Chapter 13: Behavior of Species in Aqueous Solution

Major Concept Areas: Equilibrium, Reactivity. In this chapter, the concepts of chemical equilibrium are applied to one of the important reaction types recognized in Chapter 1: acid-base reactions. The methods of application are essentially the same as for gas phase reactions in Chapter 12. The nature of acids and bases is explored in detail, and ideas are developed that enable broad understanding of the reactivity of acids and bases under a variety of circumstances in aqueous solution.

Specific Concepts in this Chapter:

• Equilibrium is dynamic with opposing processes occurring at equal rates.
• Chemical reactions achieve dynamic equilibrium similar to phase equilibrium and characterized by an equilibrium constant, K_eq, analogous to vapor pressure.
• In water, an acid transfers a proton to water to produce the hydronium ion, H₃O⁺, and the conjugate base of the acid.
• In water, a base accepts a proton from water to produce hydroxide ion, OH-, and the conjugate acid of the base.
• The ultimate origin of Bronsted acidity in aqueous solution is the Lewis acidity of cations.

13-1 Introductory Ideas . In Chapter 1, we discussed chemical reactivity in terms of three major categories of reactions: electron transfer, proton transfer, and double displacement processes. In this chapter, we will extend our knowledge of proton transfer (acid-base) reactions. We will apply the principles of chemical equilibrium to aqueous solutions of acids and bases. We will examine both qualitative and quantitative aspects of these equilibria, with emphasis on the factors (such as temperature, pressure, and the presence of other species) that influence the equilibrium position. The discussion of quantitative aspects will include an acknowledgment of the limitations of standard computational methods. Acid base equilibria are extremely important both in nature and in the commercial world. Sulfuric acid, the most-produced chemical in the United States, is used in virtually every industrial process requiring acid. And it is the presence of weak acids in carbonated beverages that gives them their interesting zest. Fluoridation of water is done to convert the principle structural component of teeth, apatite (Ca₅(OH )(PO₄)₃), to fluoroapatite (Ca₅(F)(PO₄)₃), which is both less soluble (Chapter 10) and more resistant to chemical attack by food components (particularly acids!) than is apatite.

Electrolytes. We begin by introducing a classification system for substances that can be developed strictly from experimental observation. The classification is based on whether or not an aqueous solution of a substance conducts electricity. Conductivity can be determined using the simple apparatus in Figure 13-1, consisting of a light bulb wired in series with a pair of electrodes and a switch. The electrodes are immersed in the solution to be tested, and the switch is closed. If the bulb lights, the solution conducts. If the bulb does not light, there is no conductivity. Tests like this lead to the following categories of substances.

Non-electrolytes. An aqueous solution of a non electrolyte does not conduct electricity. This implies that when the substance dissolves, it produces uncharged units in solution. We conclude that molecules of a non-electrolyte remain intact when it dissolves. An example of a non-electrolyte is sugar. It readily dissolves in water, but the resulting solution fails to conduct current. Sugar must therefore dissolve in molecular form. We represent the dissolution of sugar as follows:

Electrolytes. An electrolyte is a substance that produces a conducting solution in water. From the observation of conductivity, we conclude that when an electrolyte dissolves, it produces ions--charged species--in solution, that are capable of moving under the influence of the electric field produced by the electrodes. Although all electrolytes produce conducting solutions in water, some produce more strongly conducting solutions than others. This may be judged by the intensity of the light from the bulb. Therefore we introduce subcategories of electrolytes.

a) Strong electrolytes. These are substances that ionize or dissociate completely in water to give ions. Examples are sodium chloride (NaCl) and other similar ionic salts; hydrogen chloride (HCl), and other so-called strong acids; and sodium hydroxide (NaOH) and other so-called strong bases. We will return to the terms acid, base, and salt after we discuss what is meant by complete ionization. For now, we consider what happens when 0.1 mole of NaCl (a white, crystalline solid) dissolves in 1.0 L of water. It is readily observed that all of the NaCl dissolves; it does so in the form of its ions, Na⁺ and Cl^-. This is represented in equation 13-1-2.

The equation states that solid NaCl dissolves in water to produce aqueous sodium ions and aqueous chloride ions. The single arrow from reactants to products indicates that the process does not come to equilibrium, but instead proceeds all the way to the right. The result is a solution with [Na⁺] = 0.1 M, [Cl^-] = 0.1 M, and in which there is no un-ionized sodium chloride. This is what is meant by complete ionization.

As a second example, consider the dissolution of 0.2 mole of HCl (a colorless gas) in 1.0 L of water. Again, all of the HCl dissolves. Interestingly, despite the fact that HCl is a covalent compound, the resulting solution strongly conducts. HCl dissolves as H⁺ and Cl^- ions, rather than as HCl molecules. The following equations represent this process:

The overall process has been written as two steps. The first step is dissolution in molecular form, and the second step is the dissociation of the molecules to ions. 13-1-3b is oversimplified in that the species H⁺ is not capable of independent existence in water. Removal of the single valence electron from an atom of hydrogen leaves a bare proton, which is so tiny and consists of such a dense concentration of positive charge that it immediately attaches to the electron cloud of a lone pair of a water molecule. More accurately, then, the proton exists in water as H₃O⁺, called the hydronium ion. Association of the proton with a molecule of water is represented in 13-1-3c. Equation 13-1-4 represents the net result of 13-1-3b and 13-1-3c.

Usually, only 13-1-4 is used to represent the dissolution of HCl, with 13-1-3a being understood. As for NaCl, dissociation of HCl proceeds completely to the right to produce [H₃O⁺] = 0.2 M and [Cl^-] = 0.2 M. The concentration of dissolved but undissociated HCl molecules is essentially zero--all of the HCl that dissolves also dissociates.

Because NaCl ionizes completely when it dissolves in water, and because HCl dissociates completely, these substances are strong electrolytes. For all practical purposes, all ionic compounds are strong electrolytes (ionize completely) in water. In contrast, although some covalent compounds (like HCl) are strong electrolytes, many are either weak electrolytes or non-electrolytes. We look at weak electrolytes now.

b) Weak electrolytes. A weak electrolyte is a substance that dissociates only partially to ions in water. For weak electrolytes an equilibrium is established between undissociated molecules of substance and its constitutent ions. An example of a weak electrolyte is acetic acid, HC₂H₃O₂ , abbreviated HOAc. The structure of acetic acid is shown in Figure 13-2. The circled hydrogen atom (often called a proton, in anticipation of forming H ⁺, a proton) is the one written first in the formula. Organic chemists refer to the boxed group of atoms as the acetyl (Ac) group. The circled proton is attached to oxygen, which is in turn attached to an acetyl group. Hence the abbreviation HOAc. Acetic acid dissociates to a small extent in water according to 13-1-5:

The double arrow indicates the establishment of equilibrium. In this case, equilibrium lies far to the left. Most of the dissolved acetic acid is in the form of undissociated molecules. The conductivity of the solution is low because there are not many ions present to carry the current. Acetic acid is an example of a weak acid. Other weak acids, such as HF, HNO₂, and HCN, are also weak electrolytes.

Recognizing Acids, Bases, and Salts. In the preceding discussion, three terms have been used that come up frequently in chemistry: acid, base, and salt. We now define and discuss these terms.

Acids. For our present purposes, an acid is a substance containing a hydrogen atom attached to an electronegative atom, such as oxygen or one of the halogens (F, Cl, Br, I). Most acids fall into one of three categories.

Binary acids. These consist of hydrogen bonded to one other (electronegative) element. The common binary acids are

These have the structures expected from the electron configurations of hydrogen and the second element.

Oxoacids. Oxoacids have formulas of the type H_nEO_m, where E is the generic symbol for a third element. E is the central atom, single bonded to "n" OH groups and double bonded to "m-n" O atoms. In most oxoacids, E is in a fairly high oxidation state. Some common oxoacids are

The formulas of oxoacids are discussed in more detail at the end of the chapter.

Organic acids. Organic acids have the general formula, RCOOH, where R represents a collection of (usually) carbon, hydrogen, halogen, nitrogen, or oxygen atoms forming a single bond to the explicitly written carbon. The carbon atom shown in the formula is double bonded to one of the oxygen atoms, and single bonded to an OH group. The proton of the OH group is the acidic hydrogen. Hydrogens contained in the group R are normally not acidic. Examples of common organic acids are shown below. The presence of a -COOH group allows recognition of organic acids.

As mentioned above, the common feature of the acids discussed here is that they have a hydrogen atom attached to an electronegative atom, often oxygen.

Bases. Bases can be usefully placed in one of two categories.

Anions obtained by removing H⁺ from an acid.. Removal of a proton from an acid produces a base. Examples of bases of this type are

The anion, OH-, found in such substances as LiOH, NaOH, KOH, Mg(OH)₂, Ca(OH)₂, and Ba(OH)₂, is the most readily recognized base. It can be considered to arise by removing H⁺ from a molecule of water.

Neutral molecules containing atoms with lone pairs. This category includes ammonia, NH₃, and its relatives, such as ethylamine, CH₃CH₂NH₂. It also includes molecules like ethanol, CH₃CH₂OH, which has lone pairs on the oxygen atom.

The common feature of all bases is that they possess at least one lone pair of electrons. Thus we often represent a base generically as :B. As we will discuss more fully in the section on Lewis acids and bases, the base uses the lone electron pair to bond to H⁺.

Salt. A salt is an ionic compound consisting of a cation (other than H⁺) combined with an anion. Because they are ionic, salts are strong electrolytes. Most salts have either acidic or basic properties, usually due to the anion. We know from our discussion above that anions qualify as bases. As we will see below, however, anions exhibit a range of base strengths. Whether a salt has basic properties or not depends on how strongly the anion functions as a base. In salts of the type, NaHSO₄, the anion has a proton attached to an electronegative atom. Consequently the anion is an acid, and this salt exhibits acidic behavior. Finally, salts such as NH₄Cl function as acids due to the cation, which has a hydrogen atom attached to the electronegative nitrogen atom.

Table 13-1 shows some representative acids, bases, and salts.

13-2 Physical Properties of Acids and Bases. Acids and bases have been a part of chemistry for a long time. Ideas about their nature have changed and evolved over the years, until now we have a quite sophisticated and comprehensive understanding of the behavior of these two classes of chemical substances. Independent of the theoretical framework in which we choose to view them, however, acids and bases have some distinct and readily observable physical and chemical properties.

Acids:

• Taste sour
• Dissolve reactive metals (for example, zinc), producing hydrogen gas (H₂)
• Cause vegetable dyes to exhibit characteristic colors. For example, litmus turns red in the presence of acid
• React with bases, thereby losing their characteristic properties.

Bases:

• Taste bitter
• Feel slippery
• Cause vegetable dyes to exhibit colors different from the colors produced by acids. For example, bases turn litmus blue
• React with acids, thereby losing their characteristic properties.

The acids familiar to early chemists were mineral acids, such as H₂SO₄ and HNO₃. An early theory of acidity attributed the properties of acids to the presence of oxygen. However, the observation that HCl also produced acidic solutions, but lacks oxygen, led to the suggestion that hydrogen is the essential ingredient. Today we recognize that the second interpretation is more correct.

For the time being, we will restrict ourselves to consideration of acids and bases in water solution. This situation is by far the most commonly encountered, and an understanding of aqueous solutions is essential for an understanding of geological, biological, and environmental questions. In modern terms, we associate acidity with a substance that somehow produces hydronium ions, H₃O⁺, in water. For example, sulfuric acid, H₂SO₄, dissolves in water, then dissociates to form H₃O⁺ and HSO₄^- according to 13-2-1:

A base is a substance that somehow produces OH^-. For example, sodium hydroxide, NaOH, produces OH^- by simply dissolving. The presence of OH^- gives the solution basic properties. According to this view, when an acid and base react, H₃O⁺ combines with OH^- to form water in the process referred to as acid-base neutralization.

13-3 The Bronsted Concept. In 1923, Bronsted and Lowry independently proposed to restructure acid-base concepts in terms of proton transfer reactions such as 13-2-1. They suggested the following definitions:

Let's examine the reaction of HCl with water in terms of these definitions. The reaction is shown in 13-3-1:

All that happens is that HCl transfers a proton, H⁺, to a molecule of water, converting it to the hydronium ion and leaving a chloride ion, Cl^-, behind. In Bronsted terms, HCl is an acid, because it donates a proton; H₂O is a base, because it accepts a proton. But now think about the reverse reaction. This is of course also a proton transfer, hydronium ion serving as the donor and Cl^- as the acceptor of the proton. Thus H₃O ⁺ is also an acid, and Cl^- is a base, in Bronsted terms. Generally, we will use the formula, HA, as our generic representation for an acid. Here it is understood that A represents an atoms or group of atoms to which the acidic proton, H, is attached. An acid of formula HA reacts with water by proton transfer, like HCl does above, to produce a second acid (H₃O⁺) and a second base (A^- ):

Because the base A- is produced from the acid HA by removal of a proton, HA and A- are related species. For this reason, they are called a conjugate acid-base pair. In general a conjugate pair are related by the simple addition or removal of H⁺, as in 13-3-3:

Water and hydronium ion are also a conjugate acid base pair in both 13-3-1 and 13-3-2. The hydronium ion is the acid member of the pair (because it has one more proton than water), and water is the base member. Every proton transfer reaction involves two conjugate acid-base pairs. Such reactions may be viewed as competitions between two bases for the proton, H⁺. In general, we would expect the stronger base to win this competition. The matter of base strength will be addressed below. Table 13-2 below lists a number of common acids and their conjugate bases.

When ammonia, NH₃, is dissolved in water, the resulting solution has the characteristics of a basic solution; i.e., it contains OH^-. In the Bronsted framework this is readily explained in terms of the following process:

In this reaction, water functions as a proton donor--an acid--and ammonia as the proton acceptor, to produce the conjugate base of water, OH^-, and the conjugate acid of NH₃, NH₄⁺. Ammonia is a base, not because it contains OH^-, but because it produces OH^- by accepting a proton from a molecule of water according to 13-3-4.

Water is therefore a member of two conjugate pairs. In one case it is the basic member of the pair, and in the other case it is the acid. Which pair is active in a particular aqueous solution depends on the acidic or basic properties of the dissolved substance. If an acidic substance is dissolved, it transfers a proton to water, and the first conjugate pair prevails. If a basic substance is dissolved, it accepts a proton from water, and the second pair prevails.

It is important to understand that although we use generic symbols such as HA and A^- for the acid and base of a conjugate pair, this does not mean that in all such pairs the acid must be electrically neutral and the base a monoanion. Instead, it means that the base has a charge one unit more negative than the acid. Thus all of the pairs of species in Table 13-2 are covered by the generic HA/A- symbolism.

Example 13-1. Fill in the missing species in the table of conjugate pairs below:

Solution. Add a proton to Cl^- to produce HCl. Remove a proton from HNO₃ to produce NO₃^-; add a proton to OAc^- to produce HOAc; remove a proton from NH₄⁺ to produce NH₃.

Categorization of Acid-Base Reactions in Water. We will recognize and discuss four important and general proton transfer processes that can take place when a substance is dissolved in water.

1) The reaction of water with itself.

We begin with a consideration of pure water, which we know to be a clear colorless liquid. Careful measurements have established that water has a slight tendency to react with itself according to 13-4-1:

Thus water produces very small quantities of H₃O⁺ and OH^-. The stoichiometry of the process, however, dictates that the amounts of H₃O⁺ and OH^- produced by 13-4-1 be exactly the same, so that neither H₃O⁺ nor OH^- is present in excess in pure water. Water containing equal amounts of hydronium and hydroxide ions is said to be neutral. Neutral water does not exhibit the physical characteristics of acids or bases listed above. At 20 ^oC, reaction 13-4-1 produces only 10^-7 M H₃O⁺ and OH^-. It is no surprise that the presence of the hydronium and hydroxide ions went undetected for so long. However, since both H₃O⁺ and OH^- are produced in 13-4-1, water is behaving both as an acid and as a base toward itself. The essence of reaction 13-4-1 is the transfer of a proton from one water molecule to another.

We must keep in mind that, no matter what else may be dissolved in the water, the process in 13-4-1 always takes place. The reaction establishes an equilibrium characterized by an equilibrium constant that is usually symbolized K_w, to remind us that we are dealing with the self-reaction of water. K_w has the very small value 1.0 x 10^-14 at 20 ^oC. The equilibrium constant expression, shown following the reaction above, is the product of the concentrations of the two ions produced. According to the conventions introduced in Chapter 12, the concentration of water is not included in the expression for K_w because it is a pure liquid.

Before illustrating the use of the equilibrium constant, K_w, we offer a disclaimer that will apply to all calculations of [H₃O⁺] and [OH^-] carried out in this chapter. As a general rule, although values of equilibrium constants for proton transfer processes are often quoted to two and sometimes three significant figures, they are reliable to only one significant figure. Consequently, concentrations calculated from them can be no more reliable. Respecting this, all concentrations and concentration related quantities calculated from these equilibrium constants will be given to only one significant figure.

Example 13-2. Use the expression for K_w in 13-4-1 to show that in pure water [H₃O⁺] = [OH^-] = 10^-7 M.

Solution. According to the stoichiometry of 13-4-1, the concentrations of hydronium and hydroxide ions must be equal at equilibrium in pure water, because one of each is produced each time two molecules of water react. We may therefore substitute [H₃O⁺] for [OH^-] in the equilibrium constant expression:

Example 13-3. Show graphically how the concentrations of H₃O⁺ and OH^- are related in any water solution.

Solution. A plot of [H₃O⁺] versus [OH^-] will show this. The plot can be constructed by arbitrarily picking several values of, say, [H₃O⁺] and using the expression for K_w to calculate the corresponding [OH^-].

These points suffice to show the form of the plot, presented in Figure 13-3. The relationship between the two ion concentrations is inverse: as one goes up, the other goes down. The curve in Figure 13-3 is a hyperbola.

Example 13-4. Use the plot in Figure 13-3 to show graphically what happens to the concentrations of [H₃O⁺] and [OH^-] when 0.9 * 10^-7 moles of [H₃O⁺] is added to 1.0 L of pure water.

Solution. Pure water contains equal concentrations of H₃O⁺ and OH^-, both at 10^-7 M. Before addition of H₃O⁺ from outside, the pure water is at point 1 on the graph. The immediate effect of adding 0.9 * 10^-7 moles of [H₃O⁺] is to increase the concentration of hydronium ion along a vertical line to the value, 1 * 10^-7 + 0.9 * 10^-7 M, represented by point 2 in the diagram. However, since point 2 does not lie on the hyperbola of acceptable combinations of ion concentrations, the system is not at equilibrium at point 2. To achieve equilibrium, reaction must occur from right to left, using up H₃O⁺ and OH^- in equal increments, until point 3 on the hyperbola is reached. The return to equilibrium is represented by the diagonal line connecting points 2 and 3. The new equilibrium point corresponds to [H₃O⁺] = 9 * 10^-7 M and [OH^-] = 1 * 10^-8 M.

2) Reaction of an acid with water. This occurs generally as in 13-4-2.

This reaction, acid dissociation, occurs to varying extents, depending upon what the acid, HA, is. We have written it with a double arrow to acknowledge that often the reaction does not proceed completely to the right. Instead, an equilibrium is established between reactants and products. The magnitude of the equilibrium constant, expressed as in 13-4-3, is useful as a measure of acid strength.

The concentration of water is not included in the expression for the equilibrium constant because its value is large and essentially constant during the reaction. We thus treat it as a pure liquid and follow the convention introduced in Chapter 12 that pure solids and liquids are not written in K_eq expressions. The equilibrium constant for acid dissociation is usually labelled K_a, to remind us of the type of reaction being considered. According to 13-4-3, the larger K_a, the larger are the concentrations of the products H₃O⁺ and A^- at equilibrium relative to the concentration of undissociated acid, HA. Thus the larger K_a, the more strongly the acid donates its proton to the base, H₂O, and the stronger the acid. Table 13-3, the Bronsted Acid-Base Table, lists common acids and their conjugate bases. The acids are listed from top to bottom in order of decreasing relative strength--that is, decreasing magnitude of K_a.

Based on the Bronsted Table, strength categories for acids may be set up as follows:

• K_a > 1, strong acids. This includes H₃O⁺ and all of the acids above it. All of the strong acids have large K_a values, meaning that they react essentially to completion with water (they are strong electrolytes). Large concentrations of the products and a very small concentration of the reactant, HA, leads to a large number for K_a according to 13-4-3. From the table the value of K_a for HCl is essentially infinite, which means that in a solution of HCl in water, the concentration of undissociated HCl is zero. Since strong acids are so commonly used in the laboratory, and since there are so few of them, we recommend that you memorize them.
• K_a < 1, weak acids. The weak acids dissociate to only a small extent in water. A common weak acid is acetic acid, found in vinegar, which dissociates as follows:

  (13-4-4): HOAc(aq) + H₂O <==> H₃O⁺(aq) + OAc^-(aq)

  According to the Bronsted table, K_a for acetic acid is 1.8 * 10^-5, a very small number. Thus the major species present in a solution of acetic acid is undissociated HOAc. The products of its dissociation, H₃O⁺ and OAc^-, are present in small concentration and are thus minor species.
• K_a < 10^-14. In this text, these will be called very weak acids. Examples are NH₃ and H₂. As can be seen from the table these acids lie below and are thus less acidic than water. In water solution, therefore, the acidity of these species is not noticeable and can not be measured.

In practice, chemists frequently encounter the reverse of 13-4-2, in which hydronium ion reacts with the conjugate base species to produce the acid and water.

This process is important in acid-base titrations, which are discussed in a later section. The equilibrium constant expression for this reaction is simply the reciprocal of 13-4-3.

When HA is a weak acid (K_a < 10^-2), K_eq for reaction 13-4-5 is large, and 13-4-5 can be expected to run to completion.

3) Reaction of a base with water. This occurs generally as shown below:

This reaction is referred to as base hydrolysis. Hydrolysis is a fancy word meaning "water breaking", which is exactly what happens to the water molecule in 13-4-7. The equilibrium constant for 13-4-7 is labelled K_b to remind us that we are dealing with base hydrolysis:

In the presence of a base, water behaves as an acid, transferring a proton to the base. Bases are classified according to strength as follows:

• K_b > 1, strong bases. The large values of K_b for these species indicate that they react essentially to completion with water. An example of a strong base is the amide ion, NH₂^-, for which K_b is essentially infinite:

  (13-4-9): NH₂^-(aq) + H₂O --> NH₃(aq) + OH^-(aq)
  K_b = [NH₃][OH^-]/[NH₂^-]

  We write this reaction with a single arrow to indicate its completeness.
• K_b < 1, weak bases. These hydrolyze to only a minor extent in water. An example of a weak base is fluoride ion, F^-, which reacts with water according to

  (13-4-10): F^-(aq) + H₂O <==> HF(aq) + OH^-(aq)

  K_b for F^- is 1.54 x 10^-11, which indicates that there are only small concentrations of HF and OH^- present in a solution of F^-. The equation is written with a double arrow to indicate that equilibrium is established between reactants and products.
• K_b < 10^-14, very weak bases. Examples are Cl^- and ClO₄^- (perchlorate anion). These have no tendency to extract a proton from water.

It seems reasonable that the more readily an acid donates a proton to a specific base, the less readily the conjugate base of the acid can accept a proton. Thus the acid and base strengths of a conjugate pair are inversely related. This is clear from the values of K_a and K_b in Table 13-3. We summarize this with a commonly used slogan: The stronger an acid, the weaker its conjugate base.

Before proceeding, we should examine the significance of the K_a (and K_b) values used to divide strong from weak and weak from very weak acids (bases): 1 and 10^-14. The significance of the value, 1, is that it is the K_a for the conjugate acid of the solvent water, H₃O⁺. That this acid has K_a = 1 is readily seen by writing its reaction with water:

The products and reactants are the same! Nothing happens in this process. In the equilibrium constant expression for this reaction, concentrations of reactants and products cancel, leaving K_a = 1. Any reaction for which the products and reactants are the same must have K_eq = 1. The significance of the number, 10^-14, is that it is the value of the equilibrium constant for reaction of water with itself. This equilibrium constant is normally symbolized K_w:

Any acid below H₂O in the table is weaker than water.

Finally, you may have noticed that the product of K_a for an acid and K_b for its conjugate base is always 10^-14. This is not a coincidence. Using the equilibrium constant expressions for K_a, K_b, and K_w, you might like to prove that in general, K_a(HA)K_b(A-) = K_w.

Interestingly, the acid and base in any conjugate pair lying between the bracketed pairs H₃O⁺/H₂O and H₂O/OH^- are both weak. It is important to realize that when we say an acid is strong or weak, we mean relative to the acid H₃O⁺, the conjugate acid of the solvent water. Similarly, when we say a base is strong or weak, we mean relative to OH-. The terms "strong" and "weak" are relative, not absolute.

In practice, the reverse of 13-4-7 is encountered as frequently as is 13-4-7 itself:

13-4-13 is important in titrations of weak acids with strong bases, for example, as we will see in a later section. When K_b is small (B is a weak base), K_eq for 13-4-13 is large, and 13-4-13 can be considered to run to completion.

4) Reaction of an acid other than water with a base other than water. Water need not be involved in proton transfer reactions if more than one solute conjugate pair is represented. For example, in a solution containing HF (an acid) and NH₃(a base), HF will not react with water. Instead, it will react with NH₃ because NH₃ is a stronger base than water (see the Bronsted Table!) and will compete more effectively for the proton of the acid:

The symbol K_a is reserved for reaction of an acid with the base, H₂O; we must therefore not use the symbol K_a for the equilibrium constant in 13-4-14. The more general symbol K_eq is appropriate. You might like to show that the reaction of HF with NH₃, although not explicitly involving water, can be expressed as the sum of the dissociation of HF and the reverse of the dissociation of NH₄⁺, and that therefore K_eq = K_a(HF)/K_a(NH₄⁺). According to the Bronsted table, K_a(HF) = 6.5 x 10^-4 and K_a(NH₄⁺) = 5.6 x 10^-10. So K_eq for 13-4-14 = 1.2 x 10⁶, a very large number.

A couple of useful generalizations can be induced from the discussion of reaction between HF and NH₃ in water:

• An acid reacts with the strongest available base, and vice versa.
• An acid reacts completely with the conjugate base of any acid below it in the table; and a base reacts completely with the conjugate acid of any base above it in the table.

These two realizations give us tremendous predictive ability.

In summary, the four types of proton transfer reactions that occur commonly in water are:

We will now examine the quantitative aspects of acid-base equilibria in water.

13-5 Simple Calculations Involving Acid and Base Equilibria. It is important to have some facility with a variety of calculations involving the equilibria governing reactions of acids and bases with water and with each other in water. We devote this section to a number of examples that illustrate the methods involved.

Example 13-5. What is the [H₃O⁺] in a 0.001 M solution of the strong acid, HCl?

Solution. Strong acids all dissociate completely in aqueous solution. HCl reacts according to

From simple stoichiometry, if [HCl] is initially 0.001 M, and it all reacts, then 0.001 M H₃O⁺ is produced. [H₃O⁺] = 0.001 M = 1 x 10^-3 M. Note that the calculation carried out here would be the same for any strong acid with one proton. Thus 0.001 M solutions of HBr, HI, HNO₃, and HClO₄ would all have [H₃O⁺] = 0.001 M.

Example 13-6. What is [H₃O⁺] in a 0.10 M solution of HF?

Solution. Because HF is a weak acid, its reaction with water proceeds to only a small extent. Thus [H₃O⁺] must be calculated using the methods of Chapter 12. Briefly, the approach introduced there involved

• Writing the equation for the relevant reaction;
• Writing the initial concentrations of all species under their formulas in the equation;
• Writing the changes in concentrations that occur when reaction proceeds from the initial to the equilibrium state in terms of an algebraic unknown;
• Writing the equilibrium concentrations, expressed in terms of the unknown;
• Substituting the equilibrium concentrations into the K_eq expression, and solving for the unknown;
• Calculating the numerical values of the equilibrium concentrations.

Application of this procedure to the problem at hand proceeds as follows:

(The initial concentration of the hydronium ion is written as ~0 rather than 0 to remind us that there is always a small amount of this substance present due to reaction of water with itself.) Substituting into the equilibrium constant expression,

This expression results in a quadratic equation that may readily be solved for the value of the unknown, x. However, some common sense applied to this problem will make its solution simpler. The value of the equilibrium constant, K_a, is very small; thus the extent of reaction, x, must also be quite small relative to the amount of acid initially present. We will assume that x is small enough relative to 0.1 that it may be neglected in the denominator. The equation to be solved then simplifies to

Indeed, x is only 8% of 0.1, so the approximation that we made in neglecting it in the denominator is reasonable. Because x represents the value of [H₃O⁺], it follows that

The calculation just performed is an example of what might be called the standard approach to the calculation of [H₃O⁺] in a solution of a weak acid. Notice that, once the assumption is made that x is smaller than the initial concentration of acid, C_o, the hydronium ion concentration is calculated using the following expression

In almost all cases, this equation will suffice to estimate the hydronium ion concentration in an aqueous solution of a weak acid.

As the previous examples show, concentrations of [H₃O⁺] and equilibrium constants are frequently expressed in exponential notation. It rapidly becomes tiresome to write these long numbers over and over again. To avoid doing this, chemists have invented a quantity called pH to represent the hydronium ion concentration in shorthand form:

Thus for a solution in which [H₃O⁺] = 0.1 M, pH = 1. The pH in Example 13-5 is - log(1 x 10^-3) = 3; and in Example 13-6, pH = 2.1. These are easier numbers to write and to type than are the corresponding exponential forms. The range of hydronium ion concentrations and corresponding pH values normally encountered in aqueous solutions is shown in Table 13-4.

The same concept works fine for equilibrium constants, which are often just as tedious to write. Thus the symbol pK is taken to mean the negative logarithm (base 10) of the equilibrium constant:

Applying these ideas to 13-5-1 leads to a useful expression for estimating the pH of a solution of a weak acid in water:

Example 13-7. What is the pH of a 0.1 M solution of a strong base?

Solution. Strong bases are all ionic and therefore strong electrolytes. The concentration of OH- produced is 0.1 M multiplied by the number of hydroxide ions in the formula for the base.

Example 13-8. What is the pH in a solution containing 0.20 M CN-?

Solution. The cyanide anion, CN-, is the conjugate base of hydrocyanic acid, HCN. As a base, it will undergo a hydrolysis reaction with water to produce small amounts of HCN and OH-, governed by an equilibrium constant, K_b. The concentration of OH- at equilibrium can be calculated using the approach reviewed in Example 13-6.

Because K_b is small, x will be small and may be ignored with respect to 0.20 in the denominator. Then

At equilibrium, [OH-] = 2 * 10^-3 M. Thus [H₃O⁺] = K_w/[OH-] = 5 * 10^-12 M, and pH = 11.3.

This approach to the hydrolysis of a weak base, in which the small change in base concentration is assumed to be negligibly small when compared to the initial base concentration, C_o, might be called the standard approach. It is identical in method to the standard approach for dealing with acid dissociation in Example 13-6. Note that according to the standard approach, the concentration of hydroxide ion in a C_o molar solution of a weak base is given approximately by

Expressed in log terms,

Example 13-9. Predict the reaction that will occur when sodium acetate, NaOAc, and ammonium chloride, NH₄Cl, are dissolved in water in a beaker. Calculate K_eq for the predicted reaction.

Solution. We recognize both substances as salts. They are thus ionic and ionize completely in solution. Ionization produces the following species in substantial amount in the aqueous solution:

Of these species, we recognize that OAc^-, NH₄⁺, Cl^-, and H₂O all represent conjugate pairs from the Bronsted table. However, Cl^- is the conjugate base of the strong acid HCl, and is therefore very weak. It will not react with any acids in water. The sodium ion is not listed in the table. It, like all of the group 1 cations, is unreactive in water and can be ignored. That leaves NH₄⁺, an acid; OAc^-, a base; and H₂O, which can function as either an acid or a base. First we compare base strengths. Which of the two bases is stronger? Since OAc^- is below H₂O in the base column, OAc- is the stronger base. Now compare acid strengths. Since NH₄⁺ is above H₂O in the acid column, NH₄⁺ is the stronger acid. The reaction that takes place is

The value of K_eq is K_a(NH₄⁺)/K_a(HOAc) = 5.6 * 10^-10/1.8 * 10^-5 = 3 * 10^-5. Not much reaction will occur, and the major species at equilibrium are NH₄⁺(aq) and OAc^-(aq).

Example 13-10. Predict the reaction that will most likely occur between H₂O and NH₃ in aqueous solution.

Solution. Both H₂O and NH₃ appear in both the acid and base columns of Table 13-3. How do we know which will function as the acid and which as the base? We can write both possibilities:

H₂O is above NH₃ in the acid column, and is therefore stronger. H₂O thus functions as an acid, forcing NH₃ to be the base. Reaction 13-5-7 is correct.

Example 13-11. The concentration of H₃O⁺ in a solution of HF is measured to be 2 * 10^-3 M. What fraction of the total fluoride is present in the protonated form, as HF? (The remainder is present as F-).

Solution. HF produces H₃O⁺ by reaction 13-5-9:

The equilibrium constant expression is K_a = [H₃O⁺][F^-]/[HF]. Since [H₃O⁺] is known, we can solve the expression for the ratio of fluoride to hydrofluoric acid concentrations:

This tells us that the solution contains 32 parts F- per 100 parts HF, for 132 total parts. The fraction in the conjugate acid form is 100/132 = 0.76.

Generalization of Approach to Calculations. All calculations involving proton transfer processes in aqueous solution can been approached via a common way of thinking; indeed, the preceding calculations in this section have used this way of thinking. We now articulate this approach.

• Identify the strongest acid and the strongest base in the solution.
• Write a proton transfer reaction involving the strongest acid and base.
• Obtain the value of K_eq for this reaction.
  • If the reaction involves water, K_eq will be either K_a, K_b, or the reciprocal of one of them.
  • If the reaction does not involve water, K_eq will be K_a for the acid on the left of the equation divided by K_a for the acid on the right.
• Set up the initial, change, equilibrium table for the reaction.
  • If K_eq for the reaction is large, the values in the "change" line are determined by the limiting reagent idea.
  • If K_eq for the reaction is small, the values in the "change" line are unknown and must be determined by the usual type of equilibrium calculation.
• Calculate the equilibrium concentrations of all species in the reaction.
• If the reaction does not involve water, use K_a for one of the acids in the reaction to calculate [H₃O⁺] and [OH-].

This approach was used in each type of calculation as follows:

• Pure water. Here H₂O is both the strongest acid and the strongest base. The relevant reaction is of water with itself, according to K_a for H₂O, which is called K_w.
• Strong or weak acid HA in water. HA is the strongest acid, H₂O is the strongest base. The relevant reaction is of HA with H₂O, according to K_a for the acid.
• Strong or weak base :B in water. H₂O is the strongest acid, :B is the strongest base. The relevant reaction is of H₂O with :B, according to K_b for the base (= K_w/K_a(BH⁺).
• Strong or weak acid, HA, and strong or weak base, :B, in water. HA is the strongest acid, :B is the strongest base. The relevant reaction is of HA with :B, according to the equilibrium constant K_eq = K_a(HA)/K_a(BH⁺).

One final example will make this approach explicit.

Example. 20 mL of 0.12 M NaOCl is mixed with 15 mL of 0.20 M HCOOH. What are the concentrations of all species in the solution once proton transfer equilibrium has been attained?

Solution. We first work out the diluted concentrations after mixing:

Now identify the strongest acid and base. NaOCl is a salt with the anion OCl-, which is a weak base (thus stronger than water). HCOOH is a weak acid. The relevant reaction will involve OCl- and HCOOH:

This reaction does not involve water explicitly, so K_eq = K_a(HCOOH)/K_a(HOCl) = 1.8*10^-4/3.5*10^-8 = 5.1*10³. K_eq is large, so the limiting reagent idea will apply. Set up the initial, change, equilibrium table:

Three of the four species have known concentrations at this point: [HCOOH] = 0.017 M, [HOCl] = 0.069 M, [HCOO-] = 0.069 M. Because concentrations of both members of the HCOOH/HCOO- conjugate pair are known, we can use the K_a expression for this pair to obtain [H₃O⁺]:

Knowing [H₃O⁺] enables us to calculate [OCl-] and [OH-]:

Almost finished; but don't forget Na⁺ at 0.069 M!

13-6 Net ionic equations. In Chapter 1 we learned that the changes taking place in a chemical reaction are represented in shorthand form using a chemical equation. Several rules apply to the writing of such equations.

1. Substances that react and form are represented by their chemical formulas.
2. Reactants are written on the left and products on the right of the equation.
3. Reactants and products are separated by an arrow pointing from reactants to products.
4. The equation must be balanced to conform to the Law of Conservation of Mass.
5. Physical states of reactants and products are indicated as follows:
   
   

   (g) gas
   (l) liquid
   (s) solid
   (aq) aqueous (water) solution to indicate that the substance is dissolved in water

We will now modify these rules, beginning with rule 3. In previous chapters, we have seen two ways to connect reactants and products: a single arrow leading from reactants to products; and a double arrow between reactants and products. From now on, we must be careful in the use of these symbols. A single arrow will indicate that the reaction goes to completion; that is, that its equilibrium constant is very large in magnitude. We will consider a reaction with K_eq > 100 to go to completion. A double arrow will be used for a reaction that comes to equilibrium, and will be used in particular for reactions with small values of K_eq (K_eq < 1) to emphasize that the reaction does not proceed completely to the right. In many chemistry texts, you will find equals signs used in place of double arrows to indicate the establishment of equilibrium. The equals sign is NOT used that way in this text, because it implies an equality of reactants and products, and of all their properties, that does not exist. In choosing not to use the equals sign, we contradict common usage. However, we feel strongly enough about the misleading aspects of the equals sign to avoid using it in chemical equations.

Now, consider rule 1. There are many reactions, particularly those involving dissolved species, for which writing the formulas of the participants does not indicate clearly the nature of the reactions. For example, the equation for reaction between hydrochloric acid and sodium hydroxide in aqueous solution may be written:

This appears to be a reasonable, balanced equation for the reaction, within the framework of the five rules above. However, there is a minor problem. In our discussion of electrolytes, we acknowledged that as a strong electrolyte, HCl(aq) exists in solution as separated H₃O⁺ and Cl^- ions. Similarly, NaOH and NaCl are strong electrolytes, and exist in water as separated cations and anions. Taking this into account, we achieve a better representation of the reaction:

This equation, called an IONIC EQUATION, is an improvement over 13-6-1 because it represents the reactants and products as they exist when dissolved in water. However, 13-6-2 is still bothersome. Sodium ions, Na⁺(aq), occur on both sides of the equation in exactly the same form. Nothing happens to the sodium ions during reaction. The same is true of chloride ions, Cl^-(aq). These ions are spectators to the real action, which is combination of H₃O⁺(aq) and OH^-(aq) to form water, H₂O. For this reason, they are called spectator ions. Because they are not involved in the chemical process that occurs in solution, there is little point in including them in the equation. Reducing 13-6-2 to its essentials gives 13-6-3:

13-6-3 is called a NET IONIC EQUATION, because it shows us only the net process taking place in solution, omitting spurious species. From now on, we will use net ionic equations extensively to convey the essence of a chemical process.

In summary, our revised rules for writing chemical equations follow:

1. Chemical formulas are used to represent substances that react and form in the reaction, but the formulas are written to conform as closely as possible with the true nature of the substances in water. Species that appear in identical form on both sides are not shown.
2. Reactants are written on the left side of the equation, products on the right.
3. Reactants and products are separated by either a single or double arrow. The single arrow is used if reaction is essentially complete (K_eq > 100). The double arrow is used if an equilibrium is established (K_eq < 100).
4. The equation must be balanced to conform to the law of conservation of mass. Because a net ionic equation involves ions, we recognize also that the equation must be balanced for charge--that is, the same total charge must appear on both sides of the equation.
5. Elements and compounds that are not dissolved in water are represented by their chemical formulas followed by one of the symbols, (s), (l), or (g) to show that they participate in a reaction as pure solid, liquid, or gaseous phases, respectively. The symbol (aq) is placed after the chemical formula of a molecule or ion that is dissolved in water.

Net Ionic Equations and Bronsted Reactions. Consider reaction of fluoride ion, F-, with water to produce hydrogen fluoride (an extremely dangerous, colorless gas) and hydroxide ion:

We have discussed this reaction previously, but did not acknowledge it as a net ionic equation. To see this, we consider how we might carry out this reaction in the laboratory. First, we must put some distilled water in a beaker and add the desired amount of fluoride ion. Because charged atoms or molecules (i.e., ions) do not exist independently in bulk (large quantitities) in nature, we must deal with species that are electrically neutral; fluoride ion must be introduced as a component of a neutral compound that will ionize once dissolved to give F^-. A good choice might be either NaF or KF, both of which are soluble in water. Both compounds are (ionic) salts, and therefore strong electrolytes. Either will successfully produce F^- in solution. Both Na⁺ and K⁺ are unreactive in water, so either choice will serve. As a strong electrolyte, NaF dissolves to form ions. F- then reacts with water, producing HF and OH^-. Since for each OH^- ion produced, one F^- ion is used, there will be just the right number of Na⁺cations to balance the anions. The ionic equation representing what we just described is in 13-6-4:

Na⁺ is a spectator ion, and may be eliminated, leaving the net ionic equation 13-4-10. Almost without exception, proton transfer (Bronsted) reactions are written as net ionic equations, with only the participating species shown.

13-7 Recognition of Species. Perhaps the most difficult aspect of aqueous solution chemistry is recognizing what various species do when dissolved in water. It is important to be able to identify a species and to predict what it will do. To facilitate this, we will spend some time focusing on species identification.

As mentioned before, it is recommended that you MEMORIZE the common strong acids and bases. You will run into these frequently in the laboratory, and their properties in solution should become second nature to you. For convenient reference, the common strong acids and bases are listed in Table 13-5.

When dealing with one of these acids (or bases) you KNOW it is strong because you have memorized the list. An acid (base) not in this list is weak (dissociates only slightly) and will therefore produce only a small amount of H₃O⁺ (OH^-) when dissolved in water.

Second, most acids can be recognized from their formulas, which are often (but not always) written with the proton first. If an acid is not in the memorized strong list, it is weak and can be located in the Bronsted Table. For practice, identify which of the following are acids, and whether they are strong or weak: NaCl, NaOH, HNO₃, HCN, Na₂SO₄, H₂SeO₃. Bases are not as easy to recognize from their formulas, but we can state some general guidelines. Compounds containing nitrogen with a lone pair are usually weak bases. Anions are very often weak bases. The most commonly encountered weak bases are, of course, listed in the base column of the Bronsted Table.

Finally, many weak acids and bases come disguised as salts. Thus at first glance we would call NaCN a salt. With further thought, however, we recognize that it is ionic, will dissolve in water, and when it does so will produce Na⁺ (a spectator ion) and CN^-, the conjugate base of the acid, HCN. CN^- will therefore react with water to some extent to form HCN and OH^-. NaCN is sometimes called a basic salt, because its anion is a weak base. Similarly, NaHSO₄ is a salt and will dissolve as Na⁺ and HSO₄^- ions. The anion is found in the weak acid section of the Bronsted table. It reacts to a small extent with H₂O to produce H₃O⁺ and SO₄^2-. NaHSO₄ is called an acidic salt, because its anion is a weak acid. Similarly, NH₄Cl is an acidic salt, via its cation.

Example 13-12. 0.25 mole NaOH and 0.20 mole HOAc are added to enough water to produce 1.00 L of solution. What are the concentrations of all species present in the solution at equilibrium (i.e., after the favorable Bronsted reactions have taken place)? What are the major species present in solution?

Solution. We begin by identifying each species and deciding what it will do when dissolved:

Of these species, the strongest acid is HOAc, and the strongest base is OH-. A proton transfer reaction can occur between them. That the reaction proceeds to completion can be shown by using data from Table 13-3 to calculate K_eq for the expected reaction, given below:

This is the reverse of the reaction of a weak base with water, so its equilibrium constant is the reciprocal of K_b for OAc-. To calculate the equilibrium concentrations of the species in this reaction, we set up a table in which we list, first, the initial concentrations of all species; second, the changes in these concentrations that occur during reaction; and third, the final (equilibrium) values of the concentrations. The equilibrium constant is so large that the reaction may be considered to go to completion. The extent of reaction is thus determined by the limiting reactant concept.

The entries in the D line are obtained by realizing that, since HOAc and OH^- react in 1:1 molar ratio, HOAc is the limiting reactant. It reacts completely, using up in the process the same amount of OH^- and producing the same amount of OAc^-. Based on the equilibrium concentrations, we conclude that the major species at equilibrium are OH^-, OAc^-, and Na⁺ (the spectator), with the following concentrations:

Other species present in the solution in small amounts are H₃O⁺ and HOAc. [H₃O⁺] can be calculated using the expression for K_w:

The concentration of acetic acid can be obtained from K_a for acetic acid dissociation:

13-8 Distribution Diagrams. The standard calculations introduced in Examples 13-6 and 13-8 are found in virtually every introductory chemistry text available. Interestingly, a practicing chemist very infrequently if ever finds it necessary or useful to calculate the pH of a, say, 0.1 M solution of some weak acid or weak base. Thus one is led to question the value of the standard calculations. Of much more relevance is the ability to use equilibrium expressions and calculations to determine the manner in which a substance distributes between the conjugate forms as a function of pH. A distribution diagram is a convenient way to summarize the manner in which the concentrations of the acid and base members of a conjugate pair vary with pH. Such a diagram tells us at a glance the form in which a species exists at a specified pH. This is of great use in geological, biological, and environmental contexts. In this section, we construct the distribution diagram for acetic acid to show that the calculations involved are quite simple. We then present distribution diagrams for a number of important conjugate pairs.

The calculation flows as follows. First, we choose a pH range over which to construct the diagram. Most distribution diagrams encompass the range from 0 to 14. For each of several convenient pH values in the selected range, we calculate the value of the ratio [OAc^-]/[HOAc] using a modified form of the equilibrium constant expression for acetic acid dissociation. The modified form, shown below, simplifies calculations.

This can be obtained by taking the negative logarithm of both sides of the K_a expression, and rearranging the result to give pH on one side. The form of this equation suggests the pH values at which we should do our concentration ratio calculations. Specifically, we will calculate [OAc^-]/[HOAc] when pH = pK_a, when pH = pK_a + 1, and when pH = pK_a + 2. We then use the concentration ratio to calculate the fraction of the total acetate (the sum of the concentrations of acetate in both the conjugate acid and base forms) in each of the forms, HOAc and OAc -. Finally, we plot these fractions as percentages against pH. The calculations are outlined below.

This completes the necessary calculation. At pH values more than 2 units above pK_a for the acid, virtually all of the acetate is present in the form of the conjugate base, OAc-. At pH values more than 2 units below pK_a, the reverse is true. It should be clear that as pH increases from pK_a, the %OAc^- approaches 100 and the %HOAc approaches 0. These numbers are never actually reached, of course, because there is always some HOAc present, however tiny the amount, no matter how high the pH. A similar statement holds as we decrease pH from pK_a, except that now %HOAc approaches 100. We now plot the percentages of HOAc and OAc^- against pH as shown in Figure 13-4. Note how the amounts of conjugate acid and base are inversely related; as one goes down, the other goes up. Note also that the sum of the OAc^- and HOAc curves at any pH is exactly 100%. This says simply that the acetate has to be in one form or the other. The two curves cross at the 50% level at a pH that is numerically the same as pK_a for the acid. By choosing to begin our calculations at pH = pK_a, we began at this crossover point. Finally, note that the distribution diagram has been calculated without any knowledge of the actual amounts of OAc^- and HOAc present in the solution. We have dealt strictly with ratios of the two concentrations, which allow us to determine relative amounts. Therefore the distribution diagram applies for any desired total concentration of acetate. It is independent of total species concentration.

Using the methods illustrated in this section, it is a simple matter to calculate the distribution diagram for any conjugate acid-base pair. However, it is not necessary to do any further calculation because, although the diagrams for different acids have different crossover pH values, the relative percentages at increments of pH above and below the crossover are exactly the same as for acetic acid. To obtain the distribution diagram for any acid, all we need do is shift the acetic acid diagram right or left along the pH axis until the crossover point is at the pK_a for the desired acid. The resulting diagrams all look exactly like the one for acetate in Figure 13-4, except that the crossover points of the curves occur at pH values equal to the pK_a values of the acids concerned. Figure 13-5a shows distribution diagrams for several common acid-base pairs. The diagram for phosphoric acid, H₃PO₄, is interesting because it shows three crossovers corresponding to the pK_a values for successive loss of the three protons. The three relevant conjugate pairs are

The shape of a distribution diagram suggests that the value of K_a for a weak acid can be estimated by measuring the pH of a solution containing equal concentrations of the acid and its conjugate base. When [HA] = [A-], pK_a = pH.

A Predominance Diagram is a simplified distribution diagram. Instead of showing quantitatively the percentages of the conjugate acid-base species, it indicates the predominant species as a function of pH. To construct a predominance from a distribution diagram, draw a vertical line at each crossover pH, and write the acid member of the conjugate pair to the left of the line, the base member to the right. The predominance diagrams for acetic acid and phosphoric acid are shown in Figure 13-5b.

We can make the predominance diagram even simpler, however, by reducing it to a horizontal line representing the pH axis. A mark is placed on the line at the pK_a for the acid of interest, and it is understood that the conjugate base predominates at higher pH values to the right of the mark, the acid predominates to the left. Using this simplified form, many conjugate pairs can be shown on the same diagram, as in Figure 13-5c.

Example 13-13. The predominance diagrams for HF/F^- and NH₄⁺/NH₃ are shown in simplified form in Figure 13-6. If equal volumes of 0.1 M hydrogen fluoride and 0.15 M ammonia are mixed, will reaction occur? What species predominate at equilibrium? What is the pH at equilibrium?

Solution. The distribution diagrams show that HF and NH₃ predominate in incompatible pH ranges--that is, the pH ranges in which they exist do not overlap. When solutions of the two species are mixed, they cannot coexist but instead must react to form species that have overlapping pH ranges, and therefore can coexist. Thus reaction 13-8-2 occurs. We can be sure that it goes essentially to completion for the same reason that we are sure that reaction occurs--nonoverlap of the pH ranges of stability for HF and NH₃.

(NOTE: K_eq can be calculated from the Bronsted Table if desired.)

At equilibrium, the solution contains known amounts of both members of the NH₄⁺/NH₃ conjugate pair. If the concentrations of the two species were equal, the solution pH would be exactly the value at the dividing line in the distribution diagram for this pair. Since the acid form (NH₄⁺) has higher concentration, however, the pH will be slightly to the acid side (left) of the dividing line, or somewhat less than 9.3. We can calculate it more exactly using the K_a expression for the NH₄⁺/NH₃ pair:

It is interesting that we can calculate the pH of the final solution with certainty without explicitly knowing the pH in either solution before mixing! If desired, we can also calculate the small (unknown) concentration of HF left at equilibrium using the K_a expression for the HF/F^- pair:

13-9 pH Titration.We take up now a practical problem: How do we experimentally determine an unknown concentration of acid in an aqueous solution? This is normally done using a process called titration, involving the following steps:

• A precisely known volume of acid solution is measured out and placed in a flask. Such a precisely known volume is called an aliquot, and may be measured using a volumetric pipet, an item of glassware that, when filled with the solution of interest, delivers precisely the indicated volume of solution. Convenient pipet sizes for aliquot measurement range from 1.00 to 25.00 mL. The substance placed in the flask is called the analyte.

• To the aliquot of acid solution, a solution of base of precisely known concentration (called standard base) is added from another item of precision glassware called a buret. This process is the actual titration. The substance added from the buret, in this case the base, is called the titrant. Addition of titrant is continued until the moles of OH^- added is precisely equal to the moles of H⁺ in the acid. The point of the titration where moles OH^- added = moles H⁺ in the acid is called the equivalence point.
• From the volume of base added to reach the equivalence point and its known concentration, we calculate the concentration of acid in the original solution.

Example 13-14. 32.16 mL of 0.1042 M NaOH is needed to titrate a 25.00-mL aliquot of a solution of H₃PO₄ of unknown concentration, according to

Calculate the concentration of H₃PO₄ in the original solution.

Solution. We calculate the moles of NaOH used; divide it by 3 to obtain the moles H₃PO₄ present; and divide by 25.00 mL, the volume in which this number of moles was contained:

The Titration Curve. The discussion above has glossed over a very important practical problem: how is the equivalence point detected? Unless we can detect the equivalence point, we will not know when to stop adding titrant. Clearly, the pH of the solution must change as the titration proceeds. At the beginning, before base is added, the pH of the solution is low because it contains acid. As titration proceeds, acid is neutralized by the added base, and pH rises. Addition of base after all of the acid has been neutralized produces a basic solution, with a high pH. During the titration, then, pH runs the gamut from low to high. We detect the equivalence point from the manner in which the pH change occurs.

Two methods are commonly used to detect the equivalence point in a titration. In the first, an instrument called a pH meter is used to monitor the pH of the solution as base is added during the titration. A pH meter responds to the electrical potential of an electrode immersed in the solution being titrated. This potential is a function of [H₃O⁺] in the solution. The relationship between chemical concentration and electrical potential is the subject of Chapter 14. For now we take it on faith that we can measure pH with a pH meter. A plot of pH versus the volume of titrant added to the solution gives the so-called titration curve. The experimental curve for titration of 40.00 mL of 0.1000 M HCl with 0.1000 M NaOH is shown in Figure 13-7a. We observe that

• The curve is shaped like the letter "S". All titration curves have this characteristic shape.
• The total change in pH during the titration is about 12 units, but almost all of this pH change occurs over a very small range of base volume centered at 40 mL, which corresponds to the equivalence point. Thus the large pH change occurs right at the equivalence point! More exactly, the equivalence point corresponds with the inflection point of the curve, which is the point at which the slope of the curve is a maximum.

This provides us with the first method for determining the equivalence point: we successively add small volumes of base, measure pH after each addition, and plot the titration curve to find V_base at the inflection point (the equivalence point). Moles of acid in the original aliquot is calculated as follows:

The titration curve for a weak acid, such as acetic acid, looks different from that for a strong acid such as HCl. The curve for titration of 40.00 mL of 0.1000 M HOAc with 0.1 M NaOH is shown in Figure 13-7b. The general S-shape is preserved. However, we note several specific differences.

• The initial pH of the weak acid curve is higher than that for the strong acid, because the strong acid is completely dissociated, whereas the weak acid is dissociated to only a minor extent.
• In the weak acid titration, pH increases initially, levels off until near the equivalence point, then rises steeply over a small volume range around the equivalence point. For the strong acid, no initial rise in pH is observed.
• The sharp rise in pH at the equivalence point is larger and steeper for the strong acid.
• For the strong acid, the pH at the equivalence point is exactly 7, indicating a neutral solution. For the weak acid, pH > 7 at the equivalence point! The solution is basic when the amount of base added is precisely equal to the amount of acid initially present. This is due to the presence of OAc^-, the conjugate base of HOAc, at the equivalence point, which reacts as expected:
  OAc^-(aq) + H₂O <===> HOAc(aq) + OH^-(aq)

  The small amount of OH^- produced by this process causes the pH to be > 7.

The differences in the titration curves for strong and weak acids allow us to tell by inspection of the curve whether the titrated acid is strong or weak.

Before discussing the second method of equivalence point detection, we will consider the shape of the titration curve for a weak acid somewhat more quantitatively in terms of the equilibrium constant expression 13-9-3.

This expression must be obeyed throughout the titration. Let's recast it in a form directly involving pH by first, rearranging it to put [H₃O⁺] alone on the left side, and second, taking the negative logarithm of both sides:

This equation will be useful for calculation of pH at any point where both [OAc-] and [HOAc] have known values.

• Initial pH. Initially, we know only [HOAc] = 0.10 M. Using its tabulated pK_a of 4.74, we can estimate the initial pH readily: pH = 1/2(pK_a - log C_o) = ½ (4.7 + 1)
  pH = 2.9 This is close to the experimental initial pH in Figure 13-7b.
• pH when 10 mL NaOH has been added. At this point, the amount of OH^- added is 1/4 of the amount of acid initially present. This will convert 1/4 of the acid to its conjugate base, so the ratio of conjugate base to acid concentration is [OAc^-]/[HOAc] = (1/4) / (3/4) = 1/3
  Thus pH = pK_a + log (1/3) = 4.7 - 0.48 = 4.2
• pH when 20 mL NaOH has been added. The amount of OH^- added is half of the amount of acid initially present. The ratio of acetate to acetic acid concentration is [OAc^-]/[HOAc] = (1/2)/(1/2) = 1
  Thus pH = pK_a + log (1) = pK_a = 4.7. This is a very useful result. It tells us that halfway to the equivalence point, the pH gives a direct measure of pK_a for the acid. We have seen this idea before in our discussion of distribution diagrams.
• pH when 30 mL NaOH has been added. By reasoning similar to that above, the acetate-to-acid ratio is 3, so pH = pK_a + log(3) = 5.2
• pH when 39 mL NaOH has been added. Apply the reasoning to show that [OAc^-]/[HOAc] = 39/1 = 39, and pH = pK_a + log(39) = 6.3. Note that addition of 39 mL base has caused the pH to change by 6.3 - 2.9 = 3.4 units, and that a change of 6.3 - 5.2 = 1.1 units has been caused by the last 9 mL of the 39. Addition of only one more mL of base will put us at the equivalence point.
• pH at the equivalence point, where 40 mL NaOH solution has been added. At this point, we will have converted all but a tiny (unknown) amount of HOAc to OAc^-, so we cannot make use of 13-9-4. We can use an approach similar to the one used to calculate initial pH, however, based on the reaction of OAc^- with water:
  pOH = ½ (pKb - log C_o) = ½ (9.3 - log (0.05) = 5.3

  The concentration of OAc^- is 0.05 M rather than 0.1 M because during the titration, we have doubled the volume of solution. This dilutes the total acetate concentration by a factor of 2. This gives pH = 14 - 5.3 = 8.7. The last mL of titrant causes a pH change of 8.7 - 6.3 = 2.4 units.

  We will not make calculations for volumes of added NaOH beyond the equivalence point, because pH can be readily calculated from the excess [OH-]. You may want to attempt to calculate pH at 50 mL of added NaOH.

The form of 13-9-4 suggests that a plot of pH versus log [A-]/[HA] should be linear with slope of 1.00 and intercept of pK_a. Such a plot is in fact a variation on the distribution diagram for the HA/A- conjugate pair, discussed earlier. Figure 13-8 shows plots of pH versus log R (R = [A-]/[HA]) for the acids whose distribution diagrams are plotted in Figures 13-4, 13-5, and 6: HOAc, H₃PO₄, HF, and NH₄⁺. The advantage of the log-log plot is that several acids may be conveniently shown on the same diagram, from which pK_a values may be conveniently read from the y-intercepts. The acid form of a conjugate pair predominates at pH values less than the y-intercept (pK_a) for the acid; the base form predominates at pH values greater than the y-intercept. If the acid region for a conjugate pair overlaps the base region for another conjugate pair, the acid of the first pair and the base of the second pair are compatible and may coexist in solution. If not, they are incompatible and may not coexist. From Figure 13-8, HF and NH₃ are seen to be incompatible. They must therefore react via proton transfer, producing F- and NH₄⁺.

Acid-base indicators. The second method for flagging the equivalence point in a titration is to use an indicator. An indicator is a weak acid, usually an organic compound of complex structure, which when dissolved in water undergoes the usual dissociation:

HIn and In^- are generic symbols for the acid and base forms, respectively. The structural formula for a frequently used indicator, phenolphthalein, is shown in Figure 13-9.

An indicator is useful because its conjugate acid and base forms have different colors. For example, phenolphthalein has a colorless acid form and a pink base form; methyl red has a red acid form and a yellow base form. The color of an indicator solution therefore depends on the value of the ratio R = [In^-]/[HIn] in the solution. In practice, if R > 10, the solution shows the color of the base, In^-, because the amount of In^- is much larger than the amount of HIn. If R < 0.1, the solution shows the color of HIn. For R between these extremes, the solution color is intermediate between the HIn and In^- colors. The value of R is determined by two quantities: the value of the equilibrium constant, K_In, and the pH, according to an equation identical to 13-9-4 above:

For R = [In^-]/[HIn] to have the value 10 requires that pH have the value pK_In + 1. Similarly, R = 0.1 requires pH = pK_In - 1. Thus if the pH of the indicator solution is one or more units higher than pK_In for the indicator, the solution shows the color of In^-, the base form of the indicator. If the pH of the solution is one or more units lower than pK_In, the solution has the color of HIn, the acid form. At pH in the range pK_In ± 1, the color is in transition. If we choose an indicator so that its color change coincides with the equivalence point of the titration, the color change signals that the equivalence point has been reached. An important point arises here. The indicator is to be used only as a signal; its concentration must be low enough that ionization of the indicator by 13-9-5 does not contribute significantly to [H₃O⁺] in the solution. Ideally, the indicator should be a passive resident of the solution that responds to changes in [H₃O⁺] resulting from the titration, but that itself does not contribute appreciably to [H₃O⁺]. The concentration of indicator must be low enough that the value of the color ratio, R, is dictated by [H₃O⁺] in the solution resulting from ionization of the acid to be titrated. Obviously, if a substance is to be useful as an indicator, it must be intensely colored in one or both of its forms (i.e., it must have a large molar absorptivity). This allows it to be seen in solution when a very small amount, typically 10^-5 moles per liter, is present.

To properly choose an indicator for a particular titration requires that we have available certain information. First, we must know the pH at the equivalence point of the titration. We have seen that this is 8.7 in the titration of HOAc with NaOH. This value may be determined by first doing the titration with a pH meter and determining the equivalence point pH from the inflection point. Second, we must have available a list of pK_In values for a variety of indicators. Table 13-6 contains such a list. We then pick an indicator whose color change range (pK_In ± 1) brackets the equivalence point pH.

The advantage of indicator titrations over pH-meter titrations is speed. Using a pH meter to monitor a titration is slow because the pH must be read after each of a large number of incremental additions of titrant. In contrast, in an indicator titration, titrant can be added rapidly until near the equivalence point, slowing down only when the color change is imminent. Although it is usually necessary to perform one titration of a particular acid with a pH meter to determine the equivalence point pH, all subsequent titrations of this acid may be carried out quickly with an appropriate indicator.

We close this section by distinguishing two terms that chemists use in conjunction with titration. They are equivalence point, the point at which the number of moles of base added is exactly equal to the number of moles of acid initially present; and endpoint, the point at which the indicator changes color. In choosing an indicator, it is important that the equivalence and end points coincide.

Other Titrations. Our discussion of titration has been limited to titration of an acid with a base. It is also possible to titrate a solution containing an unknown concentration of base with an acid titrant. An example is that of the weak base, NH₃ with the strong acid HCl. The titration curve resembles those of Figure 13-7, except that the pH starts high and ends low. The calculation of pH at several points during the titration can be accomplished in a manner analogous to that used for the titration of acetic acid, and indicator choice is made according to the same criteria already discussed.

In addition, it is possible to titrate acids having more than one acidic proton. Figure 13-10 shows the curve for titration of a generic acid, H₂A, with 0.10 M NaOH. The curve has the shape that we might expect: it shows 2 equivalence points, the second occuring after addition of exactly twice the volume of base needed to reach the first. The first equivalence point is at lower pH than the second, because the first proton dissociates more readily than the second. Further, since at the first equivalence point the solution contains only HA^-, which is both an acid and a base, the pH can be either > or < 7, depending on the relative magnitudes of K_a(HA^-) and K_b(HA^-). Distinct equivalence points for titration of a 2-proton acid can be observed as long as two conditions are met:

If K_a1 and K_a2 are similar, only one equivalence point is observed, because both protons react at once according to

If the second dissociation is weak, only a single equivalence point is seen, corresponding to

In cases where both equivalence points are seen, K_a1 can be estimated from the pH at the first half-equivalence point, and K_a2 from the pH at the second half-equivalence point.

13-10 Range of Acid Strengths in Water. Is there a limitation on the range of strengths of acids that can be titrated in aqueous solution? Indeed there is, as we now show.

• Strong Acids. Strong acids are converted completely to H₃O⁺ in water, so their aqueous solutions are actually solutions of H₃O⁺ and the (conjugate base) anion. A 0.1 M solution of any of them gives a titration curve identical to Figure 13-7a because in all cases it is H₃O⁺ that is being titrated. Thus, although strong acids may be titrated in water, they may not be distinguished by titration because they all give the same curve.
• Acids with 1 > K_a > 10^-14. These acids lie between H₃O⁺ and H₂O in Table 13-3. They are weaker than H₃O⁺ but stronger than H₂O, and we have referred to them as weak acids. As K_a becomes smaller, the initial pH of a 0.1 M solution gets higher, as do the half-equivalence point and equivalence point pH values. The equivalence point pH becomes higher because A^- becomes a stronger base, and the extent of base hydrolysis increases at the equivalence point. Table 13-7 shows how these three pH values change for a 0.1 M solution of HA as K_a decreases from 10^-5 to 10^-13. Figure 13-11 shows the titration curves for 0.1 M solutions of acids having these K_a's.

Note that the initial pH of a weak acid solution is never higher than 7. We see the values approaching 7 as the acid becomes weaker. Second, the difference between the half-equivalence point and equivalence point pH's becomes smaller as the acid becomes weaker. This occurs because if K_a of the acid becomes much smaller than 10^-12, the acid no longer reacts to completion with OH^-. Addition of OH^- to a solution of such an acid does not result in complete consumption of OH^-, so pH rises rapidly at the beginning of the titration, and no clear equivalence point is detected:

For example, the pH at the equivalence point when pK_a = 13 is 12.56. Clearly this is so near the maximum possible pH (13) that no equivalence point is detected. Thus neither water itself, nor any weak acid of comparable strength, can be titrated in water. This sets the limit of weakness. Based on this, the acids HPO₄^2- and HS^-, with K_a values of 4.2 x 10^-13 and 1 x 10^-14, respectively, show no equivalence point deflection in water.

In summary, we present Table 13-8.

13-11 Buffer Solutions.Solutions containing comparable concentrations of a weak acid, HA, and its conjugate base, A^-, are called buffer solutions. In general, a buffer is something that absorbs an impact or softens the effect of a blow. When used in reference to a solution, the word buffer means something very specific:

Ordinarily, addition of even a small amount of strong acid or base to water causes a large pH change. A large pH change constitutes a substantial impact on the species present in the solution. A buffer solution is one in which this impact is softened. A solution containing comparable amounts of HA and A^- has the ability to do this. Why?

Consider a solution containing both a weak acid, HA, and its conjugate base, A^-, in comparable (i.e., roughly equal) concentrations. What factors determine the pH of such a solution? As we have done before, we obtain the answer from the K_a expression for the acid by rearranging for the hydronium ion concentration, then taking the negative logarithm of both sides.

We have seen versions of this before in 13-9-4 and 13-9-6. The solution pH depends on two quantities:

• The value of K_a, which is a constant depending on the identity of HA;
• The value of R = [A^-]/[HA]. A change in R means a change in pH; but as long as R stays nearly constant, pH does too.

What happens if a small amount of strong acid, such as HCl or HNO₃, is added? We can use the (Bronsted Table) to predict a proton transfer from H₃O⁺ (arising from the strong acid) to A^-, the conjugate base of the weak acid:

This will proceed to completion, with the result that [A-] is decreased and [HA] correspondingly increased. As long as the amount of acid added is small relative to the concentrations of HA and A^-, the ratio R decreases by only a small amount, and pH changes very little. Addition of a small amount of strong base, such as NaOH or KOH, results in the following proton transfer process, which proceeds to completion:

This causes an increase in the value of R, but as long as the amount of base added is small relative to the concentrations of HA and A^-, R and pH change little. We can thus view buffering as a shuttling back and forth between the two species HA and A^- as a result of addition of small amounts of either strong acid or base. The most effective buffering occurs when

• The concentrations of HA and A^- are large relative to the amount of H₃O⁺ or OH^- added;
• R is fairly close to 1.

The smallest percent change in R occurs when its initial value is 1. The more R deviates from this value, the more its value changes for a given amount of strong acid or base added. If R becomes larger than 10 or smaller than 0.1, buffering is no longer effective. Substituting the values 10 and 0.1, respectively, to equation 13-40, we find that a particular conjugate acid-base pair buffers best within a range of 1 pH unit to either side of its pK_a:

Thus the acetic acid/acetate pair buffers best over a range centered at 4.7, whereas the ammonium ion/ammonia pair is effective around pH 9.3. The buffer range for HOAc/OAc^- is readily visible in the titration curve of Figure 13-7b. We did not previously call your attention specifically to the pH range between about 3.7 and 5.7. Note now, however, that over this range the pH changes relatively little despite the addition of a total of 20 mL of strong base (OH^-). It is in this range that both HOAc and OAc^- are present in comparable amounts. Added OH^- is absorbed in the conversion of HOAc to OAc^-, which changes R somewhat but not dramatically. All weak acid or weak base titration curves show this buffer region centered on a pH equal to the pK_a of the acid.

Buffer action is extremely important in biological and ecological systems, which often cannot tolerate pH changes of more than a few hundredths of a pH unit. For example, human blood contains two conjugate pairs that together work to maintain the pH within 0.01 units of 7.1. These are the carbonic acid/ hydrogen carbonate pair (CO₂/HCO₃^-), with a buffering range centered at 6.37; and the dihydrogen phosphate/hydrogen phosphate pair (H₂PO₄^-/HPO₄^2-), with a range centered at 7.21. Were it not for the effective buffering action provided by these pairs, human life expectancy would be short.

Example 13-15. A solution contains 0.15 M HCOOH (formic acid) and 0.12 M NaHCOO (sodium formate). What is the pH of the solution? What is the pH following addition of 0.02 M HCl?

Solution. Equation 13-11-1 can be used to calculate the pH of the formic acid buffer system:

Addition of 0.02 M HCl, a strong acid, converts 0.02 M formate ion to formic acid. The new concentrations of the buffer components are [HCOO-] = 0.10 M and [HCOOH] = 0.17 M. The final pH is

Example 13-16. It is necessary for you to do some experiments in water buffered at physiological pH (7.1). Describe how you would prepare 500 mL of an appropriate buffer solution.

Solution. Try this yourself.

13-12 The Lewis Concept. A more general concept of acids and bases was proposed in about 1930 by GN Lewis, based on his recognition that many reactions exhibit the characteristics of acid-base reactions, even though they do not involve H⁺ or take place in aqueous solution. One such reaction is shown below.

When this reaction occurs, the reactivities of both BF₃ and NH₃ are neutralized, similar to the way in which acids and bases neutralize each other in aqueous solution. However, 13-12-1 does not involve proton transfer, and therefore cannot be classified as an acid-base process in the Bronsted framework. Lewis recognized that 13-12-1 involves the formation of a bond between N and B. The bond results from donation of the lone pair on the nitrogen atom of NH₃ to the boron atom of BF₃, which has only 6 electrons. The Lewis concept is based on the idea of electron pair donation:

A Lewis reaction is often called an adduct formation reaction.The word "adduct" is a contraction of the two words "addition product." Many reactions, including 13-12-1, that are not acid-base reactions in terms of the Bronsted concepts, are acid-base reactions in the Lewis sense. You will encounter the full application of Lewis acid-base theory if you go further with your study of chemistry. For now, we will be content to show that Lewis theory encompasses the less general Bronsted theory; that is, we will show that proton transfer reactions can be analyzed in terms of the Lewis definitions of acids and bases.

Consider the proton transfer process below.

The familiar Bronsted classifications are given below the species. Any such proton transfer can be viewed as the sum of two reactions, each involving only one conjugate pair:

When the acid HA comes apart, both electrons of the bond between H and A leave with the base, A^-. This electron pair is explicitly shown on A^-. Similarly, when the acid HB⁺ is formed, the bonding electron pair originates on the base, B. In other words, B donates an electron pair to H⁺. These two reactions are very similar in form to 13-12-1. They are in fact Lewis reactions. Equation 13-12-4 is an adduct formation process; 13-12-3 is an adduct dissociation process. Thus the Lewis interpretation of the proton transfer process, 13-12-2, is as follows.

• Bronsted bases (A^- and B) are also Lewis bases. Further, we now understand better why a Bronsted base accepts a proton. It has an available electron pair that can be used to form a chemical bond to the proton. Although the Bronsted definition of a base as a proton acceptor is fine, it does not reveal the motivation for proton acceptance by the base. In contrast, the Lewis definition reveals that the motivation is the formation of a chemical bond.
• Bronsted acids are Lewis adducts of a base and the proton, H⁺, which is a Lewis acid because it accepts an electron pair from a Lewis base. Thus HA is an adduct of the acid, H⁺, and the base A^-. HB⁺ is an adduct of H⁺ and B.

Thus the Bronsted concept is derivative of the more general Lewis concept. In the next section of the chapter, we use the Lewis concept to introduce a general view of acidity in aqueous solution.

Example 13-17. Identify the Lewis acids, bases, and adducts in the following processes:

Solution. In the first process, HC₂H₃O₂(aq) is a Lewis adduct of the Lewis acid, H⁺, and the Lewis base, C₂H₃O₂^-(aq). H₃O⁺ is a Lewis adduct of the Lewis acid, H⁺, and the Lewis base, H₂O.

Try the second one yourself.

13-13 Cations as Acids: A Unifying Concept.

The Full Bronsted Spectrum of Water. The complete series of Bronsted acid-base reactions of water can be expressed in terms of four species, three of which we have already used extensively. The Bronsted species of water are