Chemistry the central science 13e by theodore l brown 2

Chemistry the central science 13e by theodore l brown 2 Chemistry the central science 13e by theodore l brown 2 Chemistry the central science 13e by theodore l brown 2 Chemistry the central science 13e by theodore l brown 2 Chemistry the central science 13e by theodore l brown 2 Chemistry the central science 13e by theodore l brown 2 Chemistry the central science 13e by theodore l brown 2

con-and AB molecules (a) Which reaction mixture is at rium? (b) For those mixtures that are not at equilibrium, how

equilib-will the reaction proceed to reach equilibrium? [Sections 15.5 and 15.6]

15.9 The reaction A21g2 + B1g2 ∆ A1g2 + AB1g2 has an

equilibrium constant of K p= 2 The accompanying diagram shows a mixture containing A atoms (red), A2 molecules, and

AB molecules (red and blue) How many B atoms should be added to the diagram to illustrate an equilibrium mixture? [Section 15.6]

15.10 The diagram shown here represents the equilibrium state for

the reaction A21g2 + 2 B1g2 ∆ 2AB1g2 (a) Assuming

the volume is 2 L, calculate the equilibrium constant K c for the

reaction (b) If the volume of the equilibrium mixture is

decreased, will the number of AB molecules increase or crease? [Sections 15.5 and 15.7]

de-are shown here Which reaction has a larger equilibrium

con-stant? [Sections 15.1 and 15.2]

The following figures represent the concentrations at

equi-librium at the same temperature when X2 is Cl2 (green), Br2

(brown), and I2 (purple) List the equilibria from smallest to

largest equilibrium constant [Section 15.3]

(c)

15.7 When lead (IV) oxide is heated above 300 °C it

decom-poses according to the following reaction PbO21s2 ∆

PbO1s2 + O21g2 Consider the two sealed vessels of PbO2

shown here If both vessels are heated to 400 °C and allowed to

come to equilibitum which of the following statements is true?

(a) There will be less PbO2 remaining in vessel A, (b) There

will be less PbO2 remaining in vessel B, (c) The amount of

PbO2 remaining in each vessel will be the same [Section 15.4]

exercises 663

15.15 Write the expression for K c for the following reactions In each case indicate whether the reaction is homogeneous or heterogeneous

(a) 3 NO1g2 ∆ N2O1g2 + NO21g2

(b) CH41g2 + 2 H2S1g2 ∆ CS21g2 + 4 H21g2

(c) Ni1CO241g2 ∆ Ni1s2 + 4 CO1g2

(d) HF1aq2 ∆ H+1aq2 + F-1aq2

(e) 2 Ag1s2 + Zn2+1aq2 ∆ 2 Ag+1aq2 + Zn1s2

(f) H2O1l2 ∆ H+1aq2 + OH-1aq2

(g) 2 H2O1l2 ∆ 2 H+1aq2 + 2 OH-1aq2

15.16 Write the expressions for K c for the following reactions In each case indicate whether the reaction is homogeneous or heterogeneous

(a) N21g2 + O21g2 ∆ 2 NO1g2; K c = 1.5 * 10-10

(b) 2 SO21g2 + O21g2 ∆ 2 SO31g2; K p = 2.5 * 109

15.18 Which of the following reactions lies to the right, favoring the

formation of products, and which lies to the left, favoring mation of reactants?

for-(a) 2 NO1g2 + O21g2 ∆ 2 NO21g2; K p = 5.0 * 1012

(b) 2 HBr1g2 ∆ H21g2 + Br21g2; K c= 5.8 * 10-18

15.19 Which of the following statements are true and which are false?

(a) The equilibrium constant can never be a negative

num-ber (b) In reactions that we draw with a single-headed arrow,

the equilibrium constant has a value that is very close to zero

(c) As the value of the equilibrium constant increases the

speed at which a reaction reaches equilibrium increases

15.20 Which of the following statements are true and which are

false? (a) For the reaction 2 A1g2 + B1g2 ∆ A2B1g2 K c

and K p are numerically the same (b) It is possible to

distin-guish K c from K p by comparing the units used to express the

equilibrium constant (c) For the equilibrium in (a), the value

of K c increases with increasing pressure

is K c = 1.3 * 10-2 at 1000 K (a) At this temperature does

the equilibrium favor NO and Br2, or does it favor NOBr?

(b) Calculate K c for 2 NOBr1g2 ∆ 2 NO1g2 + Br21g2

(c) Calculate K c for NOBr1g2 ¡ NO1g2 + 1 Br21g2.

15.24 Consider the following equilibrium:

2 H21g2 + S21g2 ∆ 2 H2S1g2 K c = 1.08 * 107at 700 °C

15.11 The following diagrams represent equilibrium mixtures for

the reaction A2+ B ∆ A + AB at 300 K and 500 K The

A atoms are red, and the B atoms are blue Is the reaction

exo-thermic or endoexo-thermic? [Section 15.7]

15.12 The following graph represents the yield of the compound AB

at equilibrium in the reaction A1g2 + B1g2 ¡ AB1g2 at

two different pressures, x and y, as a function of temperature.

Temperature

[AB] P =x

P =y

(a) Is this reaction exothermic or endothermic? (b) Is P = x

greater or smaller than P = y? [Section 15.7]

Equilibrium; The Equilibrium Constant (sections

15.1–15.4)

15.13 Suppose that the gas-phase reactions A ¡ B and

B ¡ A are both elementary processes with rate constants

of 4.7 * 10-3 s-1 and 5.8 * 10-1 s-1, respectively (a) What

is the value of the equilibrium constant for the equilibrium

A1g2 ∆ B1g2? (b) Which is greater at equilibrium, the

partial pressure of A or the partial pressure of B?

15.14 Consider the reaction A + B ∆ C + D Assume that

both the forward reaction and the reverse reaction are

el-ementary processes and that the value of the equilibrium

constant is very large (a) Which species predominate at

equi-librium, reactants or products? (b) Which reaction has the

larger rate constant, the forward or the reverse?

15.32 Gaseous hydrogen iodide is placed in a closed container at

425 °C, where it partially decomposes to hydrogen and

io-dine: 2 HI1g2 ∆ H21g2 + I21g2 At equilibrium it is

found that 3HI4 = 3.53 * 10-3 M, 3H24 = 4.79 * 10-4 M,

and 3I24 = 4.79 * 10-4 M What is the value of K c at this temperature?

15.33 The equilibrium 2 NO1g2 + Cl21g2 ∆ 2 NOCl1g2 is

es-tablished at 500 K An equilibrium mixture of the three gases has partial pressures of 0.095 atm, 0.171 atm, and 0.28 atm for

NO, Cl2, and NOCl, respectively (a) Calculate K p for this

re-action at 500.0 K (b) If the vessel has a volume of 5.00 L,

cal-culate K c at this temperature

15.34 Phosphorus trichloride gas and chlorine gas react to form

phosphorus pentachloride gas: PCl31g2 + Cl21g2 ∆

PCl51g2 A 7.5-L gas vessel is charged with a mixture of

PCl31g2 and Cl21g2, which is allowed to equilibrate at 450

K At equilibrium the partial pressures of the three gases are

PPCl3= 0.124 atm, PCl2 = 0.157 atm, an d PPCl5 = 1.30 atm

(a) What is the value of K p at this temperature? (b) Does the equilibrium favor reactants or products? (c) Calculate K c for this reaction at 450 K

15.35 A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of

H2O is placed in a 1.0-L vessel at 300 K The following rium is established:

equilib-2 NO1gequilib-2 + equilib-2 H21g2 ∆ N21g2 + 2 H2O1g2

At equilibrium 3NO4 = 0.062 M (a) Calculate the

equilib-rium concentrations of H2, N2, and H2O (b) Calculate K c

15.36 A mixture of 1.374 g of H2 and 70.31 g of Br2 is heated in a 2.00-L vessel at 700 K These substances react according to

H21g2 + Br21g2 ∆ 2 HBr1g2

At equilibrium, the vessel is found to contain 0.566 g of H2

(a) Calculate the equilibrium concentrations of H2, Br2, and

HBr (b) Calculate K c

15.37 A mixture of 0.2000 mol of CO2, 0.1000 mol of H2, and 0.1600 mol of H2O is placed in a 2.000-L vessel The following equi-librium is established at 500 K:

CO21g2 + H21g2 ∆ CO1g2 + H2O1g2

(a) Calculate the initial partial pressures of CO2, H2, and H2O

(b) At equilibrium PH2O = 3.51 atm Calculate the rium partial pressures of CO2, H2, and CO (c) Calculate K p

equilib-for the reaction (d) Calculate K c for the reaction

15.38 A flask is charged with 1.500 atm of N2O41g2 and 1.00 atm

NO21g2 at 25 °C, and the following equilibrium is achieved:

N2O41g2 ∆ 2 NO21g2

After equilibrium is reached, the partial pressure of NO2 is

0.512 atm (a) What is the equilibrium partial pressure of

N2O4? (b) Calculate the value of K p for the reaction (c)

Calcu-late K c for the reaction

15.39 Two different proteins X and Y are dissolved in aqueous tion at 37 °C The proteins bind in a 1:1 ratio to form XY A

solu-solution that is initially 1.00 mM in each protein is allowed

to reach equilibrium At equilibrium, 0.20 mM of free X and 0.20 mM of free Y remain What is K c for the reaction?

15.40 A chemist at a pharmaceutical company is measuring

equi-librium constants for reactions in which drug candidate molecules bind to a protein involved in cancer The drug

(a) Calculate K p (b) Does the equilibrium mixture contain

mostly H2 and S2 or mostly H2S? (c) Calculate the value of K c

if you rewrote the equation H21g2 +1

2 S21g2 ∆ H2S1g2.

15.25 At 1000 K, K p= 1.85 for the reaction

SO21g2 + 1 O21g2 ∆ SO31g2

(a) What is the value of K p for the reaction SO31g2 ∆

SO21g2 + 12 O21g2? (b) What is the value of K p for the reaction

2 SO21g2 + O21g2 ∆ 2 SO31g2? (c) What is the value of

K c for the reaction in part (b)?

15.26 Consider the following equilibrium, for which K p = 0.0752

(c) What is the value of K c for the reaction in part (b)?

15.27 The following equilibria were attained at 823 K:

CoO1s2 + H21g2 ∆ Co1s2 + H2O1g2 K c = 67

CoO1s2 + CO1g2 ∆ Co1s2 + CO21g2 K c = 490

Based on these equilibria, calculate the equilibrium constant

for H21g2 + CO21g2 ∆ CO1g2 + H2O1g2 at 823 K.

15.28 Consider the equilibrium

N21g2 + O21g2 + Br21g2 ∆ 2 NOBr1g2

Calculate the equilibrium constant K p for this reaction, given

the following information (at 298 K):

2 NO1g2 + Br21g2 ∆ 2 NOBr1g2 K c = 2.0

2 NO1g2 ∆ N21g2 + O21g2 K c= 2.1* 1030

15.29 Mercury(I) oxide decomposes into elemental mercury and

elemental oxygen: 2 Hg2O1s2 ∆ 4 Hg1l2 + O21g2

(a) Write the equilibrium-constant expression for this

reac-tion in terms of partial pressures (b) Suppose you run this

reaction in a solvent that dissolves elemental mercury and

el-emental oxygen Rewrite the equilibrium-constant expression

in terms of molarities for the reaction, using (solv) to indicate

solvation

15.30 Consider the equilibrium Na2O1s2 + SO21g2 ∆ Na2SO31s2

(a) Write the equilibrium-constant expression for this

reac-tion in terms of partial pressures (b) All the compounds in

this reaction are soluble in water Rewrite the

equilibrium-constant expression in terms of molarities for the aqueous

reaction

Calculating Equilibrium Constants

(section 15.5)

15.31 Methanol 1CH3OH2 is produced commercially by the

catalyzed reaction of carbon monoxide and hydrogen:

CO1g2 + 2 H21g2 ∆ CH3OH1g2 A n e q u i l i b r i u m

mixture in a 2.00-L vessel is found to contain 0.0406 mol

CH3OH, 0.170 mol CO, and 0.302 mol H2 at 500 K Calculate

K c at this temperature

exercises 665

15.47 At 1285 °C, the equilibrium constant for the reaction

Br21g2 ∆ 2 Br1g2 is K c = 1.04 * 10-3 A 0.200-L vessel containing an equilibrium mixture of the gases has 0.245 g

Br21g2 in it What is the mass of Br1g2 in the vessel?

15.48 For the reaction H21g2 + I21g2 ∆ 2 HI1g2, K c = 55.3 at

700 K In a 2.00-L flask containing an equilibrium mixture of the three gases, there are 0.056 g H2 and 4.36 g I2 What is the mass of HI in the flask?

15.49 At 800 K, the equilibrium constant for I21g2 ∆ 2 I1g2 is

K c= 3.1 * 10-5 If an equilibrium mixture in a 10.0-L vessel contains 2.67 * 10-2 g of I(g), how many grams of I2 are in the mixture?

is K c = 2.4 * 103 If the initial concentration of NO is

0.175 M, what are the equilibrium concentrations of NO,

15.53 At 373 K, K p = 0.416 for the equilibrium

15.55 Consider the reaction

CaSO41s2 ∆ Ca2+1aq2 + SO42-1aq2

At 25 °C, the equilibrium constant is K c= 2.4 * 10-5 for

this reaction (a) If excess CaSO41s2 is mixed with water at

25 °C to produce a saturated solution of CaSO4, what are the equilibrium concentrations of Ca2+ and SO42-? (b) If the

resulting solution has a volume of 1.4 L, what is the mum mass of CaSO41s2 needed to achieve equilibrium?

15.56 At 80 °C, K c = 1.87 * 10-3 for the reaction

PH3BCl31s2 ∆ PH31g2 + BCl31g2

(a) Calculate the equilibrium concentrations of PH3 and BCl3 if

a solid sample of PH3BCl3 is placed in a closed vessel at 80 °C

and decomposes until equilibrium is reached (b) If the flask has

a volume of 0.250 L, what is the minimum mass of PH3BCl31s2

that must be added to the flask to achieve equilibrium?

15.57 For the reaction I2+ Br21g2 ∆ 2 IBr1g2, K c= 280 at

150 °C Suppose that 0.500 mol IBr in a 2.00-L flask is allowed

to reach equilibrium at 150 °C What are the equilibrium centrations of IBr, I2, and Br2?

con-molecules bind the protein in a 1:1 ratio to form a drug–

protein complex The protein concentration in aqueous

solu-tion at 25 °C is 1.50 * 10-6M Drug A is introduced into the

protein solution at an initial concentration of 2.00 * 10-6M

Drug B is introduced into a separate, identical protein

solu-tion at an initial concentrasolu-tion of 2.00 * 10-6M At

equilib-rium, the drug A-protein solution has an A-protein complex

concentration of 1.00 * 10-6M, and the drug B solution has a

B-protein complex concentration of 1.40 * 10-6M Calculate

the K c value for the A-protein binding reaction and for the

B-protein binding reaction Assuming that the drug that binds

more strongly will be more effective, which drug is the better

choice for further research?

applications of Equilibrium Constants

(section 15.6)

15.41 (a) If Q c 6 K c, in which direction will a reaction proceed in

order to reach equilibrium? (b) What condition must be

satis-fied so that Q c = K c?

15.42 (a) If Q c 7 K c, how must the reaction proceed to reach

equi-librium? (b) At the start of a certain reaction, only reactants

are present; no products have been formed What is the value

of Q c at this point in the reaction?

15.43 At 100 °C, the equilibrium constant for the reaction

COCl21g2 ∆ CO1g2 + Cl21g2 has the value K c =

2.19 * 10-10 Are the following mixtures of COCl2, CO, and

Cl2 at 100 °C at equilibrium? If not, indicate the direction that

the reaction must proceed to achieve equilibrium

(a) 3COCl24 = 2.00 * 10-3 M, 3CO4 = 3.3 * 10-6 M,

is 4.51 * 10-5 at 450 °C For each of the mixtures listed here,

indicate whether the mixture is at equilibrium at 450 °C If

it is not at equilibrium, indicate the direction (toward

prod-uct or toward reactants) in which the mixture must shift to

achieve equilibrium

(a) 98 atm NH3, 45 atm N2, 55 atm H2

(b) 57 atm NH3, 143 atm N2, no H2

(c) 13 atm NH3, 27 atm N2, 82 atm H2

15.45 At 100 °C, K c = 0.078 for the reaction

SO2Cl21g2 ∆ SO21g2 + Cl21g2

In an equilibrium mixture of the three gases, the

concentra-tions of SO2Cl2 and SO2 are 0.108 M and 0.052 M,

respec-tively What is the partial pressure of Cl2 in the equilibrium

mixture?

15.46 At 900 K, the following reaction has K p= 0.345:

2 SO21g2 + O21g2 ∆ 2 SO31g2

In an equilibrium mixture the partial pressures of SO2 and O2

are 0.135 atm and 0.455 atm, respectively What is the

equilib-rium partial pressure of SO3 in the mixture?

(d) decrease the volume of the container in which the

reac-tion occurs; (e) add a catalyst; (f) increase temperature 15.63 How do the following changes affect the value of the equi-

librium constant for a gas-phase exothermic reaction: (a) moval of a reactant, (b) removal of a product, (c) decrease in the volume, (d) decrease in the temperature, (e) addition of a

re-catalyst?

15.64 For a certain gas-phase reaction, the fraction of products in

an equilibrium mixture is increased by either increasing the temperature or by increasing the volume of the reaction ves-

sel (a) Is the reaction exothermic or endothermic? (b) Does

the balanced chemical equation have more molecules on the reactant side or product side?

15.65 Consider the following equilibrium between oxides of nitrogen

3 NO1g2 ∆ NO21g2 + N2O1g2

(a) Use data in Appendix C to calculate ∆H° for this

reaction (b) Will the equilibrium constant for the reaction increase or decrease with increasing temperature? (c) At

constant temperature, would a change in the volume of the container affect the fraction of products in the equilibrium mixture?

15.66 Methanol 1CH3OH2 can be made by the reaction of CO with H2:

CO1g2 + 2 H21g2 ∆ CH3OH1g2

(a) Use thermochemical data in Appendix C to calculate ∆H°

for this reaction (b) To maximize the equilibrium yield of methanol, would you use a high or low temperature? (c) To

maximize the equilibrium yield of methanol, would you use a high or low pressure?

15.67 Ozone, O3, decomposes to molecular oxygen in the sphere according to the reaction 2 O31g2 ¡ 3 O21g2

strato-Would an increase in pressure favor the formation of ozone or

of oxygen?

15.68 The water–gas shift reaction CO1g2 + H2O1g2 ∆

CO21g2 + H21g2 is used industrially to produce hydrogen

The reaction enthalpy is ∆H° = -41 kJ (a) To increase

the equilibrium yield of hydrogen would you use high or

low temperature? (b) Could you increase the equilibrium

yield of hydrogen by controlling the pressure of this tion? If so would high or low pressure favor formation of

reac-H21g2?

15.58 At 25 °C, the reaction

CaCrO41s2 ∆ Ca2+1aq2 + CrO42-1aq2

has an equilibrium constant K c = 7.1 * 10-4 What are the

equilibrium concentrations of Ca2 + and CrO42- in a

satu-rated solution of CaCrO4?

15.59 Methane, CH4, reacts with I2 according to the reaction

CH41g2 + l21g2 ∆ CH3l1g2 + HI1g2 At 630 K, K p for

this reaction is 2.26 * 10-4 A reaction was set up at 630 K

with initial partial pressures of methane of 105.1 torr and of

7.96 torr for I2 Calculate the pressures, in torr, of all reactants

and products at equilibrium

15.60 The reaction of an organic acid with an alcohol, in organic

sol-vent, to produce an ester and water is commonly done in the

pharmaceutical industry This reaction is catalyzed by strong

acid (usually H2SO4) A simple example is the reaction of acetic

acid with ethyl alcohol to produce ethyl acetate and water:

CH3COOH1solv2 + CH3CH2OH1solv2 ∆

CH3COOCH2CH31solv2 + H2O1solv2 where “(solv)” indicates that all reactants and products are

in solution but not an aqueous solution The equilibrium

constant for this reaction at 55 °C is 6.68 A pharmaceutical

chemist makes up 15.0 L of a solution that is initially 0.275

M in acetic acid and 3.85 M in ethanol At equilibrium, how

many grams of ethyl acetate are formed?

le Châtelier’s Principle (section 15.7)

15.61 Consider the following equilibrium for which ∆H 6 0

2 SO21g2 + O21g2 ∆ 2 SO31g2

How will each of the following changes affect an equilibrium

mixture of the three gases: (a) O21g2 is added to the system;

(b) the reaction mixture is heated; (c) the volume of the

reac-tion vessel is doubled; (d) a catalyst is added to the mixture;

(e) the total pressure of the system is increased by adding a

noble gas; (f) SO31g2 is removed from the system?

15.62 Consider the reaction

4 NH31g2 + 5 O21g2 ∆

4 NO1g2 + 6 H2O1g2, ∆H = -904.4 kJ

Does each of the following increase, decrease, or leave

unchanged the yield of NO at equilibrium? (a)

in-crease 3NH34; (b) increase 3H2O4; (c) decrease 3O24;

additional Exercises

15.69 Both the forward reaction and the reverse reaction in the

fol-lowing equilibrium are believed to be elementary steps:

CO1g2 + Cl21g2 ∆ COCl1g2 + Cl1g2

At 25 °C, the rate constants for the forward and reverse

reac-tions are 1.4 * 10-28 M-1 s-1 and 9.3 * 1010 M-1 s-1,

respec-tively (a) What is the value for the equilibrium constant at 25 °C?

(b) Are reactants or products more plentiful at equilibrium?

15.70 If K c = 1 for the equilibrium 2 A1g2 ∆ B1g2, what is the

relationship between [A] and [B] at equilibrium?

15.71 A mixture of CH4 and H2O is passed over a nickel

cata-lyst at 1000 K The emerging gas is collected in a 5.00-L

flask and is found to contain 8.62 g of CO, 2.60 g of H2, 43.0 g of CH4, and 48.4 g of H2O Assuming that equilib-rium has been reached, calculate K c and K p for the reaction

CH41g2 + H2O1g2 ∆ CO1g2 + 3 H21g2.

15.72 When 2.00 mol of SO2Cl2 is placed in a 2.00-L flask at 303 K, 56% of the SO2Cl2 decomposes to SO2 and Cl2 :

SO2Cl21g2 ∆ SO21g2 + Cl21g2

(a) Calculate K c for this reaction at this temperature

(b) Calculate K p for this reaction at 303 K (c) According to

Le Châtelier’s principle, would the percent of SO2Cl2 that composes increase, decrease or stay the same if the mixture

de-additional exercises 667

At 700 K, the equilibrium constant K p for this reaction is 0.26 Predict the behavior of each of the following mixtures at this temperature and indicate whether or not the mixtures are at equilibrium If not, state whether the mixture will need to produce more products or reactants to reach equilibrium

(a) PNO = 0.15 atm, PCl2 = 0.31 atm, PNOCl = 0.11 atm

ves-(a) 15.0 g CaCO3, 15.0 g CaO, and 4.25 g CO2

(b) 2.50 g CaCO3, 25.0 g CaO, and 5.66 g CO2

(c) 30.5 g CaCO3, 25.5 g CaO, and 6.48 g CO2

15.83 When 1.50 mol CO2 and 1.50 mol H2 are placed in a 3.00-L container at 395 °C, the following reaction occurs:

CO21g2 + H21g2 ∆ CO1g2 + H2O1g2 If K c = 0.802, what are the concentrations of each substance in the equilib-rium mixture?

15.84 The equilibrium constant K c for C1s2 + CO21g2 ∆

2 CO1g2 is 1.9 at 1000 K and 0.133 at 298 K (a) If excess C is

allowed to react with 25.0 g of CO2 in a 3.00-L vessel at 1000

K, how many grams of CO are produced? (b) How many grams of C are consumed? (c) If a smaller vessel is used for the reaction, will the yield of CO be greater or smaller? (d) Is

the reaction endothermic or exothermic?

15.85 NiO is to be reduced to nickel metal in an industrial process

by use of the reaction

NiO1s2 + CO1g2 ∆ Ni1s2 + CO21g2

At 1600 K, the equilibrium constant for the reaction is

K p= 6.0 * 102 If a CO pressure of 150 torr is to be employed in the furnace and total pressure never exceeds

760 torr, will reduction occur?

15.86 Le Châtelier noted that many industrial processes of his time

could be improved by an understanding of chemical ria For example, the reaction of iron oxide with carbon mon-oxide was used to produce elemental iron and CO2 according

equilib-to the reaction

Fe2O31s2 + 3 CO1g2 ∆ 2 Fe1s2 + 3 CO21g2

Even in Le Châtelier’s time, it was noted that a great deal of

CO was wasted, expelled through the chimneys over the naces Le Châtelier wrote, “Because this incomplete reaction was thought to be due to an insufficiently prolonged contact between carbon monoxide and the iron ore [oxide], the dimen-sions of the furnaces have been increased In England, they have been made as high as 30 m But the proportion of carbon monoxide escaping has not diminished, thus demonstrating, by

fur-an experiment costing several hundred thousfur-and frfur-ancs, that the reduction of iron oxide by carbon monoxide is a limited reaction Acquaintance with the laws of chemical equilibrium would have permitted the same conclusion to be reached more rapidly and far more economically.” What does this anecdote tell us about the equilibrium constant for this reaction?

were transferred to a 15.00-L vessel? (d) Use the equilibrium

constant you calculated above to determine the percentage of

SO2Cl2 that decomposes when 2.00 mol of SO2Cl2 is placed in

a 15.00-L vessel at 303 K

15.73 A mixture of H2, S, and H2S is held in a 1.0-L vessel at 90 °C

and reacts according to the equation:

H21g2 + S1s2 ∆ H2S1g2

At equilibrium, the mixture contains 0.46 g of H2S and 0.40

g H2 (a) Write the equilibrium-constant expression for this

reaction (b) What is the value of K c for the reaction at this

temperature?

15.74 A sample of nitrosyl bromide (NOBr) decomposes according

to the equation

2 NOBr1g2 ∆ 2 NO1g2 + Br21g2

An equilibrium mixture in a 5.00-L vessel at 100 °C contains

3.22 g of NOBr, 3.08 g of NO, and 4.19 g of Br2 (a) Calculate

K c (b) What is the total pressure exerted by the mixture of gases?

(c) What was the mass of the original sample of NOBr?

15.75 Consider the hypothetical reaction A1g2 ∆ 2 B1g2 A

flask is charged with 0.75 atm of pure A, after which it is

al-lowed to reach equilibrium at 0 °C At equilibrium, the partial

pressure of A is 0.36 atm (a) What is the total pressure in the

flask at equilibrium? (b) What is the value of K p? (c) What

could we do to maximize the yield of B?

15.76 As shown in Table 15.2, the equilibrium constant for the reaction

N21g2 + 3 H21g2 ∆ 2 NH31g2 is K p= 4.34 * 10-3 at

300 °C Pure NH3 is placed in a 1.00-L flask and allowed to reach

equilibrium at this temperature There are 1.05 g NH3 in the

equi-librium mixture (a) What are the masses of N2 and H2 in the

equilibrium mixture? (b) What was the initial mass of ammonia

placed in the vessel? (c) What is the total pressure in the vessel?

15.77 For the equilibrium

2 IBr1g2 ∆ I21g2 + Br21g2

K p= 8.5 * 10-3 at 150 °C If 0.025 atm of IBr is placed in a

2.0-L container, what is the partial pressure of all substances

after equilibrium is reached?

15.78 For the equilibrium

PH3BCl31s2 ∆ PH31g2 + BCl31g2

K p = 0.052 at 60 °C (a) Calculate K c (b) After 3.00 g of solid

PH3BCl3 is added to a closed 1.500-L vessel at 60 °C, the vessel

is charged with 0.0500 g of BCl31g2 What is the equilibrium

concentration of PH3?

[15.79]Solid NH4SH is introduced into an evacuated flask at 24 °C

The following reaction takes place:

NH4SH1s2 ∆ NH31g2 + H2S1g2

At equilibrium, the total pressure (for NH3 and H2S taken

to-gether) is 0.614 atm What is K p for this equilibrium at 24 °C?

[ 15.80 ] A 0.831-g sample of SO3 is placed in a 1.00-L container and

heated to 1100 K The SO3 decomposes to SO2 and O2:

2 SO31g2 ∆ 2 SO21g2 + O21g2

At equilibrium, the total pressure in the container is 1.300 atm

Find the values of K p and K c for this reaction at 1100 K

15.81 Nitric oxide (NO) reacts readily with chlorine gas as follows:

2 NO1g2 + Cl21g2 ∆ 2 NOCl1g2

equilibrium concentrations (in terms of x) into the equilibrium- constant expression, derive an equation that can be solved for x

(d) The equation from part (c) is a cubic equation (one that

has the form ax3+ bx2 + cx + d = 0) In general, cubic

equations cannot be solved in closed form However, you can estimate the solution by plotting the cubic equation in the al-

lowed range of x that you specified in part (b) The point at which the cubic equation crosses the x-axis is the solution

(e) From the plot in part (d), estimate the equilibrium

concen-trations of A, B, and C (Hint: You can check the accuracy of

your answer by substituting these concentrations into the librium expression.)

15.91 At 1200 K, the approximate temperature of automobile haust gases (Figure 15.15), K p for the reaction

ex-2 CO21g2 ∆ 2 CO1g2 + O21g2

is about 1 * 10-13 Assuming that the exhaust gas (total pressure 1 atm) contains 0.2% CO, 12% CO2, and 3% O2 by volume, is the system at equilibrium with respect to the CO2reaction? Based on your conclusion, would the CO concen-tration in the exhaust be decreased or increased by a catalyst that speeds up the CO2 reaction? Recall that at a fixed pres-sure and temperature, volume % = mol %

15.92 Suppose that you worked at the U.S Patent Office and a patent

application came across your desk claiming that a newly oped catalyst was much superior to the Haber catalyst for ammo-nia synthesis because the catalyst led to much greater equilibrium conversion of N2 and H2 into NH3 than the Haber catalyst under the same conditions What would be your response?

devel-[15.87]At 700 K, the equilibrium constant for the reaction

CCl41g2 ∆ C1s2 + 2 Cl21g2

is K p = 0.76 A flask is charged with 2.00 atm of CCl4, which

then reaches equilibrium at 700 K (a) What fraction of the

CCl4 is converted into C and Cl2? (b) What are the partial

pressures of CCl4 and Cl2 at equilibrium?

[15.88] The reaction PCl31g2 + Cl21g2 ∆ PCl51g2 has K p =

0.0870 at 300 °C A flask is charged with 0.50 atm PCl3, 0.50

atm Cl2, and 0.20 atm PCl5 at this temperature (a) Use the

re-action quotient to determine the direction the rere-action must

proceed to reach equilibrium (b) Calculate the equilibrium

partial pressures of the gases (c) What effect will increasing

the volume of the system have on the mole fraction of Cl2

in the equilibrium mixture? (d) The reaction is exothermic

What effect will increasing the temperature of the system

have on the mole fraction of Cl2 in the equilibrium mixture?

[15.89] An equilibrium mixture of H2, I2, and HI at 458 °C contains

0.112 mol H2, 0.112 mol I2, and 0.775 mol HI in a 5.00-L vessel

What are the equilibrium partial pressures when equilibrium is

reestablished following the addition of 0.200 mol of HI?

[15.90] Consider the hypothetical reaction A1g2 + 2 B1g2 ∆

2 C1g2, for which K c = 0.25 at a certain temperature A 1.00-L

reaction vessel is loaded with 1.00 mol of compound C, which

is allowed to reach equilibrium Let the variable x represent

the number of mol>L of compound A present at equilibrium

(a) In terms of x, what are the equilibrium concentrations of

compounds B and C? (b) What limits must be placed on the value

of x so that all concentrations are positive? (c) By putting the

integrative Exercises

15.93 Consider the reaction IO4-1aq2 + 2 H2O1l2 ∆ H4IO6-1aq2;

K c= 3.5 * 10-2 If you start with 25.0 mL of a 0.905 M

solu-tion of NaIO4, and then dilute it with water to 500.0 mL, what

is the concentration of H4IO6- at equilibrium?

[15.94] Silver chloride, AgCl(s), is an “insoluble” strong

electro-lyte (a) Write the equation for the dissolution of AgCl(s) in

H2O1l2 (b) Write the expression for K c for the reaction in

part (a) (c) Based on the thermochemical data in Appendix C

and Le Châtelier’s principle, predict whether the solubility of

AgCl in H2O increases or decreases with increasing

tempera-ture (d) The equilibrium constant for the dissolution of AgCl

in water is 1.6 * 10-10 at 25 °C In addition, Ag+1aq2 can

re-act with Cl-1aq2 according to the reaction

Ag+1aq2 + 2 Cl-1aq2 ∆ AgCl2-1aq2 where K c = 1.8 * 105 at 25 °C Although AgCl is “not solu-

ble” in water, the complex AgCl2- is soluble At 25 °C, is the

solubility of AgCl in a 0.100 M NaCl solution greater than the

solubility of AgCl in pure water, due to the formation of

sol-uble AgCl2- ions? Or is the AgCl solubility in 0.100 M NaCl

less than in pure water because of a Le Châtelier-type

argu-ment? Justify your answer with calculations (Hint: Any form

in which silver is in solution counts as “solubility.”)

[15.95]Consider the equilibrium A ∆ B in which both the

for-ward and reverse reactions are elementary (single-step)

reac-tions Assume that the only effect of a catalyst on the reaction

is to lower the activation energies of the forward and reverse

reactions, as shown in Figure 15.14 Using the Arrhenius

equation (Section 14.5), prove that the equilibrium constant is

the same for the catalyzed reaction as for the uncatalyzed one

[15.96] The phase diagram for SO2 is shown here (a) What does this

diagram tell you about the enthalpy change in the reaction

SO21l2 ¡ SO21g2? (b) Calculate the equilibrium constant

for this reaction at 100 °C and at 0 °C (c) Why is it not

pos-sible to calculate an equilibrium constant between the gas

and liquid phases in the supercritical region? (d) At which

of the three points marked in red does SO21g2 most closely

approach ideal-gas behavior? (e) At which of the three red

points does SO21g2 behave least ideally?

Gas

Critical point

Supercritical region Liquid

Design an experiment 669

15.99 Water molecules in the atmosphere can form

hydrogen-bonded dimers, 1H2O22 The presence of these dimers is thought to be important in the nucleation of ice crystals in

the atmosphere and in the formation of acid rain (a) Using

VSEPR theory, draw the structure of a water dimer, using

dashed lines to indicate intermolecular interactions (b) What

kind of intermolecular forces are involved in water dimer

for-mation? (c) The K p for water dimer formation in the gas phase

is 0.050 at 300 K and 0.020 at 350 K Is water dimer formation endothermic or exothermic?

15.100 The protein hemoglobin (Hb) transports O2 in lian blood Each Hb can bind 4 O2 molecules The equilib-rium constant for the O2 binding reaction is higher in fetal hemoglobin than in adult hemoglobin In discussing protein oxygen-binding capacity, biochemists use a measure called

mamma-the P50 value, defined as mamma-the partial pressure of oxygen at

which 50% of the protein is saturated Fetal hemoglobin has a P50 value of 19 torr, and adult hemoglobin has a P50 value

of 26.8 torr Use these data to estimate how much larger

K c is for the aqueous reaction 4 O21g2 + Hb1aq2 ¡

3Hb1O2241aq24.

[15.97]Write the equilibrium-constant expression for the equilibrium

C1s2 + CO21g2 ∆ 2 CO1g2

The table that follows shows the relative mole percentages

of CO21g2 and CO(g) at a total pressure of 1 atm for several

temperatures Calculate the value of K p at each temperature

Is the reaction exothermic or endothermic?

Temperature 1 °C2 Co 2 1mol %2 Co 1mol %2

15.98 In Section 11.5, we defined the vapor pressure of a liquid in

terms of an equilibrium (a) Write the equation representing the

equilibrium between liquid water and water vapor and the

cor-responding expression for K p (b) By using data in Appendix B,

give the value of K p for this reaction at 30 °C (c) What is the

value of K p for any liquid in equilibrium with its vapor at the

normal boiling point of the liquid?

design an Experiment

The reaction between hydrogen and iodine to form hydrogen iodide was used to

illus-trate Beer’s law in Chapter 14 (Figure 14.5) The reaction can be monitored using

visi-ble-light spectroscopy because I2 has a violet color while H2 and HI are colorless At 300 K,

the equilibrium constant for the reaction H21g2 + I21g2 ∆ 2 HI1g2 is K c = 794

To answer the following questions assume you have access to hydrogen, iodine,

hydro-gen iodide, a transparent reaction vessel, a visible-light spectrometer, and a means for

changing the temperature (a) Which gas or gases concentration could you readily

moni-tor with the spectrometer? (b) To use Beer’s law (Equation 14.5) you need to determine

the extinction coefficient, e, for the substance in question How would you determine e?

(c) Describe an experiment for determining the equilibrium constant at 600 K (d) Use the bond

enthalpies in Table 8.4 to estimate the enthalpy of this reaction (e) Based on your answer to part

(d), would you expect K c at 600 K to be larger or smaller than at 300 K?

Acid–Base Equilibria

The acids and bases that you have used so far in the laboratory are

probably solutions of relatively simple inorganic substances, such as

hydrochloric acid, sulfuric acid, sodium hydroxide, and the like

But acids and bases are important even when we are not in the lab

K w = [H3O+][OH−] defines the relationship between H3O+ and

OH− concentrations in aqueous solutions

16.4 The ph Scale We use the pH scale to describe the acidity or basicity of an aqueous solution Neutral solutions have a

pH = 7, acidic solutions have pH below 7, and basic solutions have

pH above 7

16.5 STrong acidS and BaSeS We categorize acids

and bases as being either strong or weak electrolytes Strong

acids and bases are strong electrolytes, ionizing or dissociating

completely in aqueous solution Weak acids and bases are weak

electrolytes and ionize only partially

16.1 acidS and BaSeS: a Brief review We begin by

reviewing the Arrhenius definition of acids and bases.

16.2 BrønSTed–lowry acidS and BaSeS We learn that

a Brønsted–Lowry acid is a proton donor and a Brønsted–Lowry

base is a proton acceptor Two species that differ by the presence

or absence of a proton are known as a conjugate acid–base pair.

16.3 The auToionizaTion of waTer We see that the

autoionization of water produces small quantities of H3O+

and OH− ions The equilibrium constant for autoionization,

WhAT’s

AhEAd

vinegar Grapes contain several acids that

contribute to their characteristic flavor The distinctive flavor of all vinegars is due to acetic acid Balsamic vinegar is obtained by fermenting grapes

They are ubiquitous, including in the foods we eat The characteristic flavor of the

grapes shown in the opening photograph is largely due to tartaric acid 1H2C4H4O62

and malic acid 1H2C4H4O52 (Figure 16.1), two closely related (they differ by only one

O atom) organic acids that are found in biological systems Fermentation of the sugars

in the grapes ultimately forms vinegar, the tangy, sour flavor of which is due to acetic

acid 1CH3COOH2, a substance we discussed in Section 4.3 The sour taste of oranges,

lemons, and other citrus fruits is due to citric acid 1H3C6H5O72, and, to a lesser extent,

ascorbic acid 1H2C6H6O62, better known as Vitamin C.

Acids and bases are among the most important substances in chemistry, and

they affect our daily lives in innumerable ways Not only are they present in our

foods, but acids and bases are also crucial components of living systems, such as

the amino acids that are used to synthesize proteins and the nucleic acids that code

genetic information Both citric and malic acids are among several acids involved

in the Krebs cycle (also called the citric acid cycle) that is used to generate energy

in aerobic organisms The application of acid–base chemistry has also had critical

roles in shaping modern society, including such human-driven activities as

indus-trial manufacturing, the creation of advanced pharmaceuticals, and many aspects

of the environment.

The impact of acids and bases depends not only on the type of acid or base, but

also on how much is present The time required for a metal object immersed in water

to corrode, the ability of an aquatic environment to support fish and plant life, the fate

of pollutants washed out of the air by rain, and even the rates of reactions that maintain

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

16.10 acid–BaSe Behavior and chemical STrucTure

We explore the relationship between chemical structure and acid–base behavior

16.11 lewiS acidS and BaSeS Finally, we see the most general definition of acids and bases, namely the Lewis acid–base

definition A Lewis acid is an electron-pair acceptor and a Lewis base is an electron-pair donor.

16.6 weak acidS We learn that the ionization of a weak

acid in water is an equilibrium process with an equilibrium

constant K a that can be used to calculate the pH of a weak

acid solution

16.7 weak BaSeS We learn that the ionization of a weak base

in water is an equilibrium process with equilibrium constant K b

that can be used to calculate the pH of a weak base solution

16.8 relaTionShip BeTween K a and K b We see that

K a and K b are related by the relationship K a × Kb = Kw Hence,

the stronger an acid, the weaker its conjugate base

our lives all critically depend on the acidity or basicity of solutions We will thus explore

in this chapter how we measure acidity and how the chemical reactions of acids and bases depend on their concentrations.

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

in terms of the chemical equilibria in which they participate.

From the earliest days of experimental chemistry, scientists have recognized acids and bases by their characteristic properties Acids have a sour taste and cause certain dyes to change color, whereas bases have a bitter taste and feel slippery (soap is a good exam-

ple) Use of the term base comes from the old English meaning of the word, “to bring low.” (We still use the word debase in this sense, meaning to lower the value of some-

thing.) When a base is added to an acid, the base “lowers” the amount of acid Indeed, when acids and bases are mixed in the right proportions, their characteristic properties seem to disappear altogether (Section 4.3)

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

• An acid is a substance that, when dissolved in water, increases the concentration of

H+ ions.

• A base is a substance that, when dissolved in water, increases the concentration of

OH- ions.

Hydrogen chloride gas, which is highly soluble in water, is an example of an Arrhenius

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

HCl1g2 ¡H2O H+1aq2 + Cl-1aq2 [16.1] The aqueous solution of HCl is known as hydrochloric acid Concentrated hydrochloric acid is about 37% HCl by mass and is 12 M in HCl.

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

give it some thought

Which two ions are central to the Arrhenius definitions of acids and bases?

▲ Figure 16.1 two organic acids: tartartic acid, H 2 C 4 H 4 O 6 , and malic acid, H 2 C 4 H 4 O 5

O

H OH

CCHO

H OH

O

C

secTiON 16.2 Brønsted–Lowry Acids and Bases 673

and Bases

The Arrhenius concept of acids and bases, while useful, is rather limited For one thing, it is

restricted to aqueous solutions In 1923 the Danish chemist Johannes Brønsted (1879–1947)

and the English chemist Thomas Lowry (1874–1936) independently proposed a more

gen-eral definition of acids and bases Their concept is based on the fact that acid–base reactions

involve the transfer of H+ ions from one substance to another To understand this definition

better, we need to examine the behavior of the H+ ion in water more closely.

The H+ Ion in Water

We might at first imagine that ionization of HCl in water produces just H+ and Cl-

A hydrogen ion is no more than a bare proton—a very small particle with a positive

charge As such, an H+ ion interacts strongly with any source of electron density, such

as the nonbonding electron pairs on the oxygen atoms of water molecules For

exam-ple, the interaction of a proton with water forms the hydronium ion, H3O+1aq2:

O H

H

H

O H

The behavior of H+ ions in liquid water is complex because hydronium ions interact

with additional water molecules via the formation of hydrogen bonds (Section 11.2)

For example, the H3O+ ion bonds to additional H2O molecules to generate such ions as

H5O2+ and H9O4+ (▶ Figure 16.2).

Chemists use the notations H+1aq2 and H3O+1aq2 interchangeably to represent the

hydrated proton responsible for the characteristic properties of aqueous solutions of acids

We often use the notation H+1aq2 for simplicity and convenience, as we did in Chapter 4

and Equation 16.1 The notation H3O+1aq2, however, more closely represents reality.

Proton-Transfer Reactions

In the reaction that occurs when HCl dissolves in water, the HCl molecule transfers an

H+ ion (a proton) to a water molecule Thus, we can represent the reaction as occurring

between an HCl molecule and a water molecule to form hydronium and chloride ions:

+

−

O H

H

H

O H

Notice that the reaction in Equation 16.3 involves a proton donor (HCl) and a

pro-ton acceptor 1H2O2 The notion of transfer from a proton donor to a proton acceptor is

the key idea in the Brønsted–Lowry definition of acids and bases:

• An acid is a substance (molecule or ion) that donates a proton to another substance.

• A base is a substance that accepts a proton.

H H

OHHH

OHOHH

OH

Go FiGuRE

Which type of intermolecular force

do the dotted lines in this figure represent?

Thus, when HCl dissolves in water (Equation 16.3), HCl acts as a Brønsted–Lowry acid (it donates a proton to H2O), and H2O acts as a Brønsted–Lowry base (it accepts a

proton from HCl) We see that the H2O molecule serves as a proton acceptor by using one of the nonbonding pairs of electrons on the O atom to “attach” the proton.

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

+

−

N H

H

H

N H

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

in which we have the equilibrium:

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

Base AcidAmmonia is a Brønsted–Lowry base because it accepts a proton from H2O Ammonia

is also an Arrhenius base because adding it to water leads to an increase in the tration of OH-1aq2.

concen-The transfer of a proton always involves both an acid (donor) and a base (acceptor)

In other words, a substance can function as an acid only if another substance neously behaves as a base To be a Brønsted–Lowry acid, a molecule or ion must have

simulta-a hydrogen simulta-atom it csimulta-an lose simulta-as simulta-an H+ ion To be a Brønsted–Lowry base, a molecule or ion must have a nonbonding pair of electrons it can use to bind the H+ ion.

Some substances can act as an acid in one reaction and as a base in another For ple, H2O is a Brønsted–Lowry base in Equation 16.3 and a Brønsted–Lowry acid in Equa-

exam-tion 16.5 A substance capable of acting as either an acid or a base is called amphiprotic An

amphiprotic substance acts as a base when combined with something more strongly acidic than itself and as an acid when combined with something more strongly basic than itself.

give it some thought

In the forward reaction of this equilibrium, which substance acts as the Brønsted–Lowry base?

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

Conjugate Acid–Base Pairs

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

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

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

▲ Figure 16.3 fog of NH 4Cl1s2 caused by

the reaction of HCl1g2 and NH31g2.

secTiON 16.2 Brønsted–Lowry Acids and Bases 675

donates a proton to the A- ion, so H3O+ is the acid and A- is the base When the acid

HA donates a proton, it leaves behind a substance, A-, that can act as a base Likewise,

when H2O acts as a base, it generates H3O+, which can act as an acid.

An acid and a base such as HA and A- that differ only in the presence or absence of a

proton are called a conjugate acid–base pair.* Every acid has a conjugate base, formed by

removing a proton from the acid For example, OH- is the conjugate base of H2O, and A

-is the conjugate base of HA Every base has a conjugate acid, formed by adding a proton

to the base Thus, H3O+ is the conjugate acid of H2O, and HA is the conjugate acid of A-.

In any acid–base (proton-transfer) reaction, we can identify two sets of conjugate

acid–base pairs For example, consider the reaction between nitrous acid and water:

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

Acid Base Conjugate

base Conjugateacidremove H

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

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

Acid

base

Conjugateacidremove H

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

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

(b) What is the conjugate acid of CN-, SO42 -, H2O, HCO3-?

s A m p L E

ExERcisE 16.1 identifying conjugate Acids and Bases

soLuTion

Analyze We are asked to give the conjugate base for several acids and

the conjugate acid for several bases

Plan The conjugate base of a substance is simply the parent substance

minus one proton, and the conjugate acid of a substance is the parent

substance plus one proton

Solve

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

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

CO32 -

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

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

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

as either an acid or a base

practice Exercise 1

Consider the following equilibrium reaction:

HSO4-1aq2 + OH-1aq2 ∆ SO42 -1aq2 + H2O1l2

Which substances are acting as acids in the reaction?

(a) HSO4- and OH- (b) HSO4- and H2O

Once you become proficient at identifying conjugate acid–base pairs it is not

difficult to write equations for reactions involving Brønsted–Lowry acids and bases

(proton-transfer reactions).

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

-with water (a) in which the ion acts as an acid and (b) in which the ion acts as a base In both

cases identify the conjugate acid–base pairs

s A m p L E

ExERcisE 16.2 Writing Equations for proton-Transfer Reactions

soLuTion

Analyze and Plan We are asked to write two equations representing

re-actions between HSO3- and water, one in which HSO3- should donate

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

which HSO3- should accept a proton from water, thereby acting as a

base We are also asked to identify the conjugate pairs in each equation

Solve

(a) HSO3-1aq2 + H2O1l2 ∆ SO32 -1aq2 + H3O+1aq2

The conjugate pairs in this equation are HSO3- (acid) and SO32

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

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

The conjugate pairs in this equation are H2O (acid) and OH-

(conju-gate base), and HSO3- (base) and H2SO3 (conjugate acid)

practice Exercise 1

The dihydrogen phosphate ion, H2PO4-, is amphiprotic In which

of the following reactions is this ion serving as a base?

(i) H3O+1aq2 + H2PO4-1aq2 ∆ H3PO41aq2 + H2O1l2

(ii) H3O+1aq2 + HPO42 -1aq2 ∆ H2PO4-1aq2 + H2O1l2

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

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

practice Exercise 2

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

Relative Strengths of Acids and Bases

Some acids are better proton donors than others, and some bases are better proton tors than others If we arrange acids in order of their ability to donate a proton, we find that the more easily a substance gives up a proton, the less easily its conjugate base accepts

accep-a proton Similaccep-arly, the more eaccep-asily accep-a baccep-ase accep-accepts accep-a proton, the less eaccep-asily its conjugaccep-ate accep-acid

gives up a proton In other words, the stronger an acid, the weaker its conjugate base, and the stronger a base, the weaker its conjugate acid Thus, if we know how readily an acid donates

protons, we also know something about how readily its conjugate base accepts protons The inverse relationship between the strengths of acids and their conjugate bases

is illustrated in ▼ Figure 16.4 Here we have grouped acids and bases into three broad categories based on their behavior in water:

BASE

Cl−HSO4−

HS−HPO4−

Negligible basicity

ACID

HCl

H2SO4HNO3

H3O+(aq)

HSO4−

H3PO4HF

▲ Figure 16.4 relative strengths of select conjugate acid–base pairs The two members of each

pair are listed opposite each other in the two columns

Go FiGuRE

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

secTiON 16.2 Brønsted–Lowry Acids and Bases 677

1 A strong acid completely transfers its protons to water, leaving essentially no

undis-sociated molecules in solution (Section 4.3) Its conjugate base has a negligible

tendency to accept protons in aqueous solution (The conjugate base of a strong acid

shows negligible basicity.)

2 A weak acid only partially dissociates in aqueous solution and therefore exists in

the solution as a mixture of the undissociated acid and its conjugate base The

con-jugate base of a weak acid shows a slight ability to remove protons from water

(The conjugate base of a weak acid is a weak base.)

3 A substance with negligible acidity contains hydrogen but does not demonstrate

any acidic behavior in water Its conjugate base is a strong base, reacting

com-pletely with water, to form OH- ions (The conjugate base of a substance with

neg-ligible acidity is a strong base.)

The ions H3O+1aq2 and OH-1aq2 are, respectively, the strongest possible acid and

strongest possible base that can exist at equilibrium in aqueous solution Stronger acids

react with water to produce H3O+1aq2 ions, and stronger bases react with water to

pro-duce OH-1aq2 ions, a phenomenon known as the leveling effect.

give it some thought

Given that HClO4 is a strong acid, how would you classify the basicity of ClO4-?

We can think of proton-transfer reactions as being governed by the relative

abili-ties of two bases to abstract protons For example, consider the proton transfer that

occurs when an acid HA dissolves in water:

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

If H2O (the base in the forward reaction) is a stronger base than A- (the conjugate base

of HA), it is favorable to transfer the proton from HA to H2O, producing H3O+ and

A- As a result, the equilibrium lies to the right This describes the behavior of a strong

acid in water For example, when HCl dissolves in water, the solution consists almost

entirely of H3O+ and Cl- ions with a negligible concentration of HCl molecules:

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

H2O is a stronger base than Cl- (Figure 16.4), so H2O acquires the proton to become the

hydronium ion Because the reaction lies completely to the right, we write Equation 16.10

with only an arrow to the right rather than using the double arrows for an equilibrium.

When A- is a stronger base than H2O, the equilibrium lies to the left This situation

occurs when HA is a weak acid For example, an aqueous solution of acetic acid consists

mainly of CH3COOH molecules with only a relatively few H3O+ and CH3COO- ions:

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

The CH3COO- ion is a stronger base than H2O (Figure 16.4) and therefore the reverse

reaction is favored more than the forward reaction.

From these examples, we conclude that in every acid–base reaction, equilibrium

favors transfer of the proton from the stronger acid to the stronger base to form the weaker

acid and the weaker base.

For the following proton-transfer reaction use Figure 16.4 to predict whether the equilibrium

lies to the left 1K c 6 12 or to the right 1K c7 12:

HSO4-1aq2 + CO32 -1aq2 ∆ SO42 -1aq2 + HCO3-1aq2

soLuTion

Analyze We are asked to predict whether an equilibrium lies to the right, favoring products, or to

the left, favoring reactants

s A m p L E

ExERcisE 16.3 predicting the position of a proton-Transfer

Equilibrium

16.3 | The Autoionization of Water

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

Acid Base

O H

H

H

O H

H O

We call this process the autoionization of water.

Because the forward and reverse reactions in Equation 16.12 are extremely rapid, no water molecule remains ionized for long At room temperature only about two out of every 109 water molecules are ionized at any given instant Thus, pure water consists almost entirely of H2O molecules and is an extremely poor conductor

of electricity Nevertheless, the autoionization of water is very important, as we will soon see.

Plan This is a proton-transfer reaction, and the position of the equilibrium will favor the proton going to the stronger of two bases The two bases in the equation are CO32 -, the base in the forward reaction, and SO42 -, the conjugate base of HSO4- We can find the relative positions

of these two bases in Figure 16.4 to determine which is the stronger base

Solve The CO32 - ion appears lower in the right-hand column in Figure 16.4 and is fore a stronger base than SO42 - Therefore, CO32- will get the proton preferentially to become HCO3-, while SO42 - will remain mostly unprotonated The resulting equilibrium lies to the right, favoring products (that is, K c 7 1):

there-HSO4-1aq2 + CO32-1aq2 ∆ SO42-1aq2 + HCO3-1aq2 K c 7 1

Acid Base Conjugate base Conjugate acid

Comment Of the two acids HSO4- and HCO3-, the stronger one 1HSO4-2 gives up a proton more readily, and the weaker one 1HCO3-2 tends to retain its proton Thus, the equilibrium favors the direction in which the proton moves from the stronger acid and becomes bonded

to the stronger base

practice Exercise 1

Based on information in Figure 16.4, place the following equilibria in order from smallest to largest value of K c:

(i) CH3COOH1aq2 + HS-1aq2 ∆ CH3COO-1aq2 + H2S1aq2

(ii) F-1aq2 + NH4+1aq2 ∆ HF1aq2 + NH31aq2

(iii) H2CO31aq2 + Cl-1aq2 ∆ HCO3-1aq2 + HCl1aq2

(a) i 6 ii 6 iii (b) ii 6 i 6 iii (c) iii 6 i 6 ii (d) ii 6 iii 6 i (e) iii 6 ii 6 i

practice Exercise 2

For each reaction, use Figure 16.4 to predict whether the equilibrium lies to the left or to the right:

(a) HPO42 -1aq2 + H2O1l2 ∆ H2PO4-1aq2 + OH-1aq2

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

secTiON 16.3 The Autoionization of Water 679

The Ion Product of Water

The equilibrium-constant expression for the autoionization of water is

The term [H2O] is excluded from the equilibrium-constant expression because we

exclude the concentrations of pure solids and liquids (Section 15.4) Because this

expression refers specifically to the autoionization of water, we use the symbol Kw to

denote the equilibrium constant, which we call the ion-product constant for water At

25 °C, Kw equals 1.0 * 10-14 Thus, we have

Kw = 3H3O+43OH-4 = 1.0 * 10-14 1at 25 °C2 [16.14]

Because we use H+1aq2 and H3O+1aq2 interchangeably to represent the hydrated

proton, the autoionization reaction for water can also be written as

H2O1l2 ∆ H+1aq2 + OH-1aq2 [16.15]

Likewise, the expression for Kw can be written in terms of either H3O+ or H+, and Kw

has the same value in either case:

Kw = 3H3O+43OH-4 = 3H+43OH-4 = 1.0 * 10-14 1at 25 °C2 [16.16]

This equilibrium-constant expression and the value of Kw at 25 °C are extremely

im-portant, and you should commit them to memory.

A solution in which 3H+4 = 3OH-4 is said to be neutral In most solutions,

how-ever, the H+ and OH- concentrations are not equal As the concentration of one of

these ions increases, the concentration of the other must decrease, so that the product

of their concentrations always equals 1.0 * 10-14 (▼ Figure 16.5).

Basic solution

[H+] < [OH−][H+][OH−] = 1.0 × 10−14

Suppose that equal volumes of the middle and right samples in the figure were mixed Would the resultant solution

be acidic, neutral, or basic?

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

s A m p L E

ExERcisE 16.4 calculating 3h+4 for pure Water

soLuTion

Analyze We are asked to determine the concentrations of H+ and

OH- ions in a neutral solution at 25 °C

Plan We will use Equation 16.16 and the fact that, by definition,

3H+4 = 3OH-4 in a neutral solution

Solve We will represent the concentration of H+ and OH- in neutral

solution with x This gives

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

x2 = 1.0 * 10-14

x = 1.0 * 10-7M = 3H+4 = 3OH-4

In an acid solution 3H+4 is greater than 1.0 * 10-7 M; in a basic

solu-tion 3H+4 is less than 1.0 * 10-7 M.

What makes Equation 16.16 particularly useful is that it is applicable both to pure water and to any aqueous solution Although the equilibrium between H+1aq2

and OH-1aq2 as well as other ionic equilibria are affected somewhat by the presence

of additional ions in solution, it is customary to ignore these ionic effects except in work requiring exceptional accuracy Thus, Equation 16.16 is taken to be valid for any dilute aqueous solution and can be used to calculate either 3H+4 (if 3OH-4 is known)

or 3OH-4 (if 3H+4 is known).

practice Exercise 1

In a certain acidic solution at 25 °C, 3H+4 is 100 times greater than

3OH-4 What is the value for 3OH-4 for the solution?

(a) 1.0 * 10-8 M (b) 1.0 * 10-7 M (c) 1.0 * 10-6 M

(d) 1.0 * 10-2 M (e) 1.0 * 10-9 M

practice Exercise 2

Indicate whether solutions with each of the following ion

concen-trations are neutral, acidic, or basic: (a) 3H+4 = 4 * 10-9 M;

(b) 3OH-4 = 1 * 10-7 M; (c) 3OH-4 = 1 * 10-13 M.

Calculate the concentration of H+1aq2 in (a) a solution in which 3OH-4 is 0.010 M, (b) a solution

in which 3OH-4 is 1.8 * 10-9M Note: In this problem and all that follow, we assume, unless

stated otherwise, that the temperature is 25 °C

s A m p L E

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

soLuTion

Analyze We are asked to calculate the 3H+4 concentration in an

aqueous solution where the hydroxide concentration is known Plan ionization of water and the value of We can use the equilibrium-constant expression for the auto-K w to solve for each unknown

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

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

practice Exercise 1

A solution has 3OH-4 = 4.0 * 10- 8 What is the value of 3H+4

for the solution?

The molar concentration of H+1aq2 in an aqueous solution is usually very small For

convenience, we therefore usually express 3H+4 in terms of pH, which is the negative

logarithm in base 10 of 3H+4:*

If you need to review the use of logarithms, see Appendix A.

In Sample Exercise 16.4, we saw that 3H+4 = 1.0 * 10-7 M for a neutral aqueous

solution at 25 °C We can now use Equation 16.17 to calculate the pH of a neutral solution

at 25 °C:

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

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

secTiON 16.4 The pH scale 681

Notice that the pH is reported with two decimal places We do so because only the

num-bers to the right of the decimal point are the significant figures in a logarithm Because

our original value for the concentration 11.0 * 10-7 M2 has two significant figures, the

corresponding pH has two decimal places (7.00).

What happens to the pH of a solution as we make the solution more acidic, so that

3H+4 increases? Because of the negative sign in the logarithm term of Equation 16.17,

the pH decreases as 3H+4 increases For example, when we add sufficient acid to make

3H+4 = 1.0 * 10-3 M the pH is

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

At 25 °C the pH of an acidic solution is less than 7.00.

We can also calculate the pH of a basic solution, one in which 3OH-4 7

1.0 * 10-7 M Suppose 3OH-4 = 2.0 * 10-3 M We can use Equation 16.16 to

calcu-late 3H+4 for this solution and Equation 16.17 to calculate the pH:

3H+4 = 3OH Kw-4 = 1.0 * 10-14

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

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

At 25 °C the pH of a basic solution is greater than 7.00 The relationships among

3H+4, 3OH-4, and pH are summarized in ▲ Table 16.1.

give it some thought

Is it possible for a solution to have a negative pH? If so, would that pH signify a

basic or acidic solution?

One might think that when 3H+4 is very small, as is often the case, it would be

unim-portant That reasoning is quite incorrect! Remember that many chemical processes

depend on the ratio of changes in concentration For example, if a kinetic rate law is

first order in 3H+4, doubling the H+ concentration doubles the rate even if the change

is merely from 1 * 10-7 M to 2 * 10-7 M (Section 14.3) In biological systems,

many reactions involve proton transfers and have rates that depend on 3H+4 Because

the speeds of these reactions are crucial, the pH of biological fluids must be maintained

within narrow limits For example, human blood has a normal pH range of 7.35 to 7.45

Illness and even death can result if the pH varies much from this narrow range.

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

solution Type 3h+4 1M2 3oh−4 1M2 ph

soLuTion

Analyze We are asked to determine the pH of aqueous solutions for

which we have already calculated 3H+4

Plan We can calculate pH using its defining equation, Equation 16.17

ExERcisE 16.6 calculating ph from 3h+4

Calculate the pH values for the two solutions of Sample Exercise 16.5

(b) For the second solution, 3H+4 = 5.6 * 10-6 M Before performing

the calculation, it is helpful to estimate the pH To do so, we note that 3H+4 lies between 1 * 10-6 and 1 * 10-5 Thus, we expect the pH

to lie between 6.0 and 5.0 We use Equation 16.17 to calculate the pH:

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

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

esti-pOH and Other “p” Scales

The negative logarithm is a convenient way of expressing the magnitudes of other small quantities We use the convention that the negative logarithm of a quantity is labeled

“p” (quantity) Thus, we can express the concentration of OH- as pOH:

Likewise, pKw equals -log Kw.

By taking the negative logarithm of both sides of the equilibrium-constant

expres-sion for water, Kw = 3H+43OH-4, we obtain

-log3H+4 + 1-log 3OH-42 = -log Kw [16.19] from which we obtain the useful expression

pH + pOH = 14.00 1at 25 °C2 [16.20] The pH and pOH values characteristic of a number of familiar solutions are shown in (▼ Figure 16.6 ) Notice that a change in 3H+4 by a factor of 10 causes the pH to change by 1 Thus, the concentration of H+1aq2 in a solution of pH 5 is 10 times the H+1aq2 concentra-

(a) In a sample of lemon juice, 3H+4 = 3.8 * 10-4 M What is the pH?

(b) A commonly available window-cleaning solution has

3OH-4 = 1.9 * 10-6 M What is the pH at 25 °C?

pH + pOH = 14 [H+][OH−] = 1×10 −14

pOH pH

1 (1×100)1×10−1

0.01.02.03.04.05.06.07.08.09.010.011.012.013.014.0

Human bloodSeawaterBorax

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

Go FiGuRE

Which is more acidic, black coffee or lemon juice?

secTiON 16.4 The pH scale 683

give it some thought

If the pOH for a solution is 3.00, what is the pH? Is the solution acidic or basic?

soLuTion

Analyze We need to calculate 3H+4 from pOH

Plan We will first use Equation 16.20, pH + pOH = 14.00, to

calcu-late pH from pOH Then we will use Equation 16.17 to determine the

concentration of H+

Solve From Equation 16.20, we have

pH = 14.00 - pOH

pH = 14.00 - 10.24 = 3.76Next we use Equation 16.17:

pH = -log3H+4 = 3.76Thus,

log3H+4 = -3.76

To find 3H+4, we need to determine the antilogarithm of -3.76 Your

calculator will show this command as 10x or INV log (these functions

are usually above the log key) We use this function to perform the

calculation:

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

Comment The number of significant figures in 3H+4 is two because

the number of decimal places in the pH is two

s A m p L E

ExERcisE 16.7 calculating 3h+4 from poh

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

Check Because the pH is between 3.0 and 4.0, we know that 3H+4 will

be between 1.0 * 10-3 M and 1.0 * 10-4 M Our calculated

3H+4 falls within this estimated range

practice Exercise 1

A solution at 25 °C has pOH = 10.53 Which of the following statements is or are true?

(i) The solution is acidic.

(ii) The pH of the solution is 14.00 - 10.53

(iii) For this solution, 3OH-4 = 10-10.53M.

(a) Only one of the statements is true.

(b) Statements (i) and (ii) are true.

(c) Statements (i) and (iii) are true.

(d) Statements (ii) and (iii) are true.

(e) All three statements are true.

practice Exercise 2

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

Measuring pH

The pH of a solution can be measured with a pH meter (▶ Figure 16.7) A complete

understanding of how this important device works requires a knowledge of

elec-trochemistry, a subject we take up in Chapter 20 In brief, a pH meter consists of

a pair of electrodes connected to a meter capable of measuring small voltages, on

the order of millivolts A voltage, which varies with pH, is generated when the

elec-trodes are placed in a solution This voltage is read by the meter, which is calibrated

to give pH.

Although less precise, acid–base indicators can be used to measure pH An acid–

base indicator is a colored substance that can exist in either an acid or a base form The

two forms have different colors Thus, the indicator has one color at lower pH and

another at higher pH If you know the pH at which the indicator turns from one form

to the other, you can determine whether a solution has a higher or lower pH than this

value Litmus, for example, changes color in the vicinity of pH 7 The color change,

however, is not very sharp Red litmus indicates a pH of about 5 or lower, and blue

lit-mus indicates a pH of about 8 or higher.

Some common indicators are listed in Figure 16.8 The chart tells us, for instance,

that methyl red changes color over the pH interval from about 4.5 to 6.0 Below pH 4.5

it is in the acid form, which is red In the interval between 4.5 and 6.0, it is gradually

converted to its basic form, which is yellow Once the pH rises above 6 the conversion

▲ Figure 16.7 a digital pH meter The

device is a millivoltmeter, and the electrodes immersed in a solution produce a voltage that depends on the pH of the solution

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

indi-pH values.

The chemistry of an aqueous solution often depends critically on pH It is therefore portant to examine how pH relates to acid and base concentrations The simplest cases

im-are those involving strong acids and strong bases Strong acids and bases im-are strong trolytes, existing in aqueous solution entirely as ions There are relatively few common

elec-strong acids and bases (see Table 4.2).

Strong Acids

The seven most common strong acids include six monoprotic acids (HCl, HBr,

HI, HNO3, HClO3, and HClO4), and one diprotic acid 1H2SO42 Nitric acid 1HNO32 exemplifies the behavior of the monoprotic strong acids For all practical purposes, an aqueous solution of HNO3 consists entirely of H3O+ and NO3- ions:

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

We have not used equilibrium arrows for this equation because the reaction lies entirely

to the right (Section 4.1) As noted in Section 16.3, we use H3O+1aq2 and H+1aq2

interchangeably to represent the hydrated proton in water Thus, we can simplify this acid ionization equation to

HNO31aq2 ¡ H+1aq2 + NO3-1aq2

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

pH range for color changeMethyl violet

Thymol blueMethyl orangeMethyl redBromthymol bluePhenolphthaleinAlizarin yellow R

Yellow

Yellow

VioletRed

YellowRed

▲ Figure 16.8 pH ranges for common acid–base indicators Most indicators have a useful

range of about 2 pH units

▲ Figure 16.9 solutions containing three

common acid–base indicators at various pH

values.

Go FiGuRE

Which of these indicators is best

suited to distinguish between a

solution that is slightly acidic and one

that is slightly basic?

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

secTiON 16.5 strong Acids and Bases 685

acid is straightforward because 3H+4 equals the original concentration of acid In a 0.20 M

solution of HNO31aq2, for example, 3H+4 = 3NO3-4 = 0.20 M The situation with the

diprotic acid H2SO4 is somewhat more complex, as we will see in Section 16.6.

soLuTion

Analyze and Plan Because HClO4 is a strong acid, it is completely

ionized, giving 3H+4 = 3ClO4-4 = 0.040 M.

Solve

pH = -log10.0402 = 1.40

Check Because 3H+4 lies between 1 * 10-2 and 1 * 10-1, the pH will

be between 2.0 and 1.0 Our calculated pH falls within the estimated

range Furthermore, because the concentration has two significant

figures, the pH has two decimal places

s A m p L E

ExERcisE 16.8 calculating the ph of a strong Acid

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

practice Exercise 1

Order the following three solutions from smallest to largest pH:

(i) 0.20 M HClO3 (ii) 0.0030 M HNO3 (iii) 1.50 M HCl

(a) i 6 ii 6 iii (b) ii 6 i 6 iii (c) iii 6 i 6 ii (d) ii 6 iii 6 i (e) iii 6 ii 6 i

practice Exercise 2

An aqueous solution of HNO3 has a pH of 2.34 What is the concentration of the acid?

Strong Bases

The most common soluble strong bases are the ionic hydroxides of the alkali metals,

such as NaOH, KOH, and the ionic hydroxides heavier alkaline earth metals, such as

Sr1OH22 These compounds completely dissociate into ions in aqueous solution Thus,

a solution labeled 0.30 M NaOH consists of 0.30 M Na+1aq2 and 0.30 M OH-1aq2;

there is essentially no undissociated NaOH.

give it some thought

Which solution has the higher pH, a 0.001 M solution of NaOH or a 0.001 M

solution of Ba1OH22?

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

soLuTion

Analyze We are asked to calculate the pH of two solutions of strong bases

Plan We can calculate each pH by either of two equivalent methods First, we could use Equation

16.16 to calculate 3H+4 and then use Equation 16.17 to calculate the pH Alternatively, we could

use 3OH-4 to calculate pOH and then use Equation 16.20 to calculate the pH

Solve

(a) NaOH dissociates in water to give one OH- ion per formula unit Therefore, the OH-

con-centration for the solution in (a) equals the stated concon-centration of NaOH, namely 0.028 M.

Method 1:

3H+4 = 1.0 * 10-14

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

Method 2:

pOH = -log10.0282 = 1.55 pH = 14.00 - pOH = 12.45

(b) Ca1OH22 is a strong base that dissociates in water to give two OH- ions per formula unit Thus,

the concentration of OH-1aq2 for the solution in part (b) is 2 * 10.0011 M2 = 0.0022 M.

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

Strongly basic solutions are also created by certain substances that react with water to form OH-1aq2 The most common of these contain the oxide ion Ionic metal oxides, espe-

cially Na2O and CaO, are often used in industry when a strong base is needed The O

2-reacts very exothermically with water to form OH-, leaving virtually no O2- in the solution:

O2-1aq2 + H2O1l2 ¡ 2 OH-1aq2 [16.22]

Thus, a solution formed by dissolving 0.010 mol of Na2O1s2 in enough water to form

1.0 L of solution has 3OH-4 = 0.020 M and a pH of 12.30.

give it some thought

The CH3- ion is the conjugate base of CH4, and CH4 shows no evidence of being

an acid in water Write a balanced equation for the reaction of CH3- and water

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

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

or

HA1aq2 ∆ H+1aq2 + A-1aq2 [16.24]

These equilibria are in aqueous solution, so we will use equilibrium-constant sions based on concentrations Because H2O is the solvent, it is omitted from the equilibrium-constant expression (Section 15.4) Further, we add a subscript a on

expres-the equilibrium constant to indicate that it is an equilibrium constant for expres-the ionization

of an acid Thus, we can write the equilibrium-constant expression as either:

Ka = 3H3O+43A-4

3HA4 or Ka =

3H+43A-4

Ka is called the acid-dissociation constant for acid HA.

▶ Table 16.2 shows the structural formulas, conjugate bases, and Ka values for a number of weak acids Appendix D provides a more complete list Many weak acids are organic compounds composed entirely of carbon, hydrogen, and oxygen These compounds usually contain some hydrogen atoms bonded to carbon atoms and some bonded to oxygen atoms In almost all cases, the hydrogen atoms bonded to carbon

do not ionize in water; instead, the acidic behavior of these compounds is due to the hydrogen atoms attached to oxygen atoms.

practice Exercise 1

Order the following three solutions from smallest to largest pH:

(i) 0.030 M Ba1OH22 (ii) 0.040 M KOH (iii) pure water

(a) i 6 ii 6 iii (b) ii 6 i 6 iii (c) iii 6 i 6 ii (d) ii 6 iii 6 i (e) iii 6 ii 6 i

practice Exercise 2

What is the concentration of a solution of (a) KOH for which the pH is 11.89, (b) Ca1OH22for which the pH is 11.68?

secTiON 16.6 Weak Acids 687

The magnitude of Ka indicates the tendency of the acid to ionize in water: The

larger the value of Ka, the stronger the acid Chlorous acid 1HClO22, for example, is the

strongest acid in Table 16.2, and phenol 1HOC6H52 is the weakest For most weak acids

Ka values range from 10-2 to 10-10.

give it some thought

Based on the entries in Table 16.2, which element is most commonly bonded to

the acidic hydrogen?

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

Strong acid

HA moleculescompletely dissociate

+

+

−

Table 16.2 some Weak Acids in Water at 25 °c

Acid structural Formula* conjugate Base K a

OH

OC

A student prepared a 0.10 M solution of formic acid (HCOOH) and found its pH at 25 °C to be

2.38 Calculate K a for formic acid at this temperature

a tabulation of initial and equilibrium concentrations

Solve The first step in solving any equilibrium problem is to write the equation for the rium reaction The ionization of formic acid can be written as

equilib-HCOOH1aq2 ∆ H+1aq2 + HCOO-1aq2

The equilibrium-constant expression is

K a = 3H+43HCOO-43HCOOH4From the measured pH, we can calculate 3H+4:

pH = -log 3H+4 = 2.38 log3H+4 = -2.38

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

To determine the concentrations of the species involved in the equilibrium, we imagine that the

solution is initially 0.10 M in HCOOH molecules We then consider the ionization of the acid

into H+ and HCOO- For each HCOOH molecule that ionizes, one H+ ion and one HCOO- ion are produced in solution Because the pH measurement indicates that 3H+4 = 4.2 * 10-3 M at

equilibrium, we can construct the following table:

HCOOH1aq2 ∆ H+1aq2 + HCOO-1aq2

Change in concentration (M) -4.2 * 10-3 +4.2 * 10-3 +4.2 * 10-3

Equilibrium concentration (M) 10.10 - 4.2 * 10-32 4.2 * 10-3 4.2 * 10-3Notice that we have neglected the very small concentration of H+1aq2 due to H2O autoioniza-tion Notice also that the amount of HCOOH that ionizes is very small compared with the initial concentration of the acid To the number of significant figures we are using, the subtraction

secTiON 16.6 Weak Acids 689

Percent Ionization

We have seen that the magnitude of Ka indicates the strength of a weak acid Another

measure of acid strength is percent ionization, defined as

Percent ionization = original concentration of HA * concentration of ionized HA 100% [16.26]

The stronger the acid, the greater the percent ionization.

If we assume that the autoionization of H2O is negligible, the concentration of acid

that ionizes equals the concentration of H+1aq2 that forms Thus, the percent

ioniza-tion for an acid HA can be expressed as

As calculated in Sample Exercise 16.10, a 0.10 M solution of formic acid (HCOOH) contains

4.2 * 10-3 M H+1aq2 Calculate the percentage of the acid that is ionized.

soLuTion

Analyze We are given the molar concentration of an aqueous solution of weak acid and the

equi-librium concentration of H+1aq2 and asked to determine the percent ionization of the acid.

Plan The percent ionization is given by Equation 16.27

Solve

Percent ionization = 3H

+4equilibrium3HCOOH4initial * 100% = 4.2 * 10-3 M

1 Our first step is to write the ionization equilibrium:

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

Notice that the hydrogen that ionizes is the one attached to an oxygen atom.

2 The second step is to write the equilibrium-constant expression and the value for

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

Ka = 3H+43CH3COO-4

3CH3COOH4 = 1.8 * 10-5 [16.29]

3 The third step is to express the concentrations involved in the equilibrium

reac-tion This can be done with a little accounting, as described in Sample Exercise 16.10 Because we want to find the equilibrium value for 3H+4, let’s call this quan-

tity x The concentration of acetic acid before any of it ionizes is 0.30 M The

chem-ical equation tells us that for each molecule of CH3COOH that ionizes, one H+1aq2

and one CH3COO-1aq2 are formed Consequently, if x moles per liter of H+1aq2 form at equilibrium, x moles per liter of CH3COO-1aq2 must also form and x

moles per liter of CH3COOH must be ionized:

CH3COOH1aq2 ∆ H+1aq2 + CH3COO-1aq2

4 The fourth step is to substitute the equilibrium concentrations into the

equilibrium-constant expression and solve for x:

Ka = 3H+43CH3COO-4

3CH3COOH4 =

1x21x2 0.30 - x = 1.8 * 10-5 [16.30] This expression leads to a quadratic equation in x, which we can solve by using

either an equation-solving calculator or the quadratic formula We can simplify

the problem, however, by noting that the value of Ka is quite small As a result, we

anticipate that the equilibrium lies far to the left and that x is much smaller than the initial concentration of acetic acid Thus, we assume that x is negligible relative

to 0.30, so that 0.30 - x is essentially equal to 0.30 We can (and should!) check the

validity of this assumption when we finish the problem By using this assumption, Equation 16.30 becomes

secTiON 16.6 Weak Acids 691

Now we check the validity of our simplifying assumption that 0.30 - x ≃ 0.30

The value of x we determined is so small that, for this number of significant figures, the

assumption is entirely valid We are thus satisfied that the assumption was a reasonable

one to make Because x represents the moles per liter of acetic acid that ionize, we see

that, in this particular case, less than 1% of the acetic acid molecules ionize:

Percent ionization of CH3COOH = 0.0023 M 0.30 M * 100% = 0.77%

As a general rule, if x is more than about 5% of the initial concentration value, it is better

to use the quadratic formula You should always check the validity of any simplifying

assumptions after you have finished solving a problem.

We have also made one other assumption, namely that all of the H+ in the solution

comes from ionization of CH3COOH Are we justified in neglecting the autoionization

of H2O? The answer is yes—the additional [H+4 due to water, which would be on the

order of 10-7 M, is negligible compared to the [H+4 from the acid (which in this case

is on the order of 10-3 M) In extremely precise work, or in cases involving very dilute

solutions of acids, we would need to consider the autoionization of water more fully.

give it some thought

Would a 1.0 * 10-8M solution of HCl have pH 6 7, pH = 7, or pH 7 7?

Finally, we can compare the pH value of this weak acid with the pH of a solution of

a strong acid of the same concentration The pH of the 0.30 M acetic acid is 2.64, but the

pH of a 0.30 M solution of a strong acid such as HCl is -log10.302 = 0.52 As expected,

the pH of a solution of a weak acid is higher than that of a solution of a strong acid of the

same molarity (Remember, the higher the pH value, the less acidic the solution.)

s A m p L E

ExERcisE 16.12 using Ka to calculate ph

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

soLuTion

Analyze We are given the molarity of a weak acid and are asked for

the pH From Table 16.2, K a for HCN is 4.9 * 10-10 Plan chemical equation and constructing a table of initial and equilibrium We proceed as in the example just worked in the text, writing the

concentrations in which the equilibrium concentration of H+ is our unknown

HCN1aq2 ∆ H+1aq2 + CN-1aq2

Solve Writing both the chemical equation for the ionization

reaction that forms H+1aq2 and the equilibrium-constant 1K a2

expression for the reaction:

HCN1aq2 ∆ H+1aq2 + CN-1aq2

K a= 3H+43CN-43HCN4 = 4.9 * 10-10Next, we tabulate the concentrations of the species involved in

the equilibrium reaction, letting x = 3H+4 at equilibrium:

Substituting the equilibrium concentrations into the

0.20 - x = 4.9 * 10-10

We next make the simplifying approximation that x, the amount

of acid that dissociates, is small compared with the initial

con-centration of acid, 0.20 - x ≃ 0.20 Thus,

x20.20 = 4.9 * 10-10

The properties of an acid solution that relate directly to the concentration of

H+1aq2, such as electrical conductivity and rate of reaction with an active metal, are

less evident for a solution of a weak acid than for a solution of a strong acid of the same concentration ▼ Figure 16.11 presents an experiment that demonstrates this difference

with 1 M CH3COOH and 1 M HCl The concentration of H+1aq2 in 1 M CH3COOH is

only 0.004 M, whereas the 1 M HCl solution contains 1 M H+1aq2 As a result, the

reac-tion rate with the metal is much faster in the HCl solureac-tion.

As the concentration of a weak acid increases, the equilibrium concentration of

H+1aq2 increases, as expected However, as shown in ▶ Figure 16.12, the percent tion decreases as the concentration increases Thus, the concentration of H+1aq2 is not

ioniza-directly proportional to the concentration of the weak acid For example, doubling the concentration of a weak acid does not double the concentration of H+1aq2.

▲ Figure 16.11 rates of the same reaction run in a weak acid and a strong acid The bubbles

are H2 gas, which along with metal cations, is produced when a metal is oxidized by an acid

(Section 4.4)

Reaction proceeds more rapidly

in strong acid, leading to formation of larger H2 bubbles and rapid disappearance of metal

Reaction eventually goes to completion in both acids

Reaction complete in strong acid

H2 bubbles show reaction still in progress in weak acid

1 M HCl(aq)

[H+] = 1 M 1 M CH[H+] = 0.004 M3COOH(aq)

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

x = 20.98 * 10-10 = 9.9 * 10-6 M = 3H+4

A concentration of 9.9 * 10-6 M is much smaller than 5% of

0.20, the initial HCN concentration Our simplifying

approxima-tion is therefore appropriate We now calculate the pH of the

secTiON 16.6 Weak Acids 693

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

Go FiGuRE

Is the trend observed in this graph consistent with Le Châtelier’s principle? Explain

s A m p L E

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

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

soLuTion

Analyze We are asked to calculate the percent ionization of a solution

of HF From Appendix D, we find K a= 6.8 * 10-4 Plan We write the chemical equation for the equilibrium and tabulate the We approach this problem as for previous equilibrium problems:

known and unknown concentrations of all species We then substitute the equilibrium concentrations into the equilibrium-constant expres-sion and solve for the unknown concentration of H+

K a = 3H+43F-43HF4 =

1x21x2 0.10 - x = 6.8 * 10-4

Solve The equilibrium reaction and equilibrium concentrations

The equilibrium-constant expression is

When we try solving this equation using the approximation

0.10 - x ≃ 0.10 (that is, by neglecting the concentration of acid

Because this approximation is greater than 5% of 0.10 M,

how-ever, we should work the problem in standard quadratic form

Rearranging, we have

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

x2 + 16.8 * 10-42x - 6.8 * 10-5= 0Substituting these values in the standard quadratic formula gives

x = -6.8 * 10-4 { 216.8 * 10-422 - 41-6.8 * 10-52

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

2

Polyprotic Acids

Acids that have more than one ionizable H atom are known as polyprotic acids

Sulfu-rous acid 1H2SO32, for example, can undergo two successive ionizations:

H2SO31aq2 ∆ H+1aq2 + HSO3-1aq2 Ka1= 1.7 * 10-2 [16.31] HSO3-1aq2 ∆ H+1aq2 + SO3 2-1aq2 Ka2= 6.4 * 10-8 [16.32]

Note that the acid-dissociation constants are labeled Ka1 and Ka2 The numbers on the

constants refer to the particular proton of the acid that is ionizing Thus, Ka2 always refers to the equilibrium involving removal of the second proton of a polyprotic acid.

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

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

give it some thought

What is the equilibrium associated with K a3 for H3PO4?The acid-dissociation constants for common polyprotic acids are listed in

▼ Table 16.3, and Appendix D provides a more complete list The structure of

citric acid illustrates the presence of multiple ionizable protons ◀ Figure 16.13.

Of the two solutions, only the positive value for x is chemically

rea-sonable From that value, we can determine 3H+4 and hence the pH x = 3H+4 = 3F-4 = 7.9 * 10-3 M, so pH = -log3H+4 = 2.10

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

original concentration * 100%

= 7.9 * 10-3 M 0.10 M * 100% = 7.9%

practice Exercise 1

What is the pH of a 0.010 M solution of HF?

(a) 1.58 (b) 2.10 (c) 2.30 (d) 2.58 (e) 2.64

practice Exercise 2

In Practice Exercise 2 for Sample Exercise 16.11, we found that the percent ionization of niacin

1K a= 1.5 * 10-52 in a 0.020 M solution is 2.7% Calculate the percentage of niacin molecules ionized in a

solution that is (a) 0.010 M, (b) 1.0 * 10-3 M.

Table 16.3 Acid-dissociation constants of some common polyprotic Acids

H2C

H2C

HO

Citric acid

▲ Figure 16.13 the structure of the

polyprotic acid, citric acid.

Go FiGuRE

Citric acid has four hydrogen atoms

bonded to oxygen How does the

hydrogen atom that is not an acidic

proton differ from the other three?

secTiON 16.6 Weak Acids 695

Notice in Table 16.3 that in most cases the Ka values for successive losses of protons

differ by a factor of at least 103 Notice also that the value of Ka1 for sulfuric acid is listed

simply as “large.” Sulfuric acid is a strong acid with respect to the removal of the first

proton Thus, the reaction for the first ionization step lies completely to the right:

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

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

For many polyprotic acids Ka1 is much larger than subsequent dissociation

constants, in which case the H+1aq2 in the solution comes almost entirely from

the first ionization reaction As long as successive Ka values differ by a factor of

103 or more, it is usually possible to obtain a satisfactory estimate of the pH of

polyprotic acid solutions by treating the acids as if they were monoprotic,

consid-ering only Ka1.

s A m p L E

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

The solubility of CO2 in water at 25 °C and 0.1 atm is 0.0037 M The common practice is to assume that all

the dissolved CO2 is in the form of carbonic acid 1H2CO32, which is produced in the reaction

CO21aq2 + H2O1l2 ∆ H2CO31aq2 What is the pH of a 0.0037 M solution of H2CO3?

soLuTion

Analyze We are asked to determine the pH of a 0.0037 M solution of a

polyprotic acid Plan K a1 and KH2COa23 (Table 16.3), differ by more than a factor of 10 is a diprotic acid; the two acid-dissociation constants, 3

Conse-quently, the pH can be determined by considering only K a1, thereby treating the acid as if it were a monoprotic acid

K a1 = 3H+43HCO3-4

3H2CO34 =

1x21x2 0.0037 - x = 4.3 * 10-7

Solve Proceeding as in Sample Exercises

16.12 and 16.13, we can write the

equilib-rium reaction and equilibequilib-rium

The equilibrium-constant expression is

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

Alternatively, because K a1 is small, we can

make the simplifying approximation that x is

Because we get the same value (to 2

signifi-cant figures) our simplifying assumption was

justified The pH is therefore pH = -log3H+4 = -log14.0 * 10-52 = 4.40

16.7 | Weak Bases

Many substances behave as weak bases in water Weak bases react with water, stracting protons from H2O, thereby forming the conjugate acid of the base and

ab-OH- ions:

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

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

constant Kb, the base-dissociation constant, always refers to the equilibrium in which a

base reacts with H2O to form the corresponding conjugate acid and OH-.

▶ Table 16.4 lists the Lewis structures, conjugate acids, and Kb values for a number

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

Assuming that y is small relative to

We see that the value for y is indeed very small compared with

4.0 * 10-5, showing that our assumption was justified It also shows

that the ionization of HCO3- is negligible relative to that of H2CO3,

as far as production of H+ is concerned However, it is the only

source of CO32-, which has a very low concentration in the solution

Our calculations thus tell us that in a solution of carbon dioxide in water, most of the CO2 is in the form of CO2 or H2CO3, only a small fraction ionizes to form H+ and HCO3-, and an even smaller frac-tion ionizes to give CO32- Notice also that 3CO32-4 is numerically equal to K a2

practice Exercise 1

What is the pH of a 0.28 M solution of ascorbic acid (Vitamin C)? (See Table 16.3 for K a1 and K a2.)

(a) 2.04 (b) 2.32 (c) 2.82 (d) 4.65 (e) 6.17

practice Exercise 2

(a) Calculate the pH of a 0.020 M solution of oxalic acid 1H2C2O42 (See Table 16.3 for K a1 and K a2.)

(b) Calculate the concentration of oxalate ion, 3C2O42-4, in this solution

Comment If we were asked for 3CO32-4

we would need to use K a2 Let’s illustrate

that calculation Using our calculated

values of 3HCO3-4 and 3H+4 and setting

secTiON 16.7 Weak Bases 697

Table 16.4 some Weak Bases in Water at 25 °c

Base structural Formula* conjugate Acid K b

*The atom that accepts the proton is shown in blue

HHN

s A m p L E

ExERcisE 16.15 using Kb to calculate oh

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

soLuTion

Analyze We are given the concentration of a weak base and asked to

determine the concentration of OH- Plan solving problems involving the ionization of weak acids—that is, We will use essentially the same procedure here as used in

write the chemical equation and tabulate initial and equilibrium concentrations

NH31aq2 + H2O1l2 ∆ NH4+1aq2 + OHInitial

Solve The ionization reaction and

equilibrium-constant expression are NH31aq2 + H2O1l2 ∆ NH4+1aq2 + OH-1aq2

K b= 3NH4+43OH-4

3NH34 = 1.8 * 10-5Ignoring the concentration of H2O because it

is not involved in the equilibrium-constant

expression, the equilibrium concentrations are

Inserting these quantities into the

equilibrium-constant expression gives K b= 3NH4+43OH-4

3NH34 =

1x21x2 0.15 - x = 1.8 * 10-5Because K b is small, the amount of NH3 that

reacts with water is much smaller than the

NH3 concentration, and so we can neglect x

relative to 0.15 M Then we have

x20.15 = 1.8 * 10-5

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

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

Types of Weak Bases

Weak bases fall into two general categories The first category is neutral substances that have an atom with a nonbonding pair of electrons that can accept a proton Most of these bases, including all uncharged bases in Table 16.4, contain a nitrogen atom These sub-

stances include ammonia and a related class of compounds called amines (◀ Figure 16.14)

In organic amines, at least one N ¬ H bond in NH3 is replaced with an N ¬ C bond Like

NH3, amines can abstract a proton from a water molecule by forming an N ¬ H bond, as shown here for methylamine:

N H H

CH3

CH3(aq)

N H H

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

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

Check The value obtained for x is only about 1% of the NH3 concentration, 0.15 M Therefore,

neglecting x relative to 0.15 was justified.

Comment You may be asked to find the pH of a solution of a weak base Once you have found

3OH-4, you can proceed as in Sample Exercise 16.9, where we calculated the pH of a strong base

In the present sample exercise, we have seen that the 0.15 M solution of NH3 contains 3OH-4 =

1.6 * 10-3 M Thus, pOH = -log11.6 * 10-32 = 2.80, and pH = 14.00 - 2.80 = 11.20

The pH of the solution is above 7 because we are dealing with a solution of a base

practice Exercise 1

What is the pH of a 0.65 M solution of pyridine, C5H5N? (See Table 16.4 for K b.)

(a) 4.48 (b) 8.96 (c) 9.52 (d) 9.62 (e) 9.71

practice Exercise 2

Which of the following compounds should produce the highest pH as a 0.05 M solution:

pyridine, methylamine, or nitrous acid?

▲ Figure 16.14 structures of ammonia

and two simple amines.

Go FiGuRE

When hydroxylamine acts as a base,

which atom accepts the proton?

ExERcisE 16.16 using ph to determine the concentration of a salt

A solution made by adding solid sodium hypochlorite (NaClO) to enough water to make 2.00 L of solution

has a pH of 10.50 Using the information in Equation 16.37, calculate the number of moles of NaClO added

to the water

soLuTion

Analyze NaClO is an ionic compound consisting of Na+ and ClO- ion

As such, it is a strong electrolyte that completely dissociates in solution into

Na+, a spectator ion, and ClO- ion, a weak base with K b = 3.3 * 10-7

(Equation 16.37) Given this information we must calculate the number of

moles of NaClO needed to increase the pH of 2.00-L of water to 10.50

Plan From the pH, we can determine the equilibrium concentration

of OH- We can then construct a table of initial and equilibrium concentrations in which the initial concentration of ClO- is our unknown We can calculate 3ClO-4 using the expression for K b

secTiON 16.8 relationship Between K a and K b 699

We have seen in a qualitative way that the stronger an acid, the weaker its conjugate

base To see if we can find a corresponding quantitative relationship, let’s consider the

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

NH4+, the equilibrium is

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

or written in its simpler form:

NH4+1aq2 ∆ NH31aq2 + H+1aq2 [16.38]

For the base, NH3, the equilibrium is

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

Each equilibrium is expressed by a dissociation constant:

Ka = 3NH343H+4

3NH4+4 Kb =

3NH4+43OH-4 3NH34 When we add Equations 16.38 and 16.39, the NH4+ and NH3 species cancel and we

are left with the autoionization of water:

NH4+1aq2 ∆ NH31aq2 + H+1aq2

NH31aq2 + H2O1l2 ∆ NH4+1aq2 + OH-1aq2

H2O1l2 ∆ H+1aq2 + OH-1aq2

Recall that when two equations are added to give a third, the equilibrium constant

asso-ciated with the third equation equals the product of the equilibrium constants of the

first two equations (Section 15.3)

ClO-1aq2 + H2O1l2 ∆ HClO1aq2 + OH-1aq2

concentration (M) 1x - 3.2 * 10

-42 — 3.2 * 10-4 3.2 * 10-4

Solve We can calculate 3OH-4 by using

either Equation 16.16 or Equation 16.20;

we will use the latter method here:

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

This concentration is high enough that we

can assume that Equation 16.37 is the only

source of OH-; that is, we can neglect any

OH- produced by the autoionization of H2O

We now assume a value of x for the initial

concentration of ClO- and solve the

equilib-rium problem in the usual way

We now use the expression for the base-

dissociation constant to solve for x: K b= 3HClO43OH-4

We say that the solution is 0.31 M in NaClO even though some of the ClO- ions have reacted with water

Because the solution is 0.31 M in NaClO and the total volume of solution is 2.00 L, 0.62 mol of NaClO is the

amount of the salt that was added to the water

practice Exercise 1

The benzoate ion, C6H5COO-, is a weak base with K b= 1.6 * 10-10 How many

moles of sodium benzoate are present in 0.50 L of a solution of NaC6H5COO if the

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

Kb, we obtain

Ka * Kb = a 3NH343H+4

3NH4+4 b a

3NH4+43OH-4 3NH34 b = 3H+43OH-4 = Kw

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

(Equa-us the autoionization equilibrium for water, for which the equilibrium constant

is Kw.

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

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

As the strength of an acid increases (Ka gets larger), the strength of its conjugate base

must decrease (Kb gets smaller) so that the product Ka * Kb remains 1.0 * 10-14 at

25 °C ▼ Table 16.5 demonstrates this relationship Remember, this important

rela-tionship applies only to conjugate acid–base pairs.

By using Equation 16.40, we can calculate Kb for any weak base if we know Ka

for its conjugate acid Similarly, we can calculate Ka for a weak acid if we know Kb for its conjugate base As a practical consequence, ionization constants are often listed for only one member of a conjugate acid–base pair For example, Appendix D does not

contain Kb values for the anions of weak acids because they can be readily calculated

from the tabulated Ka values for their conjugate acids.

Recall that we often express 3H+4 as pH: pH = -log 3H+4 (Section 16.4) This

“p” nomenclature is often used for other very small numbers For example, if you look

up the values for acid- or base-dissociation constants in a chemistry handbook, you

may find them expressed as pKa or pKb:

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

Using this nomenclature, Equation 16.40 can be written in terms of pKa and pKb

by taking the negative logarithm of both sides:

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

give it some thought

K a for acetic acid is 1.8 * 10-5 What is the first digit of the pK a value for acetic acid?

Table 16.5 some conjugate Acid–Base pairs

secTiON 16.8 relationship Between K a and K b 701

chemistry put to Work

amines and amine

Hydrochlorides

Many low-molecular-weight amines have a fishy odor Amines and

NH3 are produced by the anaerobic (absence of O2) decomposition of

dead animal or plant matter Two such amines with very disagreeable

aromas are H2N1CH224NH2, putrescine, and H2N1CH225NH2,

cadav-erine The names of these substances are testaments to their repugnant

odors!

Many drugs, including quinine, codeine, caffeine, and

amphet-amine, are amines Like other amines, these substances are weak

bases; the amine nitrogen is readily protonated upon treatment with

an acid The resulting products are called acid salts If we use A as

the abbreviation for an amine, the acid salt formed by reaction with

hydrochloric acid can be written AH+Cl- It can also be written as

A#HCl and referred to as a hydrochloride Amphetamine

hydro-chloride, for example, is the acid salt formed by treating

Acid salts are much less volatile, more stable, and generally more

water soluble than the corresponding amines For this reason, many

drugs that are amines are sold and administered as acid salts Some

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

Analyze We are asked to determine dissociation constants for F-, the

conjugate base of HF, and NH4+, the conjugate acid of NH3

Plan We can use the tabulated K values for HF and NH3 and the

rela-tionship between K a and K b to calculate the dissociation constants for

their conjugates, F- and NH4+

Solve

(a) For the weak acid HF, Table 16.2 and Appendix D give

K a = 6.8 * 10-4 We can use Equation 16.40 to calculate K b for

the conjugate base, F-:

K b= K w

K a = 1.0 * 10-146.8 * 10-4 = 1.5 * 10-11

(b) For NH3, Table 16.4 and in Appendix D give K b= 1.8 * 10-5, and

this value in Equation 16.40 gives us K a for the conjugate acid, NH4+:

K a= K w

K b = 1.0 * 10-141.8 * 10-5 = 5.6 * 10-10

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

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

practice Exercise 1

By using information from Appendix D, put the following three substances in order of weakest to strongest base: (i) 1CH323N, (ii) HCOO-, (iii) BrO-

(a) i 6 ii 6 iii (b) ii 6 i 6 iii (c) iii 6 i 6 ii (d) ii 6 iii 6 i (e) iii 6 ii 6 i.

practice Exercise 2

(a) Based on information in Appendix D, which of these anions

has the largest base-dissociation constant: NO2-, PO43 -, or N3-?

(b) The base quinoline has the structure

N

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