociation of a substance decreases the amount ofdissociation. In Section 11.15 we will see that the “common-ion effect” alsooccurs in so-called buffer solutions.

Because the dissociation of water shifts to the left in the presence of an acid,the concentration of H3O� and OH� ions from the dissociation of water will besmaller than the 1.0 � 10�7M concentration in pure water. The total concentra-tion of the H3O� ion will be equal to the H3O� ion from dissociation of the acid(0.010 M) plus the H3O� ion from the dissociation of water (�1.0 � 10�7M).The H3O� ion concentration from the dissociation of water, however, is so smallthat it is negligible when compared to the amount from the acid. Therefore,when the system returns to equilibrium, the H3O� ion concentration is still about0.010 M. Furthermore, when the reaction returns to equilibrium, the product ofthe H3O� and OH� ion concentrations is once again approximated by Kw.

If the concentration of the H3O� ion is 0.010 M, the concentration of the OH�

ion when the system returns to equilibrium is only 1.0 � 10�12M.

Some of the H3O� ions in the solution come from the dissociation of water;others come from the acid that has been added to the water. All of the OH� ions,on the other hand, come from the dissociation of water. In pure water, dissocia-tion of H2O molecules gives us an OH� ion concentration of 1.0 � 10�7M. In theacid solution, the OH� ion concentration is 5 orders of magnitude smaller. Thismeans that adding enough acid to water to increase the concentration of the H3O�

ion to 0.010 M decreases the dissociation of water by a factor of about 100,000.Adding an acid to water therefore has an effect on the concentration of both

the H3O� and OH� ions. Some H3O� is already present as a result of the disso-ciation of water. Adding an acid to water increases the concentration of that ion.

[OH-] =

Kw

[H3O+]

=

1.0 * 10- 14

0.010= 1.0 * 10- 12 M

[H3O+][OH-] = 1.0 * 10- 14



11.6 TO WHAT EXTENT DOES WATER DISSOCIATE TO FORM IONS? 477


However, adding an acid to water decreases the extent to which water dissociatesand therefore leads to a significant decrease in the concentration of the OH� ion.

As might be expected, the opposite effect is observed when a base is addedto water. Because we are adding a base, the OH� ion concentration increases. Oncethe system returns to equilibrium, the product of the H3O� and OH� ion concen-trations is once again equal to Kw. The only way this can be achieved, of course,is by decreasing the concentration of the H3O� ion from the dissociation of water.

11.7 pH as a Measure of the Concentration of the H3O� IonThe range of concentrations of the H3O� and OH� ions in aqueous solutions isenormous. Typical solutions in the laboratory have concentrations of the H3O� orOH� ion as large as 1.0 M (or larger) and as small as 1 � 10�14 M. Perhaps thebest way to bring a range of 14 orders of magnitude into perspective is to notethat it is comparable to the difference between 10¢ and the national debt of morethan $10 trillion, or the difference between the radius of a gold atom and a dis-tance of 8 miles.

In 1909, the Danish biochemist S. P. L. Sørenson proposed a way of con-veniently describing a large range of concentrations. Sørenson worked at a labo-ratory set up by the Carlsberg Brewery to apply scientific methods to the studyof the fermentation reactions in brewing beer. Faced with the task of construct-ing graphs of the activity of the malt versus the H3O� ion concentration, Søren-son suggested using logarithmic mathematics to condense the range of H3O� andOH� concentrations to a more convenient scale. By definition, the logarithm ofa number is the power to which a base must be raised to obtain that number. Thelogarithm to the base 10 of 10�7, for example, is �7.

Because the concentrations of the H3O� and OH� ions in aqueous solutionsare usually smaller than 1 M, the logarithms of these concentrations are negativenumbers. Because he considered positive numbers more convenient, Sørenson sug-gested that the sign of the logarithm should be changed after it had been calculated.He therefore introduced the symbol p to indicate the negative of the logarithm of anumber. Thus, pH is the negative of the logarithm of the H3O� ion concentration.

Similarly, pOH is the negative of the logarithm of the OH� ion concentration.

It should be noted that the equations for pH and pOH are valid only fordilute solutions of acid or base in pure water. Most solutions used in the labora-tory do not meet this criterion, so the above equations can only be used to approx-imate the pH and pOH of real solutions. Since values from these calculations areapproximations, they are normally reported to no more than two digits after thedecimal. A pH meter, however, can accurately determine the pH of a solution totwo digits after the decimal. If the pH is given, the H3O� ion concentration canbe calculated with the following equation.

[H3O+] = 10- pH

pOH = - log[OH-]

pH = - log[H3O+]

log(10- 7) = -7


➤ CHECKPOINTWhen an acid is added to water, theequilibrium between neutral water mol-ecules and their ions shifts to the left,decreasing the amount of H3O� fromthe dissociation of water.

Does Kw still equal 1.0 � 10�14?



The advantage of the pH scale is illustrated in Figure 11.5, which shows thepossible combinations of H3O� ion and OH� concentrations in an aqueous solu-tion. In Figure 11.5a, the concentration of the OH� ion is plotted on the verticalaxis, and the concentration of the H3O� ion is plotted on the horizontal axis. Forthe plot to be a reasonable size, it has to be restricted to only 1 order of magni-tude of concentration.

The concept of pH compresses the range of H3O� concentrations onto a scalethat is much easier to handle. As the H3O� concentration decreases from roughly1 M to 10�14M, the pH of the solution increases from 0 to 14. The relationshipbetween the H3O� concentration, OH� concentration, and the pH of a solution isshown in Table 11.2. The advantage of the pH scale is illustrated in Figure 11.5.The only way to plot the concentration of the OH� ion versus the concentrationof the H3O� ion is to restrict the data to a narrow range of concentrations, asshown in Figure 11.5a. The entire range of data in Table 11.2, however, can beplotted in a graph of pH versus pOH, as shown in Figure 11.5b. Note that not only

11.7 pH AS A MEASURE OF THE CONCENTRATION OF THE H3O� ION 479

E x e r c i s e 1 1 . 3What is the pH of Pepsi Cola if the concentration of the H3O� ion in the solu-tion is 0.0035 M?

SolutionThe pH of a solution is the negative of the logarithm of the H3O� ion con-centration. The pH of this solution is therefore 2.5.

pH = - (-2.5) = 2.5

pH = - log(3.5 * 10- 3)

pH = - log3H3O+4

Fig. 11.5 (a) A small fraction of data showing the relationship between the H3O� and OH� ionconcentrations in aqueous solutions. Every point on the solid line represents a pair of H3O� andOH� concentrations when the solution is at equilibrium. (b) We can fit the entire range of H3O� and OH�

concentrations from Table 11.2 on a single graph by plotting pH versus pOH. Any point on the solid linecorresponds to a solution at equilibrium.

0 2

14

10

12

8

6

4

2

0

pH(−log[H3O+])

pOH

(−lo

g[O

H− ])

4 6 8 10 12 14

Acidic (pH < 7, pOH > 7)

Basic (pH > 7, pOH < 7)

Neutral (pH = pOH = 7)

1 × 10−7 2 × 10−7 4 × 10−7 6 × 10−7 8 × 10−7

8 × 10−7

6 × 10−7

4 × 10−7

2 × 10−7

1 × 10−7

[H3O+] (mol/L)

(a) (b)

[OH

− ] (

mol

/L)

Acidic solution ([H3O+] > [OH−])

Basic solution ([OH−] > [H3O+])

Neutral solution ([H3O+] = [OH−])


is the scale compressed by using pH and pOH data, but the shape of the graph haschanged from a sharp curve in Figure 11.5a to a straight line in Figure 11.5b.

The concentration of the H3O� ion in pure water at 25�C is 1.0 � 10�7 M.Thus the pH of pure water is 7.

Solutions for which the concentrations of the H3O� and OH� ions are equal aresaid to be neutral. Solutions in which the concentration of the H3O� ion is largerthan 1 � 10�7M at 25�C are described as acidic. Those in which the concentra-tion of the H3O� ion is smaller than 1 � 10�7M are basic. Thus, at 25�C whenthe pH of a solution is less than 7, the solution is acidic. When the pH is morethan 7, the solution is basic.

(at 25°C)

Measurements of pH in the laboratory were historically done with acid–baseindicators, which are weak acids or weak bases that change color when they gainor lose an H� ion. Acid–base indicators are still used in the laboratory for roughpH measurements. An example of an acid–base indicator is litmus, which turnspink in solutions whose pH is below 5 and turns blue when the pH is above 8.To a large extent, these indicators have been replaced by pH meters, which aremore accurate. The actual measuring device in a pH meter is an electrode thatconsists of a resin-filled tube with a thin glass bulb at one end. When the elec-trode is immersed in a solution to be measured, it produces an electric potentialthat is related to the H3O� ion concentration in the solution.

As we have seen, adding an acid to water increases the H3O� concentrationand decreases the OH� concentration. Adding a base does the opposite. Regard-less of what is added to water, however, the product of the concentrations of theions at equilibrium is 1.0 � 10�14 at 25�C.

[H3O+][OH-] = 1.0 * 10- 14

Basic: [H3O+] 6 1 * 10- 7M pH 7 7

Acidic: [H3O+] 7 1 * 10- 7M pH 6 7

pH = - log[H3O+] = - log(1.0 * 10- 7M) = 7.0


Table 11.2 Pairs of Equilibrium Concentrations of the H3O� and OH� Ions with theCorresponding pH Values in Water at 25�C

pH

1 0123 Acidic solution4567 Neutral solution89

1011 Basic solution1213

1 14 1 * 10-141 * 10-11 * 10-131 * 10-21 * 10-121 * 10-31 * 10-111 * 10-41 * 10-101 * 10-51 * 10-91 * 10-61 * 10-81 * 10-71 * 10-71 * 10-81 * 10-61 * 10-91 * 10-51 * 10-101 * 10-41 * 10-111 * 10-31 * 10-121 * 10-21 * 10-131 * 10-11 * 10-14

[OH-] (mol/L)[H3O+] (mol/L)

u

u


The relationship between the pH and pOH of an aqueous solution can be derivedby taking the logarithm of both sides of the Kw expression.

The log of the product of two numbers is equal to the sum of their logs. Thus thesum of the logs of the H3O

� and OH� ion concentrations is equal to the log of 10�14.

Both sides of the equation can now be multiplied by �1.

Substituting the definitions of pH and pOH into the equation gives the following result.

This equation can be used to convert from pH to pOH, and vice versa, for anyaqueous solution at 25�C, regardless of how much acid or base has been added tothe solution.

pH + pOH = 14.0

- log[H3O+] - log[OH-] = 14.0

log[H3O+] + log[OH-] = -14.0

log([H3O+][OH-]) = log(1.0 * 10- 14)

11.8 RELATIVE STRENGTHS OF ACIDS AND BASES 481

➤ CHECKPOINTDescribe what happens to the pH of asolution as the concentration of theH3O� ion in the solution increases.

E x e r c i s e 1 1 . 4The pH values of samples of lemon juice (pH 2.2) and vinegar (pH 2.5) arequite similar. Calculate the H3O� and OH� concentrations for both solutionsand compare them to one another.

Solution

Lemon juice:

Vinegar:

Note that the concentration of the H3O� ion in lemon juice is twice as largeas that in vinegar even though the two solutions have similar pH values.

[OH-] = 10-11.5= 3 * 10-12 M[H3O

+] = 10-2.5= 3 * 10-3 M

pOH = - log[OH-] = 11.5pH = - log[H3O+] = 2.5

[OH-] = 10-11.8= 2 * 10-12 M[H3O

+] = 10-2.2= 6 * 10-3 M

pOH = - log[OH-] = 11.8pH = - log[H3O+] = 2.2

11.8 Relative Strengths of Acids and BasesMany hardware stores sell muriatic acid––a 6 M solution of hydrochloric acid,HCl(aq)––to clean bricks and concrete and control the pH of swimming pools.Grocery stores sell vinegar, which is a 1 M solution of acetic acid, CH3CO2H.Although both substances are acids, you wouldn’t use muriatic acid in salad dress-ing, and vinegar is ineffective in cleaning bricks or concrete.

The difference between the two acids can be explored with the apparatusshown in Figure 11.6. Two metal electrodes connected to a source of electricityare dipped into a beaker that contains a solution of one of these acids. If the solu-tion conducts an electric current because of the presence of positive and negativeions, the flow of these ions between the electrodes completes the electric circuit


and the lightbulb starts to glow. The intensity with which the bulb glows dependson the ability of the solution to conduct a current. This, in turn, depends on theconcentration of the positive and negative ions in the solution that conducts theelectric current. The larger the number of positive and negative ions in the solu-tion, the brighter the lightbulb glows.

When the electrodes of this apparatus are immersed in pure water, the light-bulb doesn’t glow. This isn’t surprising because the concentrations of H3O� andOH� ions in pure water are very small, only about 10�7M at room temperature.When the electrodes are immersed in a 1 M solution of acetic acid, the bulb glows,but only dimly. When the electrodes are immersed in a 1 M solution of hydrochlo-ric acid, the lightbulb glows very brightly. Although the concentration of the acidin both solutions is the same, 1 M, hydrochloric acid contains far more ions thanthe equivalent acetic acid solution.

The difference between these acids is a result of differences in the abilitiesof the two acids to donate a proton to water. The hydrochloric acid in muriaticacid is a strong acid because HCl it is very good at transferring an H� ion to awater molecule. Virtually all of the HCl molecules in this solution react with waterto form H3O� and Cl� ions.

In a 0.1 M solution of hydrochloric acid, very few HCl molecules remain in solu-tion. Hydrochloric acid is therefore a strong acid indeed.

Note the difference between the ideas of a strong acid and a concentrated acid.A strong acid dissociates more or less completely when dissolved in water. A con-centrated acid is one that has a relatively large value of the molarity. It is possibleto have a concentrated solution of a weak acid. Consider concentrated orange juice,for example. A dilute solution of a strong acid will dissociate essentially 100% in



Fig. 11.6 The conductivity apparatus shows that even though the solution of HCl on the left and the solution of acetic acid on the righthave the same concentrations, the acid in the solution on the left with the brightly glowing bulb has dissociated to produce more ions.


water and is therefore a strong acid even though it has a relatively small concen-tration, as shown on the left in Figure 11.7.

Acetic acid (often abbreviated as HOAc) is a weak acid because it is notvery good at transferring H� ions to water. In a 1 M solution, less than 0.4% ofthe CH3CO2H molecules react with water to form H3O� and CH3CO2

� ions.

More than 99.6% of the acetic acid molecules remain undissociated, as shown onthe right in Figure 11.7, even though this is a more concentrated solution than thestrong acid shown on the left.

It would be useful to have a quantitative measure of the relative strengthsof acids to replace the labels strong and weak. The extent to which an acid dissoci-ates in water to produce the H3O� ion is described in terms of an acid-dissociationequilibrium constant, Ka. To understand the nature of this equilibrium constant,let’s assume that the reaction between an acid and water can be represented bythe following generic equation.

In other words, we will assume that some of the HA molecules react to formH3O� and A� ions, as shown in Figure 11.8. By convention, the equilibrium con-centrations of the ions in units of moles per liter are represented by the symbols[H3O�] and [A�]. The concentration of the undissociated HA molecules thatremain in solution is represented by the symbol [HA].

The equilibrium constant expression for the reaction between HA and waterwould be written as follows.

Most acid solutions are so dilute that the concentration of H2O at equilibrium iseffectively the same as before the acid was added. The equilibrium constant forthe reaction is therefore written as follows.

The result is an equilibrium constant for the equation known as the acid dissoci-ation equilibrium constant, Ka.

When a strong acid dissolves in water, it reacts extensively with water toform H3O� and A� ions. (Essentially 100% of the strong acid dissociates, leavingonly a very small residual concentration of HA molecules in solution.) The prod-uct of the concentrations of the H3O� and A� ions is therefore much larger than

Ka =

[H3O+][A-]

[HA]= Kc * [H2O]

[H3O+][A-]

[HA]= Kc * [H2O]

Kc =

[H3O+][A-]

[HA][H2O]


CH3CO2H(aq) + H2O(l) uv H3O+(aq) + CH3CO2

-(aq)

11.8 RELATIVE STRENGTHS OF ACIDS AND BASES 483

Fig. 11.8 Some, but not all, of the HAmolecules in a typical acid react withwater to form H3O� and A� ionswhen the acid dissolves in water. In astrong acid, there are very fewundissociated HA molecules. In aweak acid, most of the HA moleculesremain.

Fig. 11.7 (a) A dilute HCl solution. Essentiallyall of the HCl that dissolves in water dissociatesto form H3O� and Cl� ions. (b) A concentratedsolution of acetic acid. Only about 0.4% of theacetic acid molecules dissociate to form H3O�

and OAc� ions when HOAc dissolves in water.

A-

HAH3O+

H3O+

A-

HA

HA

HA

H3O+

H3O+

H3O+H3O+

H3O+

HOAcHOAcHOAcOAc–

HOAc

H3O+

HOAcHOAcHOAcHOAcHOAc

HOAcHOAcHOAcHOAcHOAcHOAc

Cl–

Cl–Cl–

Cl–Cl–

Cl–

Strong acids arealmost totallydissociated intoions in dilutesolutions.

Only about 0.4% ofthe HOAc moleculesin a 1.0 M solutionof acetic aciddissociate into ions.


the concentration of the HA molecules, so Ka for a strong acid is greater than 1.Hydrochloric acid, for example, has a Ka of roughly 1 � 106.

Weak acids, on the other hand, react only slightly with water. The productof the concentrations of the H3O� and A� ions is therefore smaller than the con-centration of the residual HA molecules. As a result, Ka for a weak acid is lessthan 1. Acetic acid, for example, has a Ka of only 1.8 � 10�5.

Ka can therefore be used to distinguish between strong acids and weak acids. Asa general rule,

A list of common acids and their acid dissociation constants, Ka, is givenin Table 11.3. A more complete list can be found in Table B.8 in Appendix B.Also included in Table B.8 are pKa values (pKa � �log Ka). Commonly acceptedKa values are reported for the strong acids. Experimental determination of Ka val-ues for strong acids is difficult, resulting in considerable variation in the valuesreported in the literature.

The ionization of bases in water can be described in a similar manner. A genericbase, B, reacting with water can be described by the following chemical equation.

Strict adherence to the rules for writing equilibrium constant expressions givesthe following result.

As we saw in our discussion of acids, the concentration of water at equilibrium isessentially constant, allowing the equilibrium expression to be written as follows.

The new equilibrium constant is known as the base ionization equilibrium con-stant, Kb. Just as Ka values describe the relative strengths of acids, Kb valuesdescribe the relative strengths of bases. The values of Kb for a limited number ofbases are given in Table B.9 in Appendix B. This table also lists the pKb valuesof these bases (pKb � �log Kb).

Strong bases are defined as ions or molecules that ionize more or less com-pletely in water to produce the OH� ion. Examples of strong bases include thefollowing:

● Group IA metal hydroxides: LiOH, NaOH, KOH, RbOH, and CsOH

NaOH(s) ¡ Na+(aq) + OH-(aq)

Kb =

[BH+][OH-]

[B]= Kc * [H2O]

Kc =

[BH+][OH-]

[B][H2O]

B(aq) + H2O(l) uv BH+(aq) + OH-(aq)

Weak acids: Ka 6 1

Strong acids: Ka 7 1

[H3O+][CH3CO2

-]

[CH3CO2H]= 1.8 * 10- 5

[H3O+][Cl-]

[HCl]= 1 * 106


➤ CHECKPOINTWrite the equilibrium expressions thatdescribe the Ka and the Kw of water.Show how the two expressions aredifferent.

Table 11.3 Common Acids and TheirAcid Dissociation EquilibriumConstant for the Loss ofOne Proton

Ka

Strong AcidsHI 3 � 109

HBr 1 � 109

HClO4 1 � 108

HCl 1 � 106

H2SO4 1 � 103

H3O� 55HNO3 28H2CrO4 9.6

Weak AcidsH3PO4 7.1 � 10�3

HF 7.2 � 10�4

Citric acid 7.5 � 10�4

CH3CO2H 1.8 � 10�5

H2S 1.0 � 10�7

H2CO3 4.5 � 10�7

H3BO3 7.3 � 10�10

H2O 1.8 � 10�16


● Group IIA metal hydroxides that are soluble or slightly soluble in water:Ca(OH)2 (slightly soluble), Sr(OH)2, and Ba(OH)2

● Soluble metal oxides: Li2O, Na2O, K2O, and CaO

In Section 11.7, pH was defined as the negative of the logarithm of theH3O� ion concentration.

Similarly, pOH was defined as the negative of the logarithm of the OH� ionconcentration.

Mathematicians would describe “p” as an operator. It implies the same operationwhenever it is invoked. In each case, we take the negative of the logarithm of thevariable to which it is applied. Thus we can also calculate pKa’s from the valueof the acid-dissociation equilibrium constant for a reaction

and pKb’s from the value of the base-ionization equilibrium constant.

11.9 Relative Strengths of Conjugate Acid–Base PairsThe relationship between the strength of an acid and its conjugate base can beunderstood by considering the implications of the fact that HCl is a strong acid.If it is a strong acid, HCl must be a good proton donor. HCl can only be a goodproton donor, however, if the Cl� ion is a poor proton acceptor. Thus, the Cl�

ion must be a weak base.

Strong acid very very weak base

Because HCl is a strong acid, it dissociates essentially completely, and there islittle tendency for the reaction to go in the reverse direction. Thus the chlorideanion is such a weak conjugate base that it has essentially no base properties.

Let’s now consider the relationship between the strength of the ammoniumion (NH4

�) and its conjugate base, ammonia (NH3). The NH4� ion is a weak

acid, and its conjugate base ammonia is a reasonably good base.

Weak acid Good base

These examples can be summarized in the form of the following general rules:

The stronger the acid, the weaker the conjugate base.

The stronger the base, the weaker the conjugate acid.

NH4+(aq) + H2O(l) uv H3O

+(aq) + NH3(aq)


pKb = - log(Kb)

pKa = - log(Ka)

pOH = - log[OH-]

pH = - log[H3O+]

Li2O(s) + H2O(l) ¡ 2 Li+(aq) + 2 OH-(aq)

Ca(OH)2(aq) ¡ Ca2 +(aq) + 2 OH-(aq)

11.9 RELATIVE STRENGTHS OF CONJUGATE ACID–BASE PAIRS 485

➤ CHECKPOINTWhich solution has the smallest pH, asolution with a low concentration of anacid with a small Ka or a solution witha low concentration of an acid with ahigh Ka?


11.10 Relative Strengths of Different Acids and BasesWe can summarize the relative strengths of Brønsted acids and bases by organ-izing the reactions in which a given acid is converted into its conjugate base, asshown in Table 11.4. The strongest acids are in the upper-left corner of this table,and the strongest bases are in the bottom-right corner.

This table can be used to compare the relative strengths of a pair of acids,to decide which is stronger. Consider HCl and the H3O� ion, for example.

The Ka values suggest that both are strong acids, but HCl is a stronger acid thanthe H3O� ion.

We can now understand why such a large proportion of the HCl moleculesin an aqueous solution react with water to form H3O� and Cl� ions. The Brønsted

H3O+ : Ka = 55

HCl: Ka = 1 * 106


Table 11.4 Relative Strengths of Typical Brønsted Acids and Bases

Acid Conjugate Base Ka

Best 3 � 109

Brønsted 1 � 108

acids 1 � 106

1 � 103

5 � 102

55289.6

1.2 � 10�2

1.1 � 10�2

7.1 � 10�3

7.2 � 10�4

1.8 � 10�5

4.5 � 10�7

1.0 � 10�7

2.9 � 10�8

6.3 � 10�8

7.3 � 10�10

5.6 � 10�10

4.7 � 10�11

4.2 � 10�13

1.3 � 10�13

1.8 � 10�16

1 � 10�18

1 � 10�25

1 � 10�33

1 � 10�35 Best1 � 10�44 Brønsted1 � 10�49 BasesCH4 uv H+

+ CH3-

CH2“CH2 uv H+

+ CH2“CH-

H2 uv H+

+ H-

NH3 uv H+

+ NH2-

HC‚CH uv H+

+ HC‚C-

CH3OH uv H+

+ CH3O-

H2O uv H+

+ OH-

HS- uv H+

+ S2 -

HPO42 - uv H+

+ PO43 -

HCO3- uv H+

+ CO32 -

NH4+ uv H+

+ NH3

H3BO3 uv H+

+ H2BO3-

H2PO4- uv H+

+ HPO42 -

HClO uv H+

+ ClO-

H2S uv H+

+ HS-

H2CO3 uv H+

+ HCO3-

CH3CO2H uv H+

+ CH3CO2-

HF uv H+

+ F-

H3PO4 uv H+

+ H2PO4-

HClO2 uv H+

+ ClO2-

HSO4- uv H+

+ SO42 -

H2CrO4 uv H+

+ HCrO42 -

HNO3 uv H+

+ NO3-

H3O+ uv H+

+ H2OHClO3 uv H+

+ ClO3-

H2SO4 uv H+

+ HSO4-

HCl uv H+

+ Cl-

HClO4 uv H+

+ ClO4-

HI uv H+

+ I-


model suggests that every acid–base reaction converts an acid into its conjugatebase and a base into its conjugate acid.

Acid Base Acid Base

There are two acids and two bases in the reaction. The stronger acid, however, ison the left side of the equation, which suggests that the reaction should proceedto the right. HCl should react with water to form the H3O� and Cl� ions.

Stronger acid Weaker acid

What about the two bases: H2O and the Cl� ion? The general rules givenin the previous section suggest that the stronger of a pair of acids must form theweaker of a pair of conjugate bases. The fact that HCl is a stronger acid than theH3O� ion implies that the Cl� ion is a weaker base than water.

Thus the equation for the reaction between HCl and water can be written as follows.

Stronger Stronger Weaker Weakeracid base acid base

It isn’t surprising that almost all of the HCl molecules in an aqueous solution reactwith water to give H3O� ions and Cl� ions. The stronger of a pair of acids shouldreact with the stronger of a pair of bases to form a weaker acid and a weaker base.

As a general rule, acids with Ka values greater than the H3O� ion (Ka � 55) canbe assumed to dissociate completely (100%) in water.

Let’s now look at the relative strengths of acetic acid and the H3O� ion.

These Ka values suggest that acetic acid is a much weaker acid than the H3O�

ion, which explains why acetic acid is a weak acid in water. Once again, the reac-tion between the acid and water converts the acid into its conjugate base and thebase into its conjugate acid.

Acid Base Acid Base

But in this case, the stronger acid and the stronger base are on the right side ofthe equation.

Weaker Weaker Stronger Strongeracid base acid base


-(aq)


-(aq)

H3O+ : Ka = 55

CH3CO2H: Ka = 1.8 * 10- 5


Base strength: Cl-

6 H2O

Acid strength: HCl 7 H3O+

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


11.10 RELATIVE STRENGTHS OF DIFFERENT ACIDS AND BASES 487


As a result, only a small percentage of the CH3CO2H molecules actually donatean H� ion to a water molecule to form the H3O� and CH3CO2

� ions. At equi-librium an acid–base reaction should lie predominantly on the side of the chem-ical equation that contains the weaker acid and weaker base.

Thus, Table 11.4 can be used to predict whether certain acid–base reactionsshould occur. Each base in this table is strong enough to deprotonate the acid inany line above it. According to this table, the acetate (CH3CO2

�) ion, for exam-ple, should be a strong enough base to remove a proton from nitric acid.


This table also predicts that ammonia (NH3) should be a strong enough base toremove both protons from sulfuric acid (H2SO4) to form ammonium sulfate.


The usefulness of Table 11.4 might best be illustrated by asking whether thefollowing reaction should occur as written.

The data in Table 11.4 suggest that H2PO4� ion is a stronger base than the SO4

2�

ion. These data also suggest that the HSO4� ion is a stronger acid than H3PO4.

Because the stronger acid and the stronger base are on the left side of the reac-tion, we expect the reaction to occur as written.

Stronger Stronger Weaker Weakerbase acid acid base

The data in the table do not indicate that H2PO4� is a strong acid, in the absolute

sense. These data merely indicate that H2PO4� is a stronger acid than HSO4

�.

H2PO4-(aq) + HSO4

-(aq) uv H3PO4(aq) + SO42 -(aq)

H2PO4-(aq) + HSO4

-(aq) uv H3PO4(aq) + SO42 -(aq)

H2SO4(aq) + 2 NH3(aq) uv 2 NH4+(aq) + SO4

2 -(aq)

HNO3(aq) + CH3CO2-(aq) uv CH3CO2H(aq) + NO3

-(aq)


E x e r c i s e 1 1 . 5The data in Table 11.4 are particularly useful for predicting some of theacid–base reactions that occur among organic compounds. Which of the fol-lowing acid–base reactions would you expect to occur?

(a) The reaction between methanol (CH3OH) and sodium hydride (NaH)

(b) The reaction between acetylene (HC‚CH) and sodium amide (NaNH2)

Solution(a) Sodium hydride is a source of the H� or hydride ion. According to Table

11.4, the H� ion should be a strong enough base to remove the acidic pro-ton from methanol.

Stronger Stronger Weaker Weakeracid base base acid

CH3OH + H- uv CH3O-

+ H2


The magnitude of Ka can also be used to explain why some compounds thatare potentially Brønsted acids or bases don’t act like acids or bases when they dis-solve in water. As long as Ka for the acid is significantly larger than the value ofKa for water (1.8 � 10�16), the acid will ionize to some extent when it dissolvesin water. As the Ka value for the acid approaches the Ka for water, the compoundbecomes more like water in its acidity. Although it is still potentially a Brønstedacid, it is so weak that we may be unable to detect the acidity in aqueous solution.

Measurements of the pH of dilute solutions are good indicators of the rel-ative strengths of acids and bases. Experimental values of the pH of 0.10 Msolutions of a number of common acids and bases are given in Table 11.5. Notethat the pH of an 0.1 M solution of HCl is 1.1, not 1.0 as would be expectedfrom the concentration of the solution. That can be understood by noting thatHCl does not dissociate as completely as our model, so far, would predict.

THE LEVELING EFFECT OF WATER

Because all strong acids dissociate in water to produce the same H3O� ion, theyall seem to have the same strength when dissolved in water, regardless of the

11.10 RELATIVE STRENGTHS OF DIFFERENT ACIDS AND BASES 489

(b) Sodium amide is a source of the or amide ion, which is a strongenough base to remove a proton from acetylene to form the acetylide ion.

Stronger Stronger Weaker Weakeracid base base acid

HC‚CH + NH2- uv HC‚C-

+ NH3

NH -

2

➤ CHECKPOINTHOBr is a weaker acid than HOCl.Which conjugate base, OBr� or OCl�,is stronger?

Table 11.5 pH of 0.10 M Solutions of Common Acids and Bases

Compound pH

HCl (hydrochloric acid) 1.1H2SO4 (sulfuric acid) 1.2NaHSO4 (sodium hydrogen sulfate) 1.4H2SO3 (sulfurous acid) 1.5H3PO4 (phosphoric acid) 1.5HF (hydrofluoric acid) 2.1CH3CO2H (acetic acid) 2.9H2CO3 (carbonic acid) 3.8 (saturated solution)H2S (hydrogen sulfide) 4.1NaH2PO4 (sodium dihydrogen phosphate) 4.4NH4Cl (ammonium chloride) 4.6HCN (hydrocyanic acid) 5.1NaCl (sodium chloride) 6.4H2O (freshly boiled distilled water) 7.0NaCH3CO2 (sodium acetate) 8.4NaHCO3 (sodium hydrogen carbonate) 8.4Na2HPO4 (sodium hydrogen phosphate) 9.3Na2SO3 (sodium sulfite) 9.8NaCN (sodium cyanide) 11.0NH3 (aqueous ammonia) 11.1Na2CO3 (sodium carbonate) 11.6Na3PO4 (sodium phosphate) 12.0NaOH (sodium hydroxide, lye) 13.0


value of Ka. This phenomenon is known as the leveling effect of water––the ten-dency of water to limit the strengths of strong acids and bases. Close to 100% ofthe HCl molecules in hydrochloric acid react with water to form H3O� and Cl�

ions, for example. Thus a 0.10 M solution of HCl produces approximately 0.10 MH3O�. In the same manner, a 0.10 M solution of the strong acid HClO4 will alsoproduce 0.10 M H3O�.

As a result, there is no difference between the acidity of these solutions eventhough HClO4 (Ka � 1 � 108) is supposedly a stronger acid than HCl (Ka � 1 �106). The acidity of both solutions is limited by the strength of the acid (H3O�)formed when water molecules accept an H� ion.

A similar phenomenon occurs in solutions of strong bases. Once the basereacts with water to form the OH� ion, the solution cannot become any morebasic. The strength of a strong base is limited by the strength of the base (OH�)formed when water molecules lose an H� ion.

11.11 Relationship of Structure to Relative Strengths of Acids and BasesSeveral factors influence the probability of having a heart attack. Heart attacks occurmore often among those who smoke than among those who don’t, among the eld-erly more often than the young, among those who are overweight more often thanamong those who aren’t, and among those who never exercise more often thanamong those who exercise regularly. It is possible to sort out the relative importanceof these factors, however, by trying to keep the other factors as constant as possible.

The same approach can be used to identify the factors that control the rela-tive strengths of acids and bases. Three factors affect the acidity of the X¬ H bondin an acid: (1) the polarity of the bond, (2) the size of the atom X, and (3) thecharge on the ion or molecule. A fourth factor has to be considered to understandthe acidity of the X¬OH group in a family of compounds known as “oxyacids.”

THE POLARITY OF THE X¬H BOND

When all other factors are kept constant, acids become stronger as the X¬ H bondbecomes more polar. Consider the following electrically neutral compounds, forexample, which become more acidic as the difference between the electronega-tivities of the X and H atoms (¢EN) increases and therefore the partial charge onthe hydrogen atom in these compounds (�H) increases. HF is the strongest of thefour acids, and CH4 is the weakest.

Ka EN H

HF 7.2 � 10�4 1.9 �0.29H2O 1.8 � 10�16 1.3 �0.22NH3 1 � 10�33 0.8 �0.14CH4 1 � 10�49 0.2 �0.05

In all four of the compounds, the size of the central atom, X, to which theH is bonded is about the same. However, there are large differences in ¢EN amongthe four compounds. The molecule containing the central atom with the largest

d¢

HClO4(aq) + H2O(l) ¡ H3O+(aq) + ClO4

-(aq)




electronegativity, F, also has the H with the most positive partial charge, 0.29. Asthe central atom becomes more electronegative, electron density is pulled awayfrom the hydrogen, resulting in a more positive partial charge on the hydrogenand a more polar H¬X bond.

When these compounds act as acids, the H¬X bond is broken to form H�

and X� ions. From the preceding data, it is apparent that the more polar this bond,the easier it is to form ions, and therefore the larger the value of Ka. Thus, allelse being equal, acids become stronger as the H¬X bond becomes more polar.

The data in Table 11.5 illustrate the magnitude of this effect. A 0.10 M HFsolution is moderately acidic. Water is much less acidic, and the acidity of ammo-nia is so low that the chemistry of aqueous solutions of NH3 is dominated by theability of NH3 to act as a base.

THE SIZE OF THE X ATOM

We might expect HF, HCl, HBr, and HI to become weaker acids as we go downthe column of the periodic table because the X¬H bond becomes less polar.Experimentally, we find the opposite trend. The acids actually become strongeras we go down the column.

This can be explained by recognizing that the size of the X atom also influ-ences the acidity of the X¬H bond. Acids become stronger as the X¬H bondbecomes weaker, and bonds generally become weaker as the atoms get larger (Fig-ure 11.9). The Ka data for HF, HCl, HBr, and HI reflect the fact that the magni-tude of the enthalpy of atom combination ( ) of these compounds decreasesas the X atom becomes larger. This indicates a weaker X¬H bond, which meansit will be easier to break this bond to lose an H� ion.

Ka (kJ/molrxn) �H X¬H Bond Length (nm)

HF 7.2 � 10�4 �567.7 �0.29 0.101HCl 1 � 106 �431.6 �0.11 0.136HBr 1 � 109 �365.9 �0.08 0.151HI 3 � 109 �298.0 �0.01 0.170

THE CHARGE ON THE ACID OR BASE

The charge on a molecule or an ion can influence its ability to act as an acid ora base. This is clearly shown when the pH of 0.1 M solutions of H3PO4 and theH2PO4

�, HPO42�, and PO4

3� ions are compared.

Compounds become less acidic and more basic as the negative charge increases.

Basicity: H2PO4-

6 HPO42 -

6 PO43 -

Acidity: H3PO4 7 H2PO4-

7 HPO42 -

PO43 - pH = 12.0

HPO42 - pH = 9.3

H2PO4- pH = 4.4

H3PO4 pH = 1.5

¢H°ac

¢H°ac

0.10 M NH3 pH = 11.1

H2O pH = 7

0.10 M HF pH = 2.1

11.11 RELATIONSHIP OF STRUCTURE TO RELATIVE STRENGTHS OF ACIDS AND BASES 491

➤ CHECKPOINTWhich acid is stronger, H2O or H2S?Why?

Fig. 11.9 HI is a much stronger acidthan HF because of the relative sizesof the fluorine and iodine atoms. Thecovalent radius of iodine is more thantwice as large as that of fluorine,which means that the HI bondstrength is less than that of HF.

F IH H


The decrease in acidity with increasing negative charge occurs because it ismuch easier to remove a positive H� ion from a neutral H3PO4 molecule than itis to remove an H� ion from a negatively charged H2PO4

� ion. It is even harderto remove an H� ion from a negatively charged HPO4

2� ion.The increase in basicity with increasing negative charge can be similarly

explained. There is a strong force of attraction between the negative charge on aPO4

3� ion and the positive charge on an H� ion. As a result, PO43� is a reason-

ably good base. The force of attraction between the HPO42� ion and an H� ion

is smaller because HPO42� carries a smaller charge. The HPO4

2� ion is thereforea weaker base than PO4

3�. The charge on the H2PO4� ion is even smaller than

that of HPO42�, so this ion is an even weaker base than HPO4

2�.

RELATIVE STRENGTHS OF OXYACIDS

There is no difference in either the size of the atom bonded to hydrogen or thecharge on the acid when we compare the O¬H bonds in oxyacids of the sameelement, such as H2SO4 and H2SO3 or HNO3 and HNO2 (Figure 11.10), yet thereis a significant difference in the strengths of the acids. Consider the following Ka

data, for example.

The acidity of the oxyacids increases significantly as the number of oxygen atomsattached to the central atom increases. H2SO4 is a much stronger acid than H2SO3,and HNO3 is a much stronger acid than HNO2.

This trend is easiest to see in the four oxyacids of chlorine. At first glance,we might expect these acids to have more or less the same acidity. The sameO¬H bond is broken in each case, and the O atom in this bond is bound to a Clatom in all four compounds. Oxygen, however, is second only to fluorine in elec-tronegativity, and it tends to draw electron density toward itself. As more oxygenatoms are added to the central atom, more electron density is drawn toward theoxygens. This draws electrons away from the O¬H bond in the acids, as shownin Figure 11.11. Pulling electron density away from the hydrogen results in thepartial charge on the hydrogen, �H, becoming more positive, as can be seen in theaccompanying table.

Number of Ka Oxygen Atoms H Name of Oxyacid

HOCl 2.9 � 10�8 1 �0.24 Hypochlorous acidHOClO 1.1 � 10�2 2 �0.29 Chlorous acidHOClO2 5.0 � 102 3 �0.31 Chloric acidHOClO3 1 � 108 4 �0.33 Perchloric acid

d

H2SO3 Ka = 1.7 * 10- 2 HNO2 Ka = 5.1 * 10- 4

H2SO4 Ka = 1 * 103 HNO3 Ka = 28


Fig. 11.10 Lewis structures of HNO3

and HNO2.

N

O

O

O

HNO3

H

N O

O

HNO2

H

Fig. 11.11 The large differencebetween the Ka values for HOCl andHOClO3 is the result of the additionof oxygen atoms to the chlorine. Theelectronegative oxygen atoms drawelectron density toward themselves,resulting in electron density beingpulled away from the hydrogen andtoward the oxygen in the O¬H bond.The net result is an increase in thepolarity of the O¬H bond, whichleads to an increase in the acidity ofthe compound.

Hδ+

δ–O

OCl

O

O

E x e r c i s e 1 1 . 6For each of the following pairs, predict which compound is the stronger acidand explain why.

(a) H2O or NH3

(b) NH4� or NH3


The relative strengths of Brønsted bases can be predicted from the relativestrengths of their conjugate acids, combined with the general rule that the strongerof a pair of acids always has the weaker conjugate base.

11.11 RELATIONSHIP OF STRUCTURE TO RELATIVE STRENGTHS OF ACIDS AND BASES 493

(c) NH3 or PH3

(d) H2SeO4 or H2SeO3

Solution(a) Oxygen and nitrogen atoms are about the same size, and H2O and NH3

are both neutral molecules. The difference between these compounds liesin the polarity of the X¬H bond. Because oxygen is more electronega-tive, the O¬H bond is more polar, and H2O (Ka � 1.8 � 10�16) is a muchstronger acid than NH3 (Ka � 1 � 10�33).

(b) The difference between these compounds is the charge on the ion. TheNH4

� ion is a stronger acid than NH3 because it is easier to remove an H�

ion from a positively charged NH4� ion than from a neutral NH3 molecule.

(c) PH3 is a stronger acid than NH3 because the phosphorous atom is largerthan the nitrogen atom. As a result, the P¬H bond is longer than theN¬H bond. Longer bonds are often weaker bonds and therefore more eas-ily broken. As a result, PH3 is more acidic than NH3.

(d) These compounds are both oxyacids that differ in the number of oxygenatoms attached to the selenium atom. More electron density is drawn awayfrom the O¬H bond in H2SeO4 because of the extra oxygen atom on theSe atom in the center of the molecule. Thus H2SeO4 is the stronger acid.

➤ CHECKPOINTWhich of the following acids isstronger, HOCl or HOI? Why?

E x e r c i s e 1 1 . 7For each of the following pairs of compounds, predict which compound is theweaker base.

(a) OH� or NH2�

(b) NH3 or NH2�

(c) NH2� or PH2

�

(d) NO3� or NO2

�

Solution(a) The OH� ion is the conjugate base of water, and the NH2

� ion is the con-jugate base of ammonia. Because H2O is a stronger acid than NH3, theOH� ion must be a weaker base than the NH2

� ion.

(b) Because it carries a negative charge, the NH2� ion is a stronger base than

a neutral NH3 molecule.

(c) PH3 is a stronger acid than NH3, which means the PH2� ion must be a

weaker base than the NH2� ion.

(d) HNO3 is a stronger acid than HNO2, which means that the NO3� ion is

the weaker base.


11.12 Strong Acid pH CalculationsThe simplest acid–base equilibria are those in which a strong acid (or base) isdissolved in water. Consider the calculation of the pH of a solution formed byadding a single drop of 2 M hydrochloric acid to 100 mL of water, for example.The key to this calculation is remembering that HCl is a strong acid (Ka � 106)and that acids as strong as HCl can be assumed to dissociate more or less com-pletely in water.

The H3O� ion concentration at equilibrium can therefore be assumed to be essen-tially the same as the initial concentration of the acid. The details of this calculationare illustrated in the drawing in the margin.

A useful rule of thumb states that there are about 20 drops in each milli-liter. One drop of 2 M HCl therefore has a volume of 0.05 mL, or 5 � 10�5 L.Now that we know the volume of the acid added to the beaker of water and theconcentration of the acid, we can calculate the number of moles of HCl that wereadded to the water.

Once we know the number of moles of HCl that have been added to the beakerand the volume of water in the beaker, we can calculate the concentration of thesolution prepared by adding one drop of 2 M HCl to 100 mL of water.

According to this calculation, the initial concentration of HCl before it dissoci-ates into ions is 1 � 10�3 M.

If we assume that the acid dissociates completely, the H3O� concentrationat equilibrium is 1 � 10�3 M. We therefore expect the pH of the solution pre-pared by adding one drop of 2 M HCl to 100 mL of water to be 3.0.

Calculations such as this give results that are close to experimental values forsolutions that are reasonably dilute.

11.13 Weak Acid pH CalculationsUnlike strong acids, which dissociate almost completely, weak acids dissociateonly slightly in water. An aqueous solution of a weak acid therefore not only con-tains the dissociated (H3O� � A�) form of the acid but also significant amountsof the undissociated (HA) form of the acid. Equilibrium problems involving weakacids such as acetic acid (CH3CO2H) can be solved by applying the techniquesdeveloped in Chapter 10.

pH = - log(1 * 10- 3) = - (-3.0) = 3.0

pH = - log[H3O+]

1 * 10- 4 mol HCl0.100 L

= 1 * 10- 3M HCl

2 mol HCl1 L

* 5 * 10- 5 L = 1 * 10- 4 mol HCl



1 drop (0.05 mL) 2 M HCl (Ka = 106)

100 mL H2O

What is the pH?

HCl + H2O H3O+ + Cl–

100%

➤ CHECKPOINTWhen a strong acid is added to water,why can the pH usually be determineddirectly from the quantity of strong acidadded?


As noted in Section 11.8, the concentration of the H3O� ion in an aqueoussolution of an acid depends on two factors:

● The strength of the acid, as represented by the value of Ka for the acid.

● The concentration of the acid, in moles per liter.

These two factors were illustrated in Figure 11.7. The beaker on the left showsa strong acid in which all of the acid molecules have reacted with water to pro-duce H3O� ions. However, the concentration of this solution is much smallerthan the concentration of the weak acid shown in the beaker on the right. Thebeaker on the right shows a solution that is much more concentrated because itcontains more moles of the acid per liter of solution. However, because thissolution is a weak acid, only a small fraction of the acid molecules have com-bined with water to form H3O� ions. The importance of considering both thestrength of an acid and its concentration can be understood by comparing an0.0010 M HCl solution with a solution that is 1.0 M in acetic acid. HCl is amuch stronger acid (Ka � 106), but the concentration of the acetic acid solu-tion is so much larger that there are actually more H3O� ions at equilibrium inthe acetic acid solution.

Consider, for example, the dissociation of acetic acid, CH3CO2H, in water. Theformula of acetic acid is often abbreviated as HOAc, and that of its conjugate base,the acetate ion (CH3CO2

�), as OAc�. We can start the calculation of the H3O�,OAc�, and HOAc concentrations at equilibrium in a 0.10 M solution of acetic acidin water by building a representation of what we know about the reaction.

Initial: 0.10 M 0Equilibrium: ? ? ?

The relevant information for this calculation is summarized in the drawing shownin the margin.

We now compare the initial reaction quotient (Qa) with the equilibrium con-stant (Ka) for the reaction. In this case, the initial reaction quotient is smaller thanthe equilibrium constant.

We therefore conclude that the reaction must shift to the right to reach equilibrium.The balanced equation for the reaction states that we get one H3O� ion and

one OAc� ion each time an HOAc molecule dissociates. This allows us to writeequations for the equilibrium concentrations of the three components of the reac-tion. Let’s start by writing an expression that describes the concentration of aceticacid when the reaction comes to equilibrium.

The concentration of acetic acid at equilibrium will be equal to the initialconcentration of the acid, 0.10 M, minus the change in the concentration of thisacid, ¢C.

Because there is very little H3O� ion in pure water, we can assume that the initialconcentration of this acid is effectively zero ( 0). Because we get one H3O� ionand one OAc� ion every time an acetic acid molecule dissociates, the concentrations

L

[HOAc] = 0.10 M - ¢C

Qa =

(H3O+)(OAc-)

(HOAc)=

(0)(0)

0.10= 0 6 Ka = 1.8 * 10- 5

L0

HOAc(aq) + H2O(l) uv H3O+(aq) + OAc-(aq) Ka = 1.8 * 10- 5

11.13 WEAK ACID pH CALCULATIONS 495

A ball-and-stick model of thestructure of acetic acid, CH3CO2H.

Ka = 1.8 × 10–5

(weak acid ⇒ΔC << 0.10 M)

0.10 M HOAc

Δ(HOAc) = Δ(H3O+) = Δ(OAc–)

HOAc + H2OH3O+ + OAc–


of these ions at equilibrium are equal to the change in the concentration of aceticacid as this acid dissociates.

We can therefore represent the problem that needs to be solved as follows.

Initial: 0.10 M 0 0Equilibrium: 0.10 � ¢C ¢C ¢C

Substituting what we know about the system at equilibrium into the Ka

expression gives the following equation.

Although we could rearrange the equation and solve it with the quadratic formula,let’s test the assumption that acetic acid is a relatively weak acid. In other words,let’s assume that ¢C is very small compared with the initial concentration ofacetic acid. If this assumption is true, then subtracting ¢C from the initial con-centration of the acid (0.10 M) will not significantly reduce that value. Thisassumption gives us the following approximate equation:

Or

And therefore

We then solve the approximate equation for the value of ¢C.

We can now test our assumption. Is ¢C small enough to be ignored in thisproblem? Yes, because it is less than 5% of the initial concentration of acetic acid.

Our assumption is valid because subtracting 0.0013 from 0.10 would give us 0.10,to two significant figures. We can therefore use the value of ¢C obtained from thiscalculation to estimate the equilibrium concentrations of H3O�, OAc�, and HOAc.

[H3O+] = [OAc-] = ¢C � 0.0013 M

[HOAc] = 0.10 M - ¢C � 0.10 M

0.00130.10

* 100% = 1.3%

¢C L 0.0013 M

[¢C]2L 1.8 * 10- 6

[¢C]2

[0.10]L 1.8 * 10- 5

[¢C][¢C]

[0.10]L 1.8 * 10- 5

Ka =

[H3O+][OAc-]

[HOAc]=

[¢C][¢C]

[0.10 - ¢C]= 1.8 * 10- 5


[OAc-] = ¢C

[H3O+] = ¢C



The pH of a 0.10 M HOAc is therefore 2.9.

Two assumptions were made in the above calculation.

● We assumed that the amount of acid that dissociates is small compared withthe initial concentration of the acid.

● We assumed that enough acid dissociates to allow us to ignore the con-tribution to the total H3O� ion concentration from the dissociation ofwater.

In this case, both assumptions are valid. Only about 1.3% of the acetic acidmolecules actually dissociate at equilibrium. But this gives us an H3O� con-centration that is four orders of magnitude larger than the concentration of thision in pure water. Thus we can legitimately ignore the dissociation of water.

The assumptions made in this calculation work for acids that are neithertoo strong nor too weak. Fortunately, many acids fall into that category. If theacid is too strong to assume that ¢C is small compared with the initial con-centration, the problem can be solved with the quadratic equation. If the acidis too weak or too dilute to ignore the dissociation of water, variations in thepH of the water from one sample to another because of impurities dissolvedin the water are probably as significant as the effect on the pH of adding thevery weak acid.

To illustrate the importance of paying attention to the assumptions thatare made in pH calculations, consider the task of calculating the pH of a solu-tion prepared by dissolving 1.0 � 10�8 mol of a strong acid, such ashydrochloric acid, in a liter of water. It is tempting to assume that the acid dis-sociates completely in water to give a solution with the following H3O� ionconcentration.

If this concentration is used with the definition of pH, we get the following result.

The result of this calculation, however, does not make sense. It is impossible tocreate a basic solution (pH > 7) by adding an acid to water!

The mistake, of course, was ignoring the concentration of the H3O� ionfrom the dissociation of water. In pure water, we start with 1.0 � 10�7 M H3O�

ion. If we add 10�8 mol/L from the strong acid, the total concentration of H3O�

from both sources will be about 1.1 � 10�7M. Thus the pH of the solution isactually slightly less than 7.

Two factors control the concentration of the H3O� ion in a solution of aweak acid: (1) the acid dissociation equilibrium constant, Ka, of the acid and (2)the concentration of the solution. When solutions of equivalent concentrations arecompared, the amount of the H3O� ion at equilibrium increases as the value ofKa of the acid increases, as shown in Exercise 11.8.

pH = - log[H3O+] = - log(1.1 * 10- 7M) = 6.96

pH = - log[H3O+] = - log(1.0 * 10- 8M) = 8.0

[H3O+] = 1.0 * 10- 8M

pH = - log[H3O+] = 2.9


➤ CHECKPOINTAssume that you have two solutions.Solution 1 has a low concentration ofan acid with a large Ka. Solution 2 has ahigh concentration of an acid with asmall Ka. Do you have enough informa-tion to determine which solution hasthe largest pH? Explain your answer.



E x e r c i s e 1 1 . 8Determine the approximate pH of 0.10 M solutions of the following acids.

(a) Hypochlorous acid, HOCl, Ka � 2.9 � 10�8

(b) Hypobromous acid, HOBr, Ka � 2.4 � 10�9

(c) Hypoiodous acid, HOI, Ka � 2.3 � 10�11

SolutionThe first calculation can be set up as follows.

Initial: 0Equilibrium:

Substituting this information into the Ka expression gives the following equation.

Because the equilibrium constant for the reaction is relatively small, we cantry the assumption that ¢C is small compared with the initial concentration ofthe acid.

Solving the equation for ¢C gives the following result.

Both assumptions made in the calculation are legitimate. ¢C is small com-pared with the initial concentration of HOCl, but it is several orders of magni-tude larger than the H3O� ion concentration from the dissociation of water. Wecan therefore use this value of ¢C to calculate the pH of the solution.

Repeating the calculation with the same assumptions for hypobromous andhypoiodous acid gives the following results.

As expected, the H3O� ion concentration at equilibrium––and therefore the pHof the solution––depends on the value of Ka for the acid. The H3O� ion con-centration increases, and the pH of the solution decreases as the value of Ka

becomes larger.

HOI [H3O+] L 1.5 * 10- 6M pH L 5.8

HOBr [H3O+] L 1.5 * 10- 5M pH L 4.8

HOCl [H3O+] L 5.4 * 10- 5M pH L 4.3

pH = - log[H3O+] = - log(5.4 * 10- 5M) = 4.3

¢C = 5.4 * 10- 5M

[¢C][¢C]

[0.10]L 2.9 * 10- 8

Ka =

[H3O+][OCl-]

[HOCl]=

[¢C][¢C]

[0.10 - ¢C]= 2.9 * 10- 8

¢C¢C0.10 � ¢C�00.10 M

HOCl(aq) + H2O(l) uv H3O+(aq) + OCl-(aq)


The next exercise shows how the H3O� ion concentration at equilibriumdepends on the initial concentration of the acid.


E x e r c i s e 1 1 . 9Determine the H3O� ion concentration and pH of acetic acid solutions withthe following concentrations: 1.0 M, 0.10 M, and 0.010 M.

SolutionThe first calculation can be set up as follows.

Initial: 1.0 M L0 0Equilibrium:

Substituting this information into the Ka expression gives the following result.

We now assume that ¢C is small compared with the initial concentration ofacetic acid,

and then solve the equation for an approximate value of ¢C.

Once again, both of the characteristic assumptions of weak acid equi-librium calculations are legitimate in this case. ¢C is small compared with theinitial concentration of acid but large compared with the concentration ofH3O� ion from the dissociation of water. We can therefore use this value of¢C to calculate the pH of the solution.

Repeating the calculation for the different initial concentrations gives the fol-lowing results.

The concentration of the H3O� ion in an aqueous solution of a weak acid grad-ually increases and the pH of the solution decreases as the solution becomesmore concentrated.

0.010 M HOAc [H3O+] L 4.2 * 10- 4M pH L 3.4

0.10 M HOAc [H3O+] L 1.3 * 10- 3M pH L 2.9

1.0 M HOAc [H3O+] L 4.2 * 10- 3M pH L 2.4

pH = - log[H3O+] = - log(4.2 * 10- 3M) = 2.4

¢C L 0.0042 M

[¢C][¢C]

[1.0]L 1.8 * 10- 5

Ka =

[H3O+][OAc-]

[HOAc]=

[¢C][¢C]

[1.0 - ¢C]= 1.8 * 10- 5

¢C¢C1.0 - ¢C

HOAc(aq) + H2O(l) uv H3O+(aq) + OAc-aq Ka = 1.8 * 10- 5

The vinegar that is often mixed withvegetable oils as the basis of saladdressings is an acetic acid solutionthat has a concentration ofapproximately 1 M.

➤ CHECKPOINTWhich of the acetic acid solutions inExercise 11.9 is the most acidic?


11.14 Base pH CalculationsWith minor modifications, the techniques applied to equilibrium calculations foracids are valid for solutions of bases in water. Consider, for example, the calcu-lation of the pH of a 0.10 M NH3 solution. The details of this calculation areillustrated in the drawing in the margin. We can start by writing an equation forthe reaction between ammonia and water.

The ionization of NH3 in water can be described by the following base ioniza-tion equilibrium constant expression.

The values of Kb for a limited number of bases can be found in Table B.9 inAppendix B. For NH3, Kb is 1.8 � 10�5.

We can organize what we know about the equilibrium with the format weused for equilibria involving acids.

Initial: 0.10 M 0 0Equilibrium: 0.10 � ¢C ¢C ¢C

Substituting what we know about the concentrations of the components of thisequilibrium into the Kb expression gives the following equation.

Kb for ammonia is small enough to consider the assumption that ¢C is smallcompared with the initial concentration of the base.

Solving the approximate equation gives the following result.

The value of ¢C is small enough compared with the initial concentration of NH3

to be ignored and yet large enough compared with the OH� ion concentration inwater to ignore the dissociation of water. ¢C can therefore be used to calculatethe pOH of the solution.

This, in turn, can be used to calculate the pH of the solution.

pH = 14.0 - pOH = 11.1

pOH = - log(1.3 * 10- 3) = 2.9

¢C L 1.3 * 10- 3M

[¢C][¢C]

[0.10]L 1.8 * 10- 5

Kb =

[NH4+][OH-]

[NH3]=

[¢C][¢C]

[0.10 - ¢C]= 1.8 * 10- 5

L

NH3(aq) + H2O(l) uv NH4+(aq) + OH-(aq) Kb = 1.8 * 10- 5

Kb =

[NH4+][OH-]

[NH3]



pKb = 4.740.10 M NH3

NH3 + H2O

NH4+ + OH–

Δ(NH3) = Δ(NH4+) = Δ(OH–)

Common window cleaners are oftendilute solutions of ammonia in water.

➤ CHECKPOINTWhich is more basic, a 1 M solution ofa base that has a small Kb or a 1 Msolution of a base that has a large Kb?


THE RELATIONSHIP BETWEEN Ka AND Kb

Equilibrium problems involving bases can be solved if the value of Kb for thebase is known. Values of Kb are listed in Table B.9 in Appendix B, however, foronly a limited number of compounds. The first step in many base equilibrium cal-culations therefore involves determining the value of Kb for the reaction from thevalue of Ka for the conjugate acid.

The relationship between Ka and Kb for a conjugate acid–base pair can beunderstood by using a specific example of this relationship. Consider a solutioncontaining the acetate ion (OAc�), for example. We already know the value of Ka

for the conjugate acid, acetic acid (HOAc)

What is the value of Kb for the conjugate base, OAc�?

The Ka and Kb expressions for acetic acid and its conjugate base both con-tain the ratio of the equilibrium concentrations of the acid and its conjugate base.Ka is proportional to [OAc�] divided by [HOAc], whereas Kb is proportional to[HOAc] divided by [OAc�].

Two changes need to be made to derive the Kb expression from the Ka

expression: We need to remove the [H3O�] term and introduce an [OH�] term.We can do this by multiplying the top and bottom of the Ka expression by theOH� ion concentration.

Rearranging this equation gives the following result.

The two terms on the right side of the equation should look familiar. The first isthe inverse of the Kb expression, and the second is the expression for Kw.


According to this equation, the value of Kb for the reaction between the acetateion and water can be calculated from the Ka for acetic acid.

Kb =

Kw

Ka=

1.0 * 10- 14

1.8 * 10- 5= 5.6 * 10- 10

KaKb = Kw = 1.0 * 10- 14

Ka =

1Kb

* Kw

Ka =

[OAc-]

[HOAc][OH-]* [H3O

+][OH-]

Ka =

[H3O+][OAc-]

[HOAc]*

[OH-]

[OH-]

Ka =

[H3O+][OAc-]

[HOAc] Kb =

[HOAc][OH-]

[OAc-]

OAc-(aq) + H2O(l) uv HOAc(aq) + OH-(aq) Kb = ?


11.14 BASE pH CALCULATIONS 501


The relationship between Ka and Kb for a conjugate acid–base pair can beused to rationalize why strong acids tend to have weak conjugate bases and strongbases tend to have weak conjugate acids. The product of Ka for any Brønsted acidtimes Kb for its conjugate base is equal to Kw for water.

But Kw is a relatively small number (1.0 � 10�14). As a result, the value of Ka

and Kb for a conjugate acid–base pair can’t both be large. When Ka is relativelylarge––indicating a strong acid––Kb for the conjugate base must be relativelysmall––indicating a weak conjugate base. Conversely, if Kb is large––indicating astrong base––Ka for the conjugate acid must be small––indicating a weak acid.As noted in Section 11.9, the relative strengths of a conjugate acid–base pair canbe summarized as follows.

The stronger the acid, the weaker its conjugate base.

The stronger the base, the weaker its conjugate acid.

Ka and Kb for a conjugate acid–base pair can’t both be large, but they canboth be relatively small. Consider ammonia (NH3) and its conjugate acid, theammonium ion (NH4

�), for example. NH3 is a moderately weak base (Kb � 1.8 �10�5), and the NH4

� ion is a relatively weak acid (Ka � 5.6 � 10�10).

ACID–BASE PROPERTIES OF A SALT

The reaction of an acid with a base produces an ionic compound referred to as asalt. A salt is an ionic compound with a cation other than H� and an anion otherthan OH� or O2�. Dilute solutions of soluble salts are assumed to ionize com-pletely in water. The acid–base properties of these compounds are therefore deter-mined by the acid–base properties of the aqueous ions formed when the saltdissolves in water. As a general rule, Group I and II cations (Li�, Na�, K�, Rb�,Cs�, Ca2�, Sr2�, Ba2�) have no significant acid or base properties. The conju-gate bases of strong monoprotic acids (Cl�, Br�, I�, ClO4

�, NO3�) are so

extremely weak that they have essentially no basic properties in water.Other salts, however, can exhibit the properties of an acid or a base. Con-

sider potassium fluoride, for example.

The potassium ion is neither an acid nor a base. However, the F� ion is the con-jugate base of a weak acid, HF. The fluoride ion will therefore behave as a base.

A solution of KF in water is therefore mildly basic (0.10 M NaF, pH � 8.0), andpotassium fluoride is referred to as a basic salt.

Ammonium nitrate, NH4NO3, is an example of an acidic salt. A dilute solu-tion will ionize completely when it dissolves in water.

The nitrate anion, NO3�, is the conjugate base of the strong acid HNO3 and there-

fore has essentially no base properties. The ammonium cation, NH4�, however,

NH4NO3(aq) ¡ NH4+(aq) + NO3

-(aq)

F-(aq) + H2O(l) uv HF(aq) + OH-(aq)

KF(aq) ¡

H2OK+(aq) + F-(aq)

KaKb = Kw = 1.0 * 10- 14



is the conjugate acid of the weak base ammonia (NH3) and should thereforebehave as an acid in water. Thus an aqueous solution of NH4NO3 is slightly acidic(0.10 M NH4NO3, pH � 5.4).

Sodium chloride, NaCl, is an example of a salt that produces essentially aneutral solution (0.10 M NaCl, pH � 6.4 7). In dilute aqueous solutions, sodiumchloride ionizes completely.

The chloride ion is the conjugate base of a strong acid and is therefore much tooweak a base to be noticed. The sodium ion is neither an acid nor a base in theBrønsted model of acids and bases. Aqueous solutions of NaCl are thereforeessentially neutral.

NaCl(aq) ¡ Na+(aq) + Cl-(aq)

L


+(aq) + NH3(aq)

11.14 BASE pH CALCULATIONS 503

E x e r c i s e 1 1 . 1 0Calculate the HOAc, OAc�, and OH� concentrations and pH at equilibrium in a0.10 M NaOAc solution. (For HOAc, Ka � 1.8 � 10�5.)

SolutionWe start with a solution of an ionic salt––NaOAc––dissolved in water andassume that the salt ionizes completely.

Because the Na� ion is neither an acid nor a base, only the acetate ion willsignificantly affect the pH. We therefore start, as always, by building a repre-sentation of the information in the statement of the problem. The relevant infor-mation for this calculation is illustrated in the drawing in the margin.

Initial: 0.10 M 0 L 0Equilibrium: 0.10 � ¢C ¢C ¢C

The only other information we need to solve the problem is the value of Kb

for the reaction between the acetate ion and water, which can be calculatedfrom the value of Ka for acetic acid.

Substituting what we know into the equilibrium constant expression givesthe following result.

Kb =

[HOAc][OH-]

[OAc-]=

[¢C][¢C]

[0.10 - ¢C]= 5.6 * 10- 10

Kb =

Kw

Ka=

1.0 * 10- 14

1.8 * 10- 5= 5.6 * 10- 10

OAc-(aq) + H2O(l) uv HOAc(aq) + OH-(aq) Kb = ?

NaOAc(aq) ¡ Na+(aq) + OAc-(aq)

Kb = 5.6 × 10 –10

(very small ⇒ ΔC << 0.10 M)

0.10 M NaOAc

OAc– + H2O

HOAc + OH–


11.15 Mixtures of Acids and Bases: BuffersMixtures of a weak acid and its conjugate base are called buffers. So far, we haveexamined 0.10 M solutions of both acetic acid and sodium acetate.

What would happen if we prepared a buffer from enough sodium acetate andacetic acid solution so that the solution is simultaneously 0.10 M in both HOAcand NaOAc? The relevant information for this calculation is illustrated in thedrawing in the margin.

The first step toward answering this question is recognizing that there aretwo sources of the OAc� ion in the solution. Acetic acid, of course, dissociatesto give the H3O� and OAc� ions.

The salt, sodium acetate, however, also dissociates in water to give the OAc� ion.

NaOAc(s) ¡

H2ONa+(aq) + OAc-(aq)

HOAc(aq) + H2O(l) uv H3O+(aq) + OAc-(aq)

0.10 M NaOAc [H3O+] L 7.5 * 10- 6M pH L 8.9

0.10 M HOAc [H3O+] L 1.3 * 10- 3M pH L 2.9


➤ CHECKPOINTIf Ka for HNO2 is 5.1 � 10�4, what isKb for the NO2

� ion?

Because the value of Kb for the OAc� ion is relatively small, we can feel rea-sonably confident that ¢C is small compared with the initial concentration of base.

We can then solve the approximate equation the assumption generates.

¢C is very much smaller than the initial concentration of the base,which means that one of the key assumptions of acid–base equilibrium cal-culations is legitimate. But [OH�] is still at least 75 times larger than theconcentration of the OH� ion in pure water, which means that we can alsolegitimately ignore the dissociation of water in the calculation. We can there-fore use the value of ¢C to calculate the equilibrium concentrations of HOAc,OAc�, and OH�.

The pH of this solution is therefore 8.9.

pH = - log[H3O+] = 8.9

[H3O+] =

Kw

[OH-]=

1.0 * 10- 14

[7.5 * 10- 6]= 1.3 * 10- 9

[HOAc] = [OH-] = ¢C � 7.5 * 10- 6M

[OAc-] = 0.10M - ¢C L 0.10M

¢C L 7.5 * 10- 6M

[¢C][¢C]

[0.10]= 5.6 * 10- 10

0.10 M HOAc

0.10 M OAc–

NaOAc

Ka =1.8 × 10–5 OAc–

⇒ [HOAc] ≈ [OAc–]⇒ [H3O+] ≈ Ka


These reactions share a common ion: the OAc� ion. Le Châtelier’s principle pre-dicts that adding sodium acetate to acetic acid should shift the equilibriumbetween HOAc and the H3O� and OAc� ions to the left, another example of thecommon-ion effect. Thus adding NaOAc reduces the extent to which HOAc dis-sociates. The mixture is therefore less acidic than 0.10 M HOAc by itself.

In Exercise 11.10 we saw that the OAc� ion behaves as a base. The Kb forOAc� is several orders of magnitude smaller than Ka for HOAc. Therefore, the reac-tion of OAc� with water is small compared to the reaction of HOAc with water.

We can quantify the discussion by setting up the problem as follows.

Initial: 0.10M 0 0.10MEquilibrium: ? ? ?

We don’t have any basis for predicting whether the dissociation of water can beignored in this calculation, so let’s make that assumption and check its validity later.

If most of the H3O� ion concentration at equilibrium comes from dissoci-ation of the acetic acid, the reaction has to shift to the right to reach equilibrium.

Initial: 0.10M 0 0.10MEquilibrium: 0.10 � ¢C ¢C 0.10 � ¢C

Substituting this information into the Ka expression gives the following result.

Let’s assume that ¢C is small compared with the initial concentrations of HOAcand OAc�.

We then solve the approximate equation for ¢C.

Are our two assumptions legitimate? Is ¢C small enough compared with 0.10to be ignored? Is ¢C large enough so that the dissociation of water can be ignored?The answer to both questions is yes. We can therefore use the value of ¢C to cal-culate the H3O� ion concentration at equilibrium and the pH of the solution.

We therefore conclude that the buffer solution is acidic, but less acidic thanHOAc by itself. As we would expect from Le Châtelier’s principle, adding asource of the acetate ion shifts the equilibrium for the dissociation of aceticacid in water toward the reactants. This results in a decrease in the amount ofH3O�. In other words, the HOAc solution is less acidic due to the addition ofa base, OAc�.

pH = - log(1.8 * 10- 5) = 4.7

[H3O+] L ¢C L 1.8 * 10- 5M

¢C L 1.8 * 10- 5

[¢C][0.10]

[0.10]L 1.8 * 10- 5

Ka =

[H3O+][OAc-]

[HOAc]=

[¢C][0.10 + ¢C]

[0.10 - ¢C]= 1.8 * 10- 5

L


L


11.15 MIXTURES OF ACIDS AND BASES: BUFFERS 505


11.16 Buffers and Buffer CapacityThe term buffer means “to lessen or absorb shock.” Mixtures of weak acids andtheir conjugate bases are buffers because they lessen or absorb the drastic changein pH that occurs when small amounts of acids or bases are added to water.

In Section 11.12 we concluded that adding a single drop of 2 M HCl to 100 mL of pure water would change the pH by 4 units, from pH 7 to pH 3. Ifwe add the same amount of HCl to a buffer solution that is 0.10 M in both aceticacid (HOAc) and a salt of the acetate ion (OAc�), there is no observable changein the pH of the solution. In fact, we can add 10 times as much 2 M HCl to thebuffer solution without noticing a significant change in the pH of the solution.Thus the pH of the buffer solution is truly “buffered” against the effect of smallamounts of acid or base.

To understand how the buffer works, it is important to recognize that a bufferis made by mixing a weak acid and its conjugate base so that significant amountsof both a weak acid and weak base are present in solution.

Weak acid Conjugate base

At equilibrium, the concentrations of acetic acid and the H3O� and OAc� ionsin the solution are related by the following equation.

If we rearrange the equation, by solving for the H3O� ion concentration at equi-librium, we get the following result.

This equation suggests that the H3O� ion concentration in the solution willstay more or less constant as long as there is no significant change in the ratio ofthe concentrations of the undissociated HOAc molecules and the OAc� ions. Thebuffer solution discussed in Section 11.15 was prepared by adding enough sodiumacetate (NaOAc) to acetic acid (HOAc) to make the concentration of both the acid(HOAc) and its conjugate base (OAc�) the same. (The solution is simultaneously0.10 M HOAc and 0.10 M OAc�.) The pH of this solution as determined by apH meter is 4.74.

If a small amount of a strong acid, H3O�, is added to the solution, it willreact with the OAc� ion to form a bit more acetic acid. This is illustrated by thedrawing in the margin.

The equilibrium constant for this reaction is the inverse of the value of Ka foracetic acid because the reaction is written in the opposite direction from the equa-tion that describes the dissociation of this acid.

K¿ =

1Ka

=

1

1.8 * 10- 5= 5.6 * 104

OAc-(aq) + H3O+

¡ HOAc(aq) + H2O(l)

[H3O+] = Ka *

[HOAc]

[OAc-]

Ka =

[H3O+][OAc-]

[HOAc]

HOAc(aq) + H2O(l) uv H3O+(aq) + OAc-(aq)



The equilibrium constant is large enough that the reaction can be assumed to pro-ceed essentially to completion, forming acetic acid.

The H3O� concentration after the acid is added to the buffer solution canbe calculated as follows.

Because the extent of the reaction is so large, adding one drop of 2 M HCl to100 mL of this HOAc/OAc� buffer gives a change in the concentration, ¢C, ofonly 0.001 M and the new H3O� concentration is

The pH of the solution is therefore still 4.74The net result of adding one drop of 2 M HCl to the buffer solution is a

slight increase in the concentration of the undissociated HOAc molecules and aslight decrease in the concentration of the OAc� ion. But if only a reasonablysmall amount of acid is added to the solution, the ratio of the concentrations wouldbe roughly the same. As a result, the H3O� ion concentration in the solutionremains essentially the same, which means there is no change in the pH of thesolution. The addition of a strong acid, H3O�, to the buffer results in the H3O�

being converted into an equivalent amount of undissociated weaker acid, HOAc.If a small amount of a strong base, OH�, is added to the solution, it will

react with some of the undissociated HOAc molecules to form a bit more of theOAc� ion.

But, once again, if we add only a reasonably small amount of the base, the ratioof the concentrations of HOAc and OAc� in the solution will be more or less thesame. This means that the pH of the solution is more or less the same. The addi-tion of a strong base, OH�, to the buffer converts the OH� into an equivalentamount of weaker base, OAC�.

Because it contains significant amounts of both HOAc molecules and theOAc� ion, the solution can resist a change in pH when reasonably small amountsof either a strong acid or a strong base are added to the solution. The concentra-tions of acetic acid and the acetate ion will change slightly when the acid or baseis added to the buffer solution, but the ratio of the concentrations of these twocomponents of the equilibrium will not change by a large enough amount to pro-duce a significant change in the pH of the solution.

As long as the concentrations of HOAc and the OAc� ion in the buffer arelarger than the amount of acid or base added to the solution, the pH remains con-stant. Table 11.6 compares the effects of adding different amounts of hydrochlo-ric acid to water and two buffer solutions of different concentrations. The firstcolumn gives the number of moles of HCl added per liter. The second columnshows the effect of this much acid on the pH of water. The third column showsthe effect on a buffer solution that contains 0.10 M HOAc and 0.10 M NaOAc.The fourth column shows the result of adding the acid to a more concentratedbuffer that contains twice as much acetic acid and twice as much sodium acetate.Even when as much as 0.01 mol/L of acid is added to the first buffer, there isvery little change in the pH of the buffer solution, as shown in Figure 11.12.

HOAc(aq) + OH-(aq) ¡ OAc-(aq) + H2O(l)

[H3O+] = 1.8 * 10- 5 0.101

0.099= 1.8 * 10- 5

[H3O+] = 1.8 * 10- 5 [HOAc]

[OAc-]= 1.8 * 10- 5 [0.100 + ¢C]

[0.100 - ¢C]

11.16 BUFFERS AND BUFFER CAPACITY 507

100 mL0.10 M HOAc0.10 M OAc–

1 drop 2 M HCl≈ 1 × 10–4 mol

HCl + OAc– HOAc + Cl–

Decreasesby

0.001 M

Increasesby

0.001 M

Fig. 11.12 The pH of water is verysensitive to the addition ofhydrochloric acid. Adding as little as10�3 mol of HCl to a liter of watercan drop the pH by 4 units. The pHof a buffer, however, remainsconstant until relatively largeamounts of hydrochloric acid havebeen added. The buffer contains 0.10 M OAc� and HOAc.

10−6 10−5 10−4 10−3 10−2 10−11

2

3

4

5

6

7

Moles of HCl added per liter

pH

Buffer + HCl

Pure water + HCl


By definition, the buffer capacity of a buffer is the amount of strong acidor base that can be added before the pH of the solution changes significantly.The capacity of the second buffer solution in Table 11.6 is even larger than thatof the first. Buffer 2 protects against quantities of acid or base almost as largeas 0.1 mol/L.

The acetic acid/acetate buffer was described using the Ka expression foracetic acid.

This expression can be rearranged to give a more useful form of the equation fordescribing buffers.

Taking the negative log of both sides of the previous equation yields the following.

The term on the left side of this equation is the expression for pH. The firstterm on the right side of the equation is the expression for pKa. Substituting pHand pKa into this equation and then inverting the logarithm term gives the fol-lowing result.

This equation can be generalized, as follows, to obtain an expression known asthe Henderson–Hasselbalch equation.

If the concentrations of the conjugate acid and the conjugate base used toprepare a buffer are the same, the logarithm term in the Henderson–Hasselbalch

pH = pKa + loga [conjugate base]

[conjugate acid]b

pH = pKa + loga [OAc-]

[HOAc]b

- log[H3O+] = - logKa- loga [HOAc]

[OAc-]b

[H3O+] = Ka *

[HOAc]

[OAc-]

Ka =

[H3O+][OAc-]

[HOAc]


Table 11.6 Effect of Adding Hydrochloric Acid to Water and Two Buffer Solutions

pH of Buffer 1 pH of Buffer 2 Mol HCl/L Added to (0.10 M HOAc/ (0.20 M HOAc/Each Solution pH of Water 0.10 M OAc�) 0.20 M OAc�)

0 7 4.74 4.740.000001 6 4.74 4.740.00001 5 4.74 4.740.0001 4 4.74 4.740.001 3 4.74 4.740.01 2 4.66 4.700.1 1 2.72 4.27


equation is equal to zero. The pH of this solution is therefore equal to the pKa ofthe acid. By adjusting the ratio of the concentrations of the conjugate acid and itsconjugate base it is possible to prepare a buffer with a pH slightly above orslightly below the pKa.

Suppose a buffer with a pH of 4.74 is desired; this would suggest that anacetic acid/acetate conjugate pair with a pKa of 4.76 would be appropriate.

If a buffer with a pH of 10.0 is needed, the NaHCO3/Na2CO3 conjugateacid–base pair, which has a pKa of 9.25 could be used. When the conjugateacid–base pair has a pKa close to the desired pH, the ratio of the concentrationsof acid to base can be calculated from the previous equation.

Buffers can also be made from a weak base and its conjugate acid, such asammonia and a salt of the ammonium ion.

Conjugate acid Weak base

There is an important difference, however, between this buffer and one made bymixing sodium acetate and acetic acid. Mixtures of HOAc and OAc� ion forman acidic buffer, with a pH below 7. Mixtures of NH3 and the NH4

� ion form abasic buffer, with a pH above 7.

We can predict whether a buffer will be acidic or basic by comparing thevalues of Ka and Kb for the conjugate acid–base pair. Ka for acetic acid is sig-nificantly larger than Kb for the acetate ion. We therefore expect mixtures of theconjugate acid–base pair to be acidic.

Kb for ammonia, on the other hand, is much larger than Ka for the ammoniumion. Mixtures of this acid–base pair are therefore basic.

In general, if Ka for the acid is larger than 1 � 10�7, the buffer will be acidic.If Kb is larger than 1 � 10�7, the buffer is basic.

The choice of the conjugate acid/base pair used to prepare a buffer solu-tion can be made by comparing the desired pH with a table of values of Ka forweak acids.

NH4+ Ka = 5.6 * 10- 10

NH3 Kb = 1.8 * 10- 5

OAc- Kb = 5.6 * 10- 10

HOAc Ka = 1.8 * 10- 5


+(aq) + NH3(aq)

11.16 BUFFERS AND BUFFER CAPACITY 509

E x e r c i s e 1 1 . 1 1Describe how to prepare a pH 9.35 buffer solution.

SolutionThe data in either Table 11.4 or Table B.8 in the Appendix suggest that theNH4

�/NH3 acid–base pair could be used to prepare this buffer because the pKa

for the ammonium ion is 9.25.

NH4+(aq) + H2O(l) uv NH3(aq) + H3O

+(aq)


Buffers are used extensively in the chemistry laboratory. They can be pur-chased from chemical suppliers as solutions with known concentrations and pHvalues. They can also be purchased as packets of a mixture of a solid conjugateacid–base pair. Dissolving the packet in water yields a buffer with a pH equal tothe value stated on the package. In addition, buffers can be prepared by mixingmeasured amounts of an appropriate conjugate acid–base pair and then addingstrong acid or base to adjust the pH to the desired value. The pH of the buffer isnormally monitored with a pH meter as the strong acid or strong base is added.

11.17 Buffers in the BodyBuffers are very important in living organisms for maintaining the pH of biolog-ical fluids within the very narrow ranges necessary for the biochemical reactionsof life processes. Three primary buffer systems maintain the pH of blood in the


We can start by substituting the values of the pH of the buffer solution and thepKa value for the conjugate acid into the Henderson–Hasselbalch equation.

Solving for the ratio of the concentrations of the conjugate acid–base pair givesthe following result.

Thus if the concentration of the base is 1.3 times that of the concentration ofthe acid, the solution would be an effective buffer at a pH of 9.35.

Suppose that 1.0 L of 0.10 M NH3 was available. The equilibrium con-stant for the reaction between NH3 and water is so small that the concentra-tion of NH3 at equilibrium is essentially 0.10 M.

We can calculate the concentration of the NH4� needed to prepare a pH 9.35

buffer as follows.

Because adding 0.077 mole of solid NH4Cl to the 0.10 M NH3 solution wouldproduce a negligible change in the volume of the solution, the result would be1.0 L of a buffer with the desired pH.

[NH4+] = 0.077 M

[0.10 M]

[NH4+]

= 1.3

[NH3]

[NH4+]

= 1.3

NH3(aq) + H2O(l) uv NH4+(aq) + OH-(aq) Kb = 1.8 * 10- 5

[conjugate base]

[conjugate acid]= 1.3

loga [conjugate base]

[conjugate acid]b = 9.35 - 9.25 = 0.10

9.35 = 9.25 + loga [conjugate base]

[conjugate acid]b

pH = pKa + loga [conjugate base]

[conjugate acid]b

➤ CHECKPOINTThe pH of a buffered solution is 7.4.What happens to the pH if an acid isadded to the buffer? What happens tothe pH if a base is added to the buffer?Does the resulting pH depend on howmuch acid or base is added?

pH meters are often calibrated withbuffer solutions because the pH ofthese solutions is stable. [Copyright ofThermo Fisher Scientific Inc., 2010.]


human body within the limits 7.36 to 7.42. One of these buffers is composed ofcarbonic acid, H2CO3, and its conjugate base, HCO3

�, the hydrogen carbonate ion.Some of the CO2 that enters a red blood cell from respiring tissue reacts with waterand is converted to carbonic acid by an enzyme known as carbonic anhydrase.

The carbonic acid then partially dissociates to form an equilibrium mixture withthe hydrogen carbonate (or bicarbonate) ion.

When the concentration of the H3O� ion in the blood is too high (too lowa pH), a condition known as acidosis results. This condition can be caused bydiseases such as diabetes mellitus, in which large amounts of acids are produced.As the concentration of H3O� in blood increases, the H2CO3/HCO3

� equilibriumshifts to the left to maintain the proper pH.

A low concentration of H3O� in the blood (too high a pH) is known as alka-losis. This condition can result during hyperventilation (rapid breathing) whenexcess amounts of CO2 are lost from the body through the lungs. The loss of CO2

shifts the CO2/H2CO3 equilibrium to the left, resulting in a decrease in the con-centration of carbonic acid in the blood. This disrupts the H2CO3/HCO3

� bufferby shifting the equilibrium to the left and therefore reducing the amount of H3O�

in the blood. The resulting increase in pH is alkalosis.

11.18 Acid–Base ReactionsSTRONG ACID–STRONG BASE

The reaction between an acid and a base can produce a salt and water. Consider,for example, the reaction of the strong acid hydrochloric acid with the strong basesodium hydroxide.

The ionic equation for this reaction is

For acid and base reactions, H�(aq) is best represented by H3O�(aq), and the netionic equation for this reaction is usually written as follows.

Stronger acid Stronger base Weaker acid Weaker base

H3O� is a stronger acid than H2O, and OH� is a stronger base than H2O. Asshown in Section 11.10, the equilibrium in the above reaction lies almost entirelyto the right. The equilibrium constant for this reaction is the reciprocal of Kw, or1.0 � 1014. No matter which strong acid or strong base is reacted, the net ionicreaction is always that given by the preceding equation, with K � 1.0 � 1014.

When just enough strong acid has been added to completely react with astrong base, the resultant solution will have equal concentrations of H3O� andOH� and thus give a pH of 7.0.

H3O+(aq) + OH-(aq) ¡ H2O(l) + H2O(l)

H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) ¡ H2O(l) + Na+(aq) + Cl-(aq)

HCl(aq) + NaOH(aq) ¡ H2O(l) + NaCl(aq)

H2CO3(aq) + H2O(l) uv HCO3-(aq) + H3O

+(aq)

CO2(g) + H2O(l) uv H2CO3(aq)

11.18 ACID–BASE REACTIONS 511

➤ CHECKPOINTLactic acid, HC3H5O3, is produced inthe body during strenuous exercise.Write a chemical reaction that can oc-cur in blood to maintain the proper pHwhen lactic acid is produced.


WEAK ACID–STRONG BASE

If a weak acid is reacted with a strong base, the reaction will go almost tocompletion.

Weak acid Strong base

In contrast to the reaction of a strong acid with a strong base, the pH of a solu-tion produced from the complete reaction of a weak acid and strong base will notbe 7.0 but will depend on the interaction of the A� ion with water.

The weak acid HNO2 reacts with a sodium hydroxide solution, for exam-ple, to form water and the NO2

� ion.

Weak acid Strong base

When this reaction is essentially complete, the pH of the resulting solution is dueto the interaction of the weak base NO2

� with H2O.The equilibrium constant for this reaction can be calculated from tabulated val-

ues of the equilibrium constants for HNO2 (see Table B.8 in Appendix B) and Kw,

where K for the second reaction is the inverse of Kw. Adding the two reactions gives

The equilibrium constant is so large that the reaction is essentially complete. Ingeneral, the reaction of a weak acid with a strong base proceeds far to the right.

STRONG ACID–WEAK BASE

The weak base methyl amine, CH3NH2, reacts with a strong acid such as HIaccording to the following equation.

Assuming that HI dissociates more or less completely in water, we can write theequation for this reaction as follows.

The equilibrium constant for this reaction can be found by combining appro-priate reactions to give the desired reaction for which the equilibrium constantsare known. Tables B.8 and B.9 in Appendix B contain a tabulation of many ofthese useful equilibrium constants.

CH3NH2(aq) + H3O+(aq) uv CH3NH3

+(aq) + H2O(l) Krxn = Kb * K = 4.8 * 1010

H3O+(aq) + OH-(aq) uv 2 H2O(l) K = 1.0 * 1014

CH3NH2(aq) + H2O(l) uv CH3NH3+(aq) + OH-(aq) Kb = 4.8 * 10- 4

CH3NH2(aq) + H3O+(aq) ¡ CH3NH3

+(aq) + H2O(l)

CH3NH2(aq) + HI(aq) ¡ CH3NH3+(aq) + I-(aq)

Krxn = Ka * K = 5.1 * 10- 4* 1.0 * 1014

= 5.1 * 1010

HNO2(aq) + OH-(aq) uv H2O(l) + NO2-(aq)

H3O+(aq) + OH-(aq) uv 2 H2O(l) K = 1.0 * 1014

HNO2(aq) + H2O(l) uv H3O+(aq) + NO2

-(aq) Ka = 5.1 * 10- 4

HNO2(aq) + OH-(aq) ¡ H2O(l) + NO2-(aq)

HA(aq) + OH-(aq) uv H2O(l) + A-(aq)



The equilibrium constant for the reaction is so large that the reaction pro-ceeds nearly to completion. In general, a strong acid will react completely witha weak base. The pH of the resulting mixture will depend on the nature of theacid, in this case CH3NH3

�, produced in the reaction.

WEAK BASE–WEAK ACID

The reaction between the weak base methyl amine and the weak acid nitrous acidcan be described by the following equation.

Weak base Weak acid

This reaction will proceed to an extent that depends on the magnitude of the equi-librium constant. For such reactions in general, Krxn can be found by the follow-ing procedure.

For the CH3NH2 and HNO2 reaction, Tables B.8 and B.9 in Appendix Bgive the necessary Ka and Kb values and

In this case Krxn is quite large, and the reaction proceeds nearly to completion.However, this is not always the case. For the reaction of HCN with hydrazine,H2NNH2, the equilibrium constant is only 0.072; thus this reaction hardly pro-ceeds at all. Weak acid–weak base reactions are difficult to generalize, and in eachcase an examination of the equilibrium constants is necessary.

11.19 pH Titration CurvesWe can measure the concentration of a solution by a technique known as titra-tion. A solution of known concentration is slowly added to a known quantity of areagent with which it reacts until we observe something that tells us that exactlyequivalent numbers of moles of the reagents are present. Titrations thereforedepend on the existence of a class of compounds known as indicators. A commontype of titration is an acid–base titration in which an acid of unknown concentra-tion is titrated with a base that has an accurately known concentration. Titrationscan therefore be used to determine the concentration of an acid or base solution.

Indicators are weak acids or weak bases whose conjugate acid–base pairshave different colors in aqueous solution. A commonly used indicator is phe-nolphthalein, which is colorless in its acid form and pink in its base form. Thetwo forms of an indicator can be represented by HIn for the acid form and In�

for the base form. The ionization of the indicator can therefore be represented bythe following equation,

HIn(aq) + H2O(aq) uv H3O+(aq) + In-(aq)

Krxn = 5.1 * 10- 4* 4.8 * 10- 4

* 1.0 * 1014= 2.5 * 107

HA(aq) + B(aq) uv BH+(aq) + A-(aq) Krxn = Ka * Kb * 1.0 * 1014

H3O+(aq) + OH-(aq) uv 2 H2O(l) K = 1.0 * 1014

B(aq) + H2O(l) uv BH+(aq) + OH-(aq) Kb = ?

HA(aq) + H2O(l) uv H3O+(aq) + A-(aq) Ka = ?

CH3NH2(aq) + HNO2(aq) ¡ CH3NH3+(aq) + NO2

-(aq)

11.19 pH TITRATION CURVES 513


for which Ka could be written as follows:

This equation can be rearranged to give:

The color of the indicator solution depends on the concentration of the H3O�

ion. When the concentration of the H3O� ion is large, the value of [HIn] is largerthan [In�], and the solution has the characteristic color of the acid form of the indi-cator. If phenolphthalein is the indicator, the solution is colorless in acid. When theconcentration of the H3O� ion is small, the value of [In�] is larger than [HIn], andthe color of the base form of the indicator predominates. For phenolphthalein, thecolor of the base form is pink. For most indicators color changes occur when theratio [In�]/[HIn] changes from 1/10 to about 10. Some common indicators are listedin Table 11.7 along with the pH interval over which the indicator changes color.

[In-]

[HIn]=

Ka

[H3O+]

Ka = [H3O+]

[In-]

[HIn]


Table 11.7Common Acid–Base Indicators

Color change Indicator pH-interval (Acidform to Baseform)

Thymol blue 1.2–2.8 red-yellowMethyl orange 3.1–4.4 orange-yellowMethyl red 4.2–6.2 red-yellowBromthymol blue 6.0–7.6 yellow-bluePhenol red 6.8–8.2 yellow-redCresol red 7.2–8.8 yellow-redPhenolphthalein 8.3–10.0 colorless-pinkAlizarin yellow 10.1–12.0 yellow-red

E x e r c i s e 1 1 . 1 2Calculate the pH of a solution of methyl red (Ka � 5.0 � 10�6) for which theratio [HIn]/[In�] is 0.10, 1.0, and 10. Predict the color of the solution for eachpH if the acid form of the indicator is red and the base form is yellow.

SolutionThe H3O� ion concentration can be predicted from the [HIn]/[In�] ratio usingthe following equation.

When the [HIn]/[In�] ratio is 0.10, the pH of the solution is 6.3, and the indi-cator is yellow.

[H3O+] = 5.0 * 10- 6

*

[1]

[10]= 5.0 * 10- 7; pH = 6.3

[H3O+] = Ka

[HIn]

[In-]= 5.0 * 10- 6 [HIn]

[In-]


The end point of an acid–base titration is the point at which the indicatorturns color. The equivalence point is the point at which exactly enough base hasbeen added to neutralize the acid. Chemists try to use indicators for which theend point is as close as possible to the point at which equivalent amounts of acidand base are present.

Phenolphthalein provides an example of how chemists compensate for thefact that the indicator doesn’t always turn color exactly at the equivalence pointof the titration. Acid–base titrations that use phenolphthalein as the indicator arestopped at the point at which a pink color appears and remains for 10 seconds asthe solution is stirred. Because this is one drop before the indicator turns a per-manent color, it is as close as we can get to the equivalence point of the titration.

The selection of an indicator for the end-point determination depends on thenature of the acid and base. Consider a titration in which a strong acid such asHCl is titrated with a strong base such as NaOH.

Because neither the Na� ion nor the Cl� ion affects the pH of the water solvent,the pH will be 7.0 at the equivalence point.

If a weak acid such as acetic acid is titrated by a strong base, the solutionwill be basic at the equivalence point because of the reaction between the OAc�

ion and water.

OAc-(aq) + H2O(l) uv HOAc(aq) + OH-(aq)

NaOH(aq) + HOAc(aq) uv H2O(l) + Na+(aq) + OAc-(aq)

HCl(aq) + NaOH(aq) uv Na+(aq) + Cl-(aq) + H2O(l).


When the ratio is 1.0, the pH is 5.3 and the color of the solution should be amixture of red and yellow, and therefore orange.

When the ratio is 10, the pH is 4.3 and the indicator is red.

[H3O+] = 5.0 * 10- 6

*

[10]

[1]= 5.0 * 10- 5; pH = 4.3

[H3O+] = 5.0 * 10- 6

*

[1]

[1]= 5.0 * 10- 6; pH = 5.3

E x e r c i s e 1 1 . 1 3Select an indicator from Table 11.7 that would change color at the equivalencepoint for each of the following titrations.

(a) HNO3 and NaOH

(b) HOBr and KOH

(c) HCl and NH3

Solution(a) This titration involves the reaction between a strong acid and a strong base.

The equivalence point should therefore occur at a pH of about 7. Bromthy-mol blue would therefore be an appropriate indicator.


Let’s consider what would happen if we followed the reaction between sul-furic acid (H2SO4) and potassium hydroxide (KOH) using an appropriate titrationindicator. The reaction can be described by the following equation. Because thisis a reaction between a strong acid and a strong base, the reaction can be assumedto go to completion.

A 25.00-mL sample of sulfuric acid solution requires 42.61 mL of a 0.1500 MKOH solution to titrate to the phenolphthalein end point. What does this infor-mation tell us about the concentration of the sulfuric acid solution?

We only know the volume of the sulfuric acid solution, but we know boththe volume and the concentration of the KOH solution. It therefore seems rea-sonable to start by calculating the number of moles of KOH in this solution.

We now know the number of moles of KOH consumed in the reaction, and wehave a balanced chemical equation for the reaction that occurs when the two solu-tions are mixed. We can therefore calculate the number of moles of H2SO4 neededto consume that much KOH.

We now know the number of moles of H2SO4 in the sulfuric acid solution andthe volume of the solution. We can therefore calculate the number of moles ofsulfuric acid per liter, or the molarity of the solution.

The sulfuric acid solution therefore has a concentration of 0.1278 mole per liter.Acid–base titrations can be described by using a titration curve, a plot of

solution pH versus amount of titrant added. Buffers and buffering capacity playa major role in determining the shape of the titration curve in Figure 11.13, whichshows the pH of a solution of 25.00 mL of 0.10 M acetic acid as it is titratedwith a strong base, 0.10 M sodium hydroxide. Four points (A, B, C, and D) onthe curve will be discussed in some detail.

Point A in Figure 11.13 represents the pH at the start of the titration, thatis the pH of a 0.10 M HOAc solution.

HOAc(aq) + H2O(l) uv OAc-(aq) + H3O+(aq)

3.196 * 10- 3 mol H2SO4

0.02500 L= 0.1278M H2SO4

6.392 * 10- 3 mol KOH *

1 mol H2SO4

2 mol KOH= 3.196 * 10- 3 mol H2SO4

0.1500 mol KOH1 L

* 0.04261 L = 6.392 * 10- 3 mol KOH

H2SO4(aq) + 2 KOH(aq) ¡ 2 K+(aq) + SO42 -(aq) + 2 H2O(l)


(b) This titration involves a weak acid and a strong base, which will suggesta pH larger than 7 at the equivalence point. Cresol red would be an appro-priate indicator.

(c) This titration involves a strong acid and a weak base, which suggests a pHsmaller than 7 at the equivalence point. Methyl red is therefore an appro-priate indicator.

25.00 mLH2SO4

0.1500 M KOH42.61 mL


The pH can be calculated from

As NaOH solution is added to the HOAc solution, some of the acid is convertedto its conjugate base, the OAc� ion. The equilibrium constant for the reactionbetween a weak acid and a strong base is large, and therefore the equilibrium liesalmost entirely to the right.

As the titration proceeds, a buffer of HOAc and OAc� forms, and the pH changesslowly as the buffer capacity is exceeded with each acid addition.

Point B, corresponding to the addition of 12.50 mL of NaOH solution, isthe point at which exactly half of the HOAc molecules have been converted toOAc� ions. The concentration of the H3O� ions in the solution is controlled bythe equilibrium between HOAc and OAc�, which is described by the followingequilibrium constant expression.

At point B, the concentration of HOAc molecules is equal to the concentration ofthe OAc� ions. Because the ratio of the HOAc and OAc� concentrations is 1:1,the concentration of the H3O� ion at point B is equal to the Ka of the acetic acid.

Ka � [H3O�], and the pH is 4.7 at point B.

This provides us with a way of measuring Ka for an acid. We can titrate a sam-ple of the acid with a strong base, plot the titration curve, and then find the pointalong the curve at which exactly half of the acid has been consumed. The H3O�

ion concentration at that point will be equal to the value of Ka for the acid.Point C is the equivalence point for the titration, the point at which enough

base has been added to the solution to consume the acid present at the start ofthe titration. The goal of any titration is to find the equivalence point. What isactually observed is the end point of the titration, the point at which the indica-tor changes color.

Ka =

[H3O+][OAc-]

[HOAc]

HOAc(aq) + OH-(aq) ¡ OAc-(aq) + H2O(l)

¢C = 1.3 * 10- 3= [H3O

+], pH L 2.9

Ka = 1.8 * 10- 5=

[OAc-][H3O+]

[HOAc]L

¢C2

0.10 M


Fig. 11.13 Titration curve for the titration of 25.00 mL of a0.10 M acetic acid solution by a 0.10 M NaOH solution.

10 20 30 40 50 6000

2

4

6

8

12

10

14

Volume 0.10 M NaOH (mL)

pH

B

C

D

E

A


At point C the acetic acid has been completely reacted, and what remainsin solution are Na� ions and OAc� ions. The Na� ions have no effect on the pHof the solution, but the OAc� ions react with water to produce a basic solution.

The moles of OAc� can be found from the initial moles of HOAc.

0.025 L � 0.10 mol/L HOAc � 2.5 � 10�3 mol HOAc

At the equivalence point, 2.5 � 10�3 mol of OAc� have been formed, and thereis essentially no HOAc remaining. The total volume in the titration flask is equalto the initial 25.00 mL of HOAc plus the 25.00 ml of NaOH required to reachthe equivalence point. The concentration of the OAc� is then given by

The pH of the solution at this point can be calculated from the equilibrium con-stant, Kb, for the above reaction.

C � [OH�] � 5.3 � 10�6 M, pOH L 5.3, and the pH is 8.7 at point C.

Every effort is made to bring the end point as close as possible to the equiv-alence point of a titration. Thus in this titration an indicator whose color changeoccurs around a pH of 8.7 should be chosen. Table 11.7 shows that phenolph-thalein indicator turns from colorless to pink over a pH range of 8.3 to 10.0. Fig-ure 11.13 shows the range from points C to D over which the color change occurs.Therefore phenolphthalein would be a good indicator choice.

At point E, 50.00 mL of NaOH solution has been added. This is 25.00 mLin excess of the amount required to react will the acetic acid. Thus the solution nowhas a volume of 75.00 mL, and the excess moles of NaOH can be calculated from

0.10 mol/L � 0.02500 L � 0.0025 mol NaOH

The OH� concentration is then

[OH�] � 0.0025 mol>0.07500 L � 0.033 M

The pOH is 1.5 and the pH at point E is 12.5.Note the shape of the pH titration curve in Figure 11.13. The pH rises rap-

idly at first because we are adding a strong base to a weak acid and the base neu-tralizes some of the acid. The curve then levels off, and the pH remains more orless constant as we add base because some of the HOAc present initially is con-verted into OAc� ions to form a buffer solution. The pH of the buffer solution staysrelatively constant until most of the acid has been converted to its conjugate base.At that point the pH rises rapidly because essentially all of the HOAc in the sys-tem has been converted into OAc� ions and the buffer is exhausted. The pH thengradually levels off as the solution begins to look like a 0.10 M NaOH solution.

The titration curve for a weak base, such as ammonia, titrated with a strongacid, such as hydrochloric acid, would be analogous to the curve in Figure 11.13.

¢

Kb = 5.6 * 10- 10=

[HOAc][OH-]

[OAc-]L

¢C2

5.0 * 10- 2 M

[OAc-] =

2.5 * 10- 3 mol OAc-

0.05000 L= 5.0 * 10- 2 M OAc-

OAc-(aq) + H2O(l) uv HOAc(aq) + OH-(aq)


c11AcidsandBases.qxd 8/14/10 2:55 PM 8

The principal difference is that the pH value for this titration curve would be highat first and decrease as acid is added.


E x e r c i s e 1 1 . 1 4Figure 11.14 is the titration curve for the strong acid–weak base titration of25.00 mL of a 0.1000 M NH3 solution titrated by a 0.1000 M solution of HCl.Calculate the pH at points A, B, C, and D.

Figure 11.14 Titration curve for thetitration of 25.00 mL of a 0.1000 M NH3

solution by a 0.1000 M HCl solution.

SolutionPoint A: At start of the titration the NH3 concentration is 0.1000 M. The pHis calculated from:

C2>0.1000 M � 1.8 � 10�5, C � 1.3 � 10�3M � [OH�], pOH � 2.9 and pH � 11.1

Point B: Halfway to the equivalence point [NH4�] � [NH3] and

1.8 � 10�5 � [OH�], pOH � 4.7, pH � 9.3

Point C. At the equivalence point all of NH3 had been titrated, and only NH4�

remains in solution. The concentration of the NH4� is given by

C � [H3O�] � 5.3 � 10�6M, pH � 5.3¢

Ka = 5.6 * 10- 10=

[H3O+][NH3]

[NH4+]

=

¢C2

0.0500M


+(aq) + NH3(aq)

[NH +

4 ] =

(0.1000 M)(0.02500 L)

0.05000 L= 0.05000 M

¢¢

Kb = 1.8 * 10- 5=

[NH4+][OH-]

[NH3]


0 5 10 15 20 25 30 35 400

2

4

6

8

10

12

Volume HCI (mL)

B

A

C

D

pH



Point D. 35.00 mL of HCl solution have been added; this is 10.00 mL in excessof the end point:

0.01000 L * 0.1000 M

0.0600 L= [H3O

+] = 1.70 * 10- 2M pH = 1.8➤ CHECKPOINTDescribe what happens to the pH of anHCl solution as NaOH is slowly added.

Key TermsAcid–base indicatorAcid dissociation equilibrium

constant (Ka)Acidic bufferAcidsAlkaliesArrhenius acid/baseBase ionization equilibrium

constant (Kb)Base ionization equilibrium constant

expressionBases

Basic bufferBrønsted acid/baseBufferBuffer capacityConjugate acid–base pairEnd pointEquivalence pointHydrogen-ion acceptor/donorHydronium ionLeveling effectMonoprotic acidNeutralization reaction

pHpOHProton acceptor/donorStrong acidStrong baseTitrationWater dissociation equilibrium

constant (Kw)Weak acidWeak base

ProblemsProperties of Acids and Bases

1. Describe how acids and bases differ. Give examples ofsubstances from daily life that fit each category.

2. Many metals dissolve in acids. Write an equation for thereaction between magnesium metal in hydrochloric acid.

3. How can litmus be used to distinguish between an acidand a base?

4. What happens when acids and bases are combined?

The Arrhenius Definition of Acids and Bases5. Assume that the reaction between water and either

HBr, H2SO4, FeCl3, or Fe(OH)3 can be described bythe following equations.

Which of the following compounds are Arrhenius acids?(a) HBr (b) H2SO4 (c) FeCl3 (d) Fe(OH)3

6. Assume that the reaction between water and eitherLiOH, H2S, Na2SO4, or NH4Cl can be described by thefollowing equations.

Fe(OH)3(s) uvH2O

Fe3+(aq) + 3 OH-(aq)

FeCl3(s) ¡

H2OFe3 +(aq) + 3 Cl-(aq)

H2SO4(aq) ¡

H2OHSO4

-(aq) + H+(aq)

HBr(aq) ¡

H2OH+(aq) + Br-(aq)

Which of the following compounds are Arrhenius bases?(a) LiOH (b) H2S(c) Na2SO4 (d) NH4Cl

The Brønsted–Lowry Definition of Acids and Bases

7. Write balanced equations to show what happens whenhydrogen bromide dissolves in water to form an acidicsolution and when ammonia dissolves in water to forma basic solution.

8. Give an example of an acid whose formula carries apositive charge, an acid that is electrically neutral, andan acid that carries a negative charge.

9. The following compounds are all oxyacids. In eachcase, the acidic hydrogen atoms are bound to oxygenatoms. Write the skeleton structures for the acids.(a) H3PO4 (b) HNO3 (c) HClO4

(d) H3BO3 (e) H2CrO4

NH4Cl(s) ¡

H2ONH4

+(aq) + Cl-(aq)

Na2SO4(s) ¡

H2O2 Na+(aq) + SO4

2 -(aq)

H2S(aq) ¡

H2OH+(aq) + HS-(aq)

LiOH(s) ¡

H2OLi+(aq) + OH-(aq)


10. Describe how the Brønsted definition of an acid can beused to explain why compounds that contain hydrogenwith an oxidation number of �1 are often acids.

11. Which of the following compounds cannot be Brøn-sted bases?

(a) H3O� (b) MnO4� (c) BH4

�

(d) CN� (e) S2�

12. Which of the following compounds cannot be Brøn-sted bases?

(a) O22� (b) CH4 (c) PH3

(d) SF4 (e) CH3�

13. Label the Brønsted acids and bases in the followingreactions.

(a)

(b)

(c)

(d)

(e)

14. Write the chemical equation for the Brønsted bases, re-acting with water.

(a) NH3 (b) CH3NH2 (c) IO�

Conjugate Acid–Base Pairs15. Identify the conjugate acid–base pairs in the following

reactions.(a)(b)

(c)(d)

16. Which of the following is not an example of a conju-gate acid–base pair?

(a) NH4�/NH3 (b) H2O/OH�

(c) H3O�/OH� (d) CH4/CH3�

17. Write Lewis structures for the following Brønstedacids and their conjugate bases.

(a) formic acid, HCO2H (b) methanol, CH3OH

18. Write Lewis structures for the following Brønstedbases and their conjugate acids.

(a) methylamine, CH3NH2

(b) acetate ion, CH3CO2�

19. Identify the conjugate base of each of the followingBrønsted acids.

(a) H3O� (b) H2O (c) OH� (d) NH4�

uv Ca2 +(aq) + 2 Cl-(aq) + H2CO3(aq)CaCO3(s) + HCl(aq)NH3(aq) + H2O(l) uv NH4

+(aq) + OH-(aq)uv H2CO3(aq) + OH-(aq)

HCO3-(aq) + H2O(l)

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

uv CH4(g) + LiNH2(s)LiCH3(l) + NH3(l)

HNO3(aq) + NH3(aq) uv NH4NO3(aq)

uv CaSO4(aq) + 2 HF(aq)CaF2(s) + H2SO4(aq)

uv CH3CO2-(aq) + H2O(l)

CH3CO2H(aq) + OH-(aq)

uv H3O+(aq) + SO4

2-(aq)HSO4

-(aq) + H2O(l)

20. Identify the conjugate base of each of the followingBrønsted acids.

(a) HPO42� (b) H2PO4

�

(c) HCO3� (d) HS�

21. Identify the conjugate acid of each of the followingBrønsted bases.

(a) O2� (b) OH�

(c) H2O (d) NH2�

22. Acids (such as hydrochloric acid and sulfuric acid) re-act with bases (such as sodium hydroxide and ammo-nia) to form salts (such as sodium chloride, sodium sul-fate, ammonium chloride, and ammonium sulfate).Write balanced equations for each acid reacting witheach base.

The Role of Water in the Brønsted Model23. True or false? Water is both an acid and a base. Provide

support for your answer.

24. Write an equation that shows how a base reacts withwater. Do the same for an acid.

25. Give the Lewis structures for water and acetic acid,CH3CO2H. Write an equation that describes the inter-action between water and acetic acid using Lewisstructures.

To What Extent Does Water Dissociate to Form Ions?26. Explain why the concentrations of the H3O� ion and

the OH� ion in pure water are the same.

27. Explain why it is impossible for water to be at equilib-rium when it contains large quantities of both theH3O� and OH� ions.

28. The dissociation of water is an endothermic reaction.

Use Le Châtelier’s principle to predict what shouldhappen to the fraction of water molecules that dissoci-ate into ions as the temperature of water increases.

29. Use Le Châtelier’s principle to explain why adding ei-ther an acid or a base to water suppresses the dissocia-tion of water.

30. What is the difference between Kc and Kw for water?Calculate a value of Kc for the dissociation of H2O toH3O� and OH�.

31. When an acid is added to pure water, there are twosources for H3O�. What are they? Similarly, when abase is added to pure water, there are two sources forOH�. What are they?

32. Does the product of [H3O�] and [OH�] always have tobe 10�14 at 25�C? Explain.

¢H° = 55.84 kJ/molrxn


PROBLEMS 521


33. If the [H3O�] concentration in an aqueous solution is1.0 � 10�12, what is the [OH�]?

pH as a Measure of the Concentration of H3O� Ion34. Calculate the number of H3O� and OH� ions in 1.00

mL of pure water.35. Calculate the H3O� and OH� ion concentrations in a

solution that has a pH of 3.72.36. Explain why the pH of a solution decreases when the

H3O� ion concentration increases.37. What happens to the concentration of the H3O� ion

when a strong acid is added to water? What happens tothe concentration of the OH� ion? What happens to thepH of the solution?

38. What happens to the OH� ion concentration when astrong base is added to water? What happens to theconcentration of the H3O� ion? What happens to thepH of the solution?

39. Which of the following solutions is the most acidic?(a) 0.10 M acetic acid, pH � 2.9(b) 0.10 M hydrogen sulfide, pH � 4.1(c) 0.10 M sodium acetate, pH � 8.4(d) 0.10 M ammonia, pH � 11.1

40. Which of the following solutions is the most acidic?(a) 0.10 M H3PO4, pH � 1.5(b) 0.10 M H2PO4

�, pH � 4.4(c) 0.10 M HPO4

2�, pH � 9.3(d) 0.10 M PO4

3�, pH � 12.041. Calculate the pH and pOH of a solution for which

[H3O�] is 1.5 � 10�6 M.42. The pH of a 0.10 M solution of Na2CO3 is 11.6. What

are [H3O�] and [OH�]?43. The pH of a saturated solution of H2CO3 is 3.8. What

are [H3O�] and [OH�]?44. Calculate the pH of the following solutions

(a) 1.0 M HOAc [H3O�] � 4.2 � 10�3 M

(b) 0.10 M HOAc [H3O�] � 1.3 � 10�3 M

(c) 0.010 M HOAc [H3O�] � 4.2 � 10�4 M

45. The pH of a 0.10 M solution of ammonia is 11.1. Findthe hydroxide concentration.

Relative Strengths of Acids and Bases46. Define the following terms: weak acid, strong acid,

weak base, and strong base.47. Describe the difference between strong acids (such as

hydrochloric acid) and weak acids (such as acetic acid)and the difference between strong bases (such assodium hydroxide) and weak bases (such as ammonia).

48. Describe the relationship between the acid dissociationequilibrium constant for an acid, Ka, and the strength ofthe acid.

49. Use the table of acid dissociation equilibrium constantsin Table B.8 in Appendix B to classify the followingacids as either strong or weak.(a) acetic acid, CH3CO2H(b) boric acid, H3BO3

(c) chromic acid, H2CrO4

(d) formic acid, HCO2H(e) hydrobromic acid, HBr

50. Which of the following is the weakest Brønsted acid?(a) H2S2O3, Ka � 0.3(b) H2CrO4, Ka � 9.6(c) H3BO3, Ka � 7.3 � 10�10

(d) C6H5OH, Ka � 1.0 � 10�10

51. Which of the following solutions is the most acidic?(a) 0.10 M CH3CO2H, Ka � 1.8 � 10�5

(b) 0.10 M HCO2H, Ka � 1.8 � 10�4

(c) 0.10 M ClCH2CO2H, Ka � 1.4 � 10�3

(d) 0.10 M Cl2CHCO2H, Ka � 5.1 � 10�2

52. Which of the following compounds is the strongest base?(a) CH3CO2

� (for CH3CO2H, Ka � 1.8 � 10�5)(b) HCO2

� (for HCO2H, Ka � 1.8 � 10�4)(c) ClCH2CO2

� (for ClCH2CO2H, Ka � 1.4 � 10�3)(d) Cl2CHCO2

� (for Cl2CHCO2H, Ka � 5.1 � 10�2)53. List acetic acid, chlorous acid, hydrofluoric acid, and

nitrous acid in order of increasing strength if 0.10 Msolutions of the acids contain the following equilib-rium concentrations.

Acid [HA] [A�] [H3O�]

HOAc 0.099 M 0.0013 M 0.0013 MHOClO 0.072 M 0.028 M 0.028 MHF 0.092 M 0.0081 M 0.0081 MHNO2 0.093 M 0.0069 M 0.0069 M

Relative Strengths of Conjugate Acid–Base Pairs54. What is the relationship between the strength of a base

and its conjugate acid?55. Which of the following conjugate bases have essen-

tially no base properties in water solvent? I�, ClO4�,

F�, NO2�, NO3

�. Explain why not.56. Is the conjugate base of a weak acid a relatively strong

or weak base? Explain.57. In the Brønsted model of acid–base reactions, what

does the statement that HCl is a stronger acid than H2Omean in terms of the following reaction?

58. In the Brønsted model of acid–base reactions, whatdoes it mean to say that NH3 is a weaker acid than H2O?




59. What can you conclude from experiments suggestingthat the following reaction proceeds far to the right, aswritten?

60. What can you conclude from experiments suggestingthat the following reaction proceeds far to the right, aswritten?

61. What can you conclude from experiments suggestingthat the following reaction does not occur to an appre-ciable extent?

Relative Strengths of Different Acids and Bases62. Refer to Table 11.3 and Tables B.8 and B.9 in Appen-

dix B. For each pair of reactants, state whether the re-action proceeds nearly to completion, makes only asmall amount of product, or doesn’t proceed to a de-tectable amount.(a) H2SO4(aq) � NaOH(aq)(b) H2O(aq) � HNO3(aq)(c) HF(aq) � NaOH(aq)(d) HCl(aq) � NH3(aq)(e) C5H5NH(aq) � HNO2(aq)

63. Describe what is meant by the leveling effect of water.64. Why does a 0.10 M solution of HI contain the same

[H3O�] as a 0.10 M solution of HBr? Would a 0.10 Msolution of HI have the same [H3O�] as a 0.10 M solu-tion of HF? Explain.

65. Explain why the H3O� ion concentration in a strongacid solution depends on the concentration of the solu-tion but not on the value of Ka for the acid.

Relationship of Structure to Relative Strengthsof Acids and Bases66. Use the structures of the acids to explain the following

observations.(a) H2SO4 is a stronger acid than HSO4

�.(b) HNO3 is a stronger acid than HNO2.(c) H2S is a stronger acid than H2O.(d) H2S is a stronger acid than PH3.

67. Arrange the following in order of increasing acidstrength. Explain.(a) H3AsO4 (b) H2AsO4

�

(c) HAsO42� (d) AsO4

3�

68. Arrange the following in order of increasing acidstrength. Explain.(a) SiH4 (b) HCl (c) H2S (d) PH3

¡ HCO2H(aq) + OH-(aq)HCO2-(aq) + H2O(l)

HCl(aq) + NH3(aq) ¡ NH4+(aq) + Cl-(aq)

HBr(aq)H2O(l) ¡ H3O+(aq) + Br-(aq)

69. Which of the following is the strongest Brønsted acid?Explain.(a) H2Se (b) H2O (c) H2S (d) H2Te

70. Which of the following is the weakest Brønsted acid?Explain.(a) HBrO2 (b) HBrO(c) HBrO4 (d) HBrO3

71. Arrange the following acids in order of increasing acid-ity. Explain. HOCl, HOI, HOBr. Look up the Ka valuesin Appendix B8. Does your answer agree? Explain why.

72. Which acid, NH4� or PH4

�, has the largest Ka? Explain.73. Arrange the following acids in order of increasing Ka.

Explain your order.

Strong Acid pH Calculations74. Calculate the pH and pOH of a 0.035 M HCl solution.75. Calculate the pH and pOH of a solution that contains

0.568 g of HCl per 250 mL of solution.76. What would be the pH of a 0.010 M solution of any

strong acid? Explain.77. What would be the pH of the following solutions?

(a) 1.0 M HClO4

(b) 0.568 g of HClO4 per 250 mL of solution(c) 0.568 g of HNO3 per 250 mL of solution(d) 1.14 g of HBr per 500 mL of solution

78. Calculate the pH of a 0.056 M solution of hydroiodicacid, HI.

79. Nitric acid is often grouped with sulfuric acid and hy-drochloric acid as one of the strong acids. Calculate thepH of 0.10 M nitric acid, assuming that it is a strongacid that dissociates completely. Calculate the pH ofthe solution using the value of Ka for the acid (forHNO3, Ka � 28).

Weak Acid pH Calculations80. Describe the two assumptions that are commonly made

in weak acid equilibrium problems. Describe how youcan test whether the assumptions are valid for a partic-ular calculation.

81. Explain why the techniques used to calculate the equi-librium concentrations of the components of a weakacid solution can’t be used for either strong acids orvery weak acids.

82. Explain why the H3O� ion concentration in a weakacid solution depends on both the value of Ka for theacid and the concentration of the acid.

83. Which of the following solutions has the largest H3O�

ion concentration?(a) 0.10 M HOAc (b) 0.010 M HOAc(c) 0.0010 M HOAc

CH3COOH, CH2ClCOOH, CCl3COOH

PROBLEMS 523


84. Calculate the percentage of HOAc molecules that dis-sociate in 0.10 M, 0.010 M, and 0.0010 M solutions ofacetic acid. What happens to the percent ionization asthe solution becomes more dilute? (For HOAc, Ka �1.8 � 10�5.)

85. Formic acid (HCO2H) was first isolated by the destruc-tive distillation of ants. In fact, the name comes from theLatin word for “ants,” formi. Calculate the HCO2H,HCO2

�, and H3O� concentrations in a 0.100 M solutionof formic acid in water. (For HCO2H, Ka � 1.8 � 10�4.)

86. Hydrogen cyanide (HCN) is a gas that dissolves in wa-ter to form hydrocyanic acid. Calculate the H3O�,HCN, and CN� concentrations in a 0.174 M solutionof hydrocyanic acid. (For HCN, Ka � 6 � 10�10.)

87. The first disinfectant used by Joseph Lister was calledcarbolic acid. The substance is now known as phenol.Calculate the H3O� ion concentration in a 0.0167 Msolution of phenol (C6H5OH). (Ka � 1.0 � 10�10.)

88. Calculate the concentration of acetic acid that wouldgive an H3O� ion concentration of 2.0 � 10�3M. (ForHOAc, Ka � 1.8 � 10�5.)

89. Calculate the value of Ka for ascorbic acid (vitamin C)if 2.8% of the ascorbic acid molecules in a 0.100 Msolution dissociate.

90. Calculate the value of Ka for nitrous acid (HNO2) if a0.100 M solution is 7.1% dissociated at equilibrium.

91. Why is it wrong to assume that the pH of a 10�8M HClsolution is 8?

92. The pH of a 0.10 M solution of HOAc is 2.9. CalculateKa for acetic acid.

93. The [H3O�] of a 0.10 M solution of HOCl is 5.4 �10�5 M. Calculate the pH of this solution and the Ka forHOCl. How would the pH of this solution compare tothat of a 0.10 M solution of HNO3?

94. The pOH of a 0.10 M solution of HF is 11.9. CalculateKa for HF.

Base pH Calculations95. Which of the following equations correctly describes

the relationship between Kb for the formate ion(HCO2

�) and Ka for formic acid (HCO2H)?(a) Ka � Kw � Kb (b) Kb � Ka>Kw

(c) Kb � Kw>Ka (d) Kb � Kw � Ka

(e) Kb � Kw�Ka

96. Use the relationship between Ka for an acid and Kb forits conjugate base to explain why strong acids haveweak conjugate bases and weak acids have strong con-jugate bases.

97. Calculate the HCO2H, OH�, and HCO2� ion concen-

trations in a solution that contains 0.020 mol of sodiumformate (NaHCO2) in 250 mL of solution. (ForHCO2H, Ka � 1.8 � 10�4.)

98. Calculate the OH�, HOBr, and OBr� ion concentra-tions in a solution that contains 0.050 mol of sodiumhypobromite (NaOBr) in 500 mL of solution. (ForHOBr, Ka � 2.4 � 10�9.)

99. Calculate the pH of a 0.756 M solution of NaOAc. (ForHOAc, Ka � 1.8 � 10�5.)

100. A solution of NH3 dissolved in water is known as bothaqueous ammonia and ammonium hydroxide. Use thevalue of Kb for the following reaction to explain whyaqueous ammonia is the better name.

101. At 25�C, a 0.10 M aqueous solution of methylamine(CH3NH2) is 6.8% ionized.

Calculate Kb for methylamine. Is methylamine astronger base or a weaker base than ammonia?

102. What is the molarity of an aqueous ammonia solutionthat has an OH� ion concentration of 1.0 � 10�3M?

103. What is the pH of a 0.10 M solution of methylamine?See Problem 101.

104. Calculate Kb for hydrazine (H2NNH2) if the pH of a0.10 M aqueous solution of the rocket fuel is 10.54.

105. Calculate the pH of a 0.016 M aqueous solution of cal-cium acetate, Ca(OAc)2. (For HOAc, Ka � 1.8 � 10�5.)

106. Arrange the following 0.10 M solutions in order of in-creasing acidity.(a) KCl (b) KF (c) KOAc

107. Which of the following 0.10 M solutions will beacidic? Explain.(a) NH4Cl (b) NaF (c) RbNO3

108. Which of the following 0.10 M solutions will be basic?Explain.(a) Li2SO4 (b) KClO4 (c) NaNO2

109. Identify the conjugate acid produced by the dissocia-tion of the anion of each of the following salts.(a) Na2HPO4 (b) NaHCO3

(c) Na2SO4 (d) NaNO2

Mixtures of Acids and Bases: Buffers110. What is meant by the term common ion? If NaOAc is

added to a solution of HOAc, which way does the acidequilibrium shift? What happens to the pH?

111. What is the pH of a 0.50 M solution of HF? What is thepH of a solution containing 0.50 M HF and 0.50 MNaF? See Appendix B8 for the equilibrium constant.

112. Determine the pH of a mixture containing 0.10 Mformic acid and 0.10 M sodium formate. See AppendixB8 for the equilibrium constant.

uv CH3NH3+(aq) + OH-(aq)

CH3NH2(aq) + H2O(l)

Kb = 1.8 * 10- 5NH3(aq) + H2O(l) uv NH4

+(aq) + OH-(aq)



113. Determine the pH of a mixture containing 0.10 M am-monia and 0.10 M ammonium chloride. See AppendixB8 for the equilibrium constant.

114. Determine the pH of a solution formed by adding 15 gof benzoic acid and 10 g of sodium benzoate to waterto form 1.0 liter of solution. See Appendix B8 for theequilibrium constant.

Buffers and Buffer Capacity115. Explain how buffers resist changes in pH.116. Explain why a mixture of HOAc and NaOAc is an

acidic buffer, but a mixture of NH3 and NH4Cl is a ba-sic buffer. (For HOAc, Ka � 1.8 � 10�5; for NH4

�, Ka

� 5.6 � 10�10.)117. Which of the following mixtures in equal concentra-

tions would make the best buffer?(a) HCl and NaCl(b) NaOAc and NH3

(c) HOAc and NH4Cl(d) NaOAc and NH4Cl(e) NH3 and NH4Cl

118. Which of the following solutions is an acidic buffer?(a) 0.10 M HCl and 0.10 M NaOH(b) 0.10 M HCl and 0.10 M NaCl(c) 0.10 M HCO2H and 0.10 M NaHCO2

(d) 0.10 M NH3 and 0.10 M NH4Cl119. Which of the following solutions is a basic buffer?

(a) 0.10 M HCl and 0.10 M NaOH(b) 0.10 M HCl and 0.10 M NaCl(c) 0.10 M HCO2H and 0.10 M NaHCO2

(d) 0.10 M NH3 and 0.10 M NH4Cl120. What is the best way to increase the capacity of a

buffer made by dissolving NaHCO2 in an aqueous so-lution of HCO2H? Explain.(a) Increase the concentration of HCO2H.(b) Increase the concentration of NaHCO2.(c) Increase the concentrations of both HCO2H and

NaHCO2.(d) Increase the ratio of the concentration of HCO2H

to the concentration of NaHCO2.(e) Increase the ratio of the concentration of NaHCO2

to the concentration of HCO2H.121. In order to maintain a pH of 5.60 what must be the ra-

tio [HA]/[A�] for the following weak acids?

122. What must be the ratio of the conjugate base to that ofthe conjugate acid if the pH of a solution based on the

formic acid Ka = 1.8 * 10- 4

chlorous acid Ka = 1.1 * 10- 2

acetic acid Ka = 1.75 * 10- 5

following reaction is to maintained at 9.00? Note thatKb � 1.7 � 10�9 for this base.

123. What would be the pH of a buffer prepared from 0.50 MNa2CO3 and 0.50 M NaHCO3 solutions?

124. Calculate the pH of a solution containing 0.50 M aceticacid and 0.25 M sodium acetate. Ka � 1.75 � 10�5

Acid–Base Reactions125. Which of the following reactions would be expected to

go nearly to completion?(a) strong acid–strong base(b) weak acid–weak base(c) weak base–strong acid(d) weak acid–strong base

126. It is possible to predict the pH of the solution formed inthe following reaction by knowing only the concentra-tions and volumes of the reactants. Why can this bedone?

127. Suppose 100 mL each of 0.10 M HClO4 and KOH aremixed. What will be the pH of the solution?

128. If 50 mL of 0.10 M HClO4 is mixed with 100 mL of0.10 M KOH, what will be the resulting pH?

129. Write a net ionic equation for the reaction of thefollowing.(a) HOBr and NaOH(b) HOAc and NH3

(c) HBr and KOH(d) HCO2H and Ba(OH)2

Which of these reactions will go to completion?130. What will be the equilibrium constant for the reaction

between HOI(aq) and C5H5N(aq) (pyridine)? Whatdoes the magnitude of the equilibrium constant tell usabout this reaction?

pH Titration Curves131. Define the following: titration, indicator, end point,

and equivalence point.132. Describe in detail the experiment you would use to

measure the value of Ka for formic acid, HCO2H.133. Sketch a titration curve of 0.10 M aniline (Kb � 4.0 �

10�10) being titrated with a 0.10 M HCl solution. Labelthe equivalence point and the point at which the OH�

ion concentration is equal to Kb for the base.134. An indicator has a Ka � 1.0 � 10�7. At what pH will

the indicator change color?

HClO4(aq) + KOH(aq) ¡ H2O(l) + KClO4(aq)

uv C5H5NH+(aq) + OH-(aq)C5H5N(aq) + H2O(l)

PROBLEMS 525


135. For a solution with a pH of 10.0, what will be thecolor of the solution in the presence of the followingindicators?(a) methyl orange(b) cresol red(c) phenolphthalein(d) alizarin yellow

Integrated Problems136. Explain why pure (nonpolluted) rainwater has a pH of

5.6. Explain why boiling water to drive off the CO2

raises the pH.137. On April 10, 1974, at Pitlochry, Scotland, the rain was

found to have a pH of 2.4. Calculate the H3O� ion con-centration in the rain and compare it with the H3O� con-centration in 0.10 M acetic acid, which has a pH of 2.9.

138. Which of the following statements are true for a 0.10 Msolution of the weak acid CH3CO2H in water? Explainwhat is wrong with any incorrect answers.(a) pH � 1.00(b) [H3O�] [CH3COO�](c) [H3O�] � [CH3COO�](d) pH � 1.00

139. Which of the following statements are true for a 1.0 Msolution of the strong acid HCl in water? Explain whatis wrong with any incorrect answers.(a) [Cl�] � [H3O�] (b) pH � 0.00(c) [H3O�] � 1.0 M (d) [HCl] � 1.0 M

140. Which of the following statements are true for a 1.0 Msolution of the weak base NH3 in water? Explain whatis wrong with any incorrect answers.(a) [OH�] � [H3O�] (b) [NH4

�] � [OH�](c) pH � 7.00 (d) [NH3] � 1.0 M

141. The pH of a 0.10 M solution of formic acid is 2.37.

(a) Which of the two hydrogens in the formic acidstructure is the “acidic” hydrogen? Explain.

(b) Write a chemical equation that describes what hap-pens when formic acid is placed in water.

(c) Calculate the Ka of formic acid.(d) Estimate the pH of a 0.10 M solution of the fol-

lowing acid in water. Explain your estimate.

OBC

D GCl OOH

OS

SS

OBC Formic acid

D GH OOH

OS

SS

W

142. Methylamine, CH3NH2, is a weak base with Kb � 3.6 �10�4.(a) Write a chemical equation that describes what hap-

pens when methylamine is placed into water.(b) What is the pOH of a 0.10 M solution of methy-

lamine?(c) What is the pH of a 0.10 M solution of methy-

lamine?143. Rank the following solutions in order of increasing pH.

Explain your reasoning.(a) 0.10 M HOBrO2 (b) 0.10 M HOBrO(c) 0.10 M HBr (d) 0.10 M NaBr(e) 0.10 M NaBrO2 (f) 0.10 M NaBrO3

144. Describe a neutral (in terms of acid–base characteris-tics) solution. The Kw for the self-hydrolysis of H2O atroom temperature (25�C) is 1 � 10�14, and that atbody temperature (37�C) is 2.5 � 10�14.(a) Describe the quantitative difference in the self-

hydrolysis of pure water at 25�C and 37�C.(b) Determine the approximate pH of pure water at

25�C and 37�C.(c) Again, describe a neutral solution. Is this the same

description that you gave at the beginning of theproblem?

145. Match each of the following to its corresponding par-ticulate representation. Water molecules are not shown.Explain your reasoning.(a) a dilute solution of a strong acid(b) a concentrated solution of a weak acid(c) a good buffer


= A– = H+ = H3O+ = HA

(i) (ii)

(iii) (iv)


146. When NH4Cl is dissolved in water, the followingspecies will be present: NH4

�, Cl�, H2O, OH�, H3O�,and NH3. Using the symbols �, and arrangethe species in order of increasing concentration in thesolution.

147. The weak acid HB can be represented as � , where represents B� and ° represents H�. Match the conju-gate base of HB and the conjugate acid of HB with itsappropriate particulate representation.(a) � � (b) � (c) (d) � � (e) �

148. Which of the following compounds could dissolve inwater to give a 0.10 M solution with a pH of about 5?(a) NH3 (b) NaCl (c) HCl(d) KOH (e) NH4Cl

149. Write balanced chemical equations for the followingacid–base reactions.(a) Al2O3(s) � HCl(aq)(b) CaO(s) � H2SO4(aq)(c) Na2O(s) � H2O(l)(d) MgCO3(s) � HCl(aq)(e) NaOH(s) � H3PO4(aq)

150. Which of the following factors influences the value ofKa for the dissociation of formic acid?

(a) temperature(b) pressure(c) pH(d) the initial concentration of HCO2H(e) the initial concentration of the HCO2

� ion151. A. 1.0 mol of the acid HBr is placed in sufficient water

to make 1.0 L of solution. Ka � 1 � 109.(a) Write an equation that describes the reaction

that takes place.(b) What will be [H3O�] in the solution?(c) What is the pH?(d) What will be [OH�] in the solution?(e) What is the conjugate base of HBr?

B. 1.0 mol of the base ammonia, NH3, is placed in suf-ficient water to make 1.0 L of solution. Ka � 1.8 �10�5.(a) Write an equation that describes the reaction

that takes place.(b) What is [OH�]?(c) What is [NH3]?(d) What is the concentration of the conjugate acid

of ammonia?(e) What is the pH?

uv HCO2-(aq) + H3O

+(aq)

HCO2H(aq) + H2O(l)

¡

¡

¡

¡

¡

L ,V ,

C. 1.0 mol of NaBr is placed in sufficient water tomake 1.0 L of solution.

(a) What is [Br�] in this solution?(b) Write an equation describing the possible reaction of

Br� with water. What do you think the equilibriumconstant for this reaction might be? Explain yourreasoning. Part A of this question might be useful.

(c) What is the pH of this solution?152. The following Ka values have been experimentally de-

termined for formic and acetic acids.

(a) Which is the stronger acid? State your reasoning.(b) Write a chemical equation that describes what hap-

pens when formic acid is placed into water. UseLewis structures for products and reactants similarto those above.

(c) What is the pH of a 0.10 M solution of formic acid? Show all calculations. State and justify anyassumptions made.

(d) Which is the stronger base, HCO2� or CH3CO2

�?Explain your answer.

(e) Will the equilibrium constant for the reaction

be greater than 1 or less than 1? Explain.(f ) Which of the following acids is the strongest?

Explain.

B B

OHGD

C

H OHGD

C

Cl

B B

OHGD

C

Br OHGD

C

F

OO S OO S

OO S

OS

OO S

SOS

S

OS

SOS

S

uv CH3CO2-(aq) + HCO2H

CH3CO2H(aq) + HCO2-(aq)

B

OHGD

CH

H H

CG

GD

B

OHG

H

CD

Acetic acidKa � 1.8 � 10�5

Formic acidKa � 1.8 � 10�4

OO S

OO S

OS

S

OS

S

NaBr(s) uv Na+(aq) + Br-(aq) Kc W 1

PROBLEMS 527


153. Ka for HNO2 is 5.1 � 10�4.(a) Write an equation that describes the reaction of

HNO2 with H2O.(b) What is the pH of a 1.0 M solution of HNO2?(c) What is the pH of a 1.0 M solution of NaNO2?(d) What is the pH of a 1.0 M solution of HNO3?(e) Which of the following diagrams best describes a

1.0 M solution of HNO3? Explain.

155. Estimate the pH of each of the following solutions.Explain in each case.(a) 0.10 M HCl (b) 0.10 M NaCl(c) 0.10 M HOCl (d) 0.10 M NaF(e) 0.10 M NaOH

156. Calculate the pH of the following solutions.(a) a 2.0 M solution of HOCl; Ka � 2.9 � 10�8

(b) a 1.0 M solution of HF; Ka � 7.2 � 10�4

(c) a 0.02 M solution of HCl(d) a 1.0 M solution of NaF

157. Arrange the following 0.10 M solutions in order of in-creasing acidity. Explain.KI

KOH

158. 50.0 mL of 0.100 M HCl is titrated with 0.100 MNaOH. What is the pH at the start of the titration?What will be the pH after the addition of 20.0, 50.0,and 60.0 mL of NaOH? What is the pH at the equiva-lence point?

159. 50.0 mL of 0.100 M HOAc is titrated with 0.100 MNaOH. Calculate the pH at the beginning of the titra-tion and after the addition of 20.0, 50.0, and 60.0 mL ofNaOH.

160. Roughly sketch the titration curve for the titration of50.0 mL of 0.250 M NH3 with 0.500 M HCl. What willbe the initial pH? What volume of HCl is required toreach the equivalence point? What will be the pH at theequivalence point?

161. Which of the following ions is the conjugate base of astrong acid?(a) OH� (b) HSO4

� (c) NH2�

(d) S2� (e) H3O�

162. Many insects leave small quantities of formic acid,HCO2H, behind when they bite. Explain why the itch-ing sensation can be relieved by treating the bite withan aqueous solution of ammonia (NH3) or baking soda(NaHCO3).

(HClO4)HOClO3

NH4ClO4

(HClO3)HOClO2

KClO3


HNO3

HNO3

HNO3

HNO3

HNO3

HNO3

NO3−

NO3− NO3

−NO3

−

NO3−

NO3−

H+

H+

H+

H+

H+

H+

(a) (b)

(c)

154. Arrange the following 0.10 M solutions in order of in-creasing pH. Show clearly which has the highest andwhich has the lowest pH. Explain.CsClHOClHOClONaOClNaOClO


529

Chapter ElevenSPECIAL TOPICS

11A.1 Diprotic Acids

11A.2 Diprotic Bases

11A.3 Compounds That Could Be Either Acids or Bases


11A.1 Diprotic AcidsThe problems discussed so far that have involved acids have focused on mono-protic acids that each have a single H� ion, or proton, that can be donated whenthey act as Brønsted acids. Hydrochloric acid (HCl), acetic acid (CH3CO2H), andnitric acid (HNO3) are all monoprotic acids.

There are also important acids that can be classified as polyprotic acidsbecause they can lose more than one H� ion when they act as Brønsted acids.Diprotic acids, such as sulfuric acid (H2SO4), carbonic acid (H2CO3), hydrogensulfide (H2S), chromic acid (H2CrO4), and oxalic acid (H2C2O4), have two acidichydrogen atoms. Triprotic acids, such as phosphoric acid (H3PO4) and citric acid(C6H8O7), have three.

Naturally occurring diprotic and triprotic acids such as citric acid, fumaricacid, maleic acid, oxalic acid, and succinic acid play an important role in bio-logical systems. Furthermore, all of the amino acids used to form proteins arepolyprotic acids. Glycine, for example, is a diprotic acid, whereas other aminoacids such as aspartic acid are triprotic acids. Other diprotic acids such as adipicacid and terephthalic acid are used as starting materials from which syntheticfibers such as nylon and polyester are made. Each year more than 4 billion poundsof terephthalic acid and more than 5 billion pounds of adipic acid are used in theproduction of synthetic fibers.

Table 11A.1 gives values of Ka for some common polyprotic acids. Thereis usually a large difference in the ease with which polyprotic acids lose the firstand second (or second and third) protons. Because sulfuric acid is classified as astrong acid, it is often assumed that it loses both of its protons when it reacts withwater. That isn’t a legitimate assumption. Sulfuric acid is a strong acid becauseKa for the loss of the first proton is much larger than 1. We therefore assume thatessentially all the H2SO4 molecules in an aqueous solution lose the first protonto form the HSO4

� (hydrogen sulfate) ion.

H2SO4(aq) + H2O(l) ¡ H3O+(aq) + HSO4

-(aq) Ka1 = 1 * 103

530 CHAPTER 11 / SPECIAL TOPICS

Table 11A.1Acid Dissociation Equilibrium Constants for Common Polyprotic Acids

Acid Ka1 Ka2 Ka3

Sulfuric acid (H2SO4) 1.0 � 103 1.2 � 10�2

Chromic acid (H2CrO4) 9.6 3.2 � 10�7

Oxalic acid (HO2CCO2H) 5.4 � 10�2 5.4 � 10�5

Sulfurous acid (H2SO3) 1.7 � 10�2 6.4 � 10�8

Maleic acid (cis- HO2CCH“CHCO2H) 1.2 � 10�2 5.4 � 10�7

Phosphoric acid (H3PO4) 7.1 � 10�3 6.3 � 10�8 4.2 � 10�13

Glycine (H2NCH2CO2H) 4.6 � 10�3 2.5 � 10�10

Fumaric acid (trans-HO2CCH“CHCO2H) 9.3 � 10�4 3.6 � 10�5

Citric acid (C6H8O7) 7.5 � 10�4 1.7 � 10�5 4.0 � 10�7

Terephthalic acid (HO2CC6H4CO2H) 2.9 � 10�4 3.5 � 10�5

Adipic acid (HO2C(CH2)4CO2H) 3.7 � 10�5 3.9 � 10�6

Carbonic acid (H2CO3) 4.5 � 10�7 4.7 � 10�11

Hydrogen sulfide (H2S) 1.0 � 10�7 1.3 � 10�13


But Ka for the loss of the second proton is only 1.2 � 10�2, which means thatonly 10% of the H2SO4 molecules in a 1 M solution lose a second proton.

H2SO4 loses both H� ions only when it reacts with a base stronger than water,such as ammonia.

Table 11A.1 gives values of Ka for some common polyprotic acids. The largedifference between the values of Ka for the sequential loss of protons by apolyprotic acid is important because it means we can assume that the acids dis-sociate one step at a time, this assumption is known as stepwise dissociation.

Let’s look at the consequence of the assumption that polyprotic acids loseprotons one step at a time by examining the chemistry of a saturated solution(�0.10 M) of H2S in water. Hydrogen sulfide is the foul-smelling gas that givesrotten eggs their unpleasant odor. It is an excellent source of the S2� ion, how-ever, and is therefore commonly used in introductory chemistry laboratories. H2Sis a weak acid that dissociates in steps. Some of the H2S molecules lose a pro-ton in the first step to form the HS� (or hydrogen sulfide) ion.

A small fraction of the HS� ions formed in the reaction then go on to lose addi-tional H� ions in a second step.

Since there are two steps in the reaction, we can write two equilibrium constantexpressions.

Although each of these equations contains three terms, there are only fourunknowns––[H3O�], [H2S], [HS�], and [S2�]––because the [H3O�] and [HS�]terms appear in both equations. The [H3O�] term represents the total H3O� ionconcentration from both steps and therefore must have the same value in bothequations. Similarly, the [HS�] term, which represents the balance between theHS� ions formed in the first step and the HS� ions consumed in the second step,must have the same value for both equations.

It takes four equations to solve for four unknowns. We already have twoequations: the Ka1 and Ka2 expressions. We are going to have to find either twomore equations or a pair of assumptions that can generate two equations. We canbase one assumption on the fact that the value of Ka1 for H2S is almost a milliontimes larger than the value of Ka2.

This means that only a small fraction of the HS� ions formed in the first step goon to dissociate in the second step. If this is true, most of the H3O� ions at equi-librium come from the dissociation of H2S, and most of the HS� ions formed in

Ka1 W Ka2

Ka2 =

[H3O+][S2 -]

[HS-]= 1.3 * 10- 13

Ka1 =

[H3O+][HS-]

[H2S]= 1.0 * 10- 7

Second step: HS-(aq) + H2O(l) uv H3O+(aq) + S2 -(aq)

First step: H2S(aq) + H2O(l) uv H3O+(aq) + HS-(aq)

HSO4-(aq) + H2O(l) uv H3O

+(aq) + SO42 -(aq) Ka2 = 1.2 * 10- 2

11A.1 DIPROTIC ACIDS 531


this reaction remain in solution. As a result, we can assume that the H3O� andHS� ion concentrations at equilibrium are more or less equal.

We need one more equation, and therefore one more assumption. Note thatH2S is a weak acid (Ka1 � 1.0 � 10�7; Ka2 � 1.3 � 10�13). Thus we can assumethat most of the H2S that dissolves in water will still be present when the solu-tion reaches equilibrium. In other words, we can assume that the equilibrium con-centration of H2S is approximately equal to the initial concentration. Because asaturated solution of H2S in water has an initial concentration of about 0.10 M,we can summarize the second assumption as follows.

We now have four equations in four unknowns.

Because there is always a unique solution to four equations in four unknowns, weare now ready to calculate the H3O�, H2S, HS�, and S2� concentrations at equi-librium in a saturated solution of H2S in water. The relevant information for thiscalculation is illustrated in the drawing in the margin.

Ka1 is so much larger than Ka2 for H2S that we can assume stepwise disso-ciation. We start by assuming that we can work with the equilibrium expressionfor the first step without worrying about the second step for the moment. We there-fore start with the expression for Ka1.

We then invoke one of our assumptions.

Substituting this approximation into the Ka1 expression gives the following equation.

We then invoke the other assumption.

Substituting this approximation into the Ka1 expression gives the following result.

[¢C][¢C]

[0.10]L 1.0 * 10- 7

[H3O+] L [HS-] L ¢C

[H3O+][HS-]

[0.10]L 1.0 * 10- 7

[H2S] L 0.10 M

Ka1 =

[H3O+][HS-]

[H2S]= 1.0 * 10- 7

[H2S] L 0.10 M

[H3O+] L [HS-]

Ka2 =

[H3O+][S2 -]

[HS-]= 1.3 * 10- 13

Ka1 =

[H3O+][HS-]

[H2S]= 1.0 * 10- 7

Second assumption: [H2S] L 0.10 M

First assumption: [H3O+] L [HS-]


Saturatedsolution of

H2S

0.1 MH2S

←H2S

4 unknowns:[H3O+][H2S][HS–][S2–]

H2S + H2OH3O+ + HS–

HS– + H2OH3O+ + S2–


We now solve the approximate equation for ¢C.

If our assumptions are valid, we are three-fourths of the way to our goal.If our assumptions are legitimate, we now know the H2S, H3O�, and HS�

concentrations.

Having extracted the values of three unknowns from the first equilibrium expres-sion, we turn to the second equilibrium expression.

Substituting the known values of the H3O� and HS� ion concentrations into thisexpression gives the following equation.

Because the equilibrium concentrations of the H3O� and HS� ions are about thesame, the S2� ion concentration at equilibrium is approximately equal to the valueof Ka2 for this acid.

It is now time to check our assumptions. Is the dissociation of H2S smallcompared with the initial concentration? Yes. The HS� and H3O� ion con-centrations obtained from the calculation are 1.0 � 10�4 M, which is only0.1% of the initial concentration of H2S. The following assumption is there-fore valid.

Is the difference between the S2� and HS� ion concentrations large enough toallow us to assume that essentially all of the H3O� ions at equilibrium are formedin the first step and that essentially all of the HS� ions formed in this step remainin solution? Yes. The S2� ion concentration obtained from these calculations is109 times smaller than the HS� ion concentration. Thus our other assumption isalso valid.

We can therefore summarize the concentrations of the various components of theequilibrium as follows.

[S2 -] L 1.3 * 10- 13 M

[H3O+] L [HS-] L 1.0 * 10- 4 M

[H2S] L 0.10 M

[H3O+] L [HS-]

[H2S] L 0.10 M

[S2 -] L 1.3 * 10- 13 M

[1.0 * 10- 4][S2 -]

[1.0 * 10- 4]= 1.3 * 10- 13

Ka2 =

[H3O+][S2 -]

[HS-]= 1.3 * 10- 13

[H3S+] L [HS-] L 1.0 * 10- 4 M

[H2S] L 0.10 M

¢C L 1.0 * 10- 4

11A.1 DIPROTIC ACIDS 533


11A.2 Diprotic BasesThe techniques we have used with diprotic acids can be extended to diprotic bases.The only challenge is calculating the values of Kb for the base. Suppose we aregiven the following problem:

Calculate the H2CO3, HCO3�, CO3

2�, and OH� concentrations at equilib-rium in a solution that is initially 0.10 M in Na2CO3. (For H2CO3, Ka1 � 4.5 �10�7 and Ka2 � 4.7 � 10�11.) The relevant information for this calculation areillustrated in the drawing in the margin.

Sodium carbonate dissociates into its ions when it dissolves in water.

The carbonate ion then acts as a base toward water, picking up a pair of protons—one at a time—to form the bicarbonate ion, HCO3

�, and then eventually carbonicacid, H2CO3.

The first step in calculating the concentrations of the various components ofthis solution at equilibrium involves determining the values of Kb1 and Kb2 forthe carbonate ion. We start by writing the Kb expressions for the carbonate ionand comparing them to the Ka expressions for carbonic acid.

The expressions for Kb1 and Ka2 have something in common: They both dependon the concentrations of the HCO3

� and CO32� ions. The expressions for Kb2

and Ka1 also have something in common: They both depend on the HCO3� and

H2CO3 concentrations. We can therefore calculate Kb1 from Ka2 and Kb2 from Ka1.We start by multiplying the top and bottom of the Ka1 expression by the

OH� ion concentration to introduce the [OH�] term.

We then group the terms in the equation as follows.

The first term in the equation is the inverse of the Kb2 expression, and the sec-ond term is the Kw expression.

Ka1 =

1Kb2

* Kw

Ka1 =

[HCO3-]

[H2CO3][OH-]* [H3O

+][OH-]

Ka1 =

[H3O+][HCO -

3 ]

[H2CO3]*

[OH-]

[OH-]

Kb2 =

[H2CO3][OH-]

[HCO -

3 ] Ka1 =

[H3O+][HCO -

3 ]

[H2CO3]

Kb1 =

[HCO3-][OH-]

[CO 2 -

3 ] Ka2 =

[H3O+][CO 2 -

3 ]

[HCO -

3 ]

HCO3-(aq) + H2O(l) uv H2CO3(aq) + OH-(aq) Kb2 = ?

CO32 -(aq) + H2O(l) uv HCO -

3 (aq) + OH-(aq) Kb1 = ?

Na2CO3(s) ¡

H2O2 Na+(aq) + CO 2 -

3 (aq)


0.10 MCO3

2–CO3

2– + H2OHCO3

– + OH–

HCO3– + H2O

H2CO3 + OH–

Na2CO3



Similarly, we can multiply the top and bottom of the Ka2 expression by theOH� ion concentration.

Collecting terms gives the following equation.

The first term in the equation is the inverse of Kb1, and the second term is Kw.

This equation can therefore be rearranged as follows.

We can now calculate the values of Kb1 and Kb2 for the carbonate ion from thecorresponding values of Ka1 and Ka2 for carbonic acid.

We are now ready to do the calculations needed to determine the H2CO3,HCO3

�, CO32�, and OH� concentrations at equilibrium. We start with the Kb1

expression because the CO32� ion is the strongest base in the solution and is there-

fore the best source of the OH� ion.

The difference between Kb1 and Kb2 for the carbonate ion is large enough to sug-gest that most of the OH� ions at equilibrium come from the first step and mostof the HCO3

� ions formed in the first step remain in solution. We can thereforemake the following assumption.

The value of Kb1 is small enough to assume that ¢C should be small com-pared with the initial concentration of the carbonate ion. If this is true, the con-centration of the CO3

2� ion at equilibrium will be roughly equal to the initialconcentration of Na2CO3, which was described in the statement of the problemas 0.10 M.

[CO32 -] L 0.10 M

[OH-] L [HCO3-] L ¢C

Kb1 =

[HCO3-][OH-]

[CO32 -]

Kb2 =

Kw

Ka1=

1.0 * 10- 14

4.5 * 10- 7= 2.2 * 10- 8

Kb1 =

Kw

Ka2=

1.0 * 10- 14

4.7 * 10- 11= 2.1 * 10- 4

Ka2Kb1 = Kw

Ka2 =

1Kb1

* Kw

Ka2 =

[CO32 -]

[HCO3-][OH-]

* [H3O+][OH-]

Ka2 =

[H3O+][CO3

2 -]

[HCO3-]

*

[OH-]

[OH-]

Ka1Kb2 = Kw

11A.2 DIPROTIC BASES 535


Substituting this information into the Kb1 expression gives the following result.

This approximate equation can now be solved for ¢C.

We then use the value of ¢C to calculate the equilibrium concentrations of theOH�, HCO3

�, and CO32� ions.

We now turn to the Kb2 expression.

Substituting what we know about the OH� and HCO3� ion concentrations into

this equation gives the following result.

According to this equation, the H2CO3 concentration at equilibrium is approxi-mately equal to Kb2 for the carbonate ion.

Summarizing the results of our calculations allows us to test the assump-tions made generating the following results.

All of our assumptions are valid. The extent of the reaction between theCO3

2� ion and water to give the HCO3� ion is less than 5% of the initial con-

centration of Na2CO3. Furthermore, most of the OH� ion comes from the firststep, and most of the HCO3

� ion formed in the first step remains in solution.

11A.3 Compounds That Could Be Either Acids or BasesSometimes the hardest part of a calculation is deciding whether the compound isan acid or a base. Consider sodium bicarbonate, for example, which dissolves inwater to give the bicarbonate ion. Relevant information for this calculation is illus-trated in the drawing in the margin.

NaHCO3(s) ¡

H2ONa+(aq) + HCO -

3 (aq)

[H2CO3] L 2.2 * 10- 8 M

[OH-] L [HCO3-] L 0.0046 M

[CO32 -] L 0.10 M

[H2CO3] L 2.2 * 10- 8 M

Kb2 =

[H2CO3][0.0046]

[0.0046]L 2.2 * 10- 8

Kb2 =

[H2CO3][OH-]

[HCO -

3 ]

[OH-] L [HCO -

3 ] L 0.0046 M

[CO 2 -

3 ] L 0.10 M

¢C L 0.0046 M

[¢C][¢C]

[0.10]L 2.1 * 10- 4


NaHCO3

HCO3– + H2O

H2CO3 + OH–

Kb2=2.2 × 10–8

Kb2 >> Ka2 ⇒ HCO3– = base

HCO3– + H2O CO3

2– + H3O+

Ka2 = 4.7 × 10–11


In theory, the bicarbonate ion can act as either a Brønsted acid or a Brønsted basetoward water.

Which reaction predominates? Is the HCO3� ion more likely to act as an acid or

as a base when it dissolves in water?We can answer those questions by comparing the equilibrium constants for

the reactions. The equilibrium in which the HCO3� ion acts as a Brønsted acid

is described by Ka2 for carbonic acid.

The equilibrium in which the HCO3� acts as a Brønsted base is described by Kb2

for the carbonate ion.

Since Kb2 is significantly larger than Ka2, we conclude that the HCO3� ion is a

stronger base than it is an acid. An aqueous solution of NaHCO3 should there-fore be basic.

Kb2 =

[H3CO3][OH-]

[HCO -

3 ]= 2.2 * 10- 8

Ka2 =

[H3O+][CO 2 -

3 ]

[HCO -

3 ]= 4.7 * 10- 11

HCO3-(aq) + H2O(l) uv H3CO3(aq) + OH-(aq)

HCO3-(aq) + H2O(l) uv H3O

+(aq) + CO 2 -

3 (aq)

11A.3 COMPOUNDS THAT COULD BE EITHER ACIDS OR BASES 537

E x e r c i s e 1 1 A . 1Phosphoric acid, H3PO4, is obviously an acid. The phosphate ion, PO4

3�, is abase. Predict whether solutions of the H2PO4

� and HPO42� ions are more

likely to be acidic or basic.

SolutionLet’s start by looking at the stepwise dissociation of phosphoric acid.

We can then look at the steps by which the PO43� ion picks up protons from

water to form phosphoric acid.

Applying the procedure used for sodium carbonate in Section 11A.2 gives thefollowing relationships among the values of the six equilibrium constants.

Ka3Kb1 = Kw

Ka2Kb2 = Kw

Ka1Kb3 = Kw

H2PO4-(aq) + H2O(l) uv H3PO4(aq) + OH-(aq) Kb3 = ?

HPO42 -(aq) + H2O(l) uv H2PO4

-(aq) + OH-(aq) Kb2 = ?

PO 3 -

4 (aq) + H2O(l) uv HPO42 -(aq) + OH-(aq) Kb1 = ?

HPO42 -(aq) + H2O(l) uv H3O

+(aq) + PO 3 -

4 (aq) Ka3 = 4.2 * 10- 13


+(aq) + HPO 2 -

4 (aq) Ka2 = 6.3 * 10- 8

H3PO4(aq) + H2O(l) uv H3O+(aq) + H2PO -

4 (aq) Ka1 = 7.1 * 10- 3


In the 1880s, a variety of syrups started to appear that could be added tocarbonated water at the local drugstore. They included a root beer abstract mar-keted by Charles Hires, James Vernor’s Ginger Ale, R. S. Lazenby’s formula forDr. Pepper, John Pemberton’s formula for Coca-Cola, and Brad’s Drink whichwas later renamed Pepsi-Cola. These beverages were designed to quench thirstand provide energy during hot summer days. The energizing feeling came fromthe fact that Coca-Cola originally contained cocaine extracted from the coca plant,caffeine extracted from the kola bean, and a considerable amount of sugar. It hasalso been noted that the gentle release of CO2 gas that occurs when the liquid isswallowed has a physiologically soothing effect.

The per capita consumption of bottled carbonated beverages in the UnitedStates is almost 50 gallons per year. In order to create an acidic medium thatenhances the absorption of CO2 when these soft drinks are bottled, both phos-phoric acid (Ka1 � 7.1 � 10�3) and citric acid (Ka1 � 7.5 � 10�4) are added tosoft drinks to produce a beverage with a pH of 2.8. The high level of phosphatein soft drinks is associated with the loss of calcium from the blood, which even-tually has to be replenished by calcium from bone. The National Academy of Sci-ences recently increased the daily recommended intake of calcium from 800 mgto 1200 mg to compensate for the effect of soft drinks on the level of calcium inthe blood.

Conventional wisdom once assumed that the sugar in soft drinks could leadto tooth decay. Today we recognize that the real danger is the acid in soft drinks,which can leach enamel from the teeth, leaving a softened matrix for bacteria toenter the teeth.


We can predict whether the H2PO4� ion should be an acid or a base by

looking at the values of the Ka and Kb constants that characterize its reactionswith water.

Because Ka2 is significantly larger than Kb3, solutions of the H2PO4� ion in

water should be acidic.We can use the same method to decide whether HPO4

2� is an acid or abase when dissolved in water. We start by looking at the values of Ka and Kb

that characterize its reactions with water.

Because Kb2 is significantly larger than Ka3, solutions of the HPO42� ion in

water should be basic.Measurements of the pH of 0.10 M solutions of the species yield the fol-

lowing results, which are consistent with the predictions of this exercise.

pH = 12.0pH = 9.3pH = 4.4pH = 1.5

PO43 -HPO4

2 -H2PO4-H3PO4

HPO42 -(aq) + H2O(l) uv H2PO4

-(aq) + OH-(aq) Kb2 = 1.6 * 10- 7

HPO42 -(aq) + H2O(l) uv H3O

+(aq) + PO 3 -

4 (aq) Ka3 = 4.2 * 10- 13

H2PO4-(aq) + H2O(l) uv H3PO4(aq) + OH-(aq) Kb3 = 1.4 * 10- 12


+(aq) + HPO 2 -

4 (aq) Ka2 = 6.3 * 10- 8


PROBLEMS 539

ProblemsDiprotic AcidsA-1. Calculate the H3O�, CO3

2�, HCO3�, and H2CO3 con-

centrations at equilibrium in a solution that initiallycontained 0.100 mol of carbonic acid per liter. (ForH2CO3, Ka1 � 4.5 � 10�7 and Ka2 � 4.7 � 10�11.)

A-2. Calculate the equilibrium concentrations of the im-portant components of a 0.250 M malonic acid(HO2CCH2CO2H) solution. Use the symbol H2M asan abbreviation for malonic acid and assume stepwisedissociation of the acid. (For H2M, Ka1 � 1.4 � 10�5

and Ka2 � 2.1 � 10�8.)A-3. Check the validity of the assumption in the previous

problem that malonic acid dissociates in a stepwisefashion by comparing the concentrations of the HM�

and M2� ions obtained in the problem. Is this as-sumption valid?

A-4. Glycine, the simplest of the amino acids found inproteins, is a diprotic acid with the formulaHO2CCH2NH3

�. If we symbolize glycine as H2G�,we can write the following equations for its stepwisedissociation.

Calculate the concentrations of H3O�, G�, HG, andH2G� in a 2.00 M solution of glycine in water.

A-5. Which of the following equations accurately de-scribes a 2.0 M solution of glycine? See A-4.(a) [H3O�] [H2G�] (b) [H3O�] � [H2G�] (c) [H3O�] � [HG] (d) [H3O�] [HG](e) [H3O�] � [HG]

A-6. Which of the following equations results from thefact that glycine is a weak diprotic acid? See A-4.(a) [G�] [H2G�] (b) [G�] Ka2

(c) [G�] [HG] (d) [G�] � [HG] (e) [G�] [H3O�]

A-7. Oxalic acid (H2C2O4) has been implicated in diseasessuch as gout and kidney stones. Calculate the H3O�,H2C2O4, HC2O4

�, and C2O42� concentrations in a

1.25 M solution of oxalic acid.

Ka2 = 5.4 * 10- 5uv H3O

+(aq) + C2O42 -(aq)

HC2O4-(aq) + H2O(l)

Ka1 = 5.4 * 10- 2uv H3O

+(aq) + HC2O4-(aq)

H2C2O4(aq) + H2O(l)

L

L

LL

L

L

Ka2 = 2.5 * 10- 10HG(aq) + H2O(l) uv G-(aq) + H3O

+(aq)

Ka1 = 4.6 * 10- 3H2G

+(aq) + H2O(l) uv HG(aq) + H3O+(aq)

Diprotic BasesA-8. Which of the following sets of equations can be used

to calculate Kb1 and Kb2 for sodium oxalate(Na2C2O4) from Ka1 and Ka2 for oxalic acid(H2C2O4)?(a) Kb1 � Kw � Ka1 and Kb2 � Kw � Ka2

(b) Kb1 � Kw � Ka2 and Kb2 � Kw � Ka1

(c) Kb1 � Kw/Ka1 and Kb2 � Kw/Ka2

(d) Kb1 � Kw/Ka2 and Kb2 � Kw/Ka1

(e) Kb1 � Ka1/Kw and Kb2 � Ka2/Kw

A-9. Calculate the pH of a 0.028 M solution of sodiumoxalate (Na2C2O4) in water.

A-10. Calculate the H3O�, OH�, H2CO3, HCO3�, and

CO32� concentrations in a 0.150 M solution of

sodium carbonate (Na2CO3) in water. (For H2CO3,Ka1 � 4.5 � 10�7 and Ka2 � 4.7 � 10�11.)

Compounds That Could Be Either Acids or Bases

A-11. According to the data in Table 11.5, the pH of a 0.10 MNaHCO3 solution is 8.4. Explain why the HCO3

� ionforms solutions that are basic, not acidic.

A-12. Which of the following statements concerningsodium hydrogen sulfide (NaSH) is correct? (ForH2S, Ka1 � 1.0 � 10�7 and Ka2 � 1.3 � 10�13.)(a) NaSH is an acid because Ka1 for H2S is much

larger than Ka2.(b) NaSH is an acid because Ka1 for H2S is smaller

than Kb1 for Na2S.(c) NaSH is a base because Kb1 for Na2S is much

larger than Kb2.(d) NaSH is a base because Kb2 for Na2S is larger

than Ka2 for H2S.A-13. Calculate the pH of 0.10 M solutions of H3PO4,

NaH2PO4, Na2HPO4, and Na3PO4. (For H3PO4, Ka1 �7.1 � 10�3, Ka2 � 6.3 � 10�8, and Ka3 � 4.2 � 10�13.)

A-14. Predict whether an aqueous solution of sodium hydro-gen sulfate (NaHSO4) should be acidic, basic, or neutral.(For H2SO4, Ka1 � 1 � 103 and Ka2 � 1.2 � 10�2.)

A-15. Predict whether an aqueous solution of sodium hy-drogen sulfite (NaHSO3) should be acidic, basic, orneutral. (For H2SO3, Ka1 � 1.7 � 10�2 and Ka2 �6.4 � 10�8.)

Ka2 = 5.4 * 10- 5uv H3O

+(aq) + C2O42 -(aq)

HC2O4-(aq) + H2O(l)

Ka1 = 5.4 * 10- 2uv H3O

+(aq) + HC2O4-(aq)

H2C2O4(aq) + H2O(l)


chapter eleven - unamdepa.fquim.unam.mx/amyd/archivero/unidad_acids-bases... · 2015. 11. 17. ·...

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