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1 Acid–Base Concepts: The Brønsted–Lowry Theory

1 Acid–Base Concepts: The Brønsted–Lowry Theory

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540



Chapter 14 AQUEOUS EQUILIBRIA: ACIDS AND BASES



It follows from this equation that the products of a Brønsted–Lowry acid–base

reaction, BH + and A- , are themselves acids and bases. The species BH + produced

when the base B accepts a proton from HA can itself donate a proton back to A-,

meaning that it is a Brønsted–Lowry acid. Similarly, the species A- produced when

HA loses a proton can itself accept a proton back from BH + , meaning that it is a

Brønsted–Lowry base. Chemical species whose formulas differ only by one proton

are said to be conjugate acid–base pairs. Thus, A- is the conjugate base of the acid

HA, and HA is the conjugate acid of the base A- . Similarly, B is the conjugate base of

the acid BH +, and BH + is the conjugate acid of the base B.

To see what’s going on in an acid–base reaction, keep your eye on the proton. For

example, when a Brønsted–Lowry acid HA dissolves in water, it reacts reversibly

with water in an acid-dissociation equilibrium. The acid transfers a proton to the solvent, which acts as a base (a proton acceptor). The products are the hydronium ion,

H3O؉ (the conjugate acid of H2O), and A- (the conjugate base of HA):



+



+



H



H

+



A



H



O



H



+



O



H







+



A



+



A−(aq)

Base



H

+



HA(aq)

Acid



H2O(l)

Base



O+(aq)



H3

Acid



Conjugate acid–base pairs



In the reverse reaction, H 3O + acts as the proton donor (acid) and A- acts as the

proton acceptor (base). Typical examples of Brønsted–Lowry acids include not only

electrically neutral molecules, such as HCl, HNO3, and HF, but also cations and

anions of salts that contain transferable protons, such as NH 4 + , HSO4 - , and HCO 3 - .

When a Brønsted–Lowry base such as NH3 dissolves in water, it accepts a proton

from the solvent, which acts as an acid. The products are the hydroxide ion, OH - (the

conjugate base of water), and the ammonium ion, NH 4 + (the conjugate acid of NH3).

In the reverse reaction, NH 4 + acts as the proton donor and OH - acts as the proton

acceptor:



+

H

H



N



+



+



O



H



O



H







+



H



N



H



H



H

NH3(aq)

Base



+



H



H



H2O(l)

Acid



OH−(aq)

Base



+



NH4+(aq)

Acid



Conjugate acid–base pairs



For a molecule or ion to accept a proton, it must have at least one unshared

pair of electrons that it can use for bonding to the proton. As shown by the following



14.1 ACID–BASE CONCEPTS: THE BRØNSTED–LOWRY THEORY



electron-dot structures, all Brønsted–Lowry bases have one or more lone pairs of

electrons:

O



H



F







H







H



N



H



O



H



H

Some Brønsted–Lowry bases



WORKED EXAMPLE 14.1



EXPLAINING ACIDITY WITH THE ARRHENIUS

AND BRØNSTED–LOWRY THEORIES

Account for the acidic properties of nitrous acid (HNO2) using the Arrhenius theory

and the Brønsted–Lowry theory, and identify the conjugate base of HNO2.

STRATEGY



To account for the acidity of a substance, consider how it can produce H + ions in water

(Arrhenius theory) and how it can act as a proton donor (Brønsted–Lowry theory).

SOLUTION



HNO2 is an Arrhenius acid because it dissociates in water to produce H + ions:

HNO2(aq) Δ H + (aq) + NO 2 -(aq)

Nitrous acid is a Brønsted–Lowry acid because it acts as a proton donor when it

dissociates, transferring a proton to water to give the hydronium ion, H 3O + :

H3O+(aq) + NO2−(aq)

Acid

Base



HNO2(aq) + H2O(l)

Acid

Base



Conjugate acid–base pairs



The conjugate base of HNO2 is NO2 - , the species that remains after HNO2 has lost a

proton.

Ī PROBLEM 14.1 Write a balanced equation for the dissociation of each of the following Brønsted–Lowry acids in water:

(b) HSO4 (c) H 3O +

(d) NH 4 +

(a) H2SO4

What is the conjugate base of each acid?

Ī PROBLEM 14.2

bases?

(a) HCO 3 -



What is the conjugate acid of each of the following Brønsted–Lowry

(b) CO3 2 -



(d) H 2PO4 -



(c) OH -



WORKED EXAMPLE 14.2



IDENTIFYING BRØNSTED–LOWRY ACIDS, BASES,

AND CONJUGATE ACID–BASE PAIRS

For the following reaction in aqueous solution, identify the Brønsted–Lowry acids,

bases, and conjugate acid–base pairs:



+



+

=H



=S



=F



STRATEGY



The simplest approach is to identify the conjugate acid–base pairs, the species whose

formulas differ by just one proton.

continued on next page



Nitrous acid



541



542



Chapter 14 AQUEOUS EQUILIBRIA: ACIDS AND BASES

SOLUTION



The second reactant is HF, and the first product is its conjugate base F - . The second

product is H2S, and the first reactant is its conjugate base HS - . Therefore, the

Brønsted–Lowry acids, bases, and conjugate acid–base pairs are as follows:



+



+



HS–(aq) + HF(aq)

Base



F–(aq) + H2S(aq)



Acid



Base



Acid



Conjugate acid–base pairs



CONCEPTUAL PROBLEM 14.3 For the following reaction in aqueous solution, identify

the Brønsted–Lowry acids, bases, and conjugate acid–base pairs:



+



+



=H



=N



= Cl



14.2 ACID STRENGTH AND BASE STRENGTH

A helpful way of viewing an acid-dissociation equilibrium is to realize that the two

bases, H2O and A-, are competing for protons:

HA(aq) + H 2O(l) Δ H 3O + (aq) + A-(aq)

Acid



Base



Acid



Base



If H2O is a stronger base (a stronger proton acceptor) than A- , the H2O molecules will

get the protons and the solution will contain mainly H 3O + and A- . If A- is a stronger

base than H2O, the A- ions will get the protons and the solution will contain mainly

HA and H2O. When beginning with equal concentrations of reactants and products,

the proton is always transferred to the stronger base. This means that the direction of reaction to reach equilibrium is proton transfer from the stronger acid to the stronger

base to give the weaker acid and the weaker base:

Stronger acid + Stronger base ¡ Weaker acid + Weaker base



Remember...

Substances that dissolve in water to produce solutions that conduct electricity are

called electrolytes. Molecular substances

that dissociate into ions to a large extent

are strong electrolytes, while those that

dissociate to only a small extent are weak

electrolytes. (Section 4.2)



Different acids differ in their ability to donate protons. A strong acid is one that is

almost completely dissociated in water and is therefore a strong electrolyte (Section

4.2). Thus, the acid-dissociation equilibrium of a strong acid lies nearly 100% to the

right, and the solution contains almost entirely H 3O + and A- ions with only a negligible amount of undissociated HA molecules. Typical strong acids are perchloric acid

(HClO4), hydrochloric acid (HCl), hydrobromic acid (HBr), hydroiodic acid (HI),

nitric acid (HNO3), and sulfuric acid (H2SO4). It follows from this definition that

strong acids have very weak conjugate bases. The ions ClO4 - , Cl - , Br - , I - , NO3 - ,

and HSO4 - have only a negligible tendency to combine with a proton in aqueous

solution, and they are therefore much weaker bases than H2O.

A weak acid is one that is only partially dissociated in water and is thus a weak

electrolyte. Only a small fraction of the weak acid molecules transfer a proton to

water, and the solution therefore contains mainly undissociated HA molecules along

with small amounts of H 3O + and the conjugate base A- . Typical weak acids are

nitrous acid (HNO2), hydrofluoric acid (HF), and acetic acid (CH3CO2H). In the case

of very weak acids, such as NH3, OH -, and H2, the acid has practically no tendency

to transfer a proton to water and the acid-dissociation equilibrium lies essentially

100% to the left. It follows from this definition that very weak acids have strong



14.2 ACID STRENGTH AND BASE STRENGTH



543



conjugate bases. For example, the NH 2 - , O 2- , and H - ions are essentially 100% protonated in aqueous solution and are much stronger bases than H2O.

The equilibrium concentrations of HA, H 3O +, and A- for strong acids, weak

acids, and very weak acids are represented in Figure 14.1. The inverse relationship

between the strength of an acid and the strength of its conjugate base is illustrated in

Table 14.1.

Figure 14.1

Concentration before

dissociation



Equilibrium concentrations

after dissociation



HA



H3O+



Strong acid



Dissociation of HA involves H؉ transfer

to H2O, yielding H3O؉ and A؊ .



A–

The extent of

dissociation is

nearly 100% for

a strong acid.



~100%

dissociation

HA

HA

HA



Weak acid



Partial

dissociation



HA



Very weak acid



H3O+



A–



H3O+



A–



HA



~0%

dissociation



TABLE 14.1



Weaker

acid



The extent of

dissociation is

nearly 0% for a

very weak acid.



Relative Strengths of Conjugate Acid–Base Pairs

Base, A−



Acid, HA

Stronger

acid



The extent of

dissociation is

considerably less

than 100% for a

weak acid.



HClO4

HCl

H2SO4

HNO3

H3O+

HSO4–

H3PO4

HNO2

HF

CH3CO2H

H2CO3

H2S

NH4+

HCN

HCO3–

H2O

NH3

OH–

H2



Strong acids:

100% dissociated

in aqueous

solution.



Weak acids:

Exist in solution

as a mixture of

HA, A–, and H3O+.



Very weak acids:

Negligible tendency

to dissociate.



ClO4–

Cl–

HSO4–

NO3–

H2O

SO42–

H2PO4–

NO2–

F–

CH3CO2–

HCO3–

HS–

NH3

CN–

CO32–

OH–

NH2–

O2–

H–



Very weak bases:

Negligible tendency

to be protonated in

aqueous solution.



Weaker

base



Weak bases:

Moderate tendency

to be protonated in

aqueous solution.



Strong bases:

100% protonated in

aqueous solution.



Stronger

base



544



Chapter 14 AQUEOUS EQUILIBRIA: ACIDS AND BASES

WORKED EXAMPLE 14.3



PREDICTING THE DIRECTION OF ACID–BASE REACTIONS

If you mix equal concentrations of reactants and products, which of the following reactions proceed to the right and which proceed to the left?

(a) H 2SO4(aq) + NH 3(aq) Δ NH 4 +(aq) + HSO 4 -(aq)

(b) HCO3 -(aq) + SO 4 2-(aq) Δ HSO 4 -(aq) + CO 3 2-(aq)

STRATEGY



To predict the direction of reaction, use the balanced equation to identify the acids and

bases, and then use Table 14.1 to identify the stronger acid and the stronger base. When

equal concentrations of reactants and products are present, proton transfer always

occurs from the stronger acid to the stronger base.

SOLUTION



(a) In this reaction, H2SO4 and NH 4 + are the acids, and NH3 and HSO4 - are the bases.

According to Table 14.1, H2SO4 is a stronger acid than NH 4 + and NH3 is a stronger

base than HSO 4 -. Therefore, NH3 gets the proton and the reaction proceeds from

left to right.

H 2SO4(aq) + NH 3(aq) ¡ NH 4 +(aq) + HSO 4 -(aq)

Stronger acid Stronger base

-



-



Weaker acid



Weaker base



2-



(b) HCO3 and HSO4 are the acids, and SO4 and CO3 2- are the bases. Table 14.1

indicates that HSO4 - is the stronger acid and CO3 2- is the stronger base. Therefore,

CO 3 2- gets the proton and the reaction proceeds from right to left.

HCO3 -(aq) + SO4 2-(aq) — HSO 4 -(aq) + CO 3 2-(aq)

Weaker acid



Weaker base



Stronger acid



Stronger base



Ī PROBLEM 14.4 If you mix equal concentrations of reactants and products, which of

the following reactions proceed to the right and which proceed to the left?



(a) HF(aq) + NO 3 -(aq) Δ HNO 3(aq) + F -(aq)

(b) NH 4 +(aq) + CO3 2-(aq) Δ HCO 3 -(aq) + NH 3(aq)

CONCEPTUAL PROBLEM 14.5 The following pictures represent aqueous solutions of

two acids HA (A = X or Y); water molecules have been omitted for clarity.

= HA



HX



= H3O+



= A−



HY



(a) Which is the stronger acid, HX or HY?

(b) Which is the stronger base, X - or Y - ?

(c) If you mix equal concentrations of reactants and products, will the following

reaction proceed to the right or to the left?

HX + Y - Δ HY + X -



14.4 DISSOCIATION OF WATER



545



14.3 HYDRATED PROTONS AND

HYDRONIUM IONS

As we’ve seen, the proton is fundamental to both the Arrhenius and the

Brønsted–Lowry definitions of an acid. Dissociation of an Arrhenius acid HA gives

an aqueous hydrogen ion, or hydrated proton, written as H +(aq):

HA(aq) Δ H + (aq) + A-(aq)

As a bare proton, the positively charged H + ion is too reactive to exist in aqueous solution and so it bonds to the oxygen atom of a solvent water molecule to give

the trigonal pyramidal hydronium ion, H 3O + . The H 3O + ion, which can be

regarded as the simplest hydrate of the proton, [H(H2O)] + , can associate through

hydrogen bonding with additional water molecules to give higher hydrates with

the general formula [H(H 2O)n]+ (n = 2, 3, or 4), such as H 5O2 + , H 7O3 + , and H 9O4 + .

It’s likely that acidic aqueous solutions contain a distribution of [H(H 2O)n]+ ions

having different values of n. In this book, though, we’ll use the symbols H +(aq) and

H 3O +(aq) to mean the same thing—namely, a proton hydrated by an unspecified

number of water molecules. Ordinarily, we use H 3O + in acid–base reactions to

emphasize the proton-transfer character of those reactions.

Ī PROBLEM 14.6



H3O+ — the hydronium ion, or hydrated H+



Some of the following ions have been detected by mass spectrometry:

H 9O4 + , H 13O5 + , H 19O9 + , H 25O11 + , H 43O21 +



(a) Which should be considered as hydrates of the proton?

(b) For the ions that are hydrates, tell how many water molecules are present in

addition to the proton.



14.4 DISSOCIATION OF WATER

One of the most important properties of water is its ability to act both as an acid and

as a base. In the presence of an acid, water acts as a base, whereas in the presence of

a base, water acts as an acid. It’s not surprising, therefore, that in pure water one molecule can donate a proton to another in a reaction in which water acts as both an acid

and a base in the same reaction:

O

H

Acid



H +



O



H



H



O



H



H



Base



Acid



H



+



+



O



H







Base



Conjugate acid–base pairs



Called the dissociation of water, this reaction is characterized by the equilibrium

equation Kw = [H 3O +][OH -], where the equilibrium constant Kw is called the

ion-product constant for water.

Dissociation of Water



2 H 2O(l) Δ H 3O + (aq) + OH -(aq)



Ion-Product Constant for water



Kw = [H 3O + ][OH -]



As discussed in Section 13.4, the concentration of water is omitted from the equilibrium constant expression because water is a pure liquid.

There are two important aspects of the dynamic equilibrium in the dissociation

of water. First, the forward and reverse reactions are rapid: H2O molecules, H 3O +

ions, and OH - ions continually interconvert as protons transfer quickly from one

species to another. Second, the position of the equilibrium lies far to the left: At any

given instant, only a tiny fraction of the water molecules are dissociated into H 3O +

and OH - ions. The vast majority of the H2O molecules are undissociated.



Remember...

The concentration of a pure liquid or a

pure solid is omitted from the equilibrium

constant expression because the ratio of its

actual concentration to its concentration in

the thermodynamic standard state is equal

to 1. (Section 13.4)



546



Chapter 14 AQUEOUS EQUILIBRIA: ACIDS AND BASES



We can calculate the extent of the dissociation of the water molecules starting

from experimental measurements that show the H 3O + concentration in pure water

to be 1.0 * 10-7 M at 25 °C:

[H 3O +] = 1.0 * 10-7 M



at 25 °C



Since the dissociation reaction of water produces equal concentrations of H 3O + and

OH - ions, the OH - concentration in pure water is also 1.0 * 10-7 M at 25 °C:

[H 3O +] = [OH -] = 1.0 * 10-7 M



at 25 °C



Furthermore, we know that the molar concentration of pure water, calculated from

its density and molar mass, is 55.4 M at 25 °C:

[H 2O] = a



997 g

L



ba



1 mol

b = 55.4 mol/L

18.0 g



at 25 °C



From these facts, we conclude that the ratio of dissociated to undissociated water

molecules is about 2 in 109, a very small number indeed:

[H 2O]dissociated

1.0 * 10-7 M

=

= 1.8 * 10-9

[H 2O]undissociated

55.4 M



about 2 in 109



In addition, we can calculate that the numerical value of Kw at 25 °C is 1.0 * 10-14:

Kw = [H 3O + ][OH -] = (1.0 * 10-7)(1.0 * 10-7)

= 1.0 * 10-14

at 25 °C

In very dilute solutions, the water is almost a pure liquid and the product of the

H 3O + and OH - concentrations is unaffected by the presence of solutes. This is not

true in more concentrated solutions, but we’ll neglect that complication and assume

that the product of the H 3O + and OH - concentrations is always 1.0 * 10-14 at 25 °C

in any aqueous solution.

We can distinguish acidic, neutral, and basic aqueous solutions by the relative

values of the H 3O + and OH - concentrations:

Acidic: [H 3O + ] 7 [OH -]

Neutral: [H 3O + ] = [OH -]

Basic:

[H 3O + ] 6 [OH -]

At 25 °C, [H 3O +] 7 1.0 * 10-7 M in an acidic solution, [H 3O +] = [OH -] =

1.0 * 10-7 M in a neutral solution, and [H 3O +] 6 1.0 * 10-7 M in a basic solution

(Figure 14.2). If one of the concentrations, [H 3O +] or [OH -], is known, the other is

readily calculated:

Since



[H 3O + ][OH -] = Kw = 1.0 * 10-14



then



[H 3O + ] =



1.0 * 10-14

[OH -]



and



[OH -] =



1.0 * 10-14

[H 3O +]



Concentration



> 1.0 × 10–7 M



Figure 14.2



Values of the H3O؉ and OH؊

concentrations at 25 °C in acidic, neutral,

and basic solutions.



H3O+



OH–

H3O+ OH–



1.0 × 10–7 M

OH–



< 1.0 × 10–7 M



Acidic

solution



H3O+



Neutral

solution



Basic

solution



14.5 THE pH SCALE



In the previous discussion, we were careful to emphasize that the value of

Kw = 1.0 * 10-14 applies only at 25 °C. This is because Kw, like all equilibrium constants, is affected by temperature and the H 3O + and OH - concentrations in neutral

aqueous solutions at temperatures other than 25 °C deviate from 1.0 * 10-7 M (see

Problem 14.9). Unless otherwise indicated, we’ll always assume a temperature of 25 °C.

WORKED EXAMPLE 14.4



CALCULATING [OH ؊ ] FROM Kw AND [H3O ؉ ]



The concentration of H 3O + ions in a sample of lemon juice is 2.5 * 10-3 M. Calculate

the concentration of OH - ions, and classify the solution as acidic, neutral, or basic.

STRATEGY



When [H 3O +] is known, the OH - concentration can be found from the expression

[OH -] = Kw/[H 3O +].

SOLUTION



[OH -] =



Kw

1.0 * 10-14

=

= 4.0 * 10-12 M

[H 3O -]

2.5 * 10-3



Because [H 3O +] 7 [OH -], the solution is acidic.

BALLPARK CHECK



Because the product of the H 3O + and OH - concentrations must equal 10-14, and

because the H 3O + concentration is in the range 10-3 M to 10-2 M, the OH - concentration

must be in the range 10-11 M to 10-12 M. The ballpark check and the solution agree.

The concentration of H 3O + ions in the runoff from a coal mine is

1.4 * 10-4 M. Calculate the concentration of OH - ions, and classify the solution as acidic,

neutral, or basic.

Ī PROBLEM 14.7



The concentration of OH - in a sample of seawater is 2.0 * 10-6 M.

Calculate the concentration of H 3O + ions, and classify the solution as acidic, neutral, or

basic.

Ī PROBLEM 14.8



At 50 °C the value of Kw is 5.5 * 10-14. What are the concentrations

of H 3O and OH in a neutral solution at 50 °C?

Ī PROBLEM 14.9

+



-



14.5 THE pH SCALE

Rather than write hydronium ion concentrations in molarity, it’s more convenient to

express them on a logarithmic scale known as the pH scale. The term pH is derived

from the French puissance d’hydrogène (“power of hydrogen”) and refers to the power

of 10 (the exponent) used to express the molar H 3O + concentration. The pH of a solution is defined as the negative base-10 logarithm (log) of the molar hydronium ion

concentration:

pH = -log [H 3O + ] or [H 3O + ] = antilog (-pH) = 10-pH

Thus, an acidic solution having [H 3O +] = 10-2 M has a pH of 2, a basic solution having [OH -] = 10-2 M and [H 3O +] = 10-12 M has a pH of 12, and a neutral solution

having [H 3O +] = 10-7 M has a pH of 7. Note that we can take the log of [H 3O +]

because [H 3O +] is a dimensionless ratio of the actual concentration to the concentration (1 M) in the standard state. Although much less frequently used than pH, a pOH

can be defined in the same way as pH and used to express the molar OH - concentration. Just as pH = -log [H 3O +], so pOH = -log [OH -]. It follows from the equation

[H 3O +][OH -] = 1.0 * 10-14 that pH + pOH = 14.00.

If you use a calculator to find the pH from the H 3O + concentration, your answer

will have more decimal places than the proper number of significant figures. For example, the pH of the lemon juice in Worked Example 14.4 ([H 3O +] = 2.5 * 10-3 M) is

found on a calculator to be

pH = -log (2.5 * 10-3) = 2.602 06



547



548



Chapter 14 AQUEOUS EQUILIBRIA: ACIDS AND BASES



Acidic



Neutral



Basic



[H3O+] pH

10−14



14



10−13



13



10−12



12



10−11



11



10−10



10



10−9



9



10−8



8



10−7



7



10−6



6



10−5



5



Black coffee



10−4



4



Tomatoes



10−3



3



10−2



2



Wine

Vinegar, colas

Lemon juice



10−1



1



1



0



1.0 M NaOH



This result should be rounded to pH 2.60 (two significant figures) because [H 3O +]

has only two significant figures. Note that the only significant figures in a logarithm

are the digits to the right of the decimal point; the number to the left of the decimal

point is an exact number related to the integral power of 10 in the exponential

expression for [H 3O +]:

pH = −log(2.5 × 10−3) = −log 10−3 − log 2.5 = 3 − 0.40 = 2.60



Household

ammonia

Milk of

magnesia

Baking soda

Human blood

Pure water

Milk



2 significant

figures (2 SF’s)



Exact

number



Exact

number



2 SF’s



Exact

number



2 SF’s



Because the pH scale is logarithmic, the pH changes by 1 unit when [H 3O +]

changes by a factor of 10, by 2 units when [H 3O +] changes by a factor of 100, and by

6 units when [H 3O +] changes by a factor of 1,000,000. To appreciate the extent to

which the pH scale is a compression of the [H 3O +] scale, compare the amounts of

12 M HCl required to change the pH of the water in a backyard swimming pool:

Only about 100 mL of 12 M HCl is needed to change the pH from 7 to 6, but a 10,000 L

truckload of 12 M HCl is needed to change the pH from 7 to 1.

The pH scale and pH values for some common substances are shown in Figure 14.3.

Because the pH is the negative log of [H 3O +], the pH decreases as [H 3O +] increases.

Thus, when [H 3O +] increases from 10-7 M to 10-6 M, the pH decreases from 7 to 6. As

a result, acidic solutions have pH less than 7, and basic solutions have pH greater

than 7.

Acidic solution:

pH 6 7

Neutral solution: pH = 7

Basic solution:

pH 7 7



Stomach acid

1.0 M HCl



WORKED EXAMPLE 14.5



CALCULATING THE pH FROM THE H3O ؉ CONCENTRATION

Figure 14.3



The pH scale and pH values for some

common substances.



Calculate the pH of an aqueous ammonia solution that has an OH - concentration of

1.9 * 10-3 M.

STRATEGY



First, calculate the H 3O + concentration from the OH - concentration, and then take the

negative logarithm of [H 3O +] to convert to pH.

SOLUTION



Kw

1.0 * 10-14

=

= 5.3 * 10-12 M

[OH -]

1.9 * 10-3

pH = -log [H 3O + ] = -log (5.3 * 10-12) = 11.28



[H 3O + ] =



The pH is quoted to two significant figures (.28) because [H 3O +] is known to two significant figures (5.3).

BALLPARK CHECK



Because [OH -] is between 10-3 M and 10-2 M, [H 3O +] is between 10-11 M and 10-12 M.

Therefore, the pH is between 11 and 12, in agreement with the solution.



WORKED EXAMPLE 14.6



CALCULATING THE H3O ؉ CONCENTRATION FROM THE pH

Acid rain is a matter of serious concern because most species of fish die in waters

having a pH lower than 4.5-5.0. Calculate the H 3O + concentration in a lake that has a

pH of 4.5.

STRATEGY



Calculate the H 3O + concentration by taking the antilogarithm of the negative of

the pH.



14.6 MEASURING pH

SOLUTION



[H 3O + ] = antilog (-pH) = 10-pH = 10-4.5 = 3 * 10-5 M

[H 3O +] is reported to only one significant figure because the pH has only one digit

beyond the decimal point. (If you need help in finding the antilog of a number, see

Appendix A.2.)

BALLPARK CHECK



Because a pH of 4.5 is between 4 and 5, [H 3O +] is between 10-4 M and 10-5 M, in agreement with the solution.

Ī PROBLEM 14.10



Calculate the pH of each of the following solutions:

(a) A sample of seawater that has an OH - concentration of 1.58 * 10-6 M

(b) A sample of acid rain that has an H 3O + concentration of 6.0 * 10-5 M



Ī PROBLEM 14.11 Calculate the concentrations of H 3O + and OH - in each of the

following solutions:

(a) Human blood (pH 7.40)

(b) A cola beverage (pH 2.8)



14.6 MEASURING pH

The approximate pH of a solution can be determined by using an acid–base

indicator, a substance that changes color in a specific pH range (Figure 14.4). Indicators

(abbreviated HIn) exhibit pH-dependent color changes because they are weak acids

and have different colors in their acid (HIn) and conjugate base (In-) forms:

HIn(aq) + H 2O(l) Δ H 3O + (aq) + In-(aq)

Color A



Color B



Bromthymol blue, for example, changes color in the pH range 6.0-7.6, from yellow

in its acid form to blue in its base form.



Indicator

Methyl violet

Thymol blue

Methyl orange

Bromcresol green

Methyl red

Chlorphenol red

Bromthymol blue

Phenol red

Phenolphthalein



pH



0



1



Yellow



2



3



4



5



6



7



8



9



10



11



12



Violet

Red



Yellow (acid range)

Red



Yellow



Blue (base range)



Yellow-orange



Yellow



Blue

Red



Yellow

Yellow



Red



Yellow



Blue



Yellow



Red

Colorless



Thymolphthalein

Alizarin yellow



Figure 14.4



Some common acid–base indicators and their color changes. The color of an indicator

changes over a range of about 2 pH units.



Red

Colorless

Yellow



Blue

Violet



549



550



Chapter 14 AQUEOUS EQUILIBRIA: ACIDS AND BASES



Especially convenient for making approximate pH measurements is a commercially available mixture of indicators known as universal indicator that exhibits

various colors depending on the pH (see Figure 14.8 on page 566). More accurate

pH values can be determined with an electronic instrument called a pH meter

(Figure 14.5), a device that measures the pH-dependent electrical potential of the test

solution. We’ll learn more about pH meters in Section 17.7.



14.7 THE pH IN SOLUTIONS OF STRONG

ACIDS AND STRONG BASES

Figure 14.5



A pH meter with its electrical probe

dipping into milk of magnesia. An

accurate value of the pH (10.52) is shown

on the meter.



The commonly encountered strong acids listed in Table 14.1 include three monoprotic

acids (HClO4, HCl, and HNO3), which contain a single dissociable proton, and one

diprotic acid (H2SO4), which has two dissociable protons. Because strong monoprotic

acids are nearly 100% dissociated in aqueous solution, the H 3O + and A- concentrations are equal to the initial concentration of the acid and the concentration of

undissociated HA molecules is essentially zero.

HA(aq) + H 2O(l)



100%



" H O + (aq) + A-(aq)

3



The pH of a solution of a strong monoprotic acid is easily calculated from the

H 3O + concentration, as shown in Worked Example 14.7. Calculation of the pH of an

H2SO4 solution is more complicated because essentially 100% of the H2SO4 molecules dissociate to give H 3O + and HSO 4 - ions but much less than 100% of the

resulting HSO4 - ions dissociate to give H 3O + and SO4 2- ions. We’ll have more to say

about diprotic acids in Section 14.11.

The most familiar examples of strong bases are alkali metal hydroxides, MOH,

such as NaOH (caustic soda) and KOH (caustic potash). These compounds are watersoluble ionic solids that exist in aqueous solution as alkali metal cations (M +) and

OH -anions:

MOH(s)



H 2O



" MOH(aq)



100%



" M + (aq) + OH -(aq)



Thus, 0.10 M NaOH contains 0.10 M Na + and 0.10 M OH -, and the pH is readily calculated from the OH - concentration, as shown in Worked Example 14.8a.

The alkaline earth metal hydroxides M(OH)2 (M = Mg, Ca, Sr, or Ba) are also

strong bases ( ' 100% dissociated), but they give lower OH - concentrations because

they are less soluble. Their solubility at room temperature varies from 38 g/L for the

relatively soluble Ba(OH)2 to ' 10-2 g/L for the relatively insoluble Mg(OH)2. Aqueous suspensions of Mg(OH)2, called milk of magnesia, are used as an antacid. The most

common and least expensive alkaline earth hydroxide is Ca(OH)2, which is used in

making mortars and cements. Ca(OH)2 is called slaked lime because it is made by

treating lime (CaO) with water. Aqueous solutions of Ca(OH)2, which has a solubility

of only ' 1.3 g/L, are known as limewater.

Alkaline earth oxides, such as CaO, are even stronger bases than the corresponding hydroxides because the oxide ion (O 2-) is a stronger base than OH - (Table 14.1).

In fact, the O 2- ion can’t exist in aqueous solutions because it is immediately and

completely protonated by water, yielding OH - ions:

O 2-(aq) + H 2O(l)



100%



" OH -(aq) + OH -(aq)



Thus, when lime is dissolved in water, it gives 2 OH - ions per CaO formula unit:

CaO(s) + H 2O(l) ¡ Ca2 + (aq) + 2 OH -(aq)

Lime is the world’s most important strong base. Annual worldwide production is

around 283 million metric tons for use in steelmaking, water purification, and chemical manufacture. Lime is made by the decomposition of limestone, CaCO3, at

temperatures of 800–1000 °C:

᭡ Lime is spread on lawns and gardens to

raise the pH of acidic soils.



CaCO3(s)



Heat



" CaO(s) + CO (g)

2



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