Chapt15

Accounting

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website CHAPTER 15 Acids and Bases INTRODUCTION SOME OF THE MOST IMPORTANT PROCESSES IN CHEMICAL AND BIOLOG- ICAL SYSTEMS ARE ACID-BASE REACTIONS IN AQUEOUS SOLUTIONS. 15.1 BRØNSTED ACIDS AND BASES IN 15.2 THE ACID-BASE PROPERTIES OF WATER THIS FIRST OF TWO CHAPTERS ON THE PROPERTIES OF ACIDS AND BASES, 15.3 pH — A MEASURE OF ACIDITY WE WILL STUDY THE DEFINITIONS OF ACIDS AND BASES, THE 15.4 STRENGTH OF ACIDS AND BASES pH SCALE, THE IONIZATION OF WEAK ACIDS AND WEAK BASES, AND THE RELATIONSHIP BETWEEN ACID STRENGTH AND MOLECULAR STRUCTURE. 15.5 WEAK ACIDS AND ACID IONIZATION CONSTANTS WE 15.6 WEAK BASES AND BASE IONIZATION CONSTANTS WILL ALSO LOOK AT OXIDES THAT CAN ACT AS ACIDS OR BASES. 15.7 THE RELATIONSHIP BETWEEN THE IONIZATION CONSTANTS OF ACIDS AND THEIR CONJUGATE BASES 15.8 DIPROTIC AND POLYPROTIC ACIDS 15.9 MOLECULAR STRUCTURE AND THE STRENGTH OF ACIDS 15.10 ACID-BASE PROPERTIES OF SALTS 15.11 ACID-BASE PROPERTIES OF OXIDES AND HYDROXIDES 15.12 LEWIS ACIDS AND BASES 597 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 598 ACIDS AND BASES 15.1 Conjugate means “joined together.” BRØNSTED ACIDS AND BASES In Chapter 4 we defined a Brønsted acid as a substance capable of donating a proton, and a Brønsted base as a substance that can accept a proton. These definitions are generally suitable for a discussion of the properties and reactions of acids and bases. An extension of the Brønsted definition of acids and bases is the concept of the conjugate acid-base pair, which can be defined as an acid and its conjugate base or a base and its conjugate acid. The conjugate base of a Brønsted acid is the species that remains when one proton has been removed from the acid. Conversely, a conjugate acid results from the addition of a proton to a Brønsted base. Every Brønsted acid has a conjugate base, and every Brønsted base has a conjugate acid. For example, the chloride ion (Cl ) is the conjugate base formed from the acid HCl, and H2O is the conjugate base of the acid H3O (hydronium ion). Similarly, the ionization of acetic acid can be represented as HOOS O A H H SOS AB O HOCO COOS Q A H O HOOOH A H CH3COOH(aq) H2O(l) CH3COO (aq) H3O (aq) acid1 base2 base1 acid2 H SOS AB HOC OCOOOH O Q A H The formula of a conjugate base always has one fewer hydrogen atom and one more negative charge (or one fewer positive charge) than the formula of the corresponding acid. The subscripts 1 and 2 designate the two conjugate acid-base pairs. Thus the acetate ion (CH3COO ) is the conjugate base of CH3COOH. Both the ionization of HCl (see Section 4.3) and the ionization of CH3COOH are examples of Brønsted acid-base reactions. The Brønsted definition also allows us to classify ammonia as a base because of its ability to accept a proton: N HOOOH A H O HOOS A H H A HONOH A H NH3(aq) H2O(l) NH4(aq) OH (aq) base1 acid2 acid1 base2 O HOOS Q In this case, NH4 is the conjugate acid of the base NH3, and the hydroxide ion OH is the conjugate base of the acid H2O. Note that the atom in the Brønsted base that accepts a H ion must have a lone pair. In Example 15.1 we identify the conjugate pairs in an acid-base reaction. EXAMPLE 15.1 Identify the conjugate acid-base pairs in the reaction between ammonia and hydrofluoric acid in aqueous solution NH3(aq) HF(aq) 3:4 NH4 (aq) The conjugate acid-base pairs are (1) HF (acid) and F NH4 (acid) and NH3 (base). Answer Similar problem: 15.5. Back Forward F (aq) Main Menu TOC Study Guide TOC (base) and (2) Textbook Website MHHE Website 15.2 THE ACID-BASE PROPERTIES OF WATER 599 PRACTICE EXERCISE Identify the conjugate acid-base pairs for the reaction CN H2O 3:4 HCN OH It is acceptable to represent the proton in aqueous solution either as H or as H3O . The formula H is less cumbersome in calculations involving hydrogen ion concentrations and in calculations involving equilibrium constants, whereas H3O is more useful in a discussion of Brønsted acid-base properties. 15.2 Tap water and water from underground sources do conduct electricity because they contain many dissolved ions. THE ACID-BASE PROPERTIES OF WATER Water, as we know, is a unique solvent. One of its special properties is its ability to act either as an acid and as a base. Water functions as a base in reactions with acids such as HCl and CH3COOH, and it functions as an acid in reactions with bases such as NH3. Water is a very weak electrolyte and therefore a poor conductor of electricity, but it does undergo ionization to a small extent: H2O(l ) 3:4 H (aq) OH (aq) (15.1) This reaction is sometimes called the autoionization of water. To describe the acid-base properties of water in the Brønsted framework, we express its autoionization as follows (also shown in Figure 15.1): HOOS O A H HOOS O A H HOOOH O A H HOOS O Q or H2O H2O 3:4 H3O OH acid1 base2 base1 acid2 The acid-base conjugate pairs are (1) H2O (acid) and OH (base) and (2) H3O (acid) and H2O (base). THE ION PRODUCT OF WATER In the study of acid-base reactions in aqueous solutions, the hydrogen ion concentration is key, because it indicates the acidity or basicity of the solution. Expressing the proton as H rather than H3O , we can write the equilibrium constant for the autoionization of water, Equation (15.1), as Kc FIGURE 15.1 Reaction between two water molecules to form hydronium and hydroxide ions. Back Forward Main Menu [H ][OH ] [H2O] + TOC Study Guide TOC + Textbook Website MHHE Website 600 ACIDS AND BASES Recall that in pure water, [H2O] 55.5 M (see p. 564). Since a very small fraction of water molecules are ionized, the concentration of water [H2O] remains virtually unchanged. Therefore Kc[H2O] If you could randomly remove and examine ten particles (H2O, H , or OH ) per second from a liter of water, it would take you two years, working nonstop, to find one H ion! Kw [H ][OH ] (15.2) The equilibrium constant Kw is called the ion-product constant, which is the product of the molar concentrations of H and OH ions at a particular temperature. In pure water at 25°C, the concentrations of H and OH ions are equal and found to be [H ] 1.0 10 7 M and [OH ] 1.0 10 7 M. Thus, from Equation (15.2), at 25°C Kw (1.0 10 7)(1.0 10 7) 1.0 14 10 Whether we have pure water or a solution of dissolved species, the following relation always holds at 25°C: Kw [H ][OH ] 1.0 10 14 Whenever [H ] [OH ], the aqueous solution is said to be neutral. In an acidic solution there is an excess of H ions and [H ] [OH ]. In a basic solution there is an excess of hydroxide ions, so [H ] [OH ]. In practice we can change the concentration of either H or OH ions in solution, but we cannot vary both of them independently. If we adjust the solution so that [H ] 1.0 10 6 M, the OH concentration must change to [OH ] Kw [H ] 1.0 1.0 10 10 14 1.0 6 10 8 M Be aware that the calculations shown above, and indeed all the calculations involving solution concentrations discussed in Chapters 14 – 16, are subject to error because we have implicitly assumed ideal behavior. In reality, ion-pair formation and other types of intermolecular interactions may affect the actual concentrations of species in solution and hence also the equilibrium constant values. The situation is analogous to the relationships between ideal gas behavior and the behavior of real gases discussed in Chapter 5. Depending on temperature, volume, and amount and type of gas present, the measured gas pressure may differ from that calculated using the ideal gas equation. Similarly, the actual, or “effective,” concentration of a solute may not be what we think it should be from the amount of substance originally dissolved in solution. Just as we have the van der Waals and other equations to reconcile discrepancies between the ideal gas equation and nonideal gas behavior, we can account for nonideal behavior in solution. But for our purposes, it is acceptable to ignore deviations from ideality. In most cases this approach will give us a good approximation of the chemical processes that actually take place in the solution phase. An application of Equation (15.2) is given in Example 15.2. EXAMPLE 15.2 The concentration of OH ions in a certain household ammonia cleaning solution is 0.0025 M. Calculate the concentration of H ions. Answer Rearranging Equation (15.2), we write [H ] Many household cleaning fluids contain ammonia. Back Forward Main Menu TOC Kw [OH ] 1.0 10 0.0025 Study Guide TOC 14 4.0 10 12 M Textbook Website MHHE Website 15.3 pH—A MEASURE OF ACIDITY 601 Since [H ] [OH ], the solution is basic, as we would expect from the earlier discussion of the reaction of ammonia with water. Comment Similar problems: 15.15, 15.16. PRACTICE EXERCISE Calculate the concentration of OH ions in a HCl solution whose hydrogen ion concentration is 1.3 M. 15.3 pH — A MEASURE OF ACIDITY Because the concentrations of H and OH ions in aqueous solutions are frequently very small numbers and therefore inconvenient to work with, Soren Sorensen† in 1909 proposed a more practical measure called pH. The pH of a solution is defined as the negative logarithm of the hydrogen ion concentration (in mol/L): pH The pH of concentrated acid solutions can be negative. For example, the pH of a 2.0 M HCl solution is 0.30. log [H ] (15.3) Keep in mind that Equation (15.3) is simply a definition designed to give us convenient numbers to work with. The negative logarithm gives us a positive number for pH, which otherwise would be negative due to the small value of [H ]. Furthermore, the term [H ] in Equation (15.3) pertains only to the numerical part of the expression for hydrogen ion concentration, for we cannot take the logarithm of units. Thus, like the equilibrium constant, the pH of a solution is a dimensionless quantity. Since pH is simply a way to express hydrogen ion concentration, acidic and basic solutions at 25°C can be distinguished by their pH values, as follows: Acidic solutions: Basic solutions: Neutral solutions: [H ] [H ] [H ] 1.0 1.0 1.0 10 10 10 7 7 7 M, pH M, pH M, pH 7.00 7.00 7.00 Notice that pH increases as [H ] decreases. In the laboratory, the pH of a solution is measured with a pH meter (Figure 15.2). Table 15.1 lists the pHs of a number of common fluids. As you can see, the pH of body fluids varies greatly, depending on location and function. The low pH (high acidity) of † Soren Peer Lauritz Sorensen (1868–1939). Danish biochemist. Sorensen originally wrote the symbol as pH and called p the “hydrogen ion exponent” (Wasserstoffionexponent); it is the initial letter of Potenz (German), puissance (French), and power (English). It is now customary to write the symbol as pH. FIGURE 15.2 A pH meter is commonly used in the laboratory to determine the pH of a solution. Although many pH meters have scales marked with values from 1 to 14, pH values can, in fact, be less than 1 and greater than 14. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 602 ACIDS AND BASES TABLE 15.1 The pHs of Some Common Fluids SAMPLE pH VALUE Gastric juice in the stomach Lemon juice Vinegar Grapefruit juice Orange juice Urine Water exposed to air* Saliva Milk Pure water Blood Tears Milk of magnesia Household ammonia 1.0–2.0 2.4 3.0 3.2 3.5 4.8–7.5 5.5 6.4–6.9 6.5 7.0 7.35–7.45 7.4 10.6 11.5 *Water exposed to air for a long period of time absorbs atmospheric CO2 to form carbonic acid, H2CO3. gastric juices facilitates digestion whereas a higher pH of blood is necessary for the transport of oxygen. These pH-dependent actions will be illustrated in Chemistry in Action essays in this and the next chapter. A pOH scale analogous to the pH scale can be devised using the negative logarithm of the hydroxide ion concentration of a solution. Thus we define pOH as pOH log [OH ] Now consider again the ion-product constant for water: [H ][OH ] Kw 1.0 10 14 Taking the negative logarithm of both sides, we obtain (log [H ] log [OH ]) log [H ] log (1.0 log [OH ] 10 14 ) 14.00 From the definitions of pH and pOH we obtain pH pOH 14.00 (15.5) Equation (15.5) provides us with another way to express the relationship between the H ion concentration and the OH ion concentration. The following examples illustrate calculations involving pH. EXAMPLE 15.3 The concentration of H ions in a bottle of table wine was 3.2 10 4 M right after the cork was removed. Only half of the wine was consumed. The other half, after it had been standing open to the air for a month, was found to have a hydrogen ion concentration equal to 1.0 10 3 M. Calculate the pH of the wine on these two occasions. When the bottle was first opened, [H ] stitute in Equation (15.3). Answer In each case the pH has only two significant figures. The two figures to the right of the decimal in 3.49 tell us that there are two significant figures in the original number (see Appendix 4). (15.4) pH pH 10 4 M, which we sub- log [H ] log (3.2 On the second occasion, [H ] 3.2 1.0 10 log (1.0 10 4) 3 3.49 M, so that 10 3) 3.00 The increase in hydrogen ion concentration (or decrease in pH) is largely the result of the conversion of some of the alcohol (ethanol) to acetic acid, a reaction that takes place in the presence of molecular oxygen. Comment Similar problems: 15.17, 15.18. PRACTICE EXERCISE Nitric acid (HNO3) is used in the production of fertilizer, dyes, drugs, and explosives. Calculate the pH of a HNO3 solution having a hydrogen ion concentration of 0.76 M. EXAMPLE 15.4 The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. Calculate the H ion concentration of the rainwater. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 15.4 STRENGTHS OF ACIDS AND BASES 603 From Equation (15.3) Answer 4.82 log [H ] Taking the antilog of both sides of this equation (see Appendix 4) gives 1.5 Comment Similar problem: 15.19. 1 10 4 10 5 M [H ] Because the pH is between 4 and 5, we can expect [H ] to be between M and 1 10 5 M. Therefore, the answer is reasonable. PRACTICE EXERCISE The pH of a certain orange juice is 3.33. Calculate the H ion concentration. EXAMPLE 15.5 In a NaOH solution [OH ] is 2.9 Answer 10 4 M. Calculate the pH of the solution. We use Equation (15.4): pOH log [OH ] log (2.9 10 4) 3.54 Now we use Equation (15.5): pH pOH 14.00 pH 14.00 pOH 14.00 3.54 Similar problem: 15.18. 10.46 PRACTICE EXERCISE The OH ion concentration of a blood sample is 2.5 the blood? 15.4 10 7 M. What is the pH of STRENGTH OF ACIDS AND BASES Strong acids are strong electrolytes which, for practical purposes, are assumed to ionize completely in water (Figure 15.3). Most of the strong acids are inorganic acids: hydrochloric acid (HCl), nitric acid (HNO3), perchloric acid (HClO4), and sulfuric acid (H2SO4): HCl(aq) H2O(l ) 88n H3O (aq) Cl (aq) HNO3(aq) H2O(l ) 88n H3O (aq) NO3 (aq) HClO4(aq) H2O(l ) 88n H3O (aq) ClO4 (aq) H2SO4(aq) H2O(l ) 88n H3O (aq) HSO4 (aq) Note that H2SO4 is a diprotic acid; we show only the first stage of ionization here. At equilibrium, solutions of strong acids will not contain any nonionized acid molecules. Most acids are weak acids, which ionize only to a limited extent in water. At equilibrium, aqueous solutions of weak acids contain a mixture of nonionized acid molecules, H3O ions, and the conjugate base. Examples of weak acids are hydrofluoric acid (HF), acetic acid (CH3COOH), and the ammonium ion (NH4 ). The limited ion- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 604 ACIDS AND BASES Before ionization At equilibrium H+ A– HA (a) ization of weak acids is related to the equilibrium constant for ionization, which we will study in the next section. Like strong acids, strong bases are all strong electrolytes that ionize completely in water. Hydroxides of alkali metals and certain alkaline earth metals are strong bases. [All alkali metal hydroxides are soluble. Of the alkaline earth hydroxides, Be(OH)2 and Mg(OH)2 are insoluble; Ca(OH)2 and Sr(OH)2 are slightly soluble; and Ba(OH)2 is soluble]. Some examples of strong bases are: H2O HA NaOH(s) 88n Na (aq) HA H2O KOH(s) 88n K (aq) H2O H+ A– (b) HA Ba(OH)2(s) 88n Ba2 (aq) OH (aq) OH (aq) 2OH (aq) Strictly speaking, these metal hydroxides are not Brønsted bases because they cannot accept a proton. However, the hydroxide ion (OH ) formed when they ionize is a Brønsted base because it can accept a proton: HA H3O (aq) H+ A– (c) FIGURE 15.3 The extent of ionization of (a) a strong acid that undergoes 100 percent ionization, (b) a weak acid, and (c) a very weak acid. OH (aq) 88n 2H2O(l ) Thus, when we call NaOH or any other metal hydroxide a base, we are actually referring to the OH species derived from the hydroxide. Weak bases, like weak acids, are weak electrolytes. Ammonia is a weak base. It ionizes to a very limited extent in water: NH3(aq) H2O(l ) 3:4 NH4 (aq) OH (aq) Note that NH3 does not ionize like an acid because it does not split up to form ions the way, say, HCl does. Table 15.2 lists some important conjugate acid-base pairs, in order of their relative strengths. Conjugate acid-base pairs have the following properties: If an acid is strong, its conjugate base has no measurable strength. Thus the Cl ion, which is the conjugate base of the strong acid HCl, is an extremely weak base. • H3O is the strongest acid that can exist in aqueous solution. Acids stronger than H3O react with water to produce H3O and their conjugate bases. Thus HCl, which is a stronger acid than H3O , reacts with water completely to form H3O and Cl : • HCl(aq) H2O(l ) 88n H3O (aq) Cl (aq) Acids weaker than H3O react with water to a much smaller extent, producing H3O and their conjugate bases. For example, the following equilibrium lies primarily to the left: HF(aq) H2O(l ) 3:4 H3O (aq) F (aq) • The OH ion is the strongest base that can exist in aqueous solution. Bases stronger than OH react with water to produce OH and their conjugate acids. For example, the oxide ion (O2 ) is a stronger base than OH , so it reacts with water completely as follows: O2 (aq) H2O(l) 88n 2OH (aq) For this reason the oxide ion does not exist in aqueous solutions. The following example shows calculations of pH for a solution containing a strong acid and a solution of a strong base. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 15.4 Relative Strengths of Conjugate Acid-Base Pairs ACID CONJUGATE BASE HClO4 (perchloric acid) HI (hydroiodic acid) HBr (hydrobromic acid) HCl (hydrochloric acid) H2SO4 (sulfuric acid) HNO3 (nitric acid) H3O (hydronium ion) HSO4 (hydrogen sulfate ion) HF (hydrofluoric acid) HNO2 (nitrous acid) HCOOH (formic acid) CH3COOH (acetic acid) NH4 (ammonium ion) HCN (hydrocyanic acid) H2O (water) NH (ammonia) 3 ClO4 (perchlorate ion) I (iodide ion) Br (bromide ion) Cl (chloride ion) HSO4 (hydrogen sulfate ion) NO3 (nitrate ion) H2O (water) SO2 (sulfate ion) 4 F (fluoride ion) NO2 (nitrite ion) HCOO (formate ion) CH3COO (acetate ion) NH3 (ammonia) CN (cyanide ion) OH (hydroxide ion) NH2 (amide ion) 777777777777777777777777777777777777777777777777777777777777777777777777777777n Acid strength increases 777777777777777777777777777777777777777777777777777777777777777777777777777777n Weak acids Strong acids 605 Base strength increases TABLE 15.2 STRENGTHS OF ACIDS AND BASES EXAMPLE 15.6 Calculate the pH of (a) a 1.0 solution. Answer 3 10 M HCl solution and (b) a 0.020 M Ba(OH)2 (a) Since HCl is a strong acid, it is completely ionized in solution: HCl(aq) 88n H (aq) Cl (aq) The concentrations of all the species (HCl, H , and Cl ) before and after ionization can be represented as follows: Initial (M ): Change (M ): Final (M ): HCl(aq) 1.0 10 1.0 10 88n 3 3 88n 0.0 H (aq) 0.0 1.0 10 1.0 10 3 Cl (aq) 0.0 1.0 10 3 1.0 10 3 3 A positive ( ) change represents an increase and a negative ( ) change indicates a decrease in concentration. Thus [H ] pH 1.0 10 3 M log (1.0 10 3) 3.00 (b) Ba(OH)2 is a strong base; each Ba(OH)2 unit produces two OH ions: Ba(OH)2(aq) 88n Ba2 (aq) 2OH (aq) The changes in the concentrations of all the species can be represented as follows: Initial (M ): Change (M ): Final (M ): Back Forward Main Menu TOC Ba(OH)2(aq) 88n Ba2 (aq) 0.020 0.00 0.020 0.020 0.00 Study Guide TOC 0.020 2OH (aq) 0.00 2(0.020) 0.040 Textbook Website MHHE Website 606 ACIDS AND BASES Thus [OH ] 0.040 M pOH log 0.040 1.40 Therefore pH 14.00 pOH 14.00 1.40 12.60 Note that in both (a) and (b) we have neglected the contribution of the autoionization of water to [H ] and [OH ] because 1.0 10 7 M is so small compared with 1.0 10 3 M and 0.040 M. Comment Similar problem: 15.18. PRACTICE EXERCISE Calculate the pH of a 1.8 10 2 M Ba(OH)2 solution. If we know the relative strengths of two acids, we can predict the position of equilibrium between one of the acids and the conjugate base of the other, as illustrated in Example 15.7. EXAMPLE 15.7 Predict the direction of the following reaction in aqueous solution: HNO2(aq) CN (aq) 3:4 HCN(aq) NO2 (aq) In Table 15.2 we see that HNO2 is a stronger acid than HCN. Thus CN is a stronger base than NO2 . The net reaction will proceed from left to right as written because HNO2 is a better proton donor than HCN (and CN is a better proton acceptor than NO2 ). Answer Similar problem: 15.35. PRACTICE EXERCISE Predict whether the equilibrium constant for the following reaction is greater than or smaller than 1: CH3COOH(aq) 15.5 HCOO (aq) 3:4 CH3COO (aq) HCOOH(aq) WEAK ACIDS AND ACID IONIZATION CONSTANTS As we have seen, there are relatively few strong acids. The vast majority of acids are weak acids. Consider a weak monoprotic acid, HA. Its ionization in water is represented by HA(aq) H2O(l ) 3:4 H3O (aq) A (aq) or simply HA(aq) 3:4 H (aq) A (aq) The equilibrium expression for this ionization is All concentrations in this equation are equilibrium concentrations. Back Forward Main Menu Ka TOC [H ][A ] [HA] Study Guide TOC Textbook Website MHHE Website 15.5 607 WEAK ACIDS AND ACID IONIZATION CONSTANTS where Ka, the acid ionization constant, is the equilibrium constant for the ionization of an acid. At a given temperature, the strength of the acid HA is measured quantitatively by the magnitude of Ka. The larger Ka, the stronger the acid—that is, the greater the concentration of H ions at equilibrium due to its ionization. Keep in mind, however, that only weak acids have Ka values associated with them. Table 15.3 lists a number of weak acids and their Ka values at 25°C in order of decreasing acid strength. Although all these acids are weak, within the group there is great variation in their strengths. For example, Ka for HF (7.1 10 4) is about 1.5 million times that for HCN (4.9 10 10). Generally, we can calculate the hydrogen ion concentration or pH of an acid solution at equilibrium, given the initial concentration of the acid and its Ka value. Alternatively, if we know the pH of a weak acid solution and its initial concentration, we can determine its Ka. The basic approach for solving these problems, which deal with equilibrium concentrations, is the same one outlined in Chapter 14. However, because acid ionization represents a major category of chemical equilibrium in aqueous solution, we will develop a systematic procedure for solving this type of problem that will also help us to understand the chemistry involved. TABLE 15.3 Ionization Constants of Some Weak Acids at 25°C NAME OF ACID FORMULA STRUCTURE Ka CONJUGATE BASE Kb Hydrofluoric acid HF HOF 7.1 10 4 Nitrous acid HNO2 OPNOOOH 4.5 10 4 NO2 2.2 10 11 Acetylsalicylic acid (aspirin) C9H8O4 3.0 10 4 C9H7O4 3.3 10 11 1.7 10 4 HCOO 5.9 10 11 8.0 10 5 C6H7O6 1.3 10 10 6.5 10 5 C6H5COO 1.5 10 10 1.8 10 5 CH3COO 5.6 10 10 4.9 10 10 CN 2.0 10 5 1.3 10 10 C6H5O 7.7 10 5 O B C OOOH F 1.4 10 11 OOC OCH3 B O Formic acid HCOOH Ascorbic acid* C6H8O6 Benzoic acid O B HOC OOOH C6H5COOH OH HOO H E C C H G C C PO D CHOH O A CH2OH O B C OOOH Acetic acid CH3COOH O B CH3 OC OOOH Hydrocyanic acid HCN HOCqN Phenol C6H5OH OOH *For ascorbic acid it is the upper left hydroxyl group that is associated with this ionization constant. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 608 ACIDS AND BASES Suppose we are asked to calculate the pH of a 0.50 M HF solution at 25°C. The ionization of HF is given by HF(aq) 3:4 H (aq) F (aq) From Table 15.3 we write [H ][F ] [HF] Ka 7.1 4 10 The first step is to identify all the species present in solution that may affect its pH. Because weak acids ionize to a small extent, at equilibrium the major species present are nonionized HF and some H and F ions. Another major species is H2O, but its very small Kw (1.0 10 14) means that water is not a significant contributor to the H ion concentration. Therefore, unless otherwise stated, we will always ignore the H ions produced by the autoionization of water. Note that we need not be concerned with the OH ions that are also present in solution. The OH concentration can be determined from Equation (15.2) after we have calculated [H ]. We can summarize the changes in the concentrations of HF, H , and F according to the steps shown on p. 576 as follows: HF(aq) 3:4 H (aq) 0.50 0.00 x x Initial (M ): Change (M ): Equilibrium (M ): 0.50 x F (aq) 0.00 x x x The equilibrium concentrations of HF, H , and F , expressed in terms of the unknown x, are substituted into the ionization constant expression to give (x)(x) 0.50 x Ka 7.1 4 10 Rearranging this expression, we write x2 10 4x 7.1 3.6 10 4 0 This is a quadratic equation which can be solved using the quadratic formula (see Appendix 4). Or we can try using a shortcut to solve for x. Because HF is a weak acid and weak acids ionize only to a slight extent, we reason that x must be small compared to 0.50. Therefore we can make the approximation The sign means “approximately equal to.” An analogy of the approximation is a truck loaded with coal. Losing a few lumps of coal on a delivery trip will not significantly change the overall mass of the load. 0.50 x 0.50 Now the ionization constant expression becomes x2 0.50 x x2 0.50 7.1 10 4 Rearranging, we get x2 (0.50)(7.1 x 3.55 10 4) 10 4 3.55 10 4 0.019 M Thus we have solved for x without having to use the quadratic equation. At equilibrium, we have: [HF] [H ] Forward Main Menu TOC 0.019 M [F ] Back (0.50 0.019) M 0.019 M Study Guide TOC 0.48 M Textbook Website MHHE Website 15.5 WEAK ACIDS AND ACID IONIZATION CONSTANTS 609 and the pH of the solution is pH log (0.019) 1.72 How good is this approximation? Because Ka values for weak acids are generally known to an accuracy of only 5%, it is reasonable to require x to be less than 5% of 0.50, the number from which it is subtracted. In other words, the approximation is valid if the following expression is equal to or less than 5%: 0.019 M 0.50 M 100% 3.8% Thus the approximation we made is acceptable. Now consider a different situation. If the initial concentration of HF is 0.050 M, and we use the above procedure to solve for x, we would get 6.0 10 3 M. However, the following test shows that this answer is not a valid approximation because it is greater than 5% of 0.050 M: 6.0 10 3 M 0.050 M 100% 12% In this case there are two ways to get an accurate value for x: by solving the quadratic equation and by applying the method of successive approximation. THE QUADRATIC EQUATION We start by writing the ionization expression in terms of the unknown x: x2 0.050 x2 7.1 x 10 4x 7.1 3.6 This expression fits the quadratic equation ax2 mula, we write x b 7.1 b2 2a 10 5.6 10 3 10 bx 5 c 0 0. Using the quadratic for- 4ac 4 10 4 2 7.1 4 10 10 4)2 2 (1) (7.1 4(1)( 3.6 10 5) 0.012 M or 6.4 10 3 M The second solution (x 6.4 10 3 M) is physically impossible because the concentration of ions produced as a result of ionization cannot be negative. Choosing x 5.6 10 3 M, we can solve for [HF], [H ], [F ] as follows: [HF] (0.050 5.6 [H ] 5.6 10 3 [F ] 5.6 10 3 10 3) M 0.044 M M M The pH of the solution, then, is pH Back Forward Main Menu TOC log 5.6 Study Guide TOC 10 3 2.25 Textbook Website MHHE Website 610 ACIDS AND BASES THE METHOD OF SUCCESSIVE APPROXIMATION With this method, we first solve for x by assuming 0.050 x 0.050 as shown above. Next, we use the approximate value of x (6.0 10 3 M) to find a more exact value of the concentration for HF: [HF] (0.050 6.0 10 3) M 0.044 M Substituting this value of [HF] in the expression for Ka, we write x2 0.044 7.1 10 4 x 5.6 10 3 M 3 Using 5.6 10 M for x, we can recalculate [HF] and solve for x again. This time we find that the answer is still 5.6 10 3 M so there is no need to proceed further. In general, we apply the method of successive approximation until the value of x obtained for the last step does not differ from the value formed in the previous step. In most cases, you will need to apply this method only twice to get the correct answer. In summary, the main steps for solving weak acid ionization problems are: Identify the major species that can affect the pH of the solution. In most cases we can ignore the ionization of water. We omit the hydroxide ion because its concentration is determined by that of the H ion. 2. Express the equilibrium concentrations of these species in terms of the initial concentration of the acid and a single unknown x, which represents the change in concentration. 3. Write the acid ionization constant (Ka) in terms of the equilibrium concentrations. First solve for x by the approximate method. If the aprroximation is not valid, use the quadratic equation or the method of successive approximation to solve for x. 4. Having solved for x, calculate the equilibrium concentrations of all species and/or the pH of the solution. 1. Example 15.8 provides another illustration of the above procedure. EXAMPLE 15.8 Calculate the pH of a 0.036 M nitrous acid (HNO2) solution: HNO2(aq) 3:4 H (aq) Nitrous acid NO2 (aq) Answer Step 1: The species that can affect the pH of the solution are HNO2, H , and the conjugate base NO2 . We ignore water ’s contribution to [H ]. and NO2 ions in mol/L, we summarize: Step 2: Letting x be the equilibrium concentration of H Initial (M ): Change (M ): HNO2(aq) 3:4 H (aq) 0.036 0.00 x x Equilibrium (M ): 0.036 x x NO2 (aq) 0.00 x x Step 3: From Table 15.3 we write Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 15.5 WEAK ACIDS AND ACID IONIZATION CONSTANTS [H ][NO2 ] [HNO2] Ka x2 0.036 4.5 Applying the approximation 0.036 2 4 4.5 x 10 10 611 4 x 0.036, we obtain 2 x 0.036 x 0.036 x x2 4.5 1.62 x 4.0 4 5 10 10 10 3 M To test the approximation, 4.0 10 3 M 0.036 M 100% 11% Since this is greater than 5%, our approximation is not valid and we must solve the quadratic equation, as follows: x2 4.5 4.5 x 3.8 10 4x 10 10 3 1.62 4 10 5 0 42 (4.5 M or 10 ) 2(1) 4.3 10 3 4(1)(1.62 10 5) M The second solution is physically impossible, since the concentration of ions produced as a result of ionization cannot be negative. Therefore, the solution is given by the positive root, x 3.8 10 3 M. Step 4: At equilibrium [H ] 3.8 pH 10 log 3.8 3 M 10 3 2.42 You should check the answer by using the method of successive approximation. Comment Similar problem: 15.41. PRACTICE EXERCISE What is the pH of a 0.122 M monoprotic acid whose Ka is 5.7 10 4? One way to determine Ka of an acid is to measure the pH of the acid solution of known concentration at equilibrium. Example 15.9 shows this approach. EXAMPLE 15.9 The pH of a 0.10 M solution of formic acid (HCOOH) is 2.39. What is the Ka of the acid? Answer Step 1: The major species in solution are HCOOH, H , and the conjugate base HCOO . Formic acid Back Forward Main Menu Step 2: First we need to calculate the hydrogen ion concentration from the pH value TOC Study Guide TOC Textbook Website MHHE Website 612 ACIDS AND BASES pH log [H ] 2.39 log [H ] Taking the antilog of both sides, we get [H ] 10 2.39 4.1 10 3 M Next we summarize the changes: HCOOH(aq) 0.10 4.1 10 3 Initial (M ): Change (M ): Equilibrium (M ): (0.10 3:4 3 4.1 10 ) H (aq) 0.00 4.1 10 4.1 10 3 3 HCOO (aq) 0.00 4.1 10 3 4.1 10 3 Note that since the pH and hence the H ion concentration is known, it follows that we also know the concentrations of HCOOH and HCOO at equilibrium. Step 3: The ionization constant of formic acid is given by Ka [H ][HCOO ] [HCOOH] (4.1 10 3)(4.1 10 3) (0.10 4.1 10 3) 1.8 Similar problem: 15.43. 10 4 Comment The Ka value differs slightly from the one listed in Table 15.3 because of the rounding-off procedure we used in the calculation. PRACTICE EXERCISE The pH of a 0.060 M weak monoprotic acid is 3.44. Calculate the Ka of the acid. PERCENT IONIZATION We have seen that the magnitude of Ka indicates the strength of an acid. Another measure of the strength of an acid is its percent ionization, which is defined as We can compare the strengths of acids in terms of percent ionization only if concentrations of the acids are the same. percent ionization ionized acid concentration at equilibrium initial concentration of acid 100% (15.6) The stronger the acid, the greater the percent ionization. For a monoprotic acid HA, the concentration of the acid that undergoes ionization is equal to the concentration of the H ions or the concentration of the A ions at equilibrium. Therefore, we can write the percent ionization as percent ionization [H ] [HA]0 100% where [H ] is the concentration at equilibrium and [HA]0 is the initial concentration. Referring to Example 15.8, we see that the percent ionization of a 0.036 M HNO2 solution is percent ionization Back Forward Main Menu TOC 3.8 10 3 M 0.036 M Study Guide TOC 100% 1.1% Textbook Website MHHE Website 15.6 Strong acid % Ionization 100 Weak acid 0 Initial concentration of acid FIGURE 15.4 Dependence of percent ionization on initial concentration of acid. Note that at very low concentrations, all acids (weak and strong) are almost completely ionized. WEAK BASES AND BASE IONIZATION CONSTANTS 613 Thus only about one out of every 100 HNO2 molecules has ionized. This is consistent with the fact that HNO2 is a weak acid. The extent to which a weak acid ionizes depends on the initial concentration of the acid. The more dilute the solution, the greater the percent ionization (Figure 15.4). In qualitative terms, when an acid is diluted, the number of particles (nonionized acid molecules plus ions) per unit volume is reduced. According to Le Chatelier ’s principle (see Section 14.5), to counteract this “stress” (that is, the dilution), the equilibrium shifts from nonionized acid to H and its conjugate base to produce more particles (ions). The dependence of percent ionization on initial concentration can be illustrated by the HF case discussed on page 608: 0.50 M HF 0.019 M 0.50 M percent ionization 100% 3.8% 0.050 M HF percent ionization 5.6 10 3 M 0.050 M 100% 11.2% We see that, as expected, a more dilute HF solution has a greater percent ionization of the acid. 15.6 WEAK BASES AND BASE IONIZATION CONSTANTS The ionization of weak bases is treated in the same way as the ionization of weak acids. When ammonia dissolves in water, it undergoes the reaction NH3(aq) H2O(l ) 3:4 NH4 (aq) OH (aq) The equilibrium constant is given by K [NH4 ][OH ] [NH3][H2O] The production of hydroxide ions in this base ionization reaction means that [OH ] [H ] and therefore pH 7. Compared with the total concentration of water, very few water molecules are consumed by this reaction, so we can treat [H2O] as a constant. Thus we can write the base ionization constant (Kb), which is the equilibrium constant for the ionization reaction, as Kb K[H2O] [NH4 ][OH ] [NH3] 1.8 10 5 Table 15.4 lists a number of common weak bases and their ionization constants. Note that the basicity of all these compounds is attributable to the lone pair of electrons on the nitrogen atom. However, the ability of the lone pair to accept a H ion makes these substances Brønsted bases. In solving problems involving weak bases, we follow the same procedure we used for weak acids. The main difference is that we calculate [OH ] first, rather than [H ]. Example 15.10 shows this approach. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 614 ACIDS AND BASES TABLE 15.4 Ionization Constants of Some Common Weak Bases at 25°C NAME OF BASE FORMULA STRUCTURE Ethylamine C2H5NH2 CH3 OCH2 OOOH N A H 5.6 10 4 C2H5NH3 1.8 10 11 Methylamine CH3NH2 N CH3 OOOH A H 4.4 10 4 CH3NH3 2.3 10 11 Caffeine C8H10N4O2 O 4.1 10 4 C8H11N4O2 2.4 10 11 1.8 10 5 NH4 5.6 10 10 1.7 10 9 C5H5NH 5.9 10 6 3.8 10 C 6 H 5 NH 3 2.6 10 5 1.5 10 H3C Kb* C N C C C N Q N Ka CH3 C O CONJUGATE ACID H N CH3 Ammonia NH3 Pyridine HOOOH N A H C5H5N NS A niline C 6 H 5 NH 2 Urea N2H4CO OOOH N A H O B O HONOC OOOH N A A H H 10 14 H2NCONH3 0.67 *The nitrogen atom with the lone pair accounts for each compound’s basicity. In the case of urea, Kb can be associated with either nitrogen atom. EXAMPLE 15.10 What is the pH of a 0.40 M ammonia solution? Answer Step 1: The major species in an ammonia solution are NH3, NH4 , and OH . As in the case of weak acids, we ignore the very small contribution to OH concentration by water. Step 2: Letting x be the equilibrium concentration of NH4 and OH ions in mol/L, we summarize: Initial (M): Change (M): NH3(aq) 0.40 x Equilibrium (M): 0.40 H2O(l ) 3:4 NH4 (aq) 0.00 x x OH (aq) 0.00 x x x Step 3: Table 15.4 gives us Kb: Kb Back Forward Main Menu TOC [NH4 ][OH ] [NH3] Study Guide TOC 1.8 10 5 Textbook Website MHHE Website 15.7 THE RELATIONSHIP BETWEEN THE IONIZATION CONSTANTS OF ACIDS AND THEIR CONJUGATE BASES x2 0.40 x Applying the approximation 0.40 x2 0.40 x 1.8 x 10 615 5 0.40, we obtain x2 0.40 1.8 10 5 x2 7.2 10 6 x 2.7 10 3 M To test the approximation, we write 2.7 10 3 M 0.40 M 100% 0.68% Therefore the approximation is valid. 2.7 10 3 M. Thus Step 4: At equilibrium, [OH ] pOH 10 3) log (2.7 2.57 pH 14.00 2.57 11.43 Similar problem: 15.51. PRACTICE EXERCISE Calculate the pH of a 0.26 M methylamine solution (see Table 15.4). 15.7 THE RELATIONSHIP BETWEEN THE IONIZATION CONSTANTS OF ACIDS AND THEIR CONJUGATE BASES An important relationship between the acid ionization constant and the ionization constant of its conjugate base can be derived as follows, using acetic acid as an example: CH3COOH(aq) 3:4 H (aq) Ka CH3COO (aq) [H ][CH3COO ] [CH3COOH)] The conjugate base, CH3COO , supplied by a sodium acetate (CH3COONa) solution, reacts with water according to the equation CH3COO (aq) H2O(l ) 3:4 CH3COOH(aq) OH (aq) and we can write the base ionization constant as Kb [CH3COOH][OH ] The product of these two ionization constants is given by KaKb [H ][CH3COO ] [CH3COOH] [CH3COOH][OH ] [H ][OH ] Kw This result may seem strange at first, but if we add the two reactions we see that the sum is simply the autoionization of water. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 616 ACIDS AND BASES (1) CH3COOH(aq) 3:4 H (aq) CH3COO (aq) Ka (2) CH3COO (aq) H2O(l) 3:4 CH3COOH(aq) OH (aq) Kb (3) H2O(l) 3:4 H (aq) OH (aq) Kw This example illustrates one of the rules for chemical equilibria: When two reactions are added to give a third reaction, the equilibrium constant for the third reaction is the product of the equilibrium constants for the two added reactions (see Section 14.2). Thus, for any conjugate acid-base pair it is always true that KaKb Kw Kw Kb (15.7) Kb Expressing Equation (15.7) as Ka Kw Ka enables us to draw an important conclusion: The stronger the acid (the larger Ka), the weaker its conjugate base (the smaller Kb), and vice versa (see Tables 15.3 and 15.4). We can use Equation (15.7) to calculate the Kb of the conjugate base (CH3COO ) of CH3COOH as follows. We find the Ka value of CH3COOH in Table 15.3 and write Kb Kw Ka 1.0 1.8 14 5.6 15.8 10 10 10 10 5 DIPROTIC AND POLYPROTIC ACIDS The treatment of diprotic and polyprotic acids is more involved than that of monoprotic acids because these substances may yield more than one hydrogen ion per molecule. These acids ionize in a stepwise manner; that is, they lose one proton at a time. An ionization constant expression can be written for each ionization stage. Consequently, two or more equilibrium constant expressions must often be used to calculate the concentrations of species in the acid solution. For example, for carbonic acid, H2CO3, we write H2CO3(aq) 3:4 H (aq) HCO3 (aq) Ka1 [H ][HCO3 ] [H2CO3] HCO3 (aq) 3:4 H (aq) CO2 (aq) 3 Ka2 [H ][CO2 ] 3 [HCO3 ] Note that the conjugate base in the first ionization stage becomes the acid in the second ionization stage. Table 15.5 shows the ionization constants of several diprotic acids and one polyprotic acid. For a given acid, the first ionization constant is much larger than the second ionization constant, and so on. This trend is reasonable because it is easier to remove a H ion from a neutral molecule than to remove another H from a negatively charged ion derived from the molecule. In the following example we calculate the equilibrium concentrations of all the species of a diprotic acid in aqueous solution. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 15.8 617 DIPROTIC AND POLYPROTIC ACIDS TABLE 15.5 Ionization Constants of Some Common Diprotic and Polyprotic Acids in Water at 25°C NAME OF ACID FORMULA H2SO4 Sulfuric acid STRUCTURE O B HOOO S OOOH B O O B HOOO S OO B O Ka CONJUGATE BASE very large Kb HSO4 very small 1.3 10 2 SO2 4 7.7 10 13 C2H2O4 OO BB HOOOC OC OOOH 6.5 10 2 C2HO4 1.5 10 13 C2HO4 OO BB HOOOC OC OO 6.1 10 5 C2O2 4 1.6 10 10 H2SO3 O B HOOO S OOOH 1.3 10 2 HSO3 7.7 10 13 HSO3 O B HOOO S OO 6.3 10 8 SO2 3 1.6 10 7 H2CO3 O B HOOO C OOOH 4.2 10 7 HCO3 2.4 10 8 Hydrogen carbonate ion HCO3 O B HOOO COO 4.8 10 11 CO2 3 2.1 10 4 Hydrosulfuric acid Hydrogen sulfide ion† H2S HS HOSOH HOS 9.5 1.1 10 10 8 HS S2 1.1 1.1 10 7 105 7.5 10 3 H2PO4 1.3 10 12 6.2 10 8 HPO2 4 1.6 10 7 4.8 10 13 PO3 4 2.1 10 2 Hydrogen sulfate ion Oxalic acid Hydrogen oxalate ion Sulfurous acid* Hydrogen sulfite ion Carbonic acid Phosphoric acid HSO4 O B HOOO P OOOH A O A H H3PO4 Dihydrogen phosphate ion Hydrogen phosphate ion O B HOOO P OO A O A H H2PO4 O B HOOO P OO A O HPO2 4 19 *H2SO3 has never been isolated and exists in only minute concentration in aqueous solution of SO2. The Ka value here refers to the process SO2(g) HSO3 (aq). H2O(l ) 3 H (aq) 4 † The ionization constant of HS is very low and difficult to measure. The value listed here is only an estimate. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 618 ACIDS AND BASES EXAMPLE 15.11 Oxalic acid (C2H2O4) is a poisonous substance used chiefly as a bleaching and cleansing agent (for example, to remove bathtub rings). Calculate the concentrations of all the species present at equilibrium in a 0.10 M solution. Oxalic acid Oxalic acid is a diprotic acid (see Table 15.5). We begin with the first stage of ionization. Step 1: The major species in solution at this stage are the nonionized acid, H ions, and the conjugate base, C2HO4 . Step 2: Letting x be the equilibrium concentration of H and C2HO4 ions in mol/L, we summarize: Answer C2H2O4(aq) 3:4 H (aq) 0.10 0.00 x x Initial (M ): Change (M ): Equilibrium (M ): 0.10 x C2HO4 (aq) 0.00 x x x Step 3: Table 15.5 gives us [H ][C2HO4 ] [C2H2O4] 6.5 10 2 x2 0.10 x Ka 6.5 10 2 Applying the approximation 0.10 x 2 x 0.10 0.10, we obtain 2 x x 0.10 x2 6.5 10 3 x 8.1 10 2 6.5 10 2 M To test the approximation, 8.1 10 2 M 0.10 M 100% 81% Clearly the approximation is not valid. Therefore, we must solve the quadratic equation x2 The result is x 6.5 10 2x 6.5 10 3 0 0.054 M. Step 4: When the equilibrium for the first stage of ionization is reached, the con- centrations are [H ] 0.054 M [C2HO4 ] 0.054 M [C2H2O4] (0.10 0.054) M 0.046 M Next we consider the second stage of ionization. Step 1: At this stage, the major species are C2HO4 , which acts as the acid in the second stage of ionization, H , and the conjugate base C2O2 . 4 Step 2: Letting y be the equilibrium concentration of H and C2O2 ions in mol/L, 4 we summarize: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 15.8 619 DIPROTIC AND POLYPROTIC ACIDS Initial (M ): Change (M ): Equilibrium (M ): 0.054 C2O2 (aq) 4 0.00 y C2HO4 (aq) 3:4 H (aq) 0.054 0.054 y y y 0.054 y y Step 3: Table 15.5 gives us [H ][C2O42 ] [C2HO4 ] 6.1 10 (0.054 y)(y) (0.054 y) 6.1 10 Ka Applying the approximation 0.054 obtain (0.054)(y) (0.054) y y 5 5 0.054 and 0.054 6.1 10 5 y 0.054, we M and we test the approximation, 6.1 10 5 M 0.054 M 100% 0.11% The approximation is valid. Step 4: At equilibrium, [C2H2O4] 0.046 M [C2HO4 ] (0.054 6.1 10 5) M 0.054 M [H ] (0.054 6.1 10 5) M 0.054 M [C2O2 4 6.1 10 [OH ] Similar problem: 15.62. 5 1.0 10 14 M /0.054 1.9 10 13 M PRACTICE EXERCISE Calculate the concentrations of C2H2O4, C2HO4 , C2O2 , and H ions in a 0.20 M 4 oxalic acid solution. Phosphoric acid is largely responsible for much of the “tangy” flavor of popular cola drinks. Example 15.11 shows that for diprotic acids, if Ka1 Ka2, then we can assume that the concentration of H ions is the product of only the first stage of ionization. Furthermore, the concentration of the conjugate base for the second-stage ionization is numerically equal to Ka2. Phosphoric acid (H3PO4) is a polyprotic acid with three ionizable hydrogen atoms: H3PO4(aq) 3:4 H (aq) H2PO4 (aq) Ka1 [H ][H2PO4 ] [H3PO4] 7.5 10 3 H2PO4 (aq) 3:4 H (aq) HPO2 (aq) 4 Ka2 [H ][HPO2 ] 4 [H2PO4 ] 6.2 10 8 HPO2 (aq) 3:4 H (aq) 4 PO3 (aq) 4 Ka3 [H ][PO3 ] 4 [HPO2 ] 4 4.8 10 13 We see that phosphoric acid is a weak polyprotic acid and that its ionization constants decrease markedly for the second and third stages. Thus we can predict that, in a solution containing phosphoric acid, the concentration of the nonionized acid is the highest, and the only other species present in significant concentrations are H and H2PO4 ions. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 620 ACIDS AND BASES 15.9 MOLECULAR STRUCTURE AND THE STRENGTH OF ACIDS The strength of an acid depends on a number of factors, such as the properties of the solvent, the temperature, and, of course, the molecular structure of the acid. When we compare the strengths of two acids, we can eliminate some variables by considering their properties in the same solvent and at the same temperature and concentration. Then we can focus on the structure of the acids. Let us consider a certain acid HX. The strength of the acid is measured by its tendency to ionize: HX 88n H X Two factors influence the extent to which the acid undergoes ionization. One is the strength of the HOX bond—the stronger the bond, the more difficult it is for the HX molecule to break up and hence the weaker the acid. The other factor is the polarity of the HOX bond. The difference in the electronegativities between H and X results in a polar bond like HOX If the bond is highly polarized, that is, if there is a large accumulation of positive and negative charges on the H and X atoms, HX will tend to break up into H and X ions. So a high degree of polarity characterizes a stronger acid. Below we will consider some examples in which either bond strength or bond polarity plays a prominent role in determining acid strength. The halogens form a series of binary acids called the hydrohalic acids. The strengths of the hydrohalic acids increase in the following order: HF To review the nomenclature of inorganic acids, see Section 2.8. HCl HBr HI As Table 15.6 shows, HF has the highest bond dissociation energy of the four hydrogen halides. Since it takes 568.2 kJ/mol to break the HOF bond, HF is a weak acid. At the other extreme in the series, HI has the lowest bond energy, so HI is the strongest acid of the group. In this series of acids the polarity of the bond actually decreases from HF to HI because F is the most electronegative of the halogens (see Figure 9.5). This property should enhance the acidity of HF relative to the other hydrohalic acids, but its magnitude is not great enough to break the trend in bond dissociation energies. Now let us consider the oxoacids. Oxoacids, as we learned in Chapter 2, contain hydrogen, oxygen, and one other element Z, which occupies a central position. Figure 15.5 shows the Lewis structures of several common oxoacids. As you can see, these acids are characterized by the presence of one or more OOH bonds. The central atom Z might also have other groups attached to it: TABLE 15.6 Bond Dissociation Energies for Hydrogen Halides and Acid Strengths for Hydrohalic Acids BOND Forward ACID STRENGTH (IN WATER) HOF HOCl HOBr HOI Back BOND DISSOCIATION ENERGY (kJ/mol) 568.2 431.9 366.1 298.3 weak strong strong strong Main Menu TOC Study Guide TOC Textbook Website MHHE Website 15.9 FIGURE 15.5 Lewis structures of some common oxoacids. For simplicity, the formal charges have been omitted. MOLECULAR STRUCTURE AND THE STRENGTH OF ACIDS ðO ð B šOC OšOH HO O O HOšOšP š ONO ðO ð B šONO šð HO O O Carbonic acid Nitrous acid 621 Nitric acid ðO ð B šO P OšOH HO O O A H ðO ð B š O P O šOH HO O O A ðO ð A H ðO ð B šO S OšOH HO O O B ðO ð Phosphorous acid Phosphoric acid Sulfuric acid G O Z OOOH D As the oxidation number of an atom becomes larger, its ability to draw electrons in a bond toward itself increases. If Z is an electronegative element, or is in a high oxidation state, it will attract electrons, thus making the ZOO bond more covalent and the OOH bond more polar. Consequently, the tendency for the hydrogen to be donated as a H ion increases: G O Z OOOH D G O Z OO D H To compare their strengths, it is convenient to divide the oxoacids into two groups. 1. Oxoacids having different central atoms that are from the same group of the periodic table and that have the same oxidation number. Within this group, acid strength increases with increasing electronegativity of the central atom, as HClO3 and HBrO3 illustrate: O SOS A O O HOOO Cl OOS QQQ O SOS A O HOOO Br OOS O QQQ Cl and Br have the same oxidation number, 5. However, since Cl is more electronegative than Br, it attracts the electron pair it shares with oxygen (in the ClOOOH group) to a greater extent than Br does. Consequently, the OOH bond is more polar in chloric acid than in bromic acid and ionizes more readily. Thus the relative acid strengths are HClO3 2. HBrO3 Oxoacids having the same central atom but different numbers of attached groups. Within this group, acid strength increases as the oxidation number of the central atom increases. Consider the oxoacids of chlorine shown in Figure 15.6. In this series the ability of chlorine to draw electrons away from the OH group (thus making the OOH bond more polar) increases with the number of electronegative O atoms attached to Cl. Thus HClO4 is the strongest acid because it has the largest number of O atoms attached to Cl, and the acid strength decreases as follows: HClO4 HClO3 HClO2 HClO Example 15.12 compares the strengths of acids based on their molecular structures. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 622 ACIDS AND BASES FIGURE 15.6 Lewis structures of the oxoacids of chlorine. The oxidation number of the Cl atom is shown in parentheses. For simplicity, the formal charges have been omitted. Note that although hypochlorous acid is written as HClO, the H atom is bonded to the O atom. šš HOOO Cl ð ššO HO O O Cl Ošð Hypochlorous acid ( 1) Chlorous acid ( 3) š ðO ð A š HOO O ClO šð O ðšð O A š HO O O Cl Ošð O A ðO ð Chloric acid ( 5) Perchloric acid ( 7) EXAMPLE 15.12 Predict the relative strengths of the oxoacids in each of the following groups: (a) HClO, HBrO, and HIO; (b) HNO3 and HNO2. Answer (a) These acids all have the same structure, and the halogens all have the same oxidation number ( 1). Since the electronegativity decreases from Cl to I, the polarity of the XOO bond (where X denotes a halogen atom) increases from HClO to HIO, and the polarity of the OOH bond decreases from HClO to HIO. Thus the acid strength decreases as follows: HClO Similar problem: 15.66. HBrO HIO (b) The structures of HNO3 and HNO2 are shown in Figure 15.3. Since the oxidation number of N is 5 in HNO3 and 3 in HNO2, HNO3 is a stronger acid than HNO2. PRACTICE EXERCISE Which of the following acids is weaker: H3PO3 or H3PO4? 15.10 ACID-BASE PROPERTIES OF SALTS The word “hydrolysis” is derived from the Greek words hydro, meaning “water,” and lysis, meaning “to split apart.” As defined in Section 4.3, a salt is an ionic compound formed by the reaction between an acid and a base. Salts are strong electrolytes that completely dissociate into ions in water. The term salt hydrolysis describes the reaction of an anion or a cation of a salt, or both, with water. Salt hydrolysis usually affects the pH of a solution. SALTS THAT PRODUCE NEUTRAL SOLUTIONS It is generally true that salts containing an alkali metal ion or alkaline earth metal ion (except Be2 ) and the conjugate base of a strong acid (for example, Cl , Br , and NO3 ) do not undergo hydrolysis to an appreciable extent, and their solutions are assumed to be neutral. For instance, when NaNO3, a salt formed by the reaction of NaOH with HNO3, dissolves in water, it dissociates completely as follows: H2O NaNO3(s) 88n Na (aq) Back Forward Main Menu TOC Study Guide TOC NO3 (aq) Textbook Website MHHE Website 15.10 Metal ion hydrolysis is discussed on p. 625. ACID-BASE PROPERTIES OF SALT 623 The hydrated Na ion neither donates nor accepts H ions. The NO3 ion is the conjugate base of the strong acid HNO3, and it has no affinity for H ions. Consequently, a solution containing Na and NO3 ions is neutral, with a pH of about 7. SALTS THAT PRODUCE BASIC SOLUTIONS The solution of a salt derived from a strong base and a weak acid is basic. For example, the dissociation of sodium acetate (CH3COONa) in water is given by H2O CH3COONa(s) 88n Na (aq) CH3COO (aq) The hydrated Na ion has no acidic or basic properties. The acetate ion CH3COO , however, is the conjugate base of the weak acid CH3COOH and therefore has an affinity for H ions. The hydrolysis reaction is given by CH3COO (aq) H2O(l ) 3:4 CH3COOH(aq) OH (aq) Because this reaction produces OH ions, the sodium acetate solution will be basic. The equilibrium constant for this hydrolysis reaction is the base ionization constant expression for CH3COO , so we write (see p. 615) Kb [CH3COOH][OH ] 5.6 10 10 Since each CH3COO ion that hydrolyzes produces one OH ion, the concentration of OH at equilibrium is the same as the concentration of CH3COO that hydrolyzed. We can define the percent hydrolysis as % hydrolysis [CH3COO ]hydrolyzed [OH ]equilibrium 100% 100% A calculation based on the hydrolysis of CH3COONa is illustrated in Example 15.13. In solving salt hydrolysis problems, we follow the same procedure we used for weak acids and weak bases. EXAMPLE 15.13 Calculate the pH of a 0.15 M solution of sodium acetate (CH3COONa). What is the percent hydrolysis? Answer Step 1: The major species in a sodium acetate solution are Na and CH3COO ions. Since we started with a 0.15 M sodium acetate solution, the initial concentrations of the ions are also equal to 0.15 M: H2O CH3COONa(s) 88n Na (aq) 0.15 M CH3COO (aq) 0.15 M Of these ions, only the acetate ion will hydrolyze. We ignore the contribution to H and OH ions by water. Step 2: Let x be the equilibrium concentration of CH3COO and OH ions in mol/L, we summarize: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 624 ACIDS AND BASES Initial (M ): Change (M ): CH3COO (aq) 0.15 x Equilibrium (M ): 0.15 H2O(l ) 3:4 CH3COOH(aq) 0.00 x x x OH (aq) 0.00 x x Step 3: From the above discussion we write the equilibrium constant of hydrolysis, or the base ionization constant, as Kb [CH3COOH][OH ] 5.6 10 5.6 x2 0.15 x 10 10 10 Since Kb is very small and the initial concentration of the base is large, we can apply the approximation 0.15 x 0.15: x2 0.15 x x2 0.15 x 9.2 5.6 6 10 10 10 M Step 4: At equilibrium: [OH ] 9.2 pOH 10 6 M 10 6) log (9.2 5.04 pH 14.00 5.04 8.96 Thus the solution is basic, as we would expect. The percent hydrolysis is given by % hydrolysis 9.2 10 6 M 0.15 M 100% 0.0061% The result shows that only a very small amount of the anion undergoes hydrolysis. Note that the calculation of percent hydrolysis takes the same form as the test for the approximation, which is valid in this case. Comment Similar problem: 15.77. PRACTICE EXERCISE Calculate the pH of a 0.24 M sodium formate solution (HCOONa). SALTS THAT PRODUCE ACIDIC SOLUTIONS When a salt derived from a strong acid such as HCl and a weak base such as NH3 dissolves in water, the solution becomes acidic. For example, consider the process H2O NH4Cl(s) 88n NH4 (aq) Cl (aq) The Cl ion, being the conjugate base of a strong acid, has no affinity for H and no tendency to hydrolyze. The ammonium ion NH4 is the weak conjugate acid of the weak base NH3 and ionizes as follows: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 15.10 ACID-BASE PROPERTIES OF SALT – 3+ Al(H2O)6 FIGURE 15.7 The six H2O molecules surround the Al3 ion octahedrally. The attraction of the small Al3 ion for the lone pairs on the oxygen atoms is so great that the OOH bonds in a H2O molecule attached to the metal cation are weakened, allowing the loss of a proton (H ) to an incoming H2O molecule. By coincidence, Ka of NH4 has the same numerical value as Kb of CH3COO . + 2+ Al(OH)(H2O)5 H2O NH4 (aq) H2O(l ) 3:4 NH3(aq) + 625 + H3O+ H3O (aq) or simply NH4 (aq) 3:4 NH3(aq) H (aq) Note that this reaction also represents the hydrolysis of the NH 4 ion. Since H ions are produced, the pH of the solution decreases. The equilibrium constant (or ionization constant) for this process is given by Ka [NH3][H ] [NH4 ] Kw Kb 1.0 1.8 10 10 14 5 5.6 10 10 and we can calculate the pH of an ammonium chloride solution following the same procedure used in Example 15.13. In principle, all metal ions react with water to produce an acidic solution. However, because the extent of hydrolysis is most pronounced for the small and highly charged metal cations such as Al3 , Cr3 , Fe3 , Bi3 , and Be2 , we generally neglect the relatively small interaction of alkali metal ions and most alkaline earth metal ions with water. When aluminum chloride (AlCl3) dissolves in water, the Al3 ions take the hydrated form Al(H2O)3 (Figure 15.7). Let us consider one bond between the metal ion 6 and an oxygen atom from one of the six water molecules in Al(H2O)3 : 6 H Al O H 3 The hydrated Al3 qualifies as a proton donor and thus a Brønsted acid in this reaction. Back Forward Main Menu The positively charged Al ion draws electron density toward itself, increasing the polarity of the OOH bonds. Consequently, the H atoms have a greater tendency to ionize than those in water molecules not involved in hydration. The resulting ionization process can be written as Al(H2O)3 (aq) 6 TOC H2O(l ) 3:4 Al(OH)(H2O)2 (aq) 5 Study Guide TOC H3O (aq) Textbook Website MHHE Website 626 ACIDS AND BASES Al(H2O)3 (aq) 3:4 Al(OH)(H2O)2 (aq) 6 5 or simply H (aq) The equilibrium constant for the metal cation hydrolysis is given by Note that Al(H2O)3 is roughly as 6 strong an acid as CH3COOH. [Al(OH)(H2O)2 ][H ] 5 [Al(H2O)3 ] 6 Ka 1.3 5 10 Note that Al(OH)(H2O)2 can undergo further ionization 5 Al(OH)(H2O)2 (aq) 3:4 Al(OH)2(H2O)4 (aq) 5 H (aq) and so on. However, it is generally sufficient to take into account only the first stage of hydrolysis. The extent of hydrolysis is greatest for the smallest and most highly charged ions because a “compact” highly charged ion is more effective in polarizing the OOH bond and facilitating ionization. This is why relatively large ions of low charge such as Na and K do not undergo appreciable hydrolysis. The following example shows the calculation of the pH of an Al(NO3)3 solution. EXAMPLE 15.14 Calculate the pH of a 0.020 M Al(NO3)3 solution. Answer Step 1: Since Al(NO3)3 is a strong electrolyte, the initial dissociation is HO 2 Al(NO3)3(s) 88n Al3 (aq) 0.020 M 3NO3 (aq) 0.060 M Only the Al3 ion will hydrolyze. As in the case of NH4 , we can treat the hydrolysis of Al3 as the ionization of its hydrated ion. Step 2: Let x be the equilibrium concentration of Al(OH)(H2O)2 and of H in mol/L: 5 Al(H2O)3 (aq) 3:4 Al(OH)(H2O)2 (aq) 6 5 Initial (M): Change (M): H (aq) 0.020 0.00 x Equilibrium (M): 0.00 x x x x 0.020 x Step 3: The equilibrium constant for the ionization is Ka [Al(OH)(H2O)2 ][H ] 5 [Al(H2O)3 ] 6 x2 0.020 Applying the approximation 0.020 2 x 0.020 x x 1.3 10 5 1.3 10 5 0.020, we have 2 x 0.020 1.3 10 5 x x 5.1 10 4 M Thus the pH of the solution is pH log (5.1 10 4) 3.29 PRACTICE EXERCISE What is the pH of a 0.050 M AlCl3 solution? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 15.10 ACID-BASE PROPERTIES OF SALT 627 SALTS IN WHICH BOTH THE CATION AND THE ANION HYDROLYZE So far we have considered salts in which only one ion undergoes hydrolysis. For salts derived from a weak acid and a weak base, both the cation and the anion hydrolyze. However, whether a solution containing such a salt is acidic, basic, or neutral depends on the relative strengths of the weak acid and the weak base. Since the mathematics associated with this type of system is rather involved, we will focus on making qualitative predictions about these solutions based on the following guidelines: Kb Ka. If Kb for the anion is greater than Ka for the cation, then the solution must be basic because the anion will hydrolyze to a greater extent than the cation. At equilibrium, there will be more OH ions than H ions. • Kb Ka. Conversely, if Kb for the anion is smaller than Ka for the cation, the solution will be acidic because cation hydrolysis will be more extensive than anion hydrolysis. • Ka Kb. If Ka is approximately equal to Kb, the solution will be nearly neutral. • Table 15.7 summarizes the behavior in aqueous solution of the salts discussed in this section. Example 15.15 illustrates how to predict the acid-base properties of salt solutions. EXAMPLE 15.15 Predict whether the following solutions will be acidic, basic, or nearly neutral: (a) NH4I, (b) CaCl2, (c) KCN, (d) Fe(NO3)3. Similar problem: 15.73. Answer (a) The cation is NH4 , which will hydrolyze to produce NH3 and H . The I anion is the conjugate base of the strong acid HI. Therefore, I will not hydrolyze and so the solution is acidic. (b) The Ca2 cation does not hydrolyze. The Cl anion also does not hydrolyze. The solution will be almost neutral. (c) The K cation does not hydrolyze. CN is the conjugate base of the weak acid HCN and will hydrolyze to give OH and HCN. The solution will be basic. (d) Fe3 is a small metal ion with a high charge and hydrolyzes to produce H ions. The NO3 ion does not hydrolyze. Consequently, the solution will be acidic. PRACTICE EXERCISE Predict the pH of the following salt solutions: (a) LiClO4, (b) Na3PO4, (c) Bi(NO3)3, (d) NH4CN. TABLE 15.7 Acid-Base Properties of Salts TYPE OF SALT EXAMPLES IONS THAT UNDERGO HYDROLYSIS Cation Cation Cation Cation NaCl, KI, KNO3, RbBr, BaCl2 CH3COONa, KNO2 NH4Cl, NH4NO3 NH4NO2, CH3COONH4, NH4CN None Anion Cation Anion and cation AlCl3, Fe(NO3)3 Hydrated cation from from from from strong base; anion from strong acid strong base; anion from weak acid weak base; anion from strong acid weak base; anion from weak acid Small, highly charged cation; anion from strong acid Back Forward Main Menu TOC Study Guide TOC Textbook Website pH OF SOLUTION 7 7 7 7 if Kb 7 if Kb 7 if Kb 7 Ka Ka Ka MHHE Website 628 ACIDS AND BASES Finally we note that some anions can act either as an acid or as a base. For example, the bicarbonate ion (HCO3 ) can ionize or undergo hydrolysis as follows (see Table 15.5): HCO3 (aq) H2O(l ) 3:4 H3O (aq) CO2 (aq) 3 Ka 4.8 10 11 HCO3 (aq) H2O(l ) 3:4 H2CO3(aq) OH (aq) Kb 2.4 10 8 Since Kb Ka, we predict that the hydrolysis reaction will outweigh the ionization process. Thus, a solution of sodium bicarbonate (NaHCO3) will be basic. 15.11 ACID-BASE PROPERTIES OF OXIDES AND HYDROXIDES As we saw in Chapter 8, oxides can be classified as acidic, basic, or amphoteric. Our discussion of acid-base reactions would be incomplete if we did not examine the properties of these compounds. Figure 15.8 shows the formulas of a number of oxides of the representative elements in their highest oxidation states. Note that all alkali metal oxides and all alkaline earth metal oxides except BeO are basic. Beryllium oxide and several metallic oxides in Groups 3A and 4A are amphoteric. Nonmetallic oxides in which the oxidation number of the representative element is high are acidic (for example, N2O5, SO3, and Cl2O7), but those in which the oxidation number of the representative element is low (for example, CO and NO) show no measurable acidic properties. No nonmetallic oxides are known to have basic properties. The basic metallic oxides react with water to form metal hydroxides: H2O Na2O(s) 88n 2NaOH(aq) H2O BaO(s) 88n Ba(OH)2(aq) FIGURE 15.8 Oxides of the representative elements in their highest oxidation states. 1 1A 18 8A Basic oxide 2 2A Li2O BeO Amphoteric oxide Na2O MgO K2O 13 3A 14 4A 15 5A B2O3 CO2 N2O5 Al2O3 SiO2 P4O10 SO3 Cl2O7 CaO Ga2O3 GeO2 As2O5 SeO3 Br2O7 Rb2O SrO In2O3 SnO2 Sb2O5 TeO3 I2O7 Cs2O Back Acidic oxide BaO Tl2O3 PbO2 Bi2O5 PoO3 At2O7 3 3B Forward 4 4B 5 5B Main Menu 6 6B 7 7B TOC 8 9 8B 10 11 1B 12 2B Study Guide TOC 16 6A 17 7A OF2 Textbook Website MHHE Website 15.11 ACID-BASE PROPERTIES OF OXIDES AND HYDROXIDES 629 The reactions between acidic oxides and water are as follows: CO2(g) H2O(l ) 3:4 H2CO3(aq) SO3(g) H2O(l ) 3:4 H2SO4(aq) N2O5(g) H2O(l ) 3:4 2HNO3(aq) P4O10(s) Cl2O7(l ) We will look at the causes and effects of acid rain in Chapter 17. 6H2O(l ) 3:4 4H3PO4(aq) H2O(l ) 3:4 2HClO4(aq) The reaction between CO2 and H2O explains why when pure water is exposed to air (which contains CO2) it gradually reaches a pH of about 5.5 (Figure 15.9). The reaction between SO3 and H2O is largely responsible for acid rain (Figure 15.10). Reactions between acidic oxides and bases and those between basic oxides and acids resemble normal acid-base reactions in that the products are a salt and water: CO2(g) acidic oxide 2NaOH(aq) 88n Na2CO3(aq) base salt BaO(s) basic oxide 2HNO3(aq) 88n Ba(NO3)2(aq) acid salt H2O(l ) water H2O(l ) water As Figure 15.8 shows, aluminum oxide (Al2O3) is amphoteric. Depending on the reaction conditions, it can behave either as an acidic oxide or as a basic oxide. For example, Al2O3 acts as a base with hydrochloric acid to produce a salt (AlCl3) and water: Al2O3(s) 6HCl(aq) 88n 2AlCl3(aq) 3H2O(l ) and acts as an acid with sodium hydroxide: Al2O3(s) 2NaOH(aq) 3H2O(l ) 88n 2NaAl(OH)4(aq) FIGURE 15.9 (Left) A beaker of water to which a few drops of bromothymol blue indicator have been added. (Right) As dry ice is added to the water, the CO2 reacts to form carbonic acid, which turns the solution acidic and changes the color from blue to yellow. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 630 ACIDS AND BASES FIGURE 15.10 A forest damaged by acid rain. ” ion ers V xt -Te e in ble a ail v tA o “N The higher the oxidation number of the metal, the more covalent the compound; the lower the oxidation number, the more ionic the compound. Note that only a salt, NaAl(OH)4 [containing the Na and Al(OH)4 ions], is formed in the latter reaction; no water is produced. Nevertheless, this reaction can still be classified as an acid-base reaction because Al2O3 neutralizes NaOH. Some transition metal oxides in which the metal has a high oxidation number act as acidic oxides. Two examples are manganese(VII) oxide (Mn2O7) and chromium(VI) oxide (CrO3), both of which react with water to produce acids: Mn2O7 (l ) CrO3(s) H2O(l ) 88n 2HMnO4(aq) permanganic acid H2O(l ) 88n H2CrO4(aq) chromic acid BASIC AND AMPHOTERIC HYDROXIDES We have seen that the alkali and alkaline earth metal hydroxides [except Be(OH)2] are basic in properties. The following hydroxides are amphoteric: Be(OH)2, Al(OH)3, Sn(OH)2, Pb(OH)2, Cr(OH)3, Cu(OH)2, Zn(OH)2, and Cd(OH)2. For example, aluminum hydroxide reacts with both acids and bases: Al(OH)3(s) 3H (aq) 88777n Al3 (aq) Al(OH)3(s) OH (aq) 3:4 Al(OH)4 (aq) 3H2O(l ) All amphoteric hydroxides are insoluble. It is interesting that beryllium hydroxide, like aluminum hydroxide, exhibits amphoterism: Be(OH)2(s) Be(OH)2(s) 2H (aq) 88777n Be2 (aq) 2H2O(l ) 2OH (aq) 3:4 Be(OH)2 (aq) 4 This is another example of the diagonal relationship between beryllium and aluminum. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 15.12 LEWIS ACIDS AND BASES 631 15.12 LEWIS ACIDS AND BASES So far we have discussed acid-base properties in terms of the Brønsted theory. To behave as a Brønsted base, for example, a substance must be able to accept protons. By this definition both the hydroxide ion and ammonia are bases: O SOOH Q H H Lewis presented his theory of acids and bases in 1932, the same year in which Brønsted introduced his. All Brønsted bases are Lewis bases. H A SNOH A H HOOOH O Q H A HONOH A H In each case, the atom to which the proton becomes attached possesses at least one unshared pair of electrons. This characteristic property of OH , NH3, and other Brønsted bases suggests a more general definition of acids and bases. The American chemist G. N. Lewis formulated such a definition. He defined what we now call a Lewis base as a substance that can donate a pair of electrons. A Lewis acid is a substance that can accept a pair of electrons. For example, in the protonation of ammonia, NH3 acts as a Lewis base because it donates a pair of electrons to the proton H , which acts as a Lewis acid by accepting the pair of electrons. A Lewis acid-base reaction, therefore, is one that involves the donation of a pair of electrons from one species to another. Such a reaction does not produce a salt and water. The significance of the Lewis concept is that it is much more general than other definitions. Lewis acid-base reactions include many reactions that do not involve Brønsted acids. Consider, for example, the reaction between boron trifluoride (BF3) and ammonia to form an adduct compound (Figure 15.11): F A F OB A F acid H A SNOH A H base FH AA F O BONOH AA FH In Section 10.4 we saw that the B atom in BF3 is sp2-hybridized. The vacant, unhybridized 2p orbital accepts the pair of electrons from NH3. So BF3 functions as an acid according to the Lewis definition, even though it does not contain an ionizable proton. Note that a coordinate covalent bond is formed between the B and N atoms, as is the case in all Lewis acid-base reactions. Another Lewis acid containing boron is boric acid. Boric acid (a weak acid used in eyewash) is an oxoacid with the following structure: H A S OS A O HOQOB OOOH O O Q boric acid Boric acid does not ionize in water to produce a H ion. Its reaction with water is B(OH)3(aq) FIGURE 15.11 A Lewis acidbase reaction involving BF3 and NH3. Back Forward Main Menu H2O(l ) 3:4 B(OH)4 (aq) H (aq) In this Lewis acid-base reaction, boric acid accepts a pair of electrons from the hydroxide ion that is derived from the H2O molecule. TOC Study Guide TOC Textbook Website MHHE Website 632 ACIDS AND BASES The hydration of carbon dioxide to produce carbonic acid CO2(g) Carbon is a second-period element. It cannot expand its valence shell and can accommodate only eight valence electrons. H2O(l ) 3:4 H2CO3(aq) can be understood in the Lewis framework as follows: The first step involves donation of a lone pair on the oxygen atom in H2O to the carbon atom in CO2. An orbital is vacated on the C atom to accommodate the lone pair by removal of the electron pair in the COO pi bond. These shifts of electrons are indicated by the curved arrows. H S G O D H S O OS B C B SOS O SOS A HOOOC O AB H SOS Therefore, H2O is a Lewis base and CO2 is a Lewis acid. Next, a proton is transferred onto the O atom bearing a negative charge to form H2CO3. O SOS A OO C HOO AB H SOS O HOOS A SOO C O AB H SOS Other examples of Lewis acid-base reactions are Ag (aq) acid Cd2 (aq) acid Ni(s) acid 2NH3(aq) 3:4 Ag(NH3)2 (aq) base 4I (aq) 3:4 CdI2 (aq) 4 base 4CO(g) 3:4 Ni(CO)4(g) base It is important to note that the hydration of metal ions in solution is in itself a Lewis acid-base reaction. Thus, when copper(II) sulfate (CuSO4) dissolves in water, each Cu2 ion is associated with six water molecules as Cu(H2O)2 . In this case, the 6 Cu2 ion acts as the acid and the H2O molecules act as the base. Although the Lewis definition of acids and bases has greater significance because of its generality, we normally speak of “an acid” and “a base” in terms of the Brønsted definition. The term “Lewis acid” usually is reserved for substances that can accept a pair of electrons but do not contain ionizable hydrogen atoms. Classification of Lewis acids and bases is demonstrated in the following example. EXAMPLE 15.16 Identify the Lewis acid and Lewis base in each of the following reactions: (a) SnCl4(s) 2Cl (aq) 3:4 SnCl2 (aq) 6 (b) Hg2 (aq) 4CN (aq) 3:4 Hg(CN)2 (aq) 4 (a) In this reaction, SnCl4 accepts two pairs of electrons from the Cl ions. Therefore SnCl4 is the Lewis acid and Cl is the Lewis base. (b) Here the Hg2 ion accepts four pairs of electrons from the CN ions. Therefore Hg2 is the Lewis acid and CN is the Lewis base. Answer Similar problem: 15.90. PRACTICE EXERCISE Identify the Lewis acid and Lewis base in the reaction Co3 (aq) Back Forward Main Menu TOC 6NH3(aq) 3:4 Co(NH3)3 (aq) 6 Study Guide TOC Textbook Website MHHE Website 15.12 633 LEWIS ACIDS AND BASES Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Back Antacids and the pH Balance in Your Stomach An average adult produces between 2 and 3 L of gastric juice daily. Gastric juice is a thin, acidic digestive fluid secreted by glands in the mucous membrane that lines the stomach. It contains, among other substances, hydrochloric acid. The pH of gastric juice is about 1.5, which corresponds to a hydrochloric acid concentration of 0.03 M — a concentration strong enough to dissolve zinc metal! What is the purpose of this highly acidic medium? Where do the H ions come from? What happens when there is an excess of H ions present in the stomach? A simplified diagram of the stomach is shown on the right. The inside lining is made up of parietal cells, which are fused together to form tight junctions. The interiors of the cells are protected from the surroundings by cell membranes. These membranes allow water and neutral molecules to pass in and out of the stomach, but they usually block the movement of ions such as H , Na , K , and Cl ions. The H ions come from carbonic acid (H2CO3) formed as a result of the hydration of CO2, an end product of metabolism: CO2(g) Mucous membrane Blood Plasma HCl(aq) Blood plasma To intestines Cl– H+ (active transport) A simplified diagram of the human stomach. H2O(l ) 3:4 H2CO3(aq) H2CO2(aq) 3:4 H (aq) HCO3 (aq) These reactions take place in the blood plasma bathing the cells in the mucosa. By a process known as active transport, H ions move across the membrane into the stomach interior. (Active transport processes are aided by enzymes.) To maintain electrical balance, an equal number of Cl ions also move from the blood plasma into the stomach. Once in the stomach, most of these ions are prevented from diffusing back into the blood plasma by cell membranes. The purpose of the highly acidic medium within the stomach is to digest food and to activate certain digestive enzymes. Eating stimulates H ion secretion. A small fraction of these ions normally are reabsorbed by the mucosa, causing many tiny hemorrhages. About half a million cells are shed by the lining every minute, and a healthy stomach is completely relined every three days or so. However, if the acid content is excessively high, the constant influx of H ions Forward Food Main Menu TOC through the membrane back to the blood plasma can cause muscle contraction, pain, swelling, inflammation, and bleeding. One way to temporarily reduce the H ion concentration in the stomach is to take an antacid. The major function of antacids is to neutralize excess HCl in gastric juice. The table on p. 634 lists the active ingredients of some popular antacids. The reactions by which these antacids neutralize stomach acid are as follows: NaHCO3(aq) HCl(aq) 88n NaCl(aq) H2O(l ) CO2(g) CaCO3(s) 2HCl(aq) 88n CaCl2(aq) H2O(l ) CO2(g) MgCO3(s) 2HCl(aq) 88n MgCl2(aq) H2O(l ) CO2(g) Mg(OH)2(s) 2HCl(aq) 88n MgCl2(aq) Al(OH)2NaCO3(s) 4HCl(aq) 88n AlCl3(aq) NaCl(aq) 3H2O(l ) Study Guide TOC Textbook Website 2H2O(l ) CO2(g) MHHE Website 634 ACIDS AND BASES Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Action Some Common Commercial Antacid Preparations COMMERCIAL NAME ACTIVE INGREDIENTS Alka-2 Alka-Seltzer Calcium carbonate Aspirin, sodium bicarbonate, citric acid Aspirin, magnesium carbonate, aluminum glycinate Aspirin, magnesium carbonate, aluminum hydroxide-glycine Magnesium hydroxide Dihydroxy aluminum sodium carbonate Calcium carbonate Bufferin Buffered aspirin Milk of magnesia Rolaids Tums The CO2 released by most of these reactions increases gas pressure in the stomach, causing the person to belch. The fizzing that takes place when an AlkaSeltzer tablet dissolves in water is caused by carbon dioxide, which is released by the reaction between citric acid and sodium bicarbonate: C4H7O5(COOH)(aq) C4H7O5COONa(aq) H2O(l) CO2(g) sodium citrate This action helps to disperse the ingredients and even enhances the taste of the solution. The mucosa of the stomach is also damaged by the action of aspirin, the chemical name of which is acetylsalicylic acid. Aspirin is itself a moderately weak acid: O B OO COCH3 O B OOCOCH3 CO OH B O COO B O acetylsalicylic acid H acetylsalicylate ion In the presence of the high concentration of H ions in the stomach, this acid remains largely nonionized. SUMMARY OF KEY EQUATIONS Back NaHCO3(aq) 88n citric acid Forward • • • • Main Menu Kw [H ][OH ] pH log [H ] pOH log [OH ] pH pOH 14.00 TOC (15.2) (15.3) (15.4) (15.5) An Alka-Seltzer tablet dissolved in water gives off carbon dioxide gas. A relatively nonpolar molecule, acetylsalicylic acid has the ability to penetrate membrane barriers that are also made up of nonpolar molecules. However, inside the membrane are many small water pockets, and when an acetylsalicylic acid molecule enters such a pocket, it ionizes into H and acetylsalicylate ions. These ionic species become trapped in the interior regions of the membrane. The continued buildup of ions in this fashion weakens the structure of the membrane and eventually causes bleeding. Approximately 2 mL of blood are usually lost for every aspirin tablet taken, an amount not generally considered harmful. However, the action of aspirin can result in severe bleeding in some individuals. It is interesting to note that the presence of alcohol makes acetylsalicylic acid even more soluble in the membrane, and so further promotes the bleeding. Ion-product constant of water. Definition of pH of a solution. Definition of pOH of a solution. Another form of Equation (15.2). Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS • percent ionization • KaKb SUMMARY OF FACTS AND CONCEPTS Kw 635 ionized acid concentration at equilibrium 100% (15.6) initial concentration of acid (15.7) Relationship between the acid and base ionization constants of a conjugate acid-base pair. 1. Brønsted acids donate protons, and Brønsted bases accept protons. These are the definitions that normally underlie the use of the terms “acid” and “base.” 2. The acidity of an aqueous solution is expressed as its pH, which is defined as the negative logarithm of the hydrogen ion concentration (in mol/L). 3. At 25°C, an acidic solution has pH 7, a basic solution has pH 7, and a neutral solution has pH 7. 4. In aqueous solution, the following are classified as strong acids: HClO4, HI, HBr, HCl, H2SO4 (first stage of ionization), and HNO3. Strong bases in aqueous solution include hydroxides of alkali metals and of alkaline earth metals (except beryllium). 5. The acid ionization constant Ka increases with acid strength. Kb similarly expresses the strengths of bases. 6. Percent ionization is another measure of the strength of acids. The more dilute a solution of a weak acid, the greater the percent ionization of the acid. 7. The product of the ionization constant of an acid and the ionization constant of its conjugate base is equal to the ion-product constant of water. 8. The relative strengths of acids can be explained qualitatively in terms of their molecular structures. 9. Most salts are strong electrolytes that dissociate completely into ions in solution. The reaction of these ions with water, called salt hydrolysis, can produce acidic or basic solutions. In salt hydrolysis, the conjugate bases of weak acids yield basic solutions, and the conjugate acids of weak bases yield acidic solutions. 10. Small, highly charged metal ions, such as Al3 and Fe3 , hydrolyze to yield acidic solutions. 11. Most oxides can be classified as acidic, basic, or amphoteric. Metal hydroxides are either basic or amphoteric. 12. Lewis acids accept pairs of electrons and Lewis bases donate pairs of electrons. The term “Lewis acid” is generally reserved for substances that can accept electron pairs but do not contain ionizable hydrogen atoms. KEY WORDS Acid ionization constant (Ka), p. 607 Base ionization constant (Kb), p. 613 Conjugate acid-base pair, p. 598 Ion-product constant, p. 600 Lewis acid, p. 631 Lewis base, p. 631 Percent ionization, p. 612 pH, p. 601 Salt hydrolysis, p. 622 Strong acid, p. 603 Strong base, p. 603 Weak acid, p. 603 Weak base, p. 604 QUESTIONS AND PROBLEMS† atom in the species must possess a lone pair of electrons. Explain why this is so. BRØNSTED ACIDS AND BASES Review Questions 15.1 Define Brønsted acids and bases. Give an example of a conjugate pair in an acid-base reaction. 15.2 In order for a species to act as a Brønsted base, an † Unless otherwise stated, the temperature is assumed to be 25°C. Back Forward Main Menu TOC Problems 15.3 Classify each of the following species as a Brønsted acid or base, or both: (a) H2O, (b) OH , (c) H3O , (d) NH3, (e) NH4 , (f) NH2 , (g) NO3 , (h) CO2 , 3 (i) HBr, (j) HCN. Study Guide TOC Textbook Website MHHE Website 636 ACIDS AND BASES 15.4 Write the formulas of the conjugate bases of the following acids: (a) HNO2, (b) H2SO4, (c) H2S, (d) HCN, (e) HCOOH (formic acid). 15.5 Identify the acid-base conjugate pairs in each of the following reactions: HCN 3:4 CH3COOH CN (a) CH3COO (b) HCO3 HCO3 3:4 H2CO3 CO2 3 (c) H2PO4 NH3 3:4 HPO2 NH4 4 (d) HClO CH3NH2 3:4 CH3NH3 ClO (e) CO2 H2O 3:4 HCO3 OH 3 15.6 Write the formula for the conjugate acid of each of the following bases: (a) HS , (b) HCO3 , (c) CO2 , 3 (d) H2PO4 , (e) HPO2 , (f) PO3 , (g) HSO4 , (h) 4 4 SO2 , (i) SO2 . 4 3 15.7 Oxalic acid (C2H2O4) has the following structure: OPCO OH A OPCO OH An oxalic acid solution contains the following species in varying concentrations: C2H2O4, C2HO4 , C2O2 , and H . (a) Draw Lewis structures of 4 C2HO4 and C2O2 . (b) Which of the above four 4 species can act only as acids, which can act only as bases, and which can act as both acids and bases? 15.8 Write the formula for the conjugate base of each of the following acids: (a) CH2ClCOOH, (b) HIO4, (c) H3PO4, (d) H2PO4 , (e) HPO2 , (f) H2SO4, 4 (g) HSO4 , (h) HIO3, (i) HSO3 , ( j) NH4 , (k) H2S, (l) HS , (m) HClO. THE ACID-BASE PROPERTIES OF WATER Review Questions 15.9 What is the ion-product constant for water? 15.10 Write an equation relating [H ] and [OH ] in solution at 25°C. 15.11 The ion-product constant for water is 1.0 10 14 at 25°C and 3.8 10 14 at 40°C. Is the forward process H2O(l ) 3:4 H (aq) OH (aq) endothermic or exothermic? pH — A MEASURE OF ACIDITY Can the pH of a solution be zero or negative? If so, give examples to illustrate these values. 15.14 Define pOH. Write the equation relating pH and pOH. Problems 15.15 Calculate the concentration of OH ions in a 1.4 10 3 M HCl solution. 15.16 Calculate the concentration of H ions in a 0.62 M NaOH solution. 15.17 Calculate the pH of each of the following solutions: (a) 0.0010 M HCl, (b) 0.76 M KOH. 15.18 Calculate the pH of each of the following solutions: (a) 2.8 10 4 M Ba(OH)2, (b) 5.2 10 4 M HNO3. 15.19 Calculate the hydrogen ion concentration in mol/L for solutions with the following pH values: (a) 2.42, (b) 11.21, (c) 6.96, (d) 15.00. 15.20 Calculate the hydrogen ion concentration in mol/L for each of the following solutions: (a) a solution whose pH is 5.20, (b) a solution whose pH is 16.00, (c) a solution whose hydroxide concentration is 3.7 10 9 M. 15.21 Complete the following table for a solution: pH [H ] SOLUTION IS 7 1.0 10 7 M Neutral 15.22 Fill in the word acidic, basic, or neutral for the following solutions: (a) pOH 7; solution is (b) pOH 7; solution is (c) pOH 7; solution is 15.23 The pOH of a solution is 9.40. Calculate the hydrogen ion concentration of the solution. 15.24 Calculate the number of moles of KOH in 5.50 mL of a 0.360 M KOH solution. What is the pOH of the solution? 15.25 How much NaOH (in grams) is needed to prepare 546 mL of solution with a pH of 10.00? 15.26 A solution is made by dissolving 18.4 g of HCl in 662 mL of water. Calculate the pH of the solution. (Assume that the volume remains constant.) Review Questions 15.12 Define pH. Why do chemists normally choose to discuss the acidity of a solution in terms of pH rather than hydrogen ion concentration, [H ]? 15.13 The pH of a solution is 6.7. From this statement alone, can you conclude that the solution is acidic? If not, what additional information would you need? Back Forward Main Menu TOC STRENGTH OF ACIDS AND BASES Review Questions 15.27 Explain what is meant by the strength of an acid. 15.28 Without referring to the text, write the formulas of four strong acids and four weak acids. Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 15.29 What are the strongest acid and strongest base that can exist in water? 15.30 H2SO4 is a strong acid, but HSO4 is a weak acid. Account for the difference in strength of these two related species. Problems 15.31 Classify each of the following species as a weak or strong acid: (a) HNO3, (b) HF, (c) H2SO4, (d) HSO4 , (e) H2CO3, (f) HCO3 , (g) HCl, (h) HCN, (i) HNO2. 15.32 Classify each of the following species as a weak or strong base: (a) LiOH, (b) CN , (c) H2O, (d) ClO4 , (e) NH2 . 15.33 Which of the following statements is/are true for a 0.10 M solution of a weak acid HA? (a) The pH is 1.00. (b) [H ] [A ] (c) [H ] [A ] (d) The pH is less than 1. 15.34 Which of the following statements is/are true regarding a 1.0 M solution of a strong acid HA? (a) [A ] [H ] (b) The pH is 0.00. (c) [H ] 1.0 M (d) [HA] 1.0 M 15.35 Predict the direction that predominates in this reaction: F (aq) H2O(l ) 3:4 HF(aq) OH (aq) 15.36 Predict whether the following reaction will proceed from left to right to any measurable extent: CH3COOH(aq) Cl (aq) 88n WEAK ACIDS AND ACID IONIZATION CONSTANTS Review Questions 15.37 What does the ionization constant tell us about the strength of an acid? 15.38 List the factors on which the Ka of a weak acid depends. 15.39 Why do we normally not quote Ka values for strong acids such as HCl and HNO3? Why is it necessary to specify temperature when giving Ka values? 15.40 Which of the following solutions has the highest pH? (a) 0.40 M HCOOH, (b) 0.40 M HClO4, (c) 0.40 M CH3COOH. Problems 15.41 The Ka for benzoic acid is 6.5 10 5. Calculate the pH of a 0.10 M benzoic acid solution. Back Forward Main Menu TOC 637 15.42 A 0.0560-g quantity of acetic acid is dissolved in enough water to make 50.0 mL of solution. Calculate the concentrations of H , CH3COO , and CH3COOH at equilibrium. (Ka for acetic acid 1.8 10 5.) 15.43 The pH of an acid solution is 6.20. Calculate the Ka for the acid. The acid concentration is 0.010 M. 15.44 What is the original molarity of a solution of formic acid (HCOOH) whose pH is 3.26 at equilibrium? 15.45 Calculate the percent ionization of benzoic acid at the following concentrations: (a) 0.20 M, (b) 0.00020 M. 15.46 Calculate the percent ionization of hydrofluoric acid at the following concentrations: (a) 0.60 M, (b) 0.0046 M, (c) 0.00028 M. Comment on the trends. 15.47 A 0.040 M solution of a monoprotic acid is 14 percent ionized. Calculate the ionization constant of the acid. 15.48 (a) Calculate the percent ionization of a 0.20 M solution of the monoprotic acetylsalicylic acid (aspirin) for which Ka 3.0 10 4. (b) The pH of gastric juice in the stomach of a certain individual is 1.00. After a few aspirin tablets have been swallowed, the concentration of acetylsalicylic acid in the stomach is 0.20 M. Calculate the percent ionization of the acid under these conditions. What effect does the nonionized acid have on the membranes lining the stomach? (Hint: See the Chemistry in Action essay on p. 633.) WEAK BASES AND BASE IONIZATION CONSTANTS Review Questions 15.49 Use NH3 to illustrate what we mean by the strength of a base. 15.50 Which of the following has a higher pH? (a) 0.20 M NH3, (b) 0.20 M NaOH Problems 15.51 Calculate the pH for each of the following solutions: (a) 0.10 M NH3, (b) 0.050 M C5H5N (pyridine). 15.52 The pH of a 0.30 M solution of a weak base is 10.66. What is the Kb of the base? 15.53 What is the original molarity of a solution of ammonia whose pH is 11.22? 15.54 In a 0.080 M NH3 solution, what percent of the NH3 is present as NH4 ? THE RELATIONSHIP BETWEEN THE IONIZATION CONSTANTS OF ACIDS AND THEIR CONJUGATE BASES Review Questions 15.55 Write the equation relating Ka for a weak acid and Kb for its conjugate base. Use NH3 and its conjugate acid NH4 to derive the relationship between Ka and Kb. Study Guide TOC Textbook Website MHHE Website 638 ACIDS AND BASES 15.56 From the relationship KaKb Kw, what can you deduce about the relative strengths of a weak acid and its conjugate base. ACID-BASE PROPERTIES OF SALT SOLUTIONS DIPROTIC AND POLYPROTIC ACIDS Review Questions Review Questions 15.57 Carbonic acid is a diprotic acid. Explain what that means. 15.58 Write all the species (except water) that are present in a phosphoric acid solution. Indicate which species can act as a Brønsted acid, which as a Brønsted base, and which as both a Brønsted acid and a Brønsted base. Problems 15.59 The first and second ionization constants of a diprotic acid H2A are Ka1 and Ka2 at a certain temperature. Under what conditions will [A2 ] Ka2? 15.60 Compare the pH of a 0.040 M HCl solution with that of a 0.040 M H2SO4 solution. 15.61 What are the concentrations of HSO4 , SO2 , and 4 H in a 0.20 M KHSO4 solution? (Hint: H2SO4 is 1.3 10 2.) a strong acid; Ka for HSO4 15.62 Calculate the concentrations of H , HCO3 , and CO2 in a 0.025 M H2CO3 solution. 3 MOLECULAR STRUCTURE AND THE STRENGTH OF ACIDS Review Questions 15.63 List four factors that affect the strength of an acid. 15.64 How does the strength of an oxoacid depend on the electronegativity and oxidation number of the central atom? Problems 15.65 Predict the acid strengths of the following compounds: H2O, H2S, and H2Se. 15.66 Compare the strengths of the following pairs of acids: (a) H2SO4 and H2SeO4, (b) H3PO4 and H3AsO4. 15.67 Which of the following acids is the stronger: CH3COOH or CH2ClCOOH? Explain your choice. (Hint: Chlorine is more electronegative than hydrogen. How would its presence affect the ionization of the carboxyl group?) 15.68 Consider the following compounds: OOH phenol CH3 OOOH Forward Main Menu TOC 15.69 Define salt hydrolysis. Categorize salts according to how they affect the pH of a solution. 15.70 Explain why small, highly charged metal ions are able to undergo hydrolysis. 15.71 Al3 is not a Brønsted acid but Al(H2O)3 is. 6 Explain. 15.72 Specify which of the following salts will undergo hydrolysis: KF, NaNO3, NH4NO2, MgSO4, KCN, C6H5COONa, RbI, Na2CO3, CaCl2, HCOOK. 15.73 Predict the pH ( 7, 7, or 7) of aqueous solutions containing the following salts: (a) KBr, (b) Al(NO3)3, (c) BaCl2, (d) Bi(NO3)3. 15.74 Which ion of the alkaline earth metals is most likely to undergo hydrolysis? Problems 15.75 A certain salt, MX (containing the M and X ions), is dissolved in water, and the pH of the resulting solution is 7.0. Can you say anything about the strengths of the acid and the base from which the salt is derived? 15.76 In a certain experiment a student finds that the pHs of 0.10 M solutions of three potassium salts KX, KY, and KZ are 7.0, 9.0, and 11.0, respectively. Arrange the acids HX, HY, and HZ in the order of increasing acid strength. 15.77 Calculate the pH of a 0.36 M CH3COONa solution. 15.78 Calculate the pH of a 0.42 M NH4Cl solution. 15.79 Predict the pH ( 7, 7, ≈ 7) of a NaHCO3 solution. 15.80 Predict whether a solution containing the salt K2HPO4 will be acidic, neutral, or basic. ACIDIC AND BASIC OXIDES AND HYDROXIDES Review Questions 15.81 Classify the following oxides as acidic, basic, amphoteric, or neutral: (a) CO2, (b) K2O, (c) CaO, (d) N2O5, (e) CO, (f) NO, (g) SnO2, (h) SO3, (i) Al2O3, (j) BaO. 15.82 Write equations for the reactions between (a) CO2 and NaOH(aq), (b) Na2O and HNO3(aq). Problems methanol Experimentally, phenol is found to be a stronger acid than methanol. Explain this difference in terms of Back the structures of the conjugate bases. (Hint: A more stable conjugate base favors ionization. Only one of the conjugate bases can be stabilized by resonance.) 15.83 Explain why metal oxides tend to be basic if the oxidation number of the metal is low and acidic if the oxidation number of the metal is high. (Hint: Study Guide TOC Textbook Website MHHE Website 639 QUESTIONS AND PROBLEMS Metallic compounds in which the low oxidation numbers of the metals are low are more ionic than those in which the oxidation numbers of the metals are high.) 15.84 Arrange the oxides in each of the following groups in order of increasing basicity: (a) K2O, Al2O3, BaO, (b) CrO3, CrO, Cr2O3. 15.85 Zn(OH)2 is an amphoteric hydroxide. Write balanced ionic equations to show its reaction with (a) HCl, (b) NaOH [the product is Zn(OH)2 ]. 4 15.86 Al(OH)3 is an insoluble compound. It dissolves in excess NaOH in solution. Write a balanced ionic equation for this reaction. What type of reaction is this? 15.95 To which of the following would the addition of an equal volume of 0.60 M NaOH lead to a solution having a lower pH? (a) water, (b) 0.30 M HCl, (c) 0.70 M KOH, (d) 0.40 M NaNO3 15.96 The pH of a 0.0642 M solution of a monoprotic acid is 3.86. Is this a strong acid? 15.97 Like water, liquid ammonia undergoes autoionization: NH3 15.98 LEWIS ACIDS AND BASES Review Questions 15.87 What are the Lewis definitions of an acid and a base? In what way are they more general than the Brønsted definitions? 15.88 In terms of orbitals and electron arrangements, what must be present for a molecule or an ion to act as a Lewis acid (use H and BF3 as examples)? What must be present for a molecule or ion to act as a Lewis base (use OH and NH3 as examples)? 15.99 15.100 15.101 Problems 15.89 Classify each of the following species as a Lewis acid or a Lewis base: (a) CO2, (b) H2O, (c) I , (d) SO2, (e) NH3, (f) OH , (g) H , (h) BCl3. 15.90 Describe the following reaction in terms of the Lewis theory of acids and bases: AlCl3(s) 15.91 Which would be considered a stronger Lewis acid: (a) BF3 or BCl3, (b) Fe2 or Fe3 ? Explain. 15.92 All Brønsted acids are Lewis acids, but the reverse is not true. Give two examples of Lewis acids that are not Brønsted acids. ADDITIONAL PROBLEMS 15.93 The Ka of formic acid is 1.7 10 4 at 25°C. Will the acid become stronger or weaker at 40°C? Explain. 15.94 A typical reaction between an antacid and the hydrochloric acid in gastric juice is NaHCO3(s) HCl(aq) 3:4 NaCl(aq) H2O(l ) CO2(g) Calculate the volume (in L) of CO2 generated from 0.350 g of NaHCO3 and excess gastric juice at 1.00 atm and 37.0°C. Back Forward Main Menu TOC NH2 (a) Identify the Brønsted acids and Brønsted bases in this reaction. (b) What species correspond to H and OH and what is the condition for a neutral solution? HA and HB are both weak acids although HB is the stronger of the two. Will it take a larger volume of a 0.10 M NaOH solution to neutralize 50.0 mL of 0.10 M HB than would be needed to neutralize 50.0 mL of 0.10 M HA? A solution contains a weak monoprotic acid HA and its sodium salt NaA both at 0.1 M concentration. Show that [OH ] Kw/Ka. The three common chromium oxides are CrO, Cr2O3, and CrO3. If Cr2O3 is amphoteric, what can you say about the acid-base properties of CrO and CrO3? Use the data in Table 15.3 to calculate the equilibrium constant for the following reaction: HCOOH(aq) OH (aq) 3:4 HCOO (aq) H2O(l ) 15.102 Use the data in Table 15.3 to calculate the equilibrium constant for the following reaction: CH3COOH(aq) Cl (aq) 88n AlCl4 (aq) NH3 3:4 NH4 NO2 (aq) 3:4 CH3COO (aq) HNO2(aq) 15.103 Most of the hydrides of Group 1A and Group 2A metals are ionic (the exceptions are BeH2 and MgH2, which are covalent compounds). (a) Describe the reaction between the hydride ion (H ) and water in terms of a Brønsted acid-base reaction. (b) The same reaction can also be classified as a redox reaction. Identify the oxidizing and reducing agents. 15.104 Calculate the pH of a 0.20 M ammonium acetate (CH3COONH4) solution. 15.105 Novocaine, used as a local anesthetic by dentists, is a weak base (Kb 8.91 10 6). What is the ratio of the concentration of the base to that of its acid in the blood plasma (pH 7.40) of a patient? 15.106 Which of the following is the stronger base: NF3 or NH3? (Hint: F is more electronegative than H.) 15.107 Which of the following is a stronger base: NH3 or PH3? (Hint: The NOH bond is stronger than the POH bond.) Study Guide TOC Textbook Website MHHE Website 640 ACIDS AND BASES 15.108 The ion product of D2O is 1.35 10 15 at 25°C. (a) Calculate pD where pD log [D ]. (b) For what values of pD will a solution be acidic in D2O? (c) Derive a relation between pD and pOD. 15.109 Give an example of (a) a weak acid that contains oxygen atoms, (b) a weak acid that does not contain oxygen atoms, (c) a neutral molecule that acts as a Lewis acid, (d) a neutral molecule that acts as a Lewis base, (e) a weak acid that contains two ionizable H atoms, (f) a conjugate acid-base pair, both of which react with HCl to give carbon dioxide gas. 15.110 What is the pH of 250.0 mL of an aqueous solution containing 0.616 g of the strong acid trifluoromethane sulfonic acid (CF3SO3H)? 15.111 (a) Use VSEPR to predict the geometry of the hydronium ion, H3O . (b) The O atom in H2O has two lone pairs and in principle can accept two H ions. Explain why the species H4O2 does not exist. What would be its geometry if it did exist? 15.112 HF is a weak acid, but its strength increases with concentration. Explain. (Hint: F reacts with HF to form HF2 . The equilibrium constant for this reaction is 5.2 at 25°C.) 15.113 When chlorine reacts with water, the resulting solution is weakly acidic and reacts with AgNO3 to give a white precipitate. Write balanced equations to represent these reactions. Explain why manufacturers of household bleaches add bases such as NaOH to their products to increase their effectiveness. 15.114 When the concentration of a strong acid is not substantially higher than 1.0 10 7 M, the ionization of water must be taken into account in the calculation of the solution’s pH. (a) Derive an expression for the pH of a strong acid solution, including the contribution to [H ] from H2O. (b) Calculate the pH of a 1.0 10 7 M HCl solution. 15.115 Hemoglobin (Hb) is a blood protein that is responsible for transporting oxygen. It can exist in the protonated form as HbH . The binding of oxygen can be represented by the simplified equation HbH O2 3:4 HbO2 Forward Main Menu 2CH3COOH(g) 3:4 (CH3COOH)2(g) At 51°C the pressure of a certain acetic acid vapor system is 0.0342 atm in a 360-mL flask. The vapor is condensed and neutralized with 13.8 mL of 0.0568 M NaOH. (a) Calculate the degree of dissociation ( ) of the dimer under these conditions: (CH3COOH)2 3:4 2CH3COOH 15.118 15.119 15.120 15.121 15.122 15.123 15.124 15.125 H (a) What form of hemoglobin is favored in the lungs where oxygen concentration is highest? (b) In body tissues, where the cells release carbon dioxide produced by metabolism, the blood is more acidic due to the formation of carbonic acid. What form of hemoglobin is favored under this condition? (c) When a person hyperventilates, the concentration of CO2 in his or her blood decreases. How does this action affect the above equilibrium? Frequently a person who is hyperventilating is advised to breathe into a paper bag. Why does this action help the individual? Back 15.116 Calculate the concentrations of all species in a 0.100 M H3PO4 solution. 15.117 In the vapor phase, acetic acid molecules associate to a certain extent to form dimers: TOC 15.126 (Hint: See Problem 14.73 for general procedure.) (b) Calculate the equilibrium constant KP for the reaction in (a). Calculate the concentrations of all the species in a 0.100 M Na2CO3 solution. Henry’s law constant for CO2 at 38°C is 2.28 10 3 mol/L atm. Calculate the pH of a solution of CO2 at 38°C in equilibrium with the gas at a partial pressure of 3.20 atm. Hydrocyanic acid (HCN) is a weak acid and a deadly poisonous compound — in the gaseous form (hydrogen cyanide) it is used in gas chambers. Why is it dangerous to treat sodium cyanide with acids (such as HCl) without proper ventilation? How many grams of NaCN would you need to dissolve in enough water to make exactly 250 mL of solution with a pH of 10.00? A solution of formic acid (HCOOH) has a pH of 2.53. How many grams of formic acid are there in 100.0 mL of the solution? Calculate the pH of a 1-L solution containing 0.150 mole of CH3COOH and 0.100 mole of HCl. A 1.87-g sample of Mg reacts with 80.0 mL of a HCl solution whose pH is 0.544. What is the pH of the solution after all the Mg has reacted? Assume constant volume. You are given two beakers, one containing an aqueous solution of strong acid (HA) and the other an aqueous solution of weak acid (HB) of the same concentration. Describe how you would compare the strengths of these two acids by (a) measuring the pH, (b) measuring electrical conductance, (c) studying the rate of hydrogen gas evolution when these solutions are reacted with an active metal such as Mg or Zn. Use Le Chatelier ’s principle to predict the effect of the following changes on the extent of hydrolysis of sodium nitrite (NaNO2) solution: (a) HCl is added, (b) NaOH is added, (c) NaCl is added, (d) the solution is diluted. Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 15.127 Describe the hydration of SO2 as a Lewis acid-base reaction. (Hint: Refer to the discussion of the hydration of CO2 on p. 632.) 15.128 The disagreeable odor of fish is mainly due to organic compounds (RNH2) containing an amino group, ONH2, where R is the rest of the molecule. Amines are bases just like ammonia. Explain why putting some lemon juice on fish can greatly reduce the odor. 15.129 A solution of methylamine (CH3NH2) has a pH of 10.64. How many grams of methylamine are there in 100.0 mL of the solution? 15.130 A 0.400 M formic acid (HCOOH) solution freezes at 0.758°C. Calculate the Ka of the acid at that temperature. (Hint: Assume that molarity is equal to molality. Carry your calculations to three significant figures and round off to two for Ka.) 15.131 Both the amide ion (NH2 ) and the nitride ion (N3 ) are stronger bases than the hydroxide ion and hence do not exist in aqueous solutions. (a) Write equations showing the reactions of these ions with water, and identify the Brønsted acid and base in each case. (b) Which of the two is the stronger base? 15.132 The atmospheric sulfur dioxide (SO2) concentration over a certain region is 0.12 ppm by volume. Calculate the pH of the rainwater due to this pollutant. Assume that the dissolution of SO2 does not affect its pressure. 15.133 Calcium hypochlorite [Ca(OCl)2] is used as a disinfectant for swimming pools. When dissolved in water it produces hypochlorous acid Ca(OCl)2(s) 2H2O(l ) 3:4 2HClO(aq) Ca(OH)2(s) which ionizes as follows: HClO(aq) 3:4 H (aq) ClO (aq) Ka 3.0 10 8 As strong oxidizing agents, both HClO and ClO can kill bacteria by destroying their cellular components. However, too high a HClO concentration Back Forward Main Menu TOC 641 is irritating to the eyes of swimmers and too high a concentration of ClO will cause the ions to decompose in sunlight. The recommended pH for pool water is 7.8. Calculate the percent of these species present at this pH. 15.134 Explain the action of smelling salt, which is ammonium carbonate [(NH4)2CO3]. (Hint: The thin film of aqueous solution that lines the nasal passage is slightly basic.) 15.135 About half of the hydrochloric acid produced annually in the United States (3.0 billion pounds) is used in metal pickling. This process involves the removal of metal oxide layers from metal surfaces to prepare them for coating. (a) Write the overall and net ionic equations for the reaction between iron(III) oxide, which represents the rust layer over iron, and HCl. Identify the Brønsted acid and base. (b) Hydrochloric acid is also used to remove scale (which is mostly CaCO3) from water pipes (see p. 116). Hydrochloric acid reacts with calcium carbonate in two stages; the first stage forms the bicarbonate ion, which then reacts further to form carbon dioxide. Write equations for these two stages and for the overall reaction. (c) Hydrochloric acid is used to recover oil from the ground. It dissolves rocks (often CaCO3) so that the oil can flow more easily. In one process, a 15 percent (by mass) HCl solution is injected into an oil well to dissolve the rocks. If the density of the acid solution is 1.073 g/mL, what is the pH of the solution? Answers to Practice Exercises: 15.1 (1) H2O (acid) and OH (base); (2) HCN (acid) and CN (base). 15.2 7.7 10 15 M. 15.3 0.12. 15.4 4.7 10 4 M. 15.5 7.40. 15.6 12.56. 15.7 Smaller than 1. 15.8 2.09. 15.9 2.2 10 6. 15.10 12.03. 15.11 [C2H2O4] 0.11 M, [C2HO4 ] 0.086 M, [C2O2 ] 6.1 10 5 M, [H ] 0.086 M. 15.12 H3PO3. 4 15.13 8.58. 15.14 3.09. 15.15 (a) pH ≈ 7, (b) pH 7, (c) pH 7, (d) pH 7. 15.16 Lewis acid: Co3 ; Lewis base: NH3. Study Guide TOC Textbook Website MHHE Website C M HEMICAL YSTERY Decaying Papers L ibrarians are worried about their books. Many of the old books in their col- lections are crumbling. The situation is so bad, in fact, that about one-third of the books in the U.S. Library of Congress cannot be circulated because the pages are too brittle. Why are the books deteriorating? Until the latter part of the 18th century, practically all paper produced in the western hemisphere was made from rags of linen or cotton, which is mostly cellulose. Cellulose is a polymer comprised of glucose (C6H12O6) units joined together in a specific fashion: CH2OH HO HO H H H O H O H OH OH H H CH2OH HO O OH H H OH H H Glucose H HO O H CH2OH O H O H A portion of cellulose As the demand for paper grew, wood pulp was substituted for rags as a source of cellulose. Wood pulp also contains lignin, an organic polymer that imparts rigidity to the paper, but lignin oxidizes easily, causing the paper to discolor. Paper made from wood pulp that has not been treated to remove the lignin is used for books and newspapers for which a long life is not an important consideration. Acid-damaged paper. 642 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Another problem with paper made from wood pulp is that it is porous. Tiny holes in the surface of the paper soak up ink from a printing press, spreading it over a larger area than is intended. To prevent ink creep, a coating of aluminum sulfate [Al2(SO4)3] and rosin is applied to some paper to seal the holes. This process, called sizing, results in a smooth surface. You can readily tell the difference between papers with and without sizing by feeling the surface of a newspaper and this page. (Or try to write on them with a felt-tip pen.) Aluminum sulfate was chosen for the treatment because it is colorless and cheap. Since paper without sizing does not crumble, aluminum sulfate must be responsible for the slow decay. But how? CHEMICAL CLUES 1. When books containing “sized” paper are stored in a high-humidity environment, 2. 3. 4. 5. Al2(SO4)3 absorbs moisture, which eventually leads to the production of H ions. The H ions catalyze the hydrolysis of cellulose by attaching to the shaded O atoms in cellulose (see diagram of cellulose on p. 642). The long chain unit of glucose units breaks apart, resulting in the crumbling of the paper. Write equations for the production of H ions from Al2(SO4)3. To prevent papers from decaying, the obvious solution is to treat them with a base. However, both NaOH (a strong base) and NH3 (a weak base) are unsatisfactory choices. Suggest how you could use these substances to neutralize the acid in the paper and describe their drawbacks. After much testing, chemists developed a compound that stabilizes paper: diethylzinc [Zn(C2H5)2]. Diethylzinc is volatile so it can be sprayed onto books. It reacts with water to form zinc oxide (ZnO) and gaseous ethane (C2H6). (a) Write an equation for this reaction. (b) ZnO is an amphoteric oxide. What is its reaction with H ions? One disadvantage of diethylzinc is that it is extremely flammable in air. Therefore, oxygen must not be present when this compound is applied. How would you remove oxygen from a room before spraying diethylzinc onto stacks of books in a library? Nowadays papers are sized with titanium dioxide (TiO2), which, like ZnO, is a nontoxic white compound that will not catalyze the hydrolysis of cellulose. What advantage does TiO2 have over ZnO? 643 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ...
View Full Document

Ask a homework question - tutors are online