Chemistry the central science 13e by theodore l brown 2

662

chapter 15 Chemical Equilibrium

are shown here. Which reaction has a larger equilibrium constant? [Sections 15.1 and 15.2]
t = 10 s

20 s

30 s

A(g) + B(g)
t = 10 s

20 s

30 s

X(g) + Y(g)

40 s

50 s

AB(g)
40 s

50 s

XY(g)

15.6Ethene 1C2H42 reacts with halogens 1X22 by the following
reaction:
C2H41g2 + X21g2 ∆ C2H4X21g2

The following figures represent the concentrations at equilibrium 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]

5.0 g PbO2(g)
in both vessels

Vessel A
V = 50 mL

15.8The reaction A2 + B2 ∆ 2 AB has an equilibrium constant Kc = 1.5. The following diagrams represent reaction
mixtures containing A2 molecules (red), B2 molecules (blue),
and AB molecules. (a) Which reaction mixture is at equilibrium? (b) For those mixtures that are not at equilibrium, how
will the reaction proceed to reach equilibrium? [Sections 15.5
and 15.6]

(i)
(a)

(b)

Vessel B
V = 100 mL

(ii)

(iii)

15.9The reaction A21g2 + B1g2 ∆ A1g2 + AB1g2 has an
equilibrium constant of Kp = 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]

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

15.10The 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 Kc for the
reaction. (b) If the volume of the equilibrium mixture is
decreased, will the number of AB molecules increase or decrease? [Sections 15.5 and 15.7]


Exercises

663

15.15Write the expression for Kc 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

15.11The 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 exothermic or endothermic? [Section 15.7]

15.16Write the expressions for Kc for the following reactions. In
each case indicate whether the reaction is homogeneous or
heterogeneous.
(a) 2 O31g2 ∆ 3 O21g2

(b) Ti1s2 + 2 Cl21g2 ∆ TiCl41l2

(c) 2 C2H41g2 + 2 H2O1g2 ∆ 2 C2H61g2 + O21g2

(d) C1s2 + 2 H21g2 ∆ CH41g2

(e) 4 HCl1aq2 + O21g2 ∆ 2 H2O1l2 + 2 Cl21g2

(f) 2 C8H181l2 + 25 O21g2 ∆ 16 CO21g2 + 18 H2O1g2
(g) 2 C8H181l2 + 25 O21g2 ∆ 16 CO21g2 + 18 H2O1l2

15.17When the following reactions come to equilibrium, does
the equilibrium mixture contain mostly reactants or mostly
products?
300 K

500 K

[AB]

15.12The 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.

P=y
P=x

Temperature

(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)
15.13Suppose 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?
15.14Consider the reaction A + B ∆ C + D. Assume that
both the forward reaction and the reverse reaction are elementary processes and that the value of the equilibrium
constant is very large. (a) Which species predominate at equilibrium, reactants or products? (b) Which reaction has the
larger rate constant, the forward or the reverse?

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

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

15.18Which of the following reactions lies to the right, favoring the
formation of products, and which lies to the left, favoring formation of reactants?
(a) 2 NO1g2 + O21g2 ∆ 2 NO21g2; Kp = 5.0 * 1012
(b) 2 HBr1g2 ∆ H21g2 + Br21g2; Kc = 5.8 * 10-18

15.19Which of the following statements are true and which are false?
(a) The equilibrium constant can never be a negative number. (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.
15.20Which of the following statements are true and which are
false? (a) For the reaction 2 A1g2 + B1g2 ∆ A2B1g2 Kc
and Kp are numerically the same. (b) It is possible to distinguish Kc from Kp by comparing the units used to express the
equilibrium constant. (c) For the equilibrium in (a), the value
of Kc increases with increasing pressure.
15.21If Kc = 0.042 for PCl31g2 + Cl21g2 ∆ PCl51g2 at 500 K,
what is the value of Kp for this reaction at this temperature?

15.22Calculate Kc at 303 K for SO21g2 + Cl21g2 ∆ SO2Cl21g2

if Kp = 34.5 at this temperature.

15.23The equilibrium constant for the reaction

2 NO1g2 + Br21g2 ∆ 2 NOBr1g2

is Kc = 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 Kc for 2 NOBr1g2 ∆ 2 NO1g2 + Br21g2.
(c) Calculate Kc for NOBr1g2 ¡ NO1g2 + 12 Br21g2.
15.24Consider the following equilibrium:

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


664

chapter 15 Chemical Equilibrium


(a) Calculate Kp. (b) Does the equilibrium mixture contain
mostly H2 and S2 or mostly H2S? (c) Calculate the value of Kc
if you rewrote the equation H21g2 + 12 S21g2 ∆ H2S1g2.
15.25At 1000 K, Kp = 1.85 for the reaction
SO21g2 +

1
2

O21g2 ∆ SO31g2

(a) What is the value of Kp for the reaction SO31g2 ∆
SO21g2 + 12 O21g2? (b) What is the value of Kp for the reaction
2 SO21g2 + O21g2 ∆ 2 SO31g2? (c) What is the value of
Kc for the reaction in part (b)?
15.26Consider the following equilibrium, for which Kp = 0.0752
at 480 °C:
2 Cl21g2 + 2 H2O1g2 ∆ 4 HCl1g2 + O21g2

(a) What is the value of Kp for the reaction
4 HCl1g2 + O21g2 ∆ 2 Cl21g2 + 2 H2O1g2?
(b) What is the value of Kp for the reaction
Cl21g2 + H2O1g2 ∆ 2 HCl1g2 + 12 O21g2?

(c) What is the value of Kc for the reaction in part (b)?
15.27The following equilibria were attained at 823 K:

CoO1s2 + H21g2 ∆ Co1s2 + H2O1g2 Kc = 67

CoO1s2 + CO1g2 ∆ Co1s2 + CO21g2 Kc = 490

Based on these equilibria, calculate the equilibrium constant
for H21g2 + CO21g2 ∆ CO1g2 + H2O1g2 at 823 K.

15.28Consider the equilibrium

N21g2 + O21g2 + Br21g2 ∆ 2 NOBr1g2

Calculate the equilibrium constant Kp for this reaction, given

the following information (at 298 K):
2 NO1g2 + Br21g2 ∆ 2 NOBr1g2

Kc = 2.0

2 NO1g2 ∆ N21g2 + O21g2 Kc = 2.1 * 1030

15.29Mercury(I) oxide decomposes into elemental mercury and
e l e m e n t a l o x y g e n : 2 Hg2O1s2 ∆ 4 Hg1l2 + O21g2.
(a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) Suppose you run this
reaction in a solvent that dissolves elemental mercury and elemental oxygen. Rewrite the equilibrium-constant expression
in terms of molarities for the reaction, using (solv) to indicate
solvation.
15.30Consider the equilibrium Na2O1s2 + SO21g2 ∆ Na2SO31s2.
(a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) All the compounds in
this reaction are soluble in water. Rewrite the equilibriumconstant expression in terms of molarities for the aqueous
reaction.

Calculating Equilibrium Constants
(Section 15.5)
15.31Methanol 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
Kc at this temperature.

15.32Gaseous hydrogen iodide is placed in a closed container at
425 °C, where it partially decomposes to hydrogen and iodine: 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 Kc at this
temperature?

15.33The equilibrium 2 NO1g2 + Cl21g2 ∆ 2 NOCl1g2 is established 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 Kp for this reaction at 500.0 K. (b) If the vessel has a volume of 5.00 L, calculate Kc at this temperature.

15.34Phosphorus 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, a n d PPCl5 = 1.30 atm.
(a) What is the value of Kp at this temperature? (b) Does the
equilibrium favor reactants or products? (c) Calculate Kc for
this reaction at 450 K.
15.35A 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 equilibrium is established:
2 NO1g2 + 2 H21g2 ∆ N21g2 + 2 H2O1g2

At equilibrium 3NO4 = 0.062 M. (a) Calculate the equilibrium concentrations of H2, N2, and H2O. (b) Calculate Kc.

15.36A 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 Kc.
15.37A 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 equilibrium 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 equilibrium partial pressures of CO2, H2, and CO. (c) Calculate Kp
for the reaction. (d) Calculate Kc for the reaction.
15.38A 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 Kp for the reaction. (c) Calculate Kc for the reaction.
15.39Two different proteins X and Y are dissolved in aqueous solution at 37 °C. The proteins bind in a 1:1 ratio to form XY. A
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 Kc for the reaction?
15.40A chemist at a pharmaceutical company is measuring equilibrium constants for reactions in which drug candidate
molecules bind to a protein involved in cancer. The drug


Exercises
molecules bind the protein in a 1:1 ratio to form a drug–
protein complex. The protein concentration in aqueous solution 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 solution at an initial concentration of 2.00 * 10-6M. At equilibrium, 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 Kc value for the A-protein binding reaction and for the Bprotein 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)

15.47At 1285 °C, the equilibrium constant for the reaction
Br21g2 ∆ 2 Br1g2 is Kc = 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?
15.48For the reaction H21g2 + I21g2 ∆ 2 HI1g2, Kc = 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?

15.49At 800 K, the equilibrium constant for I21g2 ∆ 2 I1g2 is
Kc = 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?
15.50For 2 SO21g2 + O21g2 ∆ 2 SO31g2, Kp = 3.0 * 104 at
700 K. In a 2.00-L vessel, the equilibrium mixture contains
1.17 g of SO3 and 0.105 g of O2. How many grams of SO2 are
in the vessel?
15.51At 2000 °C, the equilibrium constant for the reaction

15.41(a) If Qc 6 Kc, in which direction will a reaction proceed in
order to reach equilibrium? (b) What condition must be satisfied so that Qc = Kc?
15.42(a) If Qc 7 Kc, how must the reaction proceed to reach equilibrium? (b) At the start of a certain reaction, only reactants
are present; no products have been formed. What is the value
of Qc at this point in the reaction?

15.43At 100 °C, the equilibrium constant for the reaction
COCl21g2 ∆ CO1g2 + Cl21g2 h a s t h e v a l u e Kc =
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,
3Cl24 = 6.62 * 10-6 M

(b) 3COCl24 = 4.50 * 10-2 M, 3CO4 = 1.1 * 10-7 M,
3Cl24 = 2.25 * 10-6 M

-6

(c) 3COCl24 = 0.0100 M, 3CO4 = 3Cl24 = 1.48 * 10 M

15.44As shown in Table 15.2, Kp for the equilibrium

N21g2 + 3 H21g2 ∆ 2 NH31g2

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 product 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
15.45At 100 °C, Kc = 0.078 for the reaction
SO2Cl21g2 ∆ SO21g2 + Cl21g2

In an equilibrium mixture of the three gases, the concentrations of SO2Cl2 and SO2 are 0.108 M and 0.052 M, respectively. What is the partial pressure of Cl2 in the equilibrium

mixture?
15.46At 900 K, the following reaction has Kp = 0.345:
2 SO21g2 + O21g2 ∆ 2 SO31g2

665

In an equilibrium mixture the partial pressures of SO2 and O2
are 0.135 atm and 0.455 atm, respectively. What is the equilibrium partial pressure of SO3 in the mixture?

2 NO1g2 ∆ N21g2 + O21g2

is Kc = 2.4 * 103. If the initial concentration of NO is
0.175 M, what are the equilibrium concentrations of NO,
N2, and O2?
15.52For the equilibrium
Br21g2 + Cl21g2 ∆ 2 BrCl1g2

at 400 K, Kc = 7.0. If 0.25 mol of Br2 and 0.55 mol of Cl2 are
introduced into a 3.0-L container at 400 K, what will be the
equilibrium concentrations of Br2, Cl2, and BrCl?
15.53At 373 K, Kp = 0.416 for the equilibrium
2 NOBr1g2 ∆ 2 NO1g2 + Br21g2

If the pressures of NOBr(g) and NO(g) are equal, what is the
equilibrium pressure of Br21g2?
15.54At 218 °C, Kc = 1.2 * 10-4 for the equilibrium

NH4SH1s2 ∆ NH31g2 + H2S1g2

Calculate the equilibrium concentrations of NH3 and H2S if a

sample of solid NH4SH is placed in a closed vessel at 218 °C
and decomposes until equilibrium is reached.
15.55Consider the reaction
CaSO41s2 ∆ Ca2+1aq2 + SO42-1aq2

At 25 °C, the equilibrium constant is Kc = 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 minimum mass of CaSO41s2 needed to achieve equilibrium?
15.56At 80 °C, Kc = 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?

15.57For the reaction I2 + Br21g2 ∆ 2 IBr1g2, Kc = 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 concentrations of IBr, I2, and Br2?


666

chapter 15 Chemical Equilibrium

15.58At 25 °C, the reaction
CaCrO41s2 ∆ Ca2+1aq2 + CrO42-1aq2

has an equilibrium constant Kc = 7.1 * 10-4. What are the
equilibrium concentrations of Ca2 + and CrO42- in a saturated solution of CaCrO4?
15.59Methane, CH4, reacts with I2 according to the reaction
CH41g2 + l21g2 ∆ CH3l1g2 + HI1g2. At 630 K, Kp 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.
15.60The reaction of an organic acid with an alcohol, in organic solvent, 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)
15.61Consider 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 reaction 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?
15.62Consider 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) increase 3NH34; (b) increase 3H2O4; (c) decrease 3O24;

(d) decrease the volume of the container in which the reaction occurs; (e) add a catalyst; (f) increase temperature.
15.63How do the following changes affect the value of the equilibrium constant for a gas-phase exothermic reaction: (a) removal of a reactant, (b) removal of a product, (c) decrease in
the ­volume, (d) decrease in the temperature, (e) addition of a
catalyst?
15.64For 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 vessel. (a) Is the reaction exothermic or endothermic? (b) Does
the balanced chemical equation have more molecules on the
reactant side or product side?
15.65Consider 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?
15.66Methanol 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?
15.67Ozone, O3, decomposes to molecular oxygen in the stratosphere according to the reaction 2 O31g2 ¡ 3 O21g2.
Would an increase in pressure favor the formation of ozone or
of oxygen?
15.68The 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 reaction? If so would high or low pressure favor formation of
H21g2?

Additional Exercises
15.69 Both the forward reaction and the reverse reaction in the following equilibrium are believed to be elementary steps:
CO1g2 + Cl21g2 ∆ COCl1g2 + Cl1g2

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 equilibrium has been reached, calculate Kc and Kp for the reaction
CH41g2 + H2O1g2 ∆ CO1g2 + 3 H21g2.

At 25 °C, the rate constants for the forward and reverse reactions are 1.4 * 10-28 M -1 s-1 and 9.3 * 1010 M -1 s-1, respectively. (a) What is the value for the equilibrium constant at 25 °C?
(b) Are reactants or products more plentiful at equilibrium?

15.72When 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 :

15.70If Kc = 1 for the equilibrium 2 A1g2 ∆ B1g2, what is the
relationship between [A] and [B] at equilibrium?


(a) Calculate Kc for this reaction at this temperature.
(b) Calculate Kp for this reaction at 303 K. (c) According to
Le Châtelier’s principle, would the percent of SO2Cl2 that decomposes increase, decrease or stay the same if the mixture

15.71A mixture of CH4 and H2O is passed over a nickel catalyst at 1000 K. The emerging gas is collected in a 5.00-L

SO2Cl21g2 ∆ SO21g2 + Cl21g2


Additional Exercises
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 Kc 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
Kc. (b) What is the total pressure exerted by the mixture of gases?
(c) What was the mass of the original sample of NOBr?

667

At 700 K, the equilibrium constant Kp 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
(b) PNO = 0.12 atm, PCl2 = 0.10 atm,
PNOCl = 0.050 atm
(c) PNO = 0.15 atm, PCl2 = 0.20 atm,
PNOCl = 5.10 * 10-3 atm
15.82At 900 °C, Kc = 0.0108 for the reaction
CaCO31s2 ∆ CaO1s2 + CO21g2

A mixture of CaCO3, CaO, and CO2 is placed in a 10.0-L vessel at 900 °C. For the following mixtures, will the amount of
CaCO3 increase, decrease, or remain the same as the system
approaches equilibrium?

(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

15.75 Consider the hypothetical reaction A1g2 ∆ 2 B1g2. A
flask is charged with 0.75 atm of pure A, after which it is allowed 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 Kp? (c) What
could we do to maximize the yield of B?

15.83When 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 Kc = 0.802,
what are the concentrations of each substance in the equilibrium mixture?

15.76As shown in Table 15.2, the equilibrium constant for the reaction
N21g2 + 3 H21g2 ∆ 2 NH31g2 is Kp = 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 equilibrium 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?

15.84The equilibrium constant Kc 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?

15.77For the equilibrium

15.85NiO is to be reduced to nickel metal in an industrial process
by use of the reaction

2 IBr1g2 ∆ I21g2 + Br21g2


Kp = 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?
15.78For the equilibrium
PH3BCl31s2 ∆ PH31g2 + BCl31g2


Kp = 0.052 at 60 °C. (a) Calculate Kc. (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 together) is 0.614 atm. What is Kp 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 Kp and Kc for this reaction at 1100 K.

15.81 Nitric oxide (NO) reacts readily with chlorine gas as follows:
2 NO1g2 + Cl21g2 ∆ 2 NOCl1g2

NiO1s2 + CO1g2 ∆ Ni1s2 + CO21g2

At 1600 K, the equilibrium constant for the reaction is
Kp = 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?
15.86Le Châtelier noted that many industrial processes of his time
could be improved by an understanding of chemical equilibria. For example, the reaction of iron oxide with carbon monoxide was used to produce elemental iron and CO2 according
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 furnaces. 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 dimensions 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
an experiment costing several hundred thousand francs, 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?


668

chapter 15 Chemical Equilibrium

[15.87] At 700 K, the equilibrium constant for the reaction
CCl41g2 ∆ C1s2 + 2 Cl21g2

is Kp = 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 Kp =
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 reaction quotient to determine the direction the reaction 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 Kc = 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

equilibrium concentrations (in terms of x) into the equilibriumconstant 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 allowed 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 concentrations of A, B, and C. (Hint: You can check the accuracy of
your answer by substituting these concentrations into the equilibrium expression.)
15.91At 1200 K, the approximate temperature of automobile exhaust gases (Figure 15.15), Kp for the reaction
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 CO2
reaction? Based on your conclusion, would the CO concentration in the exhaust be decreased or increased by a catalyst
that speeds up the CO2 reaction? Recall that at a fixed pressure and temperature, volume % = mol %.
15.92Suppose that you worked at the U.S. Patent Office and a patent
application came across your desk claiming that a newly developed catalyst was much superior to the Haber catalyst for ammonia 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?

Integrative Exercises
15.93Consider the reaction IO4-1aq2 + 2 H2O1l2 ∆ H4IO6-1aq2;
Kc = 3.5 * 10-2. If you start with 25.0 mL of a 0.905 M solution 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 electrolyte. (a) Write the equation for the dissolution of AgCl(s) in
H2O1l2. (b) Write the expression for Kc 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 temperature. (d) The equilibrium constant for the dissolution of AgCl
in water is 1.6 * 10-10 at 25 °C. In addition, Ag +1aq2 can react with Cl-1aq2 according to the reaction

[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 possible 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?

10

Ag +1aq2 + 2 Cl-1aq2 ∆ AgCl2-1aq2

Supercritical
region

2

Liquid

Critical point

where Kc = 1.8 * 10 at 25 °C. Although AgCl is “not soluble” 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 soluble AgCl2- ions? Or is the AgCl solubility in 0.100 M NaCl
less than in pure water because of a Le Châtelier-type argument? 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 forward and reverse reactions are elementary (single-step) reactions. 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.

Pressure (atm)

5

10

1

Gas

10−1

0

100
200
Temperature (°C)

300




Design an Experiment

[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 Kp at each temperature.
Is the reaction exothermic or endothermic?
850

Co2 1mol % 2

6.23

Co 1mol % 2

950

1.32

98.68

1050

0.37

99.63

1200

0.06

99.94

Temperature 1 °C 2

93.77

15.98In 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 corresponding expression for Kp. (b) By using data in Appendix B,
give the value of Kp for this reaction at 30 °C. (c) What is the
value of Kp for any liquid in equilibrium with its vapor at the
normal boiling point of the liquid?

669

15.99 Water molecules in the atmosphere can form hydrogenbonded 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 formation? (c) The Kp 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 mammalian blood. Each Hb can bind 4 O2 molecules. The equilibrium 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
the P50 value, defined as 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 ­e stimate how much larger
Kc is for the aqueous reaction 4 O21g2 + Hb1aq2 ¡
3Hb1O2241aq24.

Design an Experiment
The reaction between hydrogen and iodine to form hydrogen iodide was used to illustrate Beer’s law in Chapter 14 (Figure 14.5). The reaction can be monitored using visible-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 Kc = 794.
To answer the following questions assume you have access to hydrogen, iodine, hydrogen 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 monitor 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 Kc at 600 K to be larger or smaller than at 300 K?


16
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.
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 industrial 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

What’s
Ahead
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,

▶ Clusters of grapes and balsamic

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.

Kw = [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.


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

16.7 Weak Bases  We learn that the ionization of a weak base

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

16.8 Relationship between Ka and Kb  We see that

Ka and Kb are related by the relationship Ka × Kb = Kw. Hence,
the stronger an acid, the weaker its conjugate base.

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.


672

chapter 16  Acid–Base Equilibria

H OH

O
HO

C

C

C

H OH

C

H OH

O
OH

HO

O

C

C

C

H H

Tartaric acid

C

OH

O

Malic acid

▲ Figure 16.1  Two organic acids: Tartartic acid, H2C4H4O6, and malic acid, H2C4H4O5.

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 chapter we take a closer look at how acids and bases are identified and characterized. In
doing so, we consider their behavior both in terms of their structure and bonding and
in terms of the chemical equilibria in which they participate.

16.1 | Acids and Bases: A Brief Review
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 example). 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 something.) 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
 (Section 4.3)
seem to disappear altogether.
By 1830 it was evident that all acids contain hydrogen but not all hydrogencontaining 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 substances 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:


H2O

HCl1g2 ¡ 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?




673

section 16.2  Brønsted–Lowry Acids and Bases

16.2 | Brønsted–Lowry Acids

Go Figure

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 general 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.

Which type of intermolecular force
do the dotted lines in this figure
represent?
+

H

+

The H Ion in Water

O

H

H

H
+

O

H

O

H

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 example, the interaction of a proton with water forms the hydronium ion, H3O+ 1aq2:
H+ +

H

O

+

H

H
H5O2+
+

+

H

[16.2]

+

+




+

The behavior of H ions in liquid water is complex because hydronium ions interact
 (Section 11.2)
with additional water molecules via the formation of hydrogen bonds.
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:
HCl(g)

+

H2O(l)

Cl

+

O

H

H

Cl−(aq)

Cl−

+ H3O+(aq)
+

H

Acid

O

H

+

H
−

+

H

[16.3]

+

+
Base



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

H

+

O
H

H

H

O
H
H

O
H
H9O4+

▲ Figure 16.2  Ball-and-stick models and
Lewis structures for two hydrated hydronium
ions.


674

chapter 16  Acid–Base Equilibria

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:
H
Cl

H

+

N

+

H
H

Cl−

+

H

H

N

H

H
[16.4]
+

−

+
Acid

+
Base



The hazy film that forms on the windows of general chemistry laboratories and on
glassware in the laboratory (◀ Figure 16.3) is largely solid NH4Cl formed by the gasphase reaction between HCl and NH3.
Let’s consider another example that compares the relationship between the Arrhenius
and Brønsted–Lowry definitions of acids and bases—an aqueous solution of ammonia,
in which we have the equilibrium:



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

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

Acid

Ammonia is a Brønsted–Lowry base because it accepts a proton from H2O. Ammonia
is also an Arrhenius base because adding it to water leads to an increase in the concentration of OH - 1aq2.
The transfer of a proton always involves both an acid (donor) and a base (acceptor).
In other words, a substance can function as an acid only if another substance simultaneously behaves as a base. To be a Brønsted–Lowry acid, a molecule or ion must have
a hydrogen atom it can lose as an H+ ion. To be a Brønsted–Lowry base, a molecule or
ion must have a nonbonding pair of electrons it can use to bind the H+ ion.
Some substances can act as an acid in one reaction and as a base in another. For example, H2O is a Brønsted–Lowry base in Equation 16.3 and a Brønsted–Lowry acid in Equation 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




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 Ais 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:
remove H+

NO2−(aq) + H3O+(aq)

HNO2(aq) + H2O(l)
Acid

Base

Conjugate
base

[16.7]

Conjugate
acid

add H+





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

NH4+(aq) + OH−(aq)

NH3(aq) + H2O(l)
Base



Acid

Conjugate
acid

[16.8]

Conjugate
base

remove H+



Sa m p le
Exercise 16.1  Identifying Conjugate Acids and Bases
(a) What is the conjugate base of HClO4, H2S, PH4+, HCO3- ?
(b) What is the conjugate acid of CN - , SO42 - , H2O, HCO3- ?

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
(c) OH - and SO42 -   (d) SO4 2 - and H2O
(e) None of the substances are acting as acids in this reaction.

Practice Exercise 2
Write the formula for the conjugate acid of each of the following:
HSO3- , F - , PO43 - , CO.

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

(proton-transfer reactions).

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


676

chapter 16  Acid–Base Equilibria

Sa m p le
Exercise 16.2  Writing Equations for Proton-Transfer Reactions
The hydrogen sulfite ion 1HSO3- 2 is amphiprotic. Write an equation for the reaction of HSO3with 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.

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 - (conjugate 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 + HPO4 2 - 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 acceptors than others. If we arrange acids in order of their ability to donate a proton, we find
that the more easily a substance gives up a proton, the less easily its conjugate base accepts
a proton. Similarly, the more easily a base accepts a proton, the less easily its conjugate acid
gives up a proton. In other words, the stronger an acid, the weaker its conjugate base, and the
stronger a base, the weaker its conjugate acid. Thus, if we know how readily an acid donates
protons, we also know something about how readily its conjugate base accepts protons.
The inverse relationship between the strengths of acids and their conjugate bases
is illustrated in ▼ Figure 16.4. Here we have grouped acids and bases into three broad
categories based on their behavior in water:

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

Negligible acidity

Acid strength increasing

Weak acids

Base strength increasing

Strong acids

ACID
HCl
H2SO4
HNO3
H3O+(aq)
HSO4−
H3PO4
HF
CH3COOH
H2CO3
H2S
H2PO4−
NH4+
HCO3−
HPO42−
H2O

OH−
H2
CH4

BASE
Cl−
HSO4−
NO3−
H2O
SO42−
H2PO4−
F−
CH3COO−
HCO3−
HS−
HPO42−
NH3
CO32−
PO43−
OH−
O2−
H−
CH3−

Negligible basicity

Weak bases

Strong bases

▲ 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.




section 16.2  Brønsted–Lowry Acids and Bases

1.A strong acid completely transfers its protons to water, leaving essentially no undis (Section 4.3) Its conjugate base has a negligible
sociated molecules in solution.
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 conjugate 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 completely with water, to form OH - ions. (The conjugate base of a substance with negligible 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 produce 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 abilities 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.
Sa m p le
Exercise 16.3  Predicting the Position of a Proton-Transfer

Equilibrium

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

lies to the left 1Kc 6 12 or to the right 1Kc 7 12:

Solution

HSO4 - 1aq2 + CO32 - 1aq2 ∆ SO4 2 - 1aq2 + HCO3- 1aq2

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

677


678

chapter 16  Acid–Base Equilibria

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 SO4 2 - , the conjugate base of HSO4 - . We can find the relative positions
of these two bases in Figure 16.4 to determine which is the stronger base.

Solve The CO32 - ion appears lower in the right-hand column in Figure 16.4 and is there-

fore a stronger base than SO42 - . Therefore, CO32- will get the proton preferentially to become
HCO3 - , while SO4 2 - will remain mostly unprotonated. The resulting equilibrium lies to the right,
favoring products (that is, Kc 7 1):
HSO4-1aq2 + CO32-1aq2 ∆ SO42-1aq2 + HCO3-1aq2 Kc 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 Kc:
   (i) CH3COOH1aq2 + HS - 1aq2 ∆ CH3COO - 1aq2 + H2S1aq2
  (ii) F - 1aq2 + NH4+1aq2 ∆ HF1aq2 + NH31aq2

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

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

Practice Exercise 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

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:
H2O(l)

+

H2O(l)

O

+

O

H

H

H

H

OH−(aq) +

O

−

+

H



Acid

H

O

H

H

+

[16.12]
+

−

+

H3O+(aq)

+
Base

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.




section 16.3  The Autoionization of Water

679

The Ion Product of Water
The equilibrium-constant expression for the autoionization of water is
Kc = 3H3O+ 43OH -4[16.13]



The term [H2O] is excluded from the equilibrium-constant expression because we
 (Section 15.4) Because this
exclude the concentrations of pure solids and liquids.
­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 important, and you should commit them to memory.
A solution in which 3H+ 4 = 3OH - 4 is said to be neutral. In most solutions, however, 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).

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

Hydrochloric
acid
HCl(aq)

Water
H2O

Sodium

hydroxide
NaOH(aq)

Acidic solution
[H+] > [OH−]
+
[H ][OH−] = 1.0 × 10−14

Neutral solution
[H+] = [OH−]
+
[H ][OH−] = 1.0 × 10−14

Basic solution
[H+] < [OH−]
+
[H ][OH−] = 1.0 × 10−14

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

Sa m p le
Exercise 16.4  Calculating 3h + 4 for Pure Water

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

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


680

chapter 16  Acid–Base Equilibria

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?

Practice Exercise 2
Indicate whether solutions with each of the following ion concentrations 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.

(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

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).

Sa m p le
Exercise 16.5  Calculating 3h + 4 from 3oh - 4

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.

Solution

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

Plan We can use the equilibrium-constant expression for the auto-

ionization of water and the value of Kw to solve for each unknown
concentration.

aqueous solution where the hydroxide concentration is known.

Solve

(a) Using Equation 16.16, we have

This solution is basic because
(b) In this instance
This solution is acidic because

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

11.0 * 10-142
-

3OH 4

3OH - 4 7 3H+4
3H+4 =

11.0 * 10 - 142
-

3OH 4

=

=

1.0 * 10-14
= 1.0 * 10-12 M
0.010

1.0 * 10 - 14
= 5.6 * 10 - 6 M
1.8 * 10 - 9

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?
(a)2.5 * 10 - 8 M  (b) 4.0 * 10 - 8 M  (c) 2.5 * 10 - 7 M
(d)2.5 * 10 - 6 M  (e) 4.0 * 10 - 6 M

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

16.4 | The pH Scale
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:*

pH = -log[H+ ][16.17]
If you need to review the use of logarithms, see Appendix A.
In Sample Exercise 16.4, we saw that 3H+ 4 = 1.0 * 10-7 M for a neutral aqueous
solution at 25 °C. We can now use Equation 16.17 to calculate the pH of a neutral solution
at 25 °C:
pH = -log11.0 * 10-72 = -1-7.002 = 7.00
*Because 3H+ 4 and 3H3O+ 4 are used interchangeably, you might see pH defined as - log 3H3O+ 4.

section 16.4  The pH Scale

Table 16.1  Relationships among 3 h + 4 , 3 oh − 4 , and pH at 25 °c
3 h+ 4 1M 2

Solution Type

Acidic

pH

-7

6 7.00

1.0 * 10-7

1.0 * 10-7

7.00

6 1.0 * 10-7

7 1.0 * 10-7

7 7.00

7 1.0 * 10

Neutral
Basic

3oh− 4 1M 2

-7

61.0 * 10

681

Notice that the pH is reported with two decimal places. We do so because only the numbers 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 calculate 3H+ 4 for this solution and Equation 16.17 to calculate the pH:
3H+ 4 =

Kw
1.0 * 10-14

= 5.0 * 10-12 M
- =
3OH 4
2.0 * 10-3

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 unimportant. 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
 (Section 14.3) In biological systems,
is merely from 1 * 10-7 M to 2 * 10-7 M.
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.
Sa m p le
Exercise 16.6  Calculating pH from 3h + 4

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

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.
Solve

(a) In the first instance we found 3H+4 to be 1.0 * 10-12 M, so that
pH = -log11.0 * 10-122 = - 1- 12.002 = 12.00

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

(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 estimate. 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 reconsidered our calculation or estimate or both.


682

chapter 16  Acid–Base Equilibria

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

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

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:

pOH = -log 3OH- 4[16.18]
Likewise, pKw equals -log Kw.
By taking the negative logarithm of both sides of the equilibrium-constant expression 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 concentration in a solution of pH 6.

Go Figure

Increasing acid strength

Which is more acidic, black coffee or lemon juice?
pH

pOH

[OH−] (M)

1 (1×100)

0.0

14.0

1×10−14

1×10−1

1.0

13.0

1×10−13

1×10−2

2.0

12.0

1×10−12

1×10−3

3.0

11.0

1×10−11

1×10−4

4.0

10.0

1×10−10

Black coffee

1×10−5

5.0

9.0

1×10−9

Rain
Saliva
Milk

1×10−6

6.0

8.0

1×10−8

1×10−7

7.0

7.0

1×10−7
−6

Stomach acid
Lemon juice
Cola, vinegar
Wine
Tomatoes

−8

8.0

6.0

1×10

−9

1×10

9.0

5.0

1×10−5

1×10−10

10.0

4.0

1×10−4

1×10−11

11.0

3.0

1×10−3

1×10−12

12.0

2.0

1×10−2

1×10−13

13.0

1.0

1×10−1

1×10−14

14.0

0.0

1 (1×100)

1×10

Human blood
Seawater
Borax
Lime water
Household ammonia
Household bleach

pH + pOH = 14
[H+][OH−] = 1×10−14

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

Increasing base strength

[H+] (M)




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?

Sa m p le
Exercise 16.7  Calculating 3h+ 4 from pOH

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

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 = 3.76
Next we use Equation 16.17:
Thus,

pH = - log 3H+4 = 3.76
log 3H+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.

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 electrochemistry, 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 electrodes 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 litmus 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.


684

chapter 16  Acid–Base Equilibria

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

0
Methyl violet
Thymol blue
Methyl orange
Methyl red
Bromthymol blue
Phenolphthalein
Alizarin yellow R

Methyl red

Yellow
Red

2

pH range for color change
4
6
8
10

12

14

Violet
Yellow
Red

Yellow

Blue

Yellow
Red

Yellow

Yellow

Blue

Colorless

Pink
Yellow

Red

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

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

16.5 | Strong Acids and Bases

Bromthymol blue

The chemistry of an aqueous solution often depends critically on pH. It is therefore important to examine how pH relates to acid and base concentrations. The simplest cases
are those involving strong acids and strong bases. Strong acids and bases are strong electrolytes, existing in aqueous solution entirely as ions. There are relatively few common
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
 (Section 4.1) As noted in Section 16.3, we use H3O+ 1aq2 and H+ 1aq2
to the right.
interchangeably to represent the hydrated proton in water. Thus, we can simplify this
acid ionization equation to

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

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
*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

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.
Sa m p le
Exercise 16.8  Calculating the pH of a Strong Acid
What is the pH of a 0.040 M solution of HClO4?

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.

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?

Sa m p le
Exercise 16.9  Calculating the pH of a Strong Base
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 - concentration for the solution in (a) equals the stated concentration of NaOH, namely 0.028 M.
Method 1:
3H+4 =

1.0 * 10-14
= 3.57 * 10-13 M
0.028

Method 2:

pOH = - log 10.0282 = 1.55

pH = - log13.57 * 10-132 = 12.45
pH = 14.00 - pOH = 12.45

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

3H+4 =

Method 2:

1.0 * 10-14
= 4.55 * 10-12 M
0.0022
pOH = - log 10.00222 = 2.66

pH = - log14.55 * 10-122 = 11.34

pH = 14.00 - pOH = 11.34

685


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chapter 16  Acid–Base Equilibria

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) Ca1OH22

for which the pH is 11.68?

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, especially Na2O and CaO, are often used in industry when a strong base is needed. The O2reacts 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.

16.6 | Weak Acids
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 reaction 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:

or

HA1aq2 + H2O1l2 ∆ H3O+ 1aq2 + A- 1aq2[16.23]
HA1aq2 ∆ H+ 1aq2 + A- 1aq2[16.24]

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


Ka =

3H3O+ 43A- 4
3H+ 43A- 4
or Ka =
[16.25]
3HA4
3HA4

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.