OpenStax: General Chemistry
OpenStax: General Chemistry

OpenStax: General Chemistry

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14 Acid-Base Equilibria

Figure 14.1 Sinkholes such as this are the result of reactions between acidic groundwaters and basic rock formations, like limestone. (credit: modification of work by Emil Kehnel)

Chapter Outline

14.1 Brønsted-Lowry Acids and Bases

14.2 pH and pOH

14.3 Relative Strengths of Acids and Bases

14.4 Hydrolysis of Salt Solutions

14.5 Polyprotic Acids

14.6 Buffers

14.7 Acid-Base Titrations

Introduction

n our bodies, in our homes, and in our industrial society, acids and bases play key roles. Proteins, enzymes, blood, genetic material, and other components of living matter contain both acids and bases. We seem to like the sour taste of acids; we add them to soft drinks, salad dressings, and spices. Many foods, including citrus fruits and some vegetables, contain acids. Cleaners in our homes contain acids or bases. Acids and bases play important roles in the chemical industry. Currently, approximately 36 million metric tons of sulfuric acid are produced annually in the United States alone. Huge quantities of ammonia (8 million tons), urea (10 million tons), and phosphoric acid (10 million tons) are also produced annually.

This chapter will illustrate the chemistry of acid-base reactions and equilibria, and provide you with tools for quantifying the concentrations of acids and bases in solutions.

14.1 Brønsted-Lowry Acids and Bases

By the end of this section, you will be able to:

  • • Identify acids, bases, and conjugate acid-base pairs according to the Brønsted-Lowry definition
  • • Write equations for acid and base ionization reactions
  • • Use the ion-product constant for water to calculate hydronium and hydroxide ion concentrations
  • • Describe the acid-base behavior of amphiprotic substances

Acids and bases have been known for a long time. When Robert Boyle characterized them in 1680, he noted that acids dissolve many substances, change the color of certain natural dyes (for example, they change litmus from blue to red), and lose these characteristic properties after coming into contact with alkalis (bases). In the eighteenth century, it was recognized that acids have a sour taste, react with limestone to liberate a gaseous substance (now known to be CO2), and interact with alkalis to form neutral substances. In 1815, Humphry Davy contributed greatly to the development of the modern acid-base concept by demonstrating that hydrogen is the essential constituent of acids. Around that same time, Joseph Louis Gay-Lussac concluded that acids are substances that can neutralize bases and that these two classes of substances can be defined only in terms of each other. The significance of hydrogen was reemphasized in 1884 when Carl Axel Arrhenius defined an acid as a compound that dissolves in water to yield hydrogen cations (now recognized to be hydronium ions) and a base as a compound that dissolves in water to yield hydroxide anions.

In an earlier chapter on chemical reactions, we defined acids and bases as Arrhenius did: We identified an acid as a compound that dissolves in water to yield hydronium ions (H3O+) and a base as a compound that dissolves in water to yield hydroxide ions (OH). This definition is not wrong; it is simply limited.

Later, we extended the definition of an acid or a base using the more general definition proposed in 1923 by the Danish chemist Johannes Brønsted and the English chemist Thomas Lowry. Their definition centers on the proton, H+. A proton is what remains when a normal hydrogen atom, 11H, loses an electron. A compound that donates a proton to another compound is called a Brønsted-Lowry acid, and a compound that accepts a proton is called a Brønsted-Lowry base. An acid-base reaction is the transfer of a proton from a proton donor (acid) to a proton acceptor (base). In a subsequent chapter of this text we will introduce the most general model of acid-base behavior introduced by the American chemist G. N. Lewis.

Acids may be compounds such as HCl or H2SO4, organic acids like acetic acid (CH3COOH) or ascorbic acid (vitamin C), or H2O. Anions (such as HSO4, H2PO4, HS, and HCO3) and cations (such as H3O+, NH4+, and [Al(H2O)6]3+) may also act as acids. Bases fall into the same three categories. Bases may be neutral molecules (such as H2O, NH3, and CH3NH2), anions (such as OH, HS, HCO3, CO32−, F−, and PO43−), or cations (such as [Al(H2O)5OH]2+). The most familiar bases are ionic compounds such as NaOH and Ca(OH)2, which contain the hydroxide ion, OH. The hydroxide ion in these compounds accepts a proton from acids to form water:

We call the product that remains after an acid donates a proton the conjugate base of the acid. This species is a base because it can accept a proton (to re-form the acid):

We call the product that results when a base accepts a proton the base’s conjugate acid. This species is an acid because it can give up a proton (and thus re-form the base):

In these two sets of equations, the behaviors of acids as proton donors and bases as proton acceptors are represented in isolation. In reality, all acid-base reactions involve the transfer of protons between acids and bases. For example, consider the acid-base reaction that takes place when ammonia is dissolved in water. A water molecule (functioning as an acid) transfers a proton to an ammonia molecule (functioning as a base), yielding the conjugate base of water, OH, and the conjugate acid of ammonia, NH4+:

The reaction between a Brønsted-Lowry acid and water is called acid ionization. For example, when hydrogen fluoride dissolves in water and ionizes, protons are transferred from hydrogen fluoride molecules to water molecules, yielding hydronium ions and fluoride ions:

When we add a base to water, a base ionization reaction occurs in which protons are transferred from water molecules to base molecules. For example, adding ammonia to water yields hydroxide ions and ammonium ions:

Notice that both these ionization reactions are represented as equilibrium processes. The relative extent to which these acid and base ionization reactions proceed is an important topic treated in a later section of this chapter. In the preceding paragraphs we saw that water can function as either an acid or a base, depending on the nature of the solute dissolved in it. In fact, in pure water or in any aqueous solution, water acts both as an acid and a base. A very small fraction of water molecules donate protons to other water molecules to form hydronium ions and hydroxide ions:


This type of reaction, in which a substance ionizes when one molecule of the substance reacts with another molecule of the same substance, is referred to as autoionization.

Pure water undergoes autoionization to a very slight extent. Only about two out of every 109 molecules in a sample of pure water are ionized at 25 °C. The equilibrium constant for the ionization of water is called the ion-product constant for water (Kw):



The slight ionization of pure water is reflected in the small value of the equilibrium constant; at 25 °C, Kw has a value of 1.0 × 10−14. The process is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion increase with temperature. For example, at 100 °C, the value for Kw is about 5.1 × 10−13, roughly 100-times larger than the value at 25 °C.

Example 14.1

Ion Concentrations in Pure Water

What are the hydronium ion concentration and the hydroxide ion concentration in pure water at 25 °C?

Solution

The autoionization of water yields the same number of hydronium and hydroxide ions. Therefore, in pure water, [H3O+] = [OH]. At 25 °C:


So:

The hydronium ion concentration and the hydroxide ion concentration are the same, and we find that both equal 1.0 × 10−7 M.

Check Your Learning

The ion product of water at 80 °C is 2.4 × 10−13. What are the concentrations of hydronium and hydroxide ions in pure water at 80 °C?

Answer:


It is important to realize that the autoionization equilibrium for water is established in all aqueous solutions. Adding an acid or base to water will not change the position of the equilibrium. . Example 14.2 demonstrates the quantitative aspects of this relation between hydronium and hydroxide ion concentrations.

Example 14.2

The Inverse Proportionality of [H3O+] and [OH]

A solution of carbon dioxide in water has a hydronium ion concentration of 2.0 × 10−6 M. What is the concentration of hydroxide ion at 25 °C?

Solution

We know the value of the ion-product constant for water at 25 °C:

Thus, we can calculate the missing equilibrium concentration.

Rearrangement of the Kw expression yields that [OH] is directly proportional to the inverse of [H3O+]:


The hydroxide ion concentration in water is reduced to 5.0 × 10−9 M as the hydrogen ion concentration increases to 2.0 × 10−6 M. This is expected from Le Châtelier’s principle; the autoionization reaction shifts to the left to reduce the stress of the increased hydronium ion concentration and the [OH] is reduced relative to that in pure water.

A check of these concentrations confirms that our arithmetic is correct:


Check Your Learning

What is the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 M at 25 °C?

Answer:[H3O+] = 1 × 10−11 M

Amphiprotic Species

Like water, many molecules and ions may either gain or lose a proton under the appropriate conditions. Such species are said to be amphiprotic. Another term used to describe such species is amphoteric, which is a more general term for a species that may act either as an acid or a base by any definition (not just the Brønsted-Lowry one). Consider for example the bicarbonate ion, which may either donate or accept a proton as shown here:


Example 14.3

Representing the Acid-Base Behavior of an Amphoteric Substance

Write separate equations representing the reaction of HSO3

(a) as an acid with OH

(b) as a base with HI

Solution

Check Your Learning

Write separate equations representing the reaction of H2PO4

(a) as a base with HBr

(b) as an acid with OH

Answer:

Exercises

Question 14.1

14.1

Write equations that show NH3NH_3 as both a conjugate acid and a conjugate base.

Click here to see the answer to Question 14.1

Question 14.2

14.2

Write equations that show H2PO4H_{2}PO_{4} ^{−} acting both as an acid and as a base.


Question 14.3

14.3

Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid

(a)H3O+H_{3}O^+
(b) HCl (c) NH3NH_3 (d) CH3CO2HCH_{3}CO_{2}H (e) NH4+NH_{4}^{+} (f) HSO4HSO_{4}^{−}

Click here to see the answer to Question 14.3

Question 14.4

14.4

Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid:

(a) HNO3 (b) PH4+PH_{4}\:^{+} (c) H2SH_{2}S (d) CH3CH2COOHCH_{3}CH_{2}COOH (e) H2PO4H_{2} PO_{4}\:^{−} (f) HSHS^{−}


Question 14.5

14.5

Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base:

(a) H2OH_{2}O (b) OHOH^− (c) NH3NH_{3} (d) CNCN^− (e) S2S^{2−} (f) H2PO4H_{2} PO_{4}\: ^−

Click here to see the answer to Question 14.5

Question 14.6

14.6

Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base:

(a) HSHS^− (b) PO43PO_{4}\:^{3−} (c) NH2NH_{2}\: ^− (d) C2H5OHC_{2}H_{5}OH (e) O2O^{2−} (f) H2PO4H_{2} PO\:{4}\:^ −


Question 14.7

14.7

What is the conjugate acid of each of the following? What is the conjugate base of each?

(a) OHOH^− (b) H2OH_{2}O (c) HCO3HCO_{3}\:^− (d) NH3NH_{3} (e) HSO4HSO_{4}\:^− (f) H2O2H_{2}O_2 (g) HSHS^− (h)H5N2+H_{5} N_{2}^ +

Click here to see the answer to Question 14.7

Question 14.8

14.8

What is the conjugate acid of each of the following? What is the conjugate base of each?

(a) H2SH_{2}S (b) H2PO4H_{2} PO_{4}\: ^− (c) PH3PH_3 (d) HSHS^− (e) HSO3HSO_{3}\:^− (f) H3O2+H_{3} O_{2}\:^+ (g) H4N2H_{4}N_{2} (h) CH3OHCH_{3}OH


Question 14.9


14.9

Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:

(a) HNO3+H2OH3O++NO3HNO_{3} + H_{2} O \longrightarrow H_{3} O^{+} + NO_3\: {–} (b) CN+H2OHCN+OHCN^{−} + H_{2} O \longrightarrow HCN + OH^{−} (c) H2SO4+ClHCl+HSO4H_{2}SO_{4} + Cl^{−} \longrightarrow HCl + HSO_{4}\:^– (d) HSO4+OHSO42+H2OHSO_{4}\: ^− + OH^− \longrightarrow SO_{4}\:^{2−} + H_{2}O (e) O2+H2O2OHO^{2−} + H_{2}O \longrightarrow 2OH^− (f) [Cu(H2O)3(OH)]++[Al(H2O)6]3+[Cu(H2O)4]2++[Al(H2O)5(OH)]2+[Cu(H_{2} O)_{3} (OH)]^{+} + [Al(H_{2} O)_{6} ]^{3+} \longrightarrow [Cu(H_{2} O)_{4} ]^{2+} + [Al(H_{2} O)_{5} (OH)]^{2+} (g) H2S+NH2HS+NH3H_{2}S + NH_{2} − \longrightarrow HS^{−} + NH_{3}

Click here to see the answer to Question 14.9

Question 14.10

14.10

Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:

(a) NO2+H2OHNO2+OHNO_{2}^{−} + H_{2} O \longrightarrow HNO_{2} + OH^{−} (b) HBr+H2OH3O++BrHBr + H_{2} O \longrightarrow H_{3} O ^{+} + Br^{−} (c) HS+H2OH2S+OHHS^{−} + H_{2} O \longrightarrow H_{2}S + OH_{−} (d) H2PO4+OHHPO42+H2OH_{2} PO_{4}^{ −} + OH^{−} \longrightarrow HPO_{4}\:^{ 2−} + H_{2} O (e)H2PO4+HClH3PO4+Cl H_{2} PO_{4} ^{−} + HCl \longrightarrow H_{3} PO_{4} + Cl^{−} (f) [Fe(H2O)5(OH)]2++[Al(H2O)6]3+[Fe(H2O)6]3++[Al(H2O)5(OH)]2+[Fe(H2 O)5 (OH)]2+ + [Al(H2 O)6 ] 3+ \longrightarrow [Fe(H2 O)6 ] 3+ + [Al(H2 O)5 (OH)]2+ (g)CH3OH+HCH3O+H2 CH_{3} OH + H^{−} \longrightarrow CH_{3} O^{ −} + H_{2}


Question 14.11

14.11

What are amphiprotic species? Illustrate with suitable equations.

Click here to see the answer to Question 14.11

Question 14.12

14.12

State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species:

(a) H2OH_{2}O (b) H2PO4H_{2} PO_{4}\:^{ –} (c) S2S^{2−} (d) CO32CO_{3}\:^{ 2−} (e) [math]HSO_{4}^\:{ –} [/math]


Question 14.13

14.13

State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species.

(a) NH3NH_{3} (b) HPO4HPO_{4}\:^{ –} (c) ]BrBr^− (d) NH4+NH_{4}\: ^{ +} (e) ASO43ASO_{4} \:^{ 3−}

Click here to see the answer to Question 14.13

 Question 14.14

14.14

Is the self ionization of water endothermic or exothermic? The ionization constant for water (Kw)(K_w) is 2.9×10142.9 × 10^{−14} at 40 °C and 9.6×10149.6 × 10^{−14} at 60 °C.








14.2 pH and pOH

By the end of this section, you will be able to:

  • • Explain the characterization of aqueous solutions as acidic, basic, or neutral
  • • Express hydronium and hydroxide ion concentrations on the pH and pOH scales
  • • Perform calculations relating pH and pOH

As discussed earlier, hydronium and hydroxide ions are present both in pure water and in all aqueous solutions, and their concentrations are inversely proportional as determined by the ion product of water (Kw). The concentrations of these ions in a solution are often critical determinants of the solution’s properties and the chemical behaviors of its other solutes, and specific vocabulary has been developed to describe these concentrations in relative terms. A solution is neutral if it contains equal concentrations of hydronium and hydroxide ions; acidic if it contains a greater concentration of hydronium ions than hydroxide ions; and basic if it contains a lesser concentration of hydronium ions than hydroxide ions.

A common means of expressing quantities, the values of which may span many orders of magnitude, is to use a logarithmic scale. One such scale that is very popular for chemical concentrations and equilibrium constants is based on the p-function, defined as shown where “X” is the quantity of interest and “log” is the base-10 logarithm:

The pH of a solution is therefore defined as shown here, where [H3O+] is the molar concentration of hydronium ion in the solution:

Rearranging this equation to isolate the hydronium ion molarity yields the equivalent expression:

Likewise, the hydroxide ion molarity may be expressed as a p-function, or pOH:

or

Finally, the relation between these two ion concentration expressed as p-functions is easily derived from the Kw expression:



At 25 °C, the value of Kw is 1.0 × 10−14, and so:


As was shown in Example 14.1, the hydronium ion molarity in pure water (or any neutral solution) is 1.0 × 10−7 M at 25 °C. The pH and pOH of a neutral solution at this temperature are therefore:

And so, at this temperature, acidic solutions are those with hydronium ion molarities greater than 1.0 × 10−7 M and hydroxide ion molarities less than 1.0 × 10−7 M (corresponding to pH values less than 7.00 and pOH values greater than 7.00). Basic solutions are those with hydronium ion molarities less than 1.0 × 10−7 M and hydroxide ion molarities greater than 1.0 × 10−7 M (corresponding to pH values greater than 7.00 and pOH values less than 7.00).

Since the autoionization constant Kw is temperature dependent, these correlations between pH values and the acidic/neutral/basic adjectives will be different at temperatures other than 25 °C. For example, the “Check Your Learning” exercise accompanying Example 14.1 showed the hydronium molarity of pure water at 80 °C is 4.9 × 10−7 M, which corresponds to pH and pOH values of:

At this temperature, then, neutral solutions exhibit pH = pOH = 6.31, acidic solutions exhibit pH less than 6.31 and pOH greater than 6.31, whereas basic solutions exhibit pH greater than 6.31 and pOH less than 6.31. This distinction can be important when studying certain processes that occur at nonstandard temperatures, such as enzyme reactions in warm-blooded organisms. Unless otherwise noted, references to pH values are presumed to be those at standard temperature (25 °C) (Table 14.1).

Table 14.1


Figure 14.2 shows the relationships between [H3O+], [OH], pH, and pOH, and gives values for these properties at standard temperatures for some common substances.


Figure 14.2 The pH and pOH scales represent concentrations of [H3O+] and OH, respectively. The pH and pOH values of some common substances at standard temperature (25 °C) are shown in this chart.

Example 14.4

Calculation of pH from [H3O+]

What is the pH of stomach acid, a solution of HCl with a hydronium ion concentration of 1.2 × 10−3 M?

Solution


(The use of logarithms is explained in Appendix B. Recall that, as we have done here, when taking the log of a value, keep as many decimal places in the result as there are significant figures in the value.)

Check Your Learning

Water exposed to air contains carbonic acid, H2CO3, due to the reaction between carbon dioxide and water:

Air-saturated water has a hydronium ion concentration caused by the dissolved CO2 of 2.0 × 10−6 M, about 20-times larger than that of pure water. Calculate the pH of the solution at 25 °C.

Answer: 5.70

Example 14.5

Calculation of Hydronium Ion Concentration from pH

Calculate the hydronium ion concentration of blood, the pH of which is 7.3 (slightly alkaline).

Solution


On a calculator take the antilog, or the “inverse” log, of −7.3, or calculate 10−7.3.)

Check Your Learning

Calculate the hydronium ion concentration of a solution with a pH of −1.07.

Answer:12 M

How Sciences Interconnect

Environmental Science

Normal rainwater has a pH between 5 and 6 due to the presence of dissolved CO2 which forms carbonic acid:


Acid rain is rainwater that has a pH of less than 5, due to a variety of nonmetal oxides, including CO2, SO2, SO3, NO, and NO2 being dissolved in the water and reacting with it to form not only carbonic acid, but sulfuric acid and nitric acid. The formation and subsequent ionization of sulfuric acid are shown here:


Carbon dioxide is naturally present in the atmosphere because we and most other organisms produce it as a waste product of metabolism. Carbon dioxide is also formed when fires release carbon stored in vegetation or when we burn wood or fossil fuels. Sulfur trioxide in the atmosphere is naturally produced by volcanic activity, but it also stems from burning fossil fuels, which have traces of sulfur, and from the process of “roasting” ores of metal sulfides in metal-refining processes. Oxides of nitrogen are formed in internal combustion engines where the high temperatures make it possible for the nitrogen and oxygen in air to chemically combine.

Acid rain is a particular problem in industrial areas where the products of combustion and smelting are released into the air without being stripped of sulfur and nitrogen oxides. In North America and Europe until the 1980s, it was responsible for the destruction of forests and freshwater lakes, when the acidity of the rain actually killed trees, damaged soil, and made lakes uninhabitable for all but the most acid-tolerant species. Acid rain also corrodes statuary and building facades that are made of marble and limestone (Figure 14.3). Regulations limiting the amount of sulfur and nitrogen oxides that can be released into the atmosphere by industry and automobiles have reduced the severity of acid damage to both natural and manmade environments in North America and Europe. It is now a growing problem in industrial areas of China and India.

For further information on acid rain, visit this website (http://openstaxcollege.org/l/16EPA) hosted by the US Environmental Protection Agency.

Figure 14.3 (a) Acid rain makes trees more susceptible to drought and insect infestation, and depletes nutrients in the soil. (b) It also is corrodes statues that are carved from marble or limestone. (credit a: modification of work by Chris M Morris; credit b: modification of work by “Eden, Janine and Jim”/Flickr)

Example 14.6

Calculation of pOH

What are the pOH and the pH of a 0.0125-M solution of potassium hydroxide, KOH?

Solution

Potassium hydroxide is a highly soluble ionic compound and completely dissociates when dissolved in dilute solution, yielding [OH] = 0.0125 M:


The pH can be found from the pOH:


Check Your Learning

The hydronium ion concentration of vinegar is approximately 4 × 10−3 M. What are the corresponding values of pOH and pH?

Answer:pOH = 11.6, pH = 2.4

The acidity of a solution is typically assessed experimentally by measurement of its pH. The pOH of a solution is not usually measured, as it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter (Figure 14.4).


Figure 14.4 (a) A research-grade pH meter used in a laboratory can have a resolution of 0.001 pH units, an accuracy of ± 0.002 pH units, and may cost in excess of $1000. (b) A portable pH meter has lower resolution (0.01 pH units), lower accuracy (± 0.2 pH units), and a far lower price tag. (credit b: modification of work by Jacopo Werther)

The pH of a solution may also be visually estimated using colored indicators (Figure 14.5).


Figure 14.5 (a) A universal indicator assumes a different color in solutions of different pH values. Thus, it can be added to a solution to determine the pH of the solution. The eight vials each contain a universal indicator and 0.1-M solutions of progressively weaker acids: HCl (pH = l), CH3CO2H (pH = 3), and NH4Cl (pH = 5), deionized water, a neutral substance (pH = 7); and 0.1-M solutions of the progressively stronger bases: KCl (pH = 7), aniline, C6H5NH2 (pH = 9), NH3 (pH = 11), and NaOH (pH = 13). (b) pH paper contains a mixture of indicators that give different colors in solutions of differing pH values. (credit: modification of work by Sahar Atwa)

Exercises

Question 14.15

14.15

Explain why a sample of pure water at 40 °C is neutral even though [H3O+]=1.7×107M.Kwis2.9×1014at40°C.[H_{3}O^{+}] = 1.7 × 10^{−7} M. \:Kw\: is\: 2.9 × 10^{−14} at 40 °C.

Click here to see the answer to Question 14.15

Question 14.16

14.16

The ionization constant for water (Kw)(K_w) is 2.9 × 101410^{−14} at 40 °C. Calculate [H3O+],[OH],[H_{3}O^{+}], [OH^{−}], pH, and pOH for pure water at 40 °C


Question 14.17

14.17

The ionization constant for water (Kw) is 9.614×10149.614 × 10^{−14} at 60 °C. Calculate [H3O+],[OH][H_{3}O^{+}], [OH^{−}], pH, and pOH for pure water at 60 °C.

Click here to see the answer to Question 14.17

Question 14.18

14.18

Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:

(a) 0.200 M HCl (b) 0.0143 M NaOH (c) 3.0 M HNO3HNO_3 (d) 0.0031 M Ca(OH)2Ca(OH)_2


Question 14.19

17.19

Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:

(a) 0.000259 M HClO4HClO_4 (b) 0.21 M NaOH (c) 0.000071 M Ba(OH)2Ba(OH)_2 (d) 2.5 M KOH

Click here to see the answer to Question 14.19

Question 14.20

14.20

What are the pH and pOH of a solution of 2.0 M HCl, which ionizes completely?


Question 14.21

14.20

What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52?

Click here to see the answer to Question 14.21

Question 14.22

14.22

Calculate the hydrogen ion concentration and the hydroxide ion concentration in wine from its pH. See Figure for useful information.


Question 14.23

14.23

Calculate the hydronium ion concentration and the hydroxide ion concentration in lime juice from its pH. See Figure for useful information.

Click here to see the answer to Question 14.23

Question 14.24


14.24

The hydronium ion concentration in a sample of rainwater is found to be 1.7 × 10610^−6 M at 25 °C. What is the concentration of hydroxide ions in the rainwater?


Question 14.25

14.25

The hydroxide ion concentration in household ammonia is 3.2×1033.2 × 10^{−3} M at 25 °C. What is the concentration of hydronium ions in the solution?

Click here to see the answer to Question 14.25





14.3 Relative Strengths of Acids and Bases

By the end of this section, you will be able to:

  • • Assess the relative strengths of acids and bases according to their ionization constants
  • • Rationalize trends in acid–base strength in relation to molecular structure
  • • Carry out equilibrium calculations for weak acid–base systems

We can rank the strengths of acids by the extent to which they ionize in aqueous solution. The reaction of an acid with water is given by the general expression:

Water is the base that reacts with the acid HA, A is the conjugate base of the acid HA, and the hydronium ion is the conjugate acid of water. A strong acid yields 100% (or very nearly so) of H3O+ and A when the acid ionizes in water; Figure 14.6 lists several strong acids. A weak acid gives small amounts of H3O+ and A.

Figure 14.6 Some of the common strong acids and bases are listed here.

The relative strengths of acids may be determined by measuring their equlibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is called the acid-ionization constant, Ka. For the reaction of an acid HA:

we write the equation for the ionization constant as:

where the concentrations are those at equilibrium. Although water is a reactant in the reaction, it is the solvent as well, so we do not include [H2O] in the equation. The larger the Ka of an acid, the larger the concentration of H3O+ and A relative to the concentration of the nonionized acid, HA. Thus a stronger acid has a larger ionization constant than does a weaker acid. The ionization constants increase as the strengths of the acids increase. (A table of ionization constants of weak acids appears in Appendix H, with a partial listing in Table 14.2.)

The following data on acid-ionization constants indicate the order of acid strength CH3CO2H < HNO2 < HSO4:

Another measure of the strength of an acid is its percent ionization. The percent ionization of a weak acid is the ratio of the concentration of the ionized acid to the initial acid concentration, times 100:


Because the ratio includes the initial concentration, the percent ionization for a solution of a given weak acid varies depending on the original concentration of the acid, and actually decreases with increasing acid concentration.

Example 14.7

Calculation of Percent Ionization from pH

Calculate the percent ionization of a 0.125-M solution of nitrous acid (a weak acid), with a pH of 2.09.

Solution

The percent ionization for an acid is:

The chemical equation for the dissociation of the nitrous acid is:

Since 10−pH = [H3O+], we find that 10−2.09 = 8.1 × 10−3 M, so that percent ionization is:

Remember, the logarithm 2.09 indicates a hydronium ion concentration with only two significant figures.

Check Your Learning

Calculate the percent ionization of a 0.10-M solution of acetic acid with a pH of 2.89.

Answer:1.3% ionized

We can rank the strengths of bases by their tendency to form hydroxide ions in aqueous solution. The reaction of a Brønsted-Lowry base with water is given by:

Water is the acid that reacts with the base, HB+ is the conjugate acid of the base B, and the hydroxide ion is the conjugate base of water. A strong base yields 100% (or very nearly so) of OHand HB+ when it reacts with water; Figure 14.6 lists several strong bases. A weak base yields a small proportion of hydroxide ions. Soluble ionic hydroxides such as NaOH are considered strong bases because they dissociate completely when dissolved in water.

Link to Learning

View the simulation of strong and weak acids and bases at the molecular level.



As we did with acids, we can measure the relative strengths of bases by measuring their base-ionization constant (Kb) in aqueous solutions. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base has a larger ionization constant than does a weaker base. For the reaction of a base, B:

we write the equation for the ionization constant as:


where the concentrations are those at equilibrium. Again, we do not include [H2O] in the equation because water is the solvent. The chemical reactions and ionization constants of the three bases shown are:


A table of ionization constants of weak bases appears in Appendix I (with a partial list in Table 14.3). As with acids, percent ionization can be measured for basic solutions, but will vary depending on the base ionization constant and the initial concentration of the solution.

Consider the ionization reactions for a conjugate acid-base pair, HA − A−:


Adding these two chemical equations yields the equation for the autoionization for water:

As shown in the previous chapter on equilibrium, the K expression for a chemical equation derived from adding two or more other equations is the mathematical product of the other equations’ K expressions. Multiplying the mass-action expressions together and cancelling common terms, we see that:

For example, the acid ionization constant of acetic acid (CH3COOH) is 1.8 × 10−5, and the base ionization constant of its conjugate base, acetate ion (CH3COO), is 5.6 × 10−10. The product of these two constants is indeed equal to Kw:

The extent to which an acid, HA, donates protons to water molecules depends on the strength of the conjugate base, A, of the acid. If A is a strong base, any protons that are donated to water molecules are recaptured by A. Thus there is relatively little A and H3O+ in solution, and the acid, HA, is weak. If A is a weak base, water binds the protons more strongly, and the solution contains primarily A and H3O+—the acid is strong. Strong acids form very weak conjugate bases, and weak acids form stronger conjugate bases (Figure 14.7).

Figure 14.7 This diagram shows the relative strengths of conjugate acid-base pairs, as indicated by their ionization constants in aqueous solution.

Figure 14.8 lists a series of acids and bases in order of the decreasing strengths of the acids and the corresponding increasing strengths of the bases. The acid and base in a given row are conjugate to each other.

Figure 14.8 The chart shows the relative strengths of conjugate acid-base pairs.

The first six acids in Figure 14.8 are the most common strong acids. These acids are completely dissociated in aqueous solution. The conjugate bases of these acids are weaker bases than water. When one of these acids dissolves in water, their protons are completely transferred to water, the stronger base.

Those acids that lie between the hydronium ion and water in Figure 14.8 form conjugate bases that can compete with water for possession of a proton. Both hydronium ions and nonionized acid molecules are present in equilibrium in a solution of one of these acids. Compounds that are weaker acids than water (those found below water in the column of acids) in Figure 14.8 exhibit no observable acidic behavior when dissolved in water. Their conjugate bases are stronger than the hydroxide ion, and if any conjugate base were formed, it would react with water to re-form the acid.

The extent to which a base forms hydroxide ion in aqueous solution depends on the strength of the base relative to that of the hydroxide ion, as shown in the last column in Figure 14.8. A strong base, such as one of those lying below hydroxide ion, accepts protons from water to yield 100% of the conjugate acid and hydroxide ion. Those bases lying between water and hydroxide ion accept protons from water, but a mixture of the hydroxide ion and the base results. Bases that are weaker than water (those that lie above water in the column of bases) show no observable basic behavior in aqueous solution.

Example 14.8

The Product Ka × Kb = Kw

Use the Kb for the nitrite ion, NO2, to calculate the Ka for its conjugate acid.

Solution

Kb for NO2 is given in this section as 2.22 × 10−11. The conjugate acid of NO2 is HNO2; Ka for HNO2 can be calculated using the relationship:

Solving for Ka, we get:

This answer can be verified by finding the Ka for HNO2 in Appendix H.

Check Your Learning

We can determine the relative acid strengths of NH4+ and HCN by comparing their ionization constants. The ionization constant of HCN is given in Appendix H as 4 × 10−10. The ionization constant of NH4+ is not listed, but the ionization constant of its conjugate base, NH3, is listed as 1.8 × 10−5. Determine the ionization constant of NH4+, and decide which is the stronger acid, HCN or NH4+.

Answer:NH4+ is the slightly stronger acid (Ka for NH4+ = 5.6 × 10−10).


The Ionization of Weak Acids and Weak Bases

Many acids and bases are weak; that is, they do not ionize fully in aqueous solution. A solution of a weak acid in water is a mixture of the nonionized acid, hydronium ion, and the conjugate base of the acid, with the nonionized acid present in the greatest concentration. Thus, a weak acid increases the hydronium ion concentration in an aqueous solution (but not as much as the same amount of a strong acid).

Acetic acid, CH3CO2H, is a weak acid. When we add acetic acid to water, it ionizes to a small extent according to the equation:


giving an equilibrium mixture with most of the acid present in the nonionized (molecular) form. This equilibrium, like other equilibria, is dynamic; acetic acid molecules donate hydrogen ions to water molecules and form hydronium ions and acetate ions at the same rate that hydronium ions donate hydrogen ions to acetate ions to reform acetic acid molecules and water molecules. We can tell by measuring the pH of an aqueous solution of known concentration that only a fraction of the weak acid is ionized at any moment (Figure 14.9). The remaining weak acid is present in the nonionized form.

For acetic acid, at equilibrium:



Figure 14.9 pH paper indicates that a 0.l-M solution of HCl (beaker on left) has a pH of 1. The acid is fully ionized and [H3O+] = 0.1 M. A 0.1-M solution of CH3CO2H (beaker on right) is has a pH of 3 ([H3O+] = 0.001 M) because the weak acid CH3CO2H is only partially ionized. In this solution, [H3O+] < [CH3CO2H]. (credit: modification of work by Sahar Atwa)


Table 14.2

Table 14.2 gives the ionization constants for several weak acids; additional ionization constants can be found in Appendix H.

At equilibrium, a solution of a weak base in water is a mixture of the nonionized base, the conjugate acid of the weak base, and hydroxide ion with the nonionized base present in the greatest concentration. Thus, a weak base increases the hydroxide ion concentration in an aqueous solution (but not as much as the same amount of a strong base).

For example, a solution of the weak base trimethylamine, (CH3)3N, in water reacts according to the equation:


giving an equilibrium mixture with most of the base present as the nonionized amine. This equilibrium is analogous to that described for weak acids.

We can confirm by measuring the pH of an aqueous solution of a weak base of known concentration that only a fraction of the base reacts with water (Figure 14.10). The remaining weak base is present as the unreacted form. The equilibrium constant for the ionization of a weak base, Kb, is called the ionization constant of the weak base, and is equal to the reaction quotient when the reaction is at equilibrium. For trimethylamine, at equilibrium:



Figure 14.10 pH paper indicates that a 0.1-M solution of NH3 (left) is weakly basic. The solution has a pOH of 3 ([OH] = 0.001 M) because the weak base NH3 only partially reacts with water. A 0.1-M solution of NaOH (right) has a pOH of 1 because NaOH is a strong base. (credit: modification of work by Sahar Atwa)

The ionization constants of several weak bases are given in Table and in Appendix I.

Example 14.9

Determination of Ka from Equilibrium Concentrations

Acetic acid is the principal ingredient in vinegar (Figure 14.11); that's why it tastes sour. At equilibrium, a solution contains [CH3CO2H] = 0.0787 M and [H3O+]=[CH3CO2−]=0.00118M. What is the value of Ka for acetic acid?


Figure 14.1 Vinegar is a solution of acetic acid, a weak acid. (credit: modification of work by “HomeSpot HQ”/Flickr)

Solution

We are asked to calculate an equilibrium constant from equilibrium concentrations. At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:



Check Your Learning

What is the equilibrium constant for the ionization of the HSO4− ion, the weak acid used in some household cleansers:

In one mixture of NaHSO4 and Na2SO4 at equilibrium,

Answer:Ka for HSO4− = 1.2 × 10−2

Example 14.10

Determination of Kb from Equilibrium Concentrations

Caffeine, C8H10N4O2 is a weak base. What is the value of Kb for caffeine if a solution at equilibrium has [C8H10N4O2] = 0.050 M, [C8H10N4O2H+] = 5.0 × 10−3 M, and [OH−] = 2.5 × 10−3 M?

Solution

At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:



Check Your Learning

What is the equilibrium constant for the ionization of the HPO42− ion, a weak base:

In a solution containing a mixture of NaH2PO4 and Na2HPO4 at equilibrium, [OH−] = 1.3 × 10−6 M; [H2PO4]=0.042M; and [HPO42−]=0.341M.

Answer:

Example 14.11

Determination of Ka or Kb from pH

The pH of a 0.0516-M solution of nitrous acid, HNO2, is 2.34. What is its Ka?

Solution

We determine an equilibrium constant starting with the initial concentrations of HNO2, H3O+, and NO2 as well as one of the final concentrations, the concentration of hydronium ion at equilibrium. (Remember that pH is simply another way to express the concentration of hydronium ion.)

We can solve this problem with the following steps in which x is a change in concentration of a species in the reaction:

We can summarize the various concentrations and changes as shown here (the concentration of water does not appear in the expression for the equilibrium constant, so we do not need to consider its concentration):

To get the various values in the ICE (Initial, Change, Equilibrium) table, we first calculate [H3O+], the equilibrium concentration of H3O+, from the pH:


The change in concentration of H3O+, x[H3O+], is the difference between the equilibrium concentration of H3O+, which we determined from the pH, and the initial concentration, [H3O+]i. The initial concentration of H3O+ is its concentration in pure water, which is so much less than the final concentration that we approximate it as zero (~0).

The change in concentration of NO2is equal to the change in concentration of [H3O+]. For each 1 mol of H3O+ that forms, 1 mol of NO2− forms. The equilibrium concentration of HNO2 is equal to its initial concentration plus the change in its concentration.

Now we can fill in the ICE table with the concentrations at equilibrium, as shown here:

Finally, we calculate the value of the equilibrium constant using the data in the table:

Check Your Learning.

The pH of a solution of household ammonia, a 0.950-M solution of NH3, is 11.612. What is Kb for NH3.

Answer:Kb = 1.8 × 10−5

Example 14.12

Equilibrium Concentrations in a Solution of a Weak Acid

Formic acid, HCO2H, is the irritant that causes the body’s reaction to ant stings (Figure 14.12).

Figure 14.12 The pain of an ant’s sting is caused by formic acid. (credit: John Tann)

What is the concentration of hydronium ion and the pH in a 0.534-M solution of formic acid?


Solution

1. Determine x and equilibrium concentrations. The equilibrium expression is:


The concentration of water does not appear in the expression for the equilibrium constant, so we do not need to consider its change in concentration when setting up the ICE table.

The table shows initial concentrations (concentrations before the acid ionizes), changes in concentration, and equilibrium concentrations follows (the data given in the problem appear in color):

2. Solve for x and the equilibrium concentrations. At equilibrium:


Now solve for x. Because the initial concentration of acid is reasonably large and Ka is very small, we assume that x << 0.534, which permits us to simplify the denominator term as (0.534 − x) = 0.534. This gives:

Solve for x as follows:

To check the assumption that x is small compared to 0.534, we calculate:

x is less than 5% of the initial concentration; the assumption is valid.

We find the equilibrium concentration of hydronium ion in this formic acid solution from its initial concentration and the change in that concentration as indicated in the last line of the table:

The pH of the solution can be found by taking the negative log of the [H3O+], so:

Check Your Learning

Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of acetic acid in a 0.100-M solution of acetic acid, CH3CO2H?

(Hint: Determine [CH3CO2] at equilibrium.) Recall that the percent ionization is the fraction of acetic acid that is ionized × 100, or


Answer:percent ionization = 1.3%

Example 14.13

Equilibrium Concentrations in a Solution of a Weak Base

Find the concentration of hydroxide ion in a 0.25-M solution of trimethylamine, a weak base:

Solution

This problem requires that we calculate an equilibrium concentration by determining concentration changes as the ionization of a base goes to equilibrium. The solution is approached in the same way as that for the ionization of formic acid in Div. The reactants and products will be different and the numbers will be different, but the logic will be the same:

1. Determine x and equilibrium concentrations. The table shows the changes and concentrations:

2. Solve for x and the equilibrium concentrations. At equilibrium:


If we assume that x is small relative to 0.25, then we can replace (0.25 − x) in the preceding equation with 0.25. Solving the simplified equation gives:


This change is less than 5% of the initial concentration (0.25), so the assumption is justified.

Recall that, for this computation, x is equal to the equilibrium concentration of hydroxide ion in the solution (see earlier tabulation):

Then calculate pOH as follows:

Using the relation introduced in the previous section of this chapter: 

permits the computation of pH:

3. Check the work A check of our arithmetic shows that Kb = 7.4 × 10−5.

Check Your Learning

(a) Show that the calculation in Step 2 of this example gives an x of 4.3 × 10−3 and the calculation in Step 3 shows Kb = 7.4 × 10−5.

(b) Find the concentration of hydroxide ion in a 0.0325-M solution of ammonia, a weak base with a Kb of 1.76 × 10−5. Calculate the percent ionization of ammonia, the fraction ionized × 100, or 

Answer:7.56 × 10−4 M, 2.33%

Some weak acids and weak bases ionize to such an extent that the simplifying assumption that x is small relative to the initial concentration of the acid or base is inappropriate. As we solve for the equilibrium concentrations in such cases, we will see that we cannot neglect the change in the initial concentration of the acid or base, and we must solve the equilibrium equations by using the quadratic equation.

Example 14.14

Equilibrium Concentrations in a Solution of a Weak Acid

Sodium bisulfate, NaHSO4, is used in some household cleansers because it contains the HSO4 ion, a weak acid. What is the pH of a 0.50-M solution of HSO4?

Solution

We need to determine the equilibrium concentration of the hydronium ion that results from the ionization of HSO4 so that we can use [H3O+] to determine the pH. As in the previous examples, we can approach the solution by the following steps:


1. Determine x and equilibrium concentrations. This table shows the changes and concentrations:

2. Solve for x and the concentrations. As we begin solving for x, we will find this is more complicated than in previous examples. As we discuss these complications we should not lose track of the fact that it is still the purpose of this step to determine the value of x.

At equilibrium:

If we assume that x is small and approximate (0.50 − x) as 0.50, we find:

When we check the assumption, we calculate:


The value of x is not less than 5% of 0.50, so the assumption is not valid. We need the quadratic formula to find x.

The equation:

gives

or

This equation can be solved using the quadratic formula. For an equation of the form

x is given by the equation:

In this problem, a = 1, b = 1.2 × 10−3, and c = −6.0 × 10−3.

Solving for x gives a negative root (which cannot be correct since concentration cannot be negative) and a positive root:

Now determine the hydronium ion concentration and the pH:

The pH of this solution is:

Check Your Learning

(a) Show that the quadratic formula gives x = 7.2 × 10−2.

(b) Calculate the pH in a 0.010-M solution of caffeine, a weak base:


(Hint: It will be necessary to convert [OH] to [H3O+] or pOH to pH toward the end of the calculation.)

Answer:pH 11.16

The Relative Strengths of Strong Acids and Bases

Strong acids, such as HCl, HBr, and HI, all exhibit the same strength in water. The water molecule is such a strong base compared to the conjugate bases Cl, Br, and I that ionization of these strong acids is essentially complete in aqueous solutions. In solvents less basic than water, we find HCl, HBr, and HI differ markedly in their tendency to give up a proton to the solvent. For example, when dissolved in ethanol (a weaker base than water), the extent of ionization increases in the order HCl < HBr < HI, and so HI is demonstrated to be the strongest of these acids. The inability to discern differences in strength among strong acids dissolved in water is known as the leveling effect of water.

Water also exerts a leveling effect on the strengths of strong bases. For example, the oxide ion, O2−, and the amide ion, NH2, are such strong bases that they react completely with water:

Thus, O2− and NH2 appear to have the same base strength in water; they both give a 100% yield of hydroxide ion.

Effect of Molecular Structure on Acid-Base Strength

In the absence of any leveling effect, the acid strength of binary compounds of hydrogen with nonmetals (A) increases as the H-A bond strength decreases down a group in the periodic table. For group 7A, the order of increasing acidity is HF < HCl < HBr < HI. Likewise, for group 6A, the order of increasing acid strength is H2O < H2S < H2Se < H2Te.

Across a row in the periodic table, the acid strength of binary hydrogen compounds increases with increasing electronegativity of the nonmetal atom because the polarity of the H-A bond increases. Thus, the order of increasing acidity (for removal of one proton) across the second row is CH4 < NH3 < H2O < HF; across the third row, it is SiH4 < PH3 < H2S < HCl (see Figure 14.13).

Figure 14.13 As you move from left to right and down the periodic table, the acid strength increases. As you move from right to left and up, the base strength increases.

Compounds containing oxygen and one or more hydroxyl (OH) groups can be acidic, basic, or amphoteric, depending on the position in the periodic table of the central atom E, the atom bonded to the hydroxyl group. Such compounds have the general formula OnE(OH)m, and include sulfuric acid, O2S(OH)2, sulfurous acid, OS(OH)2, nitric acid, O2NOH, perchloric acid, O3ClOH, aluminum hydroxide, Al(OH)3, calcium hydroxide, Ca(OH)2, and potassium hydroxide, KOH:

If the central atom, E, has a low electronegativity, its attraction for electrons is low. Little tendency exists for the central atom to form a strong covalent bond with the oxygen atom, and bond a between the element and oxygen is more readily broken than bond b between oxygen and hydrogen. Hence bond a is ionic, hydroxide ions are released to the solution, and the material behaves as a base—this is the case with Ca(OH)2 and KOH. Lower electronegativity is characteristic of the more metallic elements; hence, the metallic elements form ionic hydroxides that are by definition basic compounds.

If, on the other hand, the atom E has a relatively high electronegativity, it strongly attracts the electrons it shares with the oxygen atom, making bond a relatively strongly covalent. The oxygen-hydrogen bond, bond b, is thereby weakened because electrons are displaced toward E. Bond b is polar and readily releases hydrogen ions to the solution, so the material behaves as an acid. High electronegativities are characteristic of the more nonmetallic elements. Thus, nonmetallic elements form covalent compounds containing acidic −OH groups that are called oxyacids.

Increasing the oxidation number of the central atom E also increases the acidity of an oxyacid because this increases the attraction of E for the electrons it shares with oxygen and thereby weakens the O-H bond. Sulfuric acid, H2SO4, or O2S(OH)2 (with a sulfur oxidation number of +6), is more acidic than sulfurous acid, H2SO3, or OS(OH)2 (with a sulfur oxidation number of +4). Likewise nitric acid, HNO3, or O2NOH (N oxidation number = +5), is more acidic than nitrous acid, HNO2, or ONOH (N oxidation number = +3). In each of these pairs, the oxidation number of the central atom is larger for the stronger acid (Figure 14.14).

Figure 14.14 As the oxidation number of the central atom E increases, the acidity also increases.

Hydroxy compounds of elements with intermediate electronegativities and relatively high oxidation numbers (for example, elements near the diagonal line separating the metals from the nonmetals in the periodic table) are usually amphoteric. This means that the hydroxy compounds act as acids when they react with strong bases and as bases when they react with strong acids. The amphoterism of aluminum hydroxide, which commonly exists as the hydrate Al(H2O)3(OH)3, is reflected in its solubility in both strong acids and strong bases. In strong bases, the relatively insoluble hydrated aluminum hydroxide, Al(H2O)3(OH)3, is converted into the soluble ion, [Al(H2O)2(OH)4]−, by reaction with hydroxide ion:

In this reaction, a proton is transferred from one of the aluminum-bound H2O molecules to a hydroxide ion in solution. The Al(H2O)3(OH)3 compound thus acts as an acid under these conditions. On the other hand, when dissolved in strong acids, it is converted to the soluble ion [Al(H2O)6]3+ by reaction with hydronium ion:


In this case, protons are transferred from hydronium ions in solution to Al(H2O)3(OH)3, and the compound functions as a base.

Exercises

Question 14.26

14.26

Explain why the neutralization reaction of a strong acid and a weak base gives a weakly acidic solution.


Question 14.27

14.27

Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basic solution.

Click here to see the answer to Question 14.27

Question 14.28


14.28

Use this list of important industrial compounds (and Figure) to answer the following questions regarding: CaO,Ca(OH)2,CH3CO2H,CO2,HCl,H2CO3,HF,HNO2,HNO3,H3PO4,H2SO4,NH3,NaOH,Na2CO3.CaO, Ca(OH)_{2}, CH_{3}CO_{2}H, CO_{2}, HCl, H_{2}CO_{3}, HF, HNO_{2}, HNO_{3}, H_{3}PO_{4}, H_{2}SO_{4}, NH_{3}, NaOH, Na_{2}CO_{3}.

(a) Identify the strong Brønsted-Lowry acids and strong Brønsted-Lowry bases.

(b) List those compounds in (a) that can behave as Brønsted-Lowry acids with strengths lying between those of H3O+andH2O.H_{3}O^{+} and H_{2}O.

(c) List those compounds in (a) that can behave as Brønsted-Lowry bases with strengths lying between those of H2OandOHH_{2}O and OH^{−}.


Question 14.29


14.29

The odor of vinegar is due to the presence of acetic acid, CH3CO2HCH_{3}CO_{2}H, a weak acid. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this acid.

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Question 14.30


14.30

Household ammonia is a solution of the weak base NH3NH_3 in water. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this base.


 Question 14.31


14.31

Explain why the ionization constant, KaK_a, for H2SO4H_{2}SO_{4} is larger than the ionization constant for H2SO3H_{2}SO_{3}.

Click here to see the answer to Question 14.31

Question 14.32


14.32

Explain why the ionization constant, KaK_a, for HI is larger than the ionization constant for HF.


Question 14.33

14.33

Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, HCl. Milk of Magnesia, a suspension of solid Mg(OH)2Mg(OH)_2 in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction, and identify the conjugate acid-base pairs.


Question 14.34


14.34

Nitric acid reacts with insoluble copper(II) oxide to form soluble copper(II) nitrate, Cu(NO3)2Cu(NO_{3})_2, a compound that has been used to prevent the growth of algae in swimming pools. Write the balanced chemical equation for the reaction of an aqueous solution of HNO3HNO_{3} with CuO.


Question 14.35

14.35

What is the ionization constant at 25 °C for the weak acid CH3NH3+,CH_{3}NH_{3}\:^{+}, the conjugate acid of the weak base CH3NH2,Kb=4.4×104.CH_{3}NH_{2}, Kb = 4.4 × 10^{−4}.

Click here to see the answer to Question 14.35

Question 14.36

14.36

What is the ionization constant at 25 °C for the weak acid (CH3)2NH2+(CH_{3})_{2}NH_{2}\:^{+}, the conjugate acid of the weak base (CH3)2NH,(CH_{3})2NH, Kb = 7.4 × 104?10^{−4}?


Question 14.37

14.37

Which base, CH3NH2CH_{3}NH_{2} or (CH3)2NH(CH_{3})_{2}NH, is the strongest base? Which conjugate acid, (CH3)2NH2+(CH_{3})_{2}NH_{2}\:^{+} or (CH3)2NH(CH_{3})_{2}NH, is the strongest acid?

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Question 14.38

14.38

Which is the stronger acid, NH4+NH_4\:^{+} or HBrO?


Question 14.39

14.39

Which is the stronger base, (CH3)3N(CH_{3})_{3}N or H2BO3?H_{2}BO_{3}\:^{−}?

Click here to see the answer to Question 14.39

Question 14.40

14.40

Predict which acid in each of the following pairs is the stronger and explain your reasoning for each.

(a) H2OorHFH_{2}O or HF

(b) B(OH)3orAl(OH)3B(OH)_{3} or Al(OH)_{3}

(c) HSO3orHSO4HSO_{3}− or HSO_{4}−

(d) NH3orH2SNH_{3} or H_{2}S

(e)H2OorH2Te H_{2}O or H_{2}Te



Question 14.41

14.41

Predict which compound in each of the following pairs of compounds is more acidic and explain your reasoning for each.

(a)HSO4orHSeO4 HSO_{4}\:^{−} or HSeO_{4}\:^{−}

(b) NH3orH2ONH_{3} or H_{2}O

(c) PH3PH_{3} or HI

(d) NH3 NH_{3} or PH3 PH_{3}

(e)H2S H_{2}S or HBr

Click here to see the answer to Question 14.41

Question 14.42

14.42

Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.

(a) acidity: HCl,HBr,HIHCl, HBr, HI

(b) basicity: H2O,OH,H,ClH_{2}O, OH^{−}, H^{−}, Cl^{−}

(c) basicity: Mg(OH)2,Si(OH)4,ClO3(OH)Mg(OH)_{2}, Si(OH)_{4}, ClO_{3}(OH) (Hint: Formula could also be written as HClO4) HClO_{4})

(d) acidity: HF,H2O,NH3,CH4HF, H_{2}O, NH_{3}, CH_{4}


Question 14.43

14.43

Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.

(a) acidity: NaHSO3,NaHSeO3,NaHSO4NaHSO_{3}, NaHSeO_{3}, NaHSO_{4}

(b) basicity: BrO2,ClO2,IO2BrO_{2}\:^{−}, ClO_{2}\:^{−}, IO_{2}\:^{−}

(c) acidity: HOCl, HOBr, HOI

(d) acidity: HOCl,HOClO,HOClO2,HOClO3HOCl, HOClO, HOClO_{2}, HOClO_{3}

(e) basicity: NH2,HS,HTe,PH2NH_{2}^{−}, HS^{−}, HTe^{−}, PH_{2}^{−}

(f) basicity: BrO,BrO2,BrO3,BrO4BrO^{−}, BrO_{2}\:^{−}, BrO_{3}\:^{−}, BrO_{4}\:^{−}

Click here to see the answer to Question 14.43

Question 14.44

14.44

Both HF and HCN ionize in water to a limited extent. Which of the conjugate bases, FF^{−} or CNCN^{−}, is the stronger base? Table 14.3.


Question 14.45

14.45

The active ingredient formed by aspirin in the body is salicylic acid, C6H4OH(CO2H).C_{6}H_{4}OH(CO_{2}H). The carboxyl group (CO2H)(−CO_{2}H) acts as a weak acid. The phenol group (an OH group bonded to an aromatic ring) also acts as an acid but a much weaker acid. List, in order of descending concentration, all of the ionic and molecular species present in a 0.001-M aqueous solution of C6H4OH(CO2H).C_{6}H_{4}OH(CO_{2}H).

Click here to see the answer to Question 14.45

 Question 14.46

14.46

What do we represent when we write:

CH3CO2H(aq)+H2O(l)H3O+(aq)+CH3CO2(aq)?CH_{3}CO_{2}H(aq)+H_{2}O(l)⇌H_{3}O\:^{+}(aq)+CH_{3}CO_{2}\:^{−}(aq)?


Question 14.47

14.47

Explain why equilibrium calculations are not necessary to determine ionic concentrations in solutions of certain strong electrolytes such as NaOH and HCl. Under what conditions are equilibrium calculations necessary as part of the determination of the concentrations of all ions of some other strong electrolytes in solution?

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Question 14.48

14.48

Are the concentrations of hydronium ion and hydroxide ion in a solution of an acid or a base in water directly proportional or inversely proportional? Explain your answer.


Question 14.49

14.49

What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak acid?

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Question 14.50

1450

What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak base?


Question 14.51

14.51

Which of the following will increase the percent of NH3NH_{3} that is converted to the ammonium ion in water (Hint: Use LeChâtelier’s principle.)?

(a) addition of NaOH

(b) addition of HCl

(c) addition of NH4ClNH_{4}Cl

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Question 14.52

14.52

Which of the following will increase the percent of HF that is converted to the fluoride ion in water?

(a) addition of NaOH

(b) addition of HCl

(c) addition of NaF


Question 14.53

14.53

What is the effect on the concentrations of NO2NO_2\:^{−}, HNO2HNO_2, and OHOH^− when the following are added to a solution of KNO2KNO^{2} in water:

(a) HCl

(b) HNO2HNO^2

(c) NaOH

(d) NaCl

(e) KNO The equation for the equilibrium is:

NO2(aq)+H2O(l)HNO2(aq)+OH(aq)NO_{2}\:^{−}(aq)+H_{2}O(l)⇌HNO_{2}(aq)+OH^{−}(aq)

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Question 14.54

14.54

What is the effect on the concentration of hydrofluoric acid, hydronium ion, and fluoride ion when the following are added to separate solutions of hydrofluoric acid?

(a) HCl

(b) KF

(c) NaCl

(d) KOH

(e) HF The equation for the equilibrium is:

HF(aq)+H2O(l)H3O+(aq)+F(aq)HF(aq)+H_{2}O(l)⇌H_{3}O+(aq)+F^{−}(aq)



Question 14.55

14.55

Why is the hydronium ion concentration in a solution that is 0.10 M in HCl and 0.10 M in HCOOH determined by the concentration of HCl?

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Question 14.56

14.56

From the equilibrium concentrations given, calculate Ka for each of the weak acids and Kb for each of the weak bases.

(a) CH3CO2H:[H3O+]=1.34×103M;CH_{3}CO_{2}H: [H_{3}O^{+}] = 1.34 × 10^{−3} M;

[CH3CO2]=1.34×103M;[CH_{3}CO_{2}\:^{−}] = 1.34 × 10^{−3} M; [CH3CO2H]=9.866×102M;[CH_{3}CO_{2}H] = 9.866 × 10^{−2} M; (b)ClO:[OH]=4.0×104M; ClO^{−}: [OH^{−}] = 4.0 × 10^{−4} M; [HClO]=2.38×105M;[HClO] = 2.38 × 10^{−5} M; [ClO][ClO^{−}] = 0.273 M; (c) HCO2H:[HCO2H]=0.524M;HCO_{2}H: [HCO_{2}H] = 0.524 M;

[H3O+]=9.8×103M;[H_{3}O\:^{+}] = 9.8 × 10^{−3} M;

[HCO2]=9.8×103M;[HCO_{2}]\:^{−} = 9.8 × 10^{−3} M;

(d)C6H5NH3+:[C6H5NH3+]=0.233M;(d) C_{6}H_{5}NH_{3}\:^{+}: [C_{6}H_{5}NH_{3}\:^{+}] = 0.233 M;

[C6H5NH2]=2.3×103M;[C_{6}H_{5}NH_{2}] = 2.3 × 10^{−3} M;

[H3O+]=2.3×103M[H_{3}O\:^{+}] = 2.3 × 10^{−3} M


Question 14.57

14.57

From the equilibrium concentrations given, calculate Ka for each of the weak acids and Kb for each of the weak bases. (a)NH3:[OH]=3.1×103 NH_3: [OH^−] = 3.1 × 10^{−3} M;

[math][NH_{4}\:^{+}] = 3.1 × 10^{−3} M; [NH3] = 0.533 M; [/math]

(b) HNO2:[H3O+]=0.011M;HNO_{2}: [H_{3}O^{+}] = 0.011 M;

[math][NO_{2}\:^{−}] = 0.0438 M; [HNO_{2}] = 1.07 M; [/math]

(c) (CH3)3N:[(CH3)3N]=0.25M;(CH_{3})_{3}N: [(CH_{3})_{3}N] = 0.25 M;

[math][(CH_{3})_{3}NH^{+}] = 4.3 × 10^{−3} M; [OH^{−}] = 4.3 × 10^{−3} M; [/math]

[math](d) NH_{4}\:{+}: [NH_{4}\:^{+}] = 0.100 M; [NH_{3}] = 7.5 × 10^{−6} M; [/math]

[H3O+]=7.5×106M[H_{3}O^{+}] = 7.5 × 10^{−6} M

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Question 14.58

14.57

Determine KbK_b for the nitrite ion, NO2NO_{2}\:^{−} . In a 0.10-M solution this base is 0.0015% ionized.


Question 14.59

14.59

Determine Ka for hydrogen sulfate ion, [math]HSO_{4}^\:{−}[/math]. In a 0.10-M solution the acid is 29% ionized.

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Question 14.60

14.60

Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:

(a) FF^{−}

(b) NH4+NH_{4}\:^+

(c) AsO43AsO^{4}3−

(d) (CH3)2NH2+ (CH_{3})2NH_{2}\:^{+}

(e) NO2 NO_{2}\:{−}

(f) HC2O4HC_{2}O_{4}\:^{−} (as a base)



 Question 14.61

14.61

Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:

(a) HTeHTe^{−} (as a base)

(b) (CH3)3NH+(CH_{3})_3NH^{+}

(c) HAsO43HAsO_{4}\:^{3−} (as a base)

(d) HO2HO_{2}\:^{−} as a base)

(e) C6H5NH3+C_{6}H_{5}NH_{3}\:^{+}

(f) HSO3(asabase)HSO_{3}\:^{−} (as a base)

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Question 14.62

14.62

For which of the following solutions must we consider the ionization of water when calculating the pH or pOH?

(a) 3×1083 × 10^{−8} M HNO3

(b) 0.10 g HCl in 1.0 L of solution

(c) 0.00080 g NaOH in 0.50 L of solution

(d) 1×107MCa(OH)21 × 10^{−7} \:M \:Ca(OH)_{2}

(e) 0.0245 M KNO3KNO_{3}


Question 14.63

14.63

Even though both NH3NH_{3} and C6H5NH2C_{6}H_{5}NH_{2} are weak bases, NH3NH_3 is a much stronger acid than C6H5NH2C_{6}H_{5}NH_{2} . Which of the following is correct at equilibrium for a solution that is initially 0.10 M in NH3NH_3 and 0.10 M in C6H5NH2C_{6}H_{5}NH_{2} ?

A

[OH]=[NH4+][OH^{−}]=[NH_{4}\:^{+}]

B

[NH4+]=C6H5NH2[NH_{4}\:^{+}]=C_{6}H_{5}NH_{2}

C

[OH]=[C6H5NH2][OH^{−}]=[C_{6}H_{5}NH_{2}]

D

[NH3]=[C6H5NH2][NH_{3}] = [C_{6}H_{5}NH_{2}]

E

both a and b are correct


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 Question 14.64

16.64

Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.25 M in HCO2HHCO_{2}H and 0.10 M in HClO.


Question 14.65

14.65

Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.134 M in HNO2HNO_{2} and 0.120 M in HBrO.

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Question 14.66

14.66

Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.25 M in CH3NH2CH_{3}NH_{2} and 0.10 M in C5H5NC_{5}H_{5}N (Kb = 1.7 × 109).10^{−9}).


Question 14.67

14.67

Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.115 M in NH3NH^{3} and 0.100 M in C6H5NH2.C_{6}H_{5}NH_{2}.

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Question 14.68

14.68

Using the KaK_a values in Appendix H, place Al(H2O)63+Al(H_{2} O)_{6}\:^{3+} in the correct location in Figure 14.8


Question 14.69

14.69

Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected. Ionization constants can be found in Appendix H and Appendix I.

(a) 0.0092 M HClO, a weak acid

(b) 0.0784 M C6H5NH2C_{6}H_{5}NH_{2}, a weak base

(c) 0.0810 M HCN, a weak acid

(d) 0.11 M (CH3)3N,(CH_{3})_{3}N, a weak base

(e) 0.120 M Fe (H2O)62+(H_{2} O)_{6}\:^{2+} a weak acid, KaK_a = 1.6 × 10710^{−7}


Question 14.70

14.70

Propionic acid, C2H5CO2HC_{2}H_{5}CO_{2}H (Ka = 1.34 × 105),10^{−5}), is used in the manufacture of calcium propionate, a food preservative. What is the hydronium ion concentration in a 0.698-M solution of C2H5CO2H?C_{2}H_{5}CO_{2}H?


Question 14.71


14.71

White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white vinegar is 1.007 g/cm3 ,what is the pH?

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Question 14.72

14.72

The ionization constant of lactic acid, CH3CH(OH)CO2H, an acid found in the blood after strenuous exercise, is 1.36 × 10−4. If 20.0 g of lactic acid is used to make a solution with a volume of 1.00 L, what is the concentration of hydronium ion in the solution?


Question 14.73


14.73

Nicotine, C10H14N2C_{10}H_{14}N_{2}, is a base that will accept two protons (K1K_1 = 7 × 10−7 , K2K_2 = 1.4 × 1011)10^{−11}). What is the concentration of each species present in a 0.050-M solution of nicotine?

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Question 14.74


14.74

The pH of a 0.20-M solution of HF is 1.92. Determine Ka for HF from these data.


Question 14.75

14.75

The pH of a 0.15-M solution of HSO4HSO_{4}\:^{−} is 1.43. Determine KaK_a for HSO4HSO_{4}\:^{−} from these data.

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Question 14.76

14.76

The pH of a 0.10-M solution of caffeine is 11.16. Determine Kb for caffeine from these data:

C8H10N4O2(aq)+H2O(l)C8H10N4O2H+(aq)+OH(aq)C_{8} H_{10} N_{4} O_{2} (aq) + H_{2} O(l) ⇌ C_{8} H_{10} N_{4} O_{2}H^{+}(aq) + OH^{−}(aq)


Question 14.77

14.77

The pH of a solution of household ammonia, a 0.950 M solution of NH3NH_3 , is 11.612. Determine KbK_{b} for NH3NH_3 from these data.

Click here to see the answer to Question 14.77




14.4 Hydrolysis of Salt Solutions

By the end of this section, you will be able to:

  • • Predict whether a salt solution will be acidic, basic, or neutral
  • • Calculate the concentrations of the various species in a salt solution
  • • Describe the process that causes solutions of certain metal ions to be acidic

As we have seen in the section on chemical reactions, when an acid and base are mixed, they undergo a neutralization reaction. The word “neutralization” seems to imply that a stoichiometrically equivalent solution of an acid and a base would be neutral. This is sometimes true, but the salts that are formed in these reactions may have acidic or basic properties of their own, as we shall now see.

Acid-Base Neutralization

A solution is neutral when it contains equal concentrations of hydronium and hydroxide ions. When we mix solutions of an acid and a base, an acid-base neutralization reaction occurs. However, even if we mix stoichiometrically equivalent quantities, we may find that the resulting solution is not neutral. It could contain either an excess of hydronium ions or an excess of hydroxide ions because the nature of the salt formed determines whether the solution is acidic, neutral, or basic. The following four situations illustrate how solutions with various pH values can arise following a neutralization reaction using stoichiometrically equivalent quantities:

1. A strong acid and a strong base, such as HCl(aq) and NaOH(aq) will react to form a neutral solution since the conjugate partners produced are of negligible strength (see Figure 14.8):

2. A strong acid and a weak base yield a weakly acidic solution, not because of the strong acid involved, but because of the conjugate acid of the weak base.

3. A weak acid and a strong base yield a weakly basic solution. A solution of a weak acid reacts with a solution of a strong base to form the conjugate base of the weak acid and the conjugate acid of the strong base. The conjugate acid of the strong base is a weaker acid than water and has no effect on the acidity of the resulting solution. However, the conjugate base of the weak acid is a weak base and ionizes slightly in water. This increases the amount of hydroxide ion in the solution produced in the reaction and renders it slightly basic.

4. A weak acid plus a weak base can yield either an acidic, basic, or neutral solution. This is the most complex of the four types of reactions. When the conjugate acid and the conjugate base are of unequal strengths, the solution can be either acidic or basic, depending on the relative strengths of the two conjugates. Occasionally the weak acid and the weak base will have the same strength, so their respective conjugate base and acid will have the same strength, and the solution will be neutral. To predict whether a particular combination will be acidic, basic or neutral, tabulated K values of the conjugates must be compared.

Chemistry in Everyday Life

Stomach Antacids

Our stomachs contain a solution of roughly 0.03 M HCl, which helps us digest the food we eat. The burning sensation associated with heartburn is a result of the acid of the stomach leaking through the muscular valve at the top of the stomach into the lower reaches of the esophagus. The lining of the esophagus is not protected from the corrosive effects of stomach acid the way the lining of the stomach is, and the results can be very painful. When we have heartburn, it feels better if we reduce the excess acid in the esophagus by taking an antacid. As you may have guessed, antacids are bases. One of the most common antacids is calcium carbonate, CaCO3. The reaction,

not only neutralizes stomach acid, it also produces CO2(g), which may result in a satisfying belch.

Milk of Magnesia is a suspension of the sparingly soluble base magnesium hydroxide, Mg(OH)2. It works according to the reaction:

The hydroxide ions generated in this equilibrium then go on to react with the hydronium ions from the stomach acid, so that :

his reaction does not produce carbon dioxide, but magnesium-containing antacids can have a laxative effect.

Several antacids have aluminum hydroxide, Al(OH)3, as an active ingredient. The aluminum hydroxide tends to cause constipation, and some antacids use aluminum hydroxide in concert with magnesium hydroxide to balance the side effects of the two substances.

Chemistry in Everyday Life

Culinary Aspects of Chemistry

Cooking is essentially synthetic chemistry that happens to be safe to eat. There are a number of examples of acid-base chemistry in the culinary world. One example is the use of baking soda, or sodium bicarbonate in baking. NaHCO3 is a base. When it reacts with an acid such as lemon juice, buttermilk, or sour cream in a batter, bubbles of carbon dioxide gas are formed from decomposition of the resulting carbonic acid, and the batter “rises.” Baking powder is a combination of sodium bicarbonate, and one or more acid salts that react when the two chemicals come in contact with water in the batter.

Many people like to put lemon juice or vinegar, both of which are acids, on cooked fish (Figure 14.15). It turns out that fish have volatile amines (bases) in their systems, which are neutralized by the acids to yield involatile ammonium salts. This reduces the odor of the fish, and also adds a “sour” taste that we seem to enjoy.


Figure 14.15 A neutralization reaction takes place between citric acid in lemons or acetic acid in vinegar, and the bases in the flesh of fish.

Pickling is a method used to preserve vegetables using a naturally produced acidic environment. The vegetable, such as a cucumber, is placed in a sealed jar submerged in a brine solution. The brine solution favors the growth of beneficial bacteria and suppresses the growth of harmful bacteria. The beneficial bacteria feed on starches in the cucumber and produce lactic acid as a waste product in a process called fermentation. The lactic acid eventually increases the acidity of the brine to a level that kills any harmful bacteria, which require a basic environment. Without the harmful bacteria consuming the cucumbers they are able to last much longer than if they were unprotected. A byproduct of the pickling process changes the flavor of the vegetables with the acid making them taste sour.

Salts of Weak Bases and Strong Acids

When we neutralize a weak base with a strong acid, the product is a salt containing the conjugate acid of the weak base. This conjugate acid is a weak acid. For example, ammonium chloride, NH4Cl, is a salt formed by the reaction of the weak base ammonia with the strong acid HCl:

A solution of this salt contains ammonium ions and chloride ions. The chloride ion has no effect on the acidity of the solution since HCl is a strong acid. Chloride is a very weak base and will not accept a proton to a measurable extent. However, the ammonium ion, the conjugate acid of ammonia, reacts with water and increases the hydronium ion concentration:

The equilibrium equation for this reaction is simply the ionization constant. Ka, for the acid NH4+:

We will not find a value of Ka for the ammonium ion in Appendix H. However, it is not difficult to determine Ka for NH4+ from the value of the ionization constant of water, Kw, and Kb, the ionization constant of its conjugate base, NH3, using the following relationship:

This relation holds for any base and its conjugate acid or for any acid and its conjugate base.

Example 14.15

The pH of a Solution of a Salt of a Weak Base and a Strong Acid

Aniline is an amine that is used to manufacture dyes. It is isolated as aniline hydrochloride, [C6H5NH3+]Cl, a salt prepared by the reaction of the weak base aniline and hydrochloric acid. What is the pH of a 0.233 M solution of aniline hydrochloride?


Solution

The new step in this example is to determine Ka for the C6H5NH3+ ion. The C6H5NH3+ ion is the conjugate acid of a weak base. The value of Ka for this acid is not listed in Appendix H, but we can determine it from the value of Kb for aniline, C6H5NH2, which is given as 4.6 × 10−10 ([link] and Appendix I):

Now we have the ionization constant and the initial concentration of the weak acid, the information necessary to determine the equilibrium concentration of H3O+, and the pH:

With these steps we find [H3O+] = 2.3 × 10−3 M and pH = 2.64

Check Your Learning

(a) Do the calculations and show that the hydronium ion concentration for a 0.233-M solution of C6H5NH3+ is 2.3 × 10−3 and the pH is 2.64.

(b) What is the hydronium ion concentration in a 0.100-M solution of ammonium nitrate, NH4NO3, a salt composed of the ions NH4+ and NO3. Use the data in Table 14.3  to determine Kb for the ammonium ion. Which is the stronger acid C6H5NH3+ or NH4+?

Answer:(a) Ka(forNH4+)=5.6×10−10, [H3O+] = 7.5 × 10−6 M; (b) C6H5NH3+ is the stronger acid.

Salts of Weak Acids and Strong Bases

When we neutralize a weak acid with a strong base, we get a salt that contains the conjugate base of the weak acid. This conjugate base is usually a weak base. For example, sodium acetate, NaCH3CO2, is a salt formed by the reaction of the weak acid acetic acid with the strong base sodium hydroxide:


A solution of this salt contains sodium ions and acetate ions. The sodium ion, as the conjugate acid of a strong base, has no effect on the acidity of the solution. However, the acetate ion, the conjugate base of acetic acid, reacts with water and increases the concentration of hydroxide ion:

The equilibrium equation for this reaction is the ionization constant, Kb, for the base CH3CO2. The value of Kb can be calculated from the value of the ionization constant of water, Kw, and Ka, the ionization constant of the conjugate acid of the anion using the equation:

For the acetate ion and its conjugate acid we have:


Some handbooks do not report values of Kb. They only report ionization constants for acids. If we want to determine a Kb value using one of these handbooks, we must look up the value of Ka for the conjugate acid and convert it to a Kb value.

Example 14.16

Equilibrium in a Solution of a Salt of a Weak Acid and a Strong Base

Determine the acetic acid concentration in a solution with [CH3CO2]=0.050M and [OH−] = 2.5 × 10−6 M at equilibrium. The reaction is:


Solution

We are given two of three equilibrium concentrations and asked to find the missing concentration. If we can find the equilibrium constant for the reaction, the process is straightforward.

The acetate ion behaves as a base in this reaction; hydroxide ions are a product. We determine Kb as follows:


Now find the missing concentration:


Solving this equation we get [CH3CO2H] = 1.1 × 10−5 M.

Check Your Learning

What is the pH of a 0.083-M solution of CN−? Use 4.0 × 10−10 as Ka for HCN. Hint: We will probably need to convert pOH to pH or find [H3O+] using [OH] in the final stages of this problem.

Answer:11.16

Equilibrium in a Solution of a Salt of a Weak Acid and a Weak Base

In a solution of a salt formed by the reaction of a weak acid and a weak base, to predict the pH, we must know both the Ka of the weak acid and the Kb of the weak base. If Ka > Kb, the solution is acidic, and if Kb > Ka, the solution is basic.

Example 14.17

Determining the Acidic or Basic Nature of Salts

Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:

(a) KBr

(b) NaHCO3

(c) NH4Cl

(d) Na2HPO4

(e) NH4F

Solution

Consider each of the ions separately in terms of its effect on the pH of the solution, as shown here:

(a) The K+ cation and the Branion are both spectators, since they are the cation of a strong base (KOH) and the anion of a strong acid (HBr), respectively. The solution is neutral.

(b) The Na+ cation is a spectator, and will not affect the pH of the solution; while the HCO3anion is amphiprotic, it could either behave as an acid or a base. The Ka of HCO3 is 4.7 × 10−11, so the Kb of its conjugate base is 

Since Kb >> Ka, the solution is basic.

(c) The NH4+ ion is acidic and the Cl− ion is a spectator. The solution will be acidic.

(d) The Na+ ion is a spectator, while the HPO4 ion is amphiprotic, with a Ka of 3.6 × 10−13

so that the Kb of its conjugate base is

 Because Kb >> Ka, the solution is basic.

(e) The NH4+ ion is listed as being acidic, and the F− ion is listed as a base, so we must directly compare the Ka and the Kb of the two ions. Ka of NH4+ is 5.6 × 10−10, which seems very small, yet the Kb of F is 1.4 × 10−11, so the solution is acidic, since Ka > Kb.

Check Your Learning

Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:

(a) K2CO3

(b) CaCl2

(c) KH2PO4

(d) (NH4)2CO3

(e) AlBr3

Answer:(a) basic; (b) neutral; (c) basic; (d) basic; (e) acidic

The Ionization of Hydrated Metal Ions

If we measure the pH of the solutions of a variety of metal ions we will find that these ions act as weak acids when in solution. The aluminum ion is an example. When aluminum nitrate dissolves in water, the aluminum ion reacts with water to give a hydrated aluminum ion, Al(H2O)63+, dissolved in bulk water. What this means is that the aluminum ion has the strongest interactions with the six closest water molecules (the so-called first solvation shell), even though it does interact with the other water molecules surrounding this Al(H2O)63+ cluster as well:

We frequently see the formula of this ion simply as “Al3+(aq)”, without explicitly noting the six water molecules that are the closest ones to the aluminum ion and just describing the ion as being solvated in water (hydrated). This is similar to the simplification of the formula of the hydronium ion, H3O+ to H+. However, in this case, the hydrated aluminum ion is a weak acid (Figure 14.16) and donates a proton to a water molecule. Thus, the hydration becomes important and we may use formulas that show the extent of hydration:

As with other polyprotic acids, the hydrated aluminum ion ionizes in stages, as shown by:

Note that some of these aluminum species are exhibiting amphiprotic behavior, since they are acting as acids when they appear on the right side of the equilibrium expressions and as bases when they appear on the left side.

Figure 14.16 When an aluminum ion reacts with water, the hydrated aluminum ion becomes a weak acid.

However, the ionization of a cation carrying more than one charge is usually not extensive beyond the first stage. Additional examples of the first stage in the ionization of hydrated metal ions are:

Example 14.18

Hydrolysis of [Al(H2O)6]3+

Calculate the pH of a 0.10-M solution of aluminum chloride, which dissolves completely to give the hydrated aluminum ion [Al(H2O)6]3+ in solution.

Solution

In spite of the unusual appearance of the acid, this is a typical acid ionization problem.

1. Determine the direction of change. The equation for the reaction and Ka are:


The reaction shifts to the right to reach equilibrium.