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2 Proton Transfer—The Reaction of a Brønsted–Lowry Acid with a Brønsted–Lowry Base

2 Proton Transfer—The Reaction of a Brønsted–Lowry Acid with a Brønsted–Lowry Base

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264



ACIDS AND BASES



This electron

pair stays on A.



This electron pair forms

a new bond to H+.

H A

acid



+



gain of H+

+

A−

H B+

conjugate base conjugate acid



B

base

loss of H+



• The product formed by loss of a proton from an acid is called its conjugate base.

• The product formed by gain of a proton by a base is called its conjugate acid.



Thus, the conjugate base of the acid HA is A:–. The conjugate acid of the base B: is HB+.

• Two species that differ by the presence of a proton are called a conjugate acid–base

pair.



Thus, in an acid–base reaction, the acid and the base on the left side of the equation (HA and B:)

form two products that are also an acid and a base (HB+ and A:–). Equilibrium arrows (

)

are often used to separate reactants and products because the reaction can proceed in either the

forward or the reverse directions. In some reactions, the products are greatly favored, as discussed in Section 9.3.

When HBr is dissolved in water, for example, the acid HBr loses a proton to form its conjugate

base Br–, and the base H2O gains a proton to form H3O+.

gain of H+

H Br

acid



+



H2O

base



+

Br−

H3O+

conjugate base conjugate acid



loss of H+



Thus, HBr and Br– are a conjugate acid–base pair since these two species differ by the presence

of a proton (H+). H2O and H3O+ are also a conjugate acid–base pair because these two species

differ by the presence of a proton as well.

The net charge must be the same on both sides of the equation. In this example, the two

reactants are neutral (zero net charge), and the sum of the –1 and +1 charges in the products is

also zero.

Take particular note of what happens to the charges in each conjugate acid–base pair. When a

species gains a proton (H+), it gains a +1 charge. Thus, if a reactant is neutral to begin with, it

ends up with a +1 charge. When a species loses a proton (H+), it effectively gains a –1 charge

since the product has one fewer proton (+1 charge) than it started with. Thus, if a reactant is

neutral to begin with, it ends up with a –1 charge.

0 charge

H2O

base

0 charge



When HCl donates a proton to NH3

in the absence of water, NH4+ and

Cl– are formed, which combine

to form solid ammonium chloride,

NH4Cl.



smi26573_ch09.indd 264



H Br

acid



add H+



lose H+



+1 charge

H3O+



Br−



−1 charge



Take away +1 charge.



The reaction of ammonia (NH3) with HCl is also a Brønsted–Lowry acid–base reaction. In this

example, NH3 is the base since it gains a proton to form its conjugate acid, NH4+. HCl is the acid

since it donates a proton, forming its conjugate base, Cl–.



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PROTON TRANSFER—THE REACTION OF A BRØNSTED–LOWRY ACID WITH A BRØNSTED–LOWRY BASE



265



gain of H+

+



H

H



N

H

base



H



+



HCl(g)

acid



H



N



H



H

conjugate acid



+ Cl−(aq)

conjugate base



loss of H+



• A Brønsted–Lowry acid–base reaction is a proton transfer reaction since it always results

in the transfer of a proton from an acid to a base.



The ability to identify and draw a conjugate acid or base from a given starting material is a necessary skill, illustrated in Sample Problems 9.4 and 9.5.



SAMPLE PROBLEM 9.4

ANALYSIS



SOLUTION



PROBLEM 9.4

SAMPLE PROBLEM 9.5

ANALYSIS



SOLUTION



PROBLEM 9.5



Draw the conjugate acid of each base: (a) F–; (b) NO3–.

Conjugate acid–base pairs differ by the presence of a proton. To draw a conjugate acid from

a base, add a proton, H+. This adds +1 to the charge of the base to give the charge on the

conjugate acid.

a. F– + H+ gives HF as the conjugate acid. HF has no charge since a proton with a +1 charge

is added to an anion with a –1 charge.

b. NO3– + H+ gives HNO3 (nitric acid) as the conjugate acid. HNO3 has no charge since a

proton with a +1 charge is added to an anion with a –1 charge.

Draw the conjugate acid of each species: (a) H2O; (b) I–; (c) HCO3–.

Draw the conjugate base of each acid: (a) H2O; (b) HCO3–.

Conjugate acid–base pairs differ by the presence of a proton. To draw a conjugate base from

an acid, remove a proton, H+. This adds –1 to the charge of the acid to give the charge on the

conjugate base.

a. Remove H+ from H2O to form –OH, the conjugate base. –OH has a –1 charge since –1 is

added to a molecule that was neutral to begin with.

b. Remove H+ from HCO3– to form CO32–, the conjugate base. CO32– has a –2 charge since

–1 is added to an anion that had a –1 charge to begin with.

Draw the conjugate base of each species: (a) H2S; (b) HCN; (c) HSO4–.



A compound that contains both a hydrogen atom and a lone pair of electrons can be either

an acid or a base, depending on the particular reaction. Such a compound is said to be amphoteric. For example, when H2O acts as a base it gains a proton, forming H3O+. Thus, H2O and

H3O+ are a conjugate acid–base pair. When H2O acts as an acid it loses a proton, forming –OH.

H2O and –OH are also a conjugate acid–base pair.



H2O as a base



H2O as an acid



smi26573_ch09.indd 265



H



H



O H

base



O H

acid



add H+



+



H

H



O



H



conjugate acid

remove H+







H



O



conjugate base



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266



ACIDS AND BASES



SAMPLE PROBLEM 9.6



Label the acid and the base and the conjugate acid and the conjugate base in the following

reaction.

NH4+(aq) + –OH(aq)



NH3(g) + H2O(l )



ANALYSIS



The Brønsted–Lowry acid loses a proton to form its conjugate base. The Brønsted–Lowry base

gains a proton to form its conjugate acid.



SOLUTION



NH4+ is the acid since it loses a proton to form NH3, its conjugate base. –OH is the base since it

gains a proton to form its conjugate acid, H2O.

gain of H+

NH4+(aq)

acid



+



−OH(aq)



+

H2O(l)

NH3(g)

conjugate base conjugate acid



base

loss of H+



PROBLEM 9.6



Label the acid and the base and the conjugate acid and the conjugate base in each reaction.

a. H2O(l) + HI(g)

I–(aq) + H3O+(aq)

b. CH3COOH(l) + NH3(g)

CH3COO–(aq) + NH4+(aq)

HBr(aq) + NO3–(aq)

c. Br–(aq) + HNO3(aq)



PROBLEM 9.7



Ammonia, NH3, is amphoteric. (a) Draw the conjugate acid of NH3. (b) Draw the conjugate

base of NH3.



PROBLEM 9.8



HSO3– can act as an acid or a base. (a) Draw the conjugate base of HSO3–. (b) Draw the

conjugate acid of HSO3–.



9.3 ACID AND BASE STRENGTH

Although all Brønsted–Lowry acids contain protons, some acids readily donate protons while

others do not. Similarly, some Brønsted–Lowry bases accept a proton much more readily than

others. How readily proton transfer occurs is determined by the strength of the acid and base.



9.3A RELATING ACID AND BASE STRENGTH

When a covalent acid dissolves in water, proton transfer forms H3O+ and an anion. This process

is called dissociation. Acids differ in their tendency to donate a proton; that is, acids differ in the

extent to which they dissociate in water.

• A strong acid readily donates a proton. When a strong acid dissolves in water, essentially

100% of the acid dissociates into ions.

• A weak acid less readily donates a proton. When a weak acid dissolves in water, only a

small fraction of the acid dissociates into ions.



Common strong acids include HI, HBr, HCl, H2SO4, and HNO3 (Table 9.1). When each acid is

dissolved in water, 100% of the acid dissociates, forming H3O+ and the conjugate base.

• Use a single reaction arrow.

• The product is greatly favored at equilibrium.



smi26573_ch09.indd 266



+

HCl(g)

strong acid



H2O(l)



H3O+(aq)



+



Cl−(aq)

conjugate base



H2SO4(l) +

strong acid



H2O(l)



H3O+(aq)



+



HSO4−(aq)

conjugate base



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ACID AND BASE STRENGTH



267







FIGURE 9.3



Focus on the Human Body: Hydrochloric Acid in the Stomach



H2O

H3O+

Cl−



mucous



stomach

gastric pit

stomach lining

The thick mucous layer protects the stomach lining.



Although HCl is a corrosive acid secreted in the stomach, a thick layer of mucous covering the

stomach wall protects it from damage by the strong acid.



HCl, hydrochloric acid, is secreted by the stomach to digest food (Figure 9.3), and H2SO4, sulfuric acid, is an important industrial starting material in the synthesis of phosphate fertilizers. A

single reaction arrow (

) is drawn to show that essentially all of the reactants are converted

to products.



TABLE 9.1



Relative Strength of Acids and Their Conjugate Bases

Acid



Conjugate Base



smi26573_ch09.indd 267



Hydroiodic acid



HI



I–



Hydrobromic acid



HBr



Br–



Bromide ion



Hydrochloric acid



HCl







Cl



Chloride ion



Sulfuric acid



H2SO4



HSO4–



Iodide ion







Hydrogen sulfate ion



Nitric acid



HNO3



NO3



Nitrate ion



Hydronium ion



H3O+



H2O



Water



H3PO4



H2PO4–



Dihydrogen phosphate ion



Weak Acids

Phosphoric acid







Hydrofluoric acid



HF



F



Acetic acid



CH3COOH



CH3COO–



Fluoride ion







Increasing base strength



Increasing acid strength



Strong Acids



Acetate ion



Carbonic acid



H2CO3



HCO3



Bicarbonate ion



Ammonium ion



NH4+



NH3



Ammonia



Hydrocyanic acid



HCN







CN



Cyanide ion



Water



H2O







OH



Hydroxide ion



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268



ACIDS AND BASES



Acetic acid, CH3COOH, is a weak acid. When acetic acid dissolves in water, only a small fraction of acetic acid molecules donate a proton to water to form H3O+ and the conjugate base,

CH3COO–. The major species at equilibrium is the undissociated acid, CH3COOH. Equilibrium

arrows that are unequal in length (

) are used to show that the equilibrium lies to the left.

Other weak acids and their conjugate bases are listed in Table 9.1.

• Use unequal reaction arrows.

• The reactants are favored at equilibrium.

CH3COOH(l)

weak acid



+



H2O(l)



H3O+(aq)



+



CH3COO−(aq)

conjugate base



Figure 9.4 illustrates the difference between an aqueous solution of a strong acid that is completely dissociated and a weak acid that contains much undissociated acid.

Bases also differ in their ability to accept a proton.

• A strong base readily accepts a proton. When a strong base dissolves in water,

essentially 100% of the base dissociates into ions.

• A weak base less readily accepts a proton. When a weak base dissolves in water, only a

small fraction of the base forms ions.



The most common strong base is hydroxide, –OH, used as a variety of metal salts, including

NaOH and KOH. Solid NaOH dissolves in water to form solvated Na+ cations and –OH anions.

In contrast, when NH3, a weak base, dissolves in water, only a small fraction of NH3 molecules

react to form NH4+ and –OH. The major species at equilibrium is the undissociated molecule,

NH3. Figure 9.5 illustrates the difference between aqueous solutions of strong and weak bases.

Table 9.1 lists common bases.





FIGURE 9.4



A Strong and Weak Acid Dissolved in Water



hydrochloric acid



vinegar



Cl−



CH3COO−



H3O+



H3O+

CH3COOH

A strong acid is

completely dissociated.



A weak acid contains mostly

undissociated acid, CH3COOH.



• The strong acid HCl completely dissociates into H3O+ and Cl– in water.

• Vinegar contains CH3COOH dissolved in H2O. The weak acid CH3COOH is only slightly

dissociated into H3O+ and CH3COO–, so mostly CH3COOH is present at equilibrium.



smi26573_ch09.indd 268



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ACID AND BASE STRENGTH



269







FIGURE 9.5



A Strong and Weak Base Dissolved in Water



sodium hydroxide



ammonia

NH3



Na+



NH4+



−OH



−OH



A strong base is

completely dissociated.



A weak base contains mostly

undissociated base, NH3.



• The strong base NaOH completely dissociates into Na+ and –OH in water.

• The weak base NH3 is only slightly dissociated into NH4+ and –OH, so mostly NH3 is

present at equilibrium.



• Use a single reaction arrow.

• The products are greatly favored at equilibrium.

NaOH(s) +

strong base



H2O(l)



Na+(aq)



+



−OH(aq)



NH3(g) +

weak base



H2O(l)



NH4+(aq)



+



−OH(aq)



• Use unequal reaction arrows.

• The reactants are favored at equilibrium.



An inverse relationship exists between acid and base strength.

• A strong acid readily donates a proton, forming a weak conjugate base.

• A strong base readily accepts a proton, forming a weak conjugate acid.



Why does this inverse relationship exist? Since a strong acid readily donates a proton, it forms

a conjugate base that has little ability to accept a proton. Since a strong base readily accepts a

proton, it forms a conjugate acid that tightly holds onto its proton, making it a weak acid.

Thus, a strong acid like HCl forms a weak conjugate base (Cl–), and a strong base like –OH forms

a weak conjugate acid (H2O). The entries in Table 9.1 are arranged in order of decreasing acid

strength. This means that Table 9.1 is also arranged in order of increasing strength of the resulting conjugate bases. Knowing the relative strength of two acids makes it possible to predict the

relative strength of their conjugate bases.



smi26573_ch09.indd 269



12/3/08 5:07:23 PM



270



ACIDS AND BASES



SAMPLE PROBLEM 9.7

ANALYSIS

SOLUTION



Using Table 9.1: (a) Is H3PO4 or HF the stronger acid? (b) Draw the conjugate base of each

acid and predict which base is stronger.

The stronger the acid, the weaker the conjugate base.

a. H3PO4 is located above HF in Table 9.1, making it the stronger acid.

b. To draw each conjugate base, remove a proton (H+). Since each acid is neutral, both

conjugate bases have a –1 charge. Since HF is the weaker acid, F– is the stronger

conjugate base.

H3PO4

stronger acid



HF

weaker acid



PROBLEM 9.9



lose H+



lose H+



H2PO4−



weaker base



F−



stronger base



Label the stronger acid in each pair. Which acid has the stronger conjugate base?

a. H2SO4 or H3PO4



b. HF or HCl



c. H2CO3 or NH4+



d. HCN or HF



PROBLEM 9.10



If HCOOH is a stronger acid than CH3COOH, which compound forms the stronger conjugate

base?



PROBLEM 9.11



(a) Draw the conjugate acids of NO2– and NO3–. (b) If NO2– is the stronger base, which acid is

stronger?



9.3B USING ACID STRENGTH TO PREDICT THE DIRECTION

OF EQUILIBRIUM

A Brønsted–Lowry acid–base reaction represents an equilibrium. Since an acid donates a proton

to a base, forming a conjugate acid and conjugate base, there are always two acids and two bases

in the reaction mixture. Which pair of acids and bases is favored at equilibrium? The position of

the equilibrium depends upon the strength of the acids and bases.

• The stronger acid reacts with the stronger base to form the weaker acid and weaker base.



Since a strong acid readily donates a proton and a strong base readily accepts one, these two

species react to form a weaker conjugate acid and base that do not donate or accept a proton as

readily. Thus, when the stronger acid and base are the reactants on the left side, the reaction

readily occurs and the reaction proceeds to the right.

A larger forward arrow means that products are favored.

H



A



+



stronger acid



B

stronger base



+

A−

weaker base



B+

weaker acid

H



Products are favored.



On the other hand, if an acid–base reaction would form the stronger acid and base, equilibrium favors the reactants and little product forms.

A larger reverse arrow means that reactants are favored.

H A

weaker acid



+



B

weaker base



+

A−

stronger base



H B+

stronger acid



Reactants are favored.



smi26573_ch09.indd 270



12/3/08 5:07:25 PM



EQUILIBRIUM AND ACID DISSOCIATION CONSTANTS



271



Predicting the direction of equilibrium using the information in Table 9.1 is illustrated in the

accompanying stepwise How To procedure.



HOW TO

EXAMPLE



Predict the Direction of Equilibrium in an Acid–Base Reaction

Are the reactants or products favored in the following acid–base reaction?

HCN(g) + –OH(aq)



Step [1]







CN(aq) + H2O(l)



Identify the acid in the reactants and the conjugate acid in the products.

gain of H+

HCN(g)

acid



+



−OH(aq)



−CN(aq)



base



conjugate base



+



H2O(l)

conjugate acid



loss of H+



• HCN is the acid since it donates a proton.

• H2O is the conjugate acid formed from the hydroxide base.



Step [2]



Determine the relative strength of the acid and the conjugate acid.

• According to Table 9.1, HCN is a stronger acid than H2O.



Step [3]



Equilibrium favors the formation of the weaker acid.

• Since the stronger acid HCN is a reactant, the reaction proceeds to the right as written, to form the weaker acid,

H2O. Unequal equilibrium arrows should be drawn with the larger arrow pointing towards the product side on

the right.

+

HCN(g)

stronger acid



−OH(aq)



−CN(aq)



+



H2O(l)

weaker acid



Products are favored.



PROBLEM 9.12



Are the reactants or products favored at equilibrium in each reaction?

a. HF(g) + –OH(aq)

F–(aq) + H2O(l)

b. NH4+(aq) + Cl–(aq)

NH3(g) + HCl(aq)



+

c. HCO3 (aq) + H3O (aq)

H2CO3(aq) + H2O(l)



PROBLEM 9.13



If lactic acid is similar in strength to acetic acid (Table 9.1), predict whether reactants or

products are favored in each reaction.

H



HEALTH NOTE



CH3



C



O

C



OH OH

C3H6O3

lactic acid



a. C3H6O3(aq)

lactic acid



+



H2O(l)



C3H5O3−(aq)



+



H3O+(aq)



b. C3H6O3(aq)

lactic acid



+



HCO3−(aq)



C3H5O3−(aq)



+



H2CO3(aq)



Lactic acid accumulates in tissues

during vigorous exercise, making

muscles feel tired and sore. The

formation of lactic acid is discussed

in greater detail in Section 24.4.



smi26573_ch09.indd 271



12/3/08 5:07:26 PM



272



ACIDS AND BASES



9.4 EQUILIBRIUM AND ACID DISSOCIATION CONSTANTS

Like all equilibria, we can write an expression for the equilibrium constant for the acid–base

reaction that takes place when an acid HA dissolves in water. The equilibrium constant (K) shows

the ratio of the concentrations of the products to the concentrations of the reactants.

Reaction



HA(aq)



Equilibrium constant



K



+



H3O+(aq)



H2O(l)



=



+



A: −



[H3O+][A:−]



concentrations of the products



[HA][H2O]



concentrations of the reactants



Water serves as both the base and the solvent. Since its concentration is essentially constant, the

equation can be rearranged by multiplying both sides by [H2O]. This forms a new constant called

the acid dissociation constant, Ka.

Ka



=



K[H2O]



=



[H3O+][A:–]

[HA]



acid dissociation

constant



How is Ka related to acid strength? The stronger the acid, the higher the concentration of the products of an acid–base reaction, and the larger the numerator in the expression for Ka. As a result:

• The stronger the acid, the larger the value of Ka.



The strong acids listed in Table 9.1 all have Ka values much greater than 1. Weak acids have Ka

values less than 1. The Ka’s for several weak acids are listed in Table 9.2.



TABLE 9.2



Increasing acidity



HEALTH NOTE



Citrus fruits (oranges, grapefruit, and

lemons) are well known sources of

vitamin C (ascorbic acid, Sample

Problem 9.9), but guava, kiwifruit, and

rose hips are excellent sources, too.



SAMPLE PROBLEM 9.8

ANALYSIS

SOLUTION



smi26573_ch09.indd 272



Acid Dissociation Constants (Ka) for Common Weak Acids



Acid



Structure



Ka



Hydrogen sulfate ion



HSO4–



1.2 × 10–2



Phosphoric acid



H3PO4



7.5 × 10–3



Hydrofluoric acid



HF



7.2 × 10–4



Acetic acid



CH3COOH



1.8 × 10–5



Carbonic acid



H2CO3



4.3 × 10–7



Dihydrogen phosphate ion



H2PO4–



6.2 × 10–8



Ammonium ion



NH4+



5.6 × 10–10



Hydrocyanic acid



HCN



4.9 × 10–10



Bicarbonate ion



HCO3–



5.6 × 10–11



Hydrogen phosphate ion



HPO42–



2.2 × 10–13



Which acid in each pair is stronger: (a) HCN or HSO4–; (b) CH3COOH or NH4+?

Use Table 9.2 to find the Ka for each acid. The acid with the larger Ka is the stronger acid.

a.



HCN

Ka = 4.9 × 10–10



HSO4–

Ka = 1.2 × 10–2

larger Ka

stronger acid



b. CH3COOH

1.8 × 10–5

larger Ka

stronger acid



NH4+

5.6 × 10–10



12/3/08 5:07:28 PM



EQUILIBRIUM AND ACID DISSOCIATION CONSTANTS



PROBLEM 9.14



273



Rank the acids in each group in order of increasing acid strength.

a. H3PO4, H2PO4–, HPO42–



PROBLEM 9.15



b. HCN, HF, CH3COOH



(a) Which compound is the stronger acid, H3PO4 or CH3COOH? (b) Draw the conjugate base

of each compound and predict which base is stronger.

Because Ka values tell us the relative strength of two acids, we can use Ka’s to predict the direction of equilibrium in an acid–base reaction, as shown in Sample Problem 9.9.

• Equilibrium favors the formation of the weaker acid—that is, the acid with the smaller Ka

value.



SAMPLE PROBLEM 9.9



Ascorbic acid, vitamin C, is needed for the formation of collagen, a common protein in

connective tissues in muscles and blood vessels. If vitamin C has a Ka of 7.9 × 10–5, are the

reactants or products favored in the following acid–base reaction?



HO



H



H



H



C



C



C



O

C



O



C6H8O6(aq)



H OH C C

HO

OH

vitamin C

ascorbic acid



+



C6H7O6−(aq)



NH3(aq)



vitamin C



+



NH4+(aq)



conjugate base

of vitamin C



C6H8O6



ANALYSIS



To determine the direction of equilibrium, we must identify the acid in the reactants and the

conjugate acid in the products. Then compare their Ka’s. Equilibrium favors the formation of

the acid with the smaller Ka value.



SOLUTION



Vitamin C is the acid and NH3 is the base on the reactant side. NH3 gains a proton to form its

conjugate acid, NH4+, which has a Ka of 5.6 × 10–10 (Table 9.2). The conjugate acid, therefore,

has a smaller Ka than vitamin C (7.9 × 10–5), making it the weaker acid. Thus, the products are

favored at equilibrium.

products favored

C6H8O6(aq)



+



NH3(aq)



vitamin C

Ka = 7.9 ×

larger Ka

stronger acid



Ka = 5.6 × 10−10

smaller Ka

weaker acid



CO32–(aq) + NH4+(aq)



Consider the weak acids, HCN and H2CO3.

a.

b.

c.

d.

e.



smi26573_ch09.indd 273



NH4+(aq)



Use the acid dissociation constants in Table 9.2 to determine whether the reactants or products

are favored in the following reaction.

HCO3–(aq) + NH3(aq)



PROBLEM 9.17



+



conjugate acid



10−5



PROBLEM 9.16



C6H7O6−(aq)



Which acid has the larger Ka?

Which acid is stronger?

Which acid has the stronger conjugate base?

Which acid has the weaker conjugate base?

When each acid is dissolved in water, for which acid does the equilibrium lie further to the

right?



12/3/08 5:07:31 PM



274



ACIDS AND BASES



9.5 DISSOCIATION OF WATER

In Section 9.2 we learned that water can behave as both a Brønsted–Lowry acid and a Brønsted–

Lowry base. As a result, two molecules of water can react together in an acid–base reaction.

loss of H+



+



H

H



O



+



H



acid



H



O







H



O



+



H



H



conjugate base



base



O



H



conjugate acid



gain of H+



• One molecule of H2O donates a proton (H+), forming its conjugate base –OH.

• One molecule of H2O accepts a proton, forming its conjugate acid H3O+.



Equilibrium favors the starting materials in this reaction, since the reactant acid, H2O, is much

weaker than the conjugate acid, H3O+. Thus, pure water contains an exceedingly low concentration of ions, H3O+ and –OH.

As usual, an expression for the equilibrium constant can be written that shows the ratio of the

concentrations of the products, H3O+ and –OH, to the concentration of the reactants, two molecules of H2O. Since the concentration of water is essentially constant, this equation can be

rearranged by multiplying by [H2O]2 to afford a new equilibrium constant, Kw, the ion–product

constant for water.

K



=



[H3O+][−OH]

[H2O][H2O]



Kw =

ion–product

constant



K[H2O]2



Kw



=



[H3O+][−OH]

[H2O]2



=



[H3O+][−OH]



=



[H3O+][−OH]



Multiply both sides by [H2O]2.



Since one H3O+ ion and one –OH ion are formed in each reaction, the concentration of H3O+

and –OH are equal in pure water. Experimentally it can be shown that the [H3O+] = [–OH] =

1.0 × 10–7 M at 25 °C. Thus,

Kw = [H3O+][–OH]

Kw = (1.0 × 10–7) × (1.0 × 10–7)

Kw = 1.0 × 1014

The product, [H3O+][OH], is a constant, 1.0 ì 1014, for all aqueous solutions at 25 °C.



Thus, the value of Kw applies to any aqueous solution, not just pure water. If we know the

concentration of one ion, H3O+ or –OH, we can find the concentration of the other by rearranging

the expression for Kw.

To calculate [−OH] when [H3O+] is known:

Kw



smi26573_ch09.indd 274



[H3O+][−OH]



=



[−OH]



=



[−OH]



=



Kw

[H3



O+]



1.0 × 10−14

[H3O+]



To calculate [H3O+] when [−OH] is known:

Kw



[H3O+][−OH]



=



[H3O+]



=



[H3O+]



=



Kw

[



−OH]



1.0 × 10−14

[−OH]



12/3/08 5:07:31 PM



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