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Many organic acids occur in the vegetable kingdom. Lemons, oranges, and tomatoes contain ascorbic acid and citric acid and rhubarbs and spinach contain oxalic acid.
Acids and Bases
E SSENTIAL C ONCEPTS
Br ø nsted Acids and Bases A Brønsted acid can donate a proton and a Brønsted base can accept a proton. For every Brønsted acid, there exists a conjugate Brønsted base and vice versa. Acid-Base Properties of Water and the pH Scale Water acts both as a Brønsted acid and as a Brønsted base. At 25C, the concen- trations of H^ and OH^ ions are both at 10^7 M. The pH scale is established to express the acidity of a solution—the smaller the pH, the higher the H^ concentration and the greater the acidity. Acid and Base Ionization Constants Strong acids and strong bases are assumed to ionize completely. Most weak acids and bases ionize to a small extent. The concentrations of the acid, conjugate base, and H^ ion at equilibrium can be calculated from the acid ionization constant, which is the equilibrium constant for the reaction. Molecular Structure and Acid Strength The strength of a series of structurally similar acids can be compared using parameters such as bond enthalpy, bond polarity, and oxidation number. Acid-Base Properties of Salts and Oxides Many salts react with water in a process called hydrolysis. From the nature of the cation and anion present in the salt, it is possible to predict the pH of the resulting solution. Most oxides also react with water to produce acidic or basic solutions. Lewis Acids and Bases A more general definition of acids and bases characterizes an acid as a substance that can accept a pair of electrons and a base as a substance that can donate a pair of electrons. All Brønsted acids and bases are Lewis acids and bases.
C H A P T E R
Activity Summary
- Animation: Acid Ionization (16.5)
- Interactivity: Calculating pH of Acid Solution (16.5)
- Animation: Base Ionization (16.6)
- Interactivity: Calculating pH of Base Solution (16.6)
- Interactivity: Molecular Structure and Acid Strength (16.8)
- Interactivity: Acid-Base Properties of Salts (16.9)
C HAPTER O UTLINE
16.1 Brønsted Acids and Bases 530
Conjugate Acid-Base Pairs
16.2 The Acid-Base Properties of Water 531
The Ion-Product of Water
16.3 pH—A Measure of Acidity 533
16.4 Strength of Acids and Bases 536
16.5 Weak Acids and Acid Ionization Constants 540
Percent Ionization • Diprotic and Polyprotic Acids
16.6 Weak Bases and Base Ionization
Constants 551
16.7 The Relationship Between Conjugate
Acid-Base Ionization Constants 553
16.8 Molecular Structure and the Strength
of Acids 554
Hydrohalic Acids • Oxoacids
16.9 Acid-Base Properties of Salts 557
Salts That Produce Neutral Solutions • Salts That Produce Basic Solutions • Salts That Produce Acidic Solutions • Metal Ion Hydrolysis • Salts in Which Both the Cation and Anion Hydrolyze
16.10 Acidic, Basic, and Amphoteric Oxides 563
16.11 Lewis Acids and Bases 565
16.1 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 capable of accepting a proton. These definitions are generally suitable for discussion of the properties and reactions of acids and bases.
Conjugate Acid-Base Pairs 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 conju- gate 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 conju- gate acid. For example, the chloride ion (Cl) is the conjugate base formed from the acid HCl, and H 2 O is the conjugate base of the acid H 3 O. Similarly, the ionization of acetic acid can be represented as
The subscripts 1 and 2 designate the two conjugate acid-base pairs. Thus, the acetate ion (CH 3 COO) is the conjugate base of the acid CH 3 COOH. Both the ionization of HCl (see Section 4.3) and the ionization of CH 3 COOH are examples of Brønsted acid- base reactions. The Brønsted definition also enables us to classify ammonia as a base because of its ability to accept a proton:
In this case, NH 4 is the conjugate acid of the base NH 3 , and OH^ is the conjugate base of the acid H 2 O. Note that the atom in the Brønsted base that accepts a H^ ion must have a lone pair.
HONOH HOOS 34 HONOH HOOS
H
A
A
H
A
H
A
H
O O OQ
NH 3 ( aq ) H 2 O( l ) 34 NH 4 ( aq ) OH( aq ) base 1 acid 2 acid 1 base (^2)
CH 3 COOH( aq ) H 2 O( l ) 34 CH 3 COO( aq ) H 3 O( aq ) acid 1 base 2 base 1 acid 2
HOCOCOOOH HOOS 34 HOCOCOOS^ HOOOH
H
A
A
H
H
A
A
H
A
H
A
H
SOS
B
SOS
B
O O O
Q
O
Q
Conjugate means “joined together.”
Electrostatic potential map of the hydronium ion. The proton is always associated with water molecules in aqueous solution. The H 3 O^ ion is the simplest formula of a hydrated proton.
Identify the conjugate acid-base pairs in the reaction between ammonia and hydrofluoric acid in aqueous solution
( Continued )
NH 3 ( aq ) HF( aq ) (^) Δ NH 4 ( aq ) F( aq )
Example 16.
530 CHAPTER 16 Acids and Bases
water molecules are ionized, the concentration of water, [H 2 O], remains virtually unchanged. Therefore, the equilibrium constant for the autoionization of water, according to Equation (16.2), is
Because we use H( aq ) and H 3 O( aq ) interchangeably to represent the hydrated pro- ton, the equilibrium constant can also be expressed as
To indicate that the equilibrium constant refers to the autoionization of water, we replace K c by K w
(16.3)
where K w is called the ion-product constant, which is the product of the molar con- centrations of H ^ and OH^ ^ ions at a particular temperature. In pure water at 25C, 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 Equa- tion (16.3), at 25C
Whether we have pure water or an aqueous solution of dissolved species, the fol- lowing relation always holds at 25C:
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^ con- centration must change to
[OH]
K w [H]
1.0 10 ^14
1.0 10 ^6
1.0 10 ^8 M
K w [H][OH] 1.0 10 ^14
K w (1.0 10 ^7 )(1.0 10 ^7 ) 1.0 10 ^14
K w [H 3 O][OH] [H][OH]
K c [H][OH]
K c [H 3 O][OH]
532 CHAPTER 16 Acids and Bases
Recall that in pure water, [H 2 O] 55.5M (see p. 502).
If you could randomly remove and examine 10 particles (H 2 O, H , or OH ) per second from a liter of water, it would take you 2 years, working nonstop, to find one H ^ ion!
The concentration of OH^ ions in a certain household ammonia cleaning solution is 0.0025 M. Calculate the concentration of H^ ions. Strategy We are given the concentration of the OH^ ions and asked to calculate [H]. The relationship between [H] and [OH] in water or an aqueous solution is given by the ion-product of water, K w [Equation (16.4)]. Solution Rearranging Equation (16.4), we write
( Continued )
[H]
K w [OH]
1.0 10 ^14
4.0 10 ^12 M
Example 16.
16.3 pH—A Measure of Acidity 533
16.3 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, the Danish chemist 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):
or (16.5)
Keep in mind that Equation (16.5) is simply a definition designed to give us conven- ient 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 (16.5) 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. Because pH is simply a way to express hydrogen ion concentration, acidic and basic solutions at 25C can be distinguished by their pH values, as follows:
Acidic solutions: [H] 1.0 10 ^7 M , pH 7. Basic solutions: [H] 1.0 10 ^7 M , pH 7. Neutral solutions: [H] 1.0 10 ^7 M , pH 7.
Notice that pH increases as [H] decreases. Sometimes we may be given the pH value of a solution and asked to calculate the H^ ion concentration. In that case, we need to take the antilog of Equation (16.5) as follows:
Be aware that the definition of pH just shown, and indeed all the calculations involving solution concentrations (expressed either as molarity or molality) discussed in previous chapters, 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. 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 pres- ent, the measured gas pressure may differ from that calculated using the ideal gas equa- tion. Similarly, the actual or “effective” concentration of a solute may not be what we think it is, knowing 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 and nonideal gas behavior, we can account for nonideal behavior in solution. One way is to replace the concentration term with activity, which is the effective concentration. Strictly speaking, then, the pH of solution should be defined as
pH log a H (16.7)
[H] 10 pH
pH log [H 3 O] pH log [H]
Check Because [H] [OH], the solution is basic, as we would expect from the earlier discussion of the reaction of ammonia with water. Practice Exercise Calculate the concentration of OH^ ions in a HCl solution whose hydrogen ion concentration is 1.3 M.
Similar problem: 16.16(c).
Note that a unit pH change corresponds to a 10-fold change in [H ].
The pH of concentrated acid solutions can be negative. For example, the pH of a 2.0M HCl solution is 0.30.
16.3 pH—A Measure of Acidity 535
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. Strategy We are given the H^ ion concentration and asked to calculate the pH of the solution. What is the definition of pH? Solution According to Equation (16.5),. When the bottle was first opened, [H] 3.2 10 ^4 M , which we substitute in Equation (16.5)
]
On the second occasion, [H] 1.0 10 ^3 M , so that
Comment 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. Practice Exercise Nitric acid (HNO 3 ) is used in the production of fertilizer, dyes, drugs, and explosives. Calculate the pH of a HNO 3 solution having a hydrogen ion concentration of 0.76 M.
pH log (1.0 10 ^3 ) 3.
log (3.2 10 ^4 ) 3.
pH log [H
pH log [H]
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. Strategy Here we are given the pH of a solution and asked to calculate [H]. Because pH is defined as pH log [H], we can solve for [H] by taking the antilog of the pH; that is, [H] 10 pH, as shown in Equation (16.6). Solution From Equation (16.5)
Therefore,
To calculate [H], we need to take the antilog of 4.
Check Because the pH is between 4 and 5, we can expect [H] to be between 1 10 ^4 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.
[H] 10 4.82^ 1.5 10 ^5 M
log [H] 4.
pH log [H] 4.
In each case the pH has only two significant figures. The two digits to the right of the decimal in 3.49 tell us that there are two significant figures in the original number (see Appendix 3).
Similar problems: 16.17(a), (d).
Scientific calculators have an antilog function that is sometimes labeled INV log or 10x.
Similar problems: 16.16(a), (b).
Example 16.
Example 16.
16.4 Strength of Acids and Bases
Strong acids are strong electrolytes which, for practical purposes, are assumed to ion- ize completely in water (Figure 16.3). Most of the strong acids are inorganic acids: hydrochloric acid (HCl), nitric acid (HNO 3 ), perchloric acid (HClO 4 ), and sulfuric acid (H 2 SO 4 ):
Note that H 2 SO 4 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 equi- librium, aqueous solutions of weak acids contain a mixture of nonionized acid mol- ecules, H 3 O^ ions, and the conjugate base. Examples of weak acids are hydrofluoric acid (HF), acetic acid (CH 3 COOH), and the ammonium ion (NH 4 ). The limited ion- 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,
H 2 SO 4 ( aq ) H 2 O( l ) (^) ¡ H 3 O( aq ) HSO 4 ( aq )
HClO 4 ( aq ) H 2 O( l ) (^) ¡ H 3 O( aq ) ClO 4 ( aq )
HNO 3 ( aq ) H 2 O( l ) (^) ¡ H 3 O( aq ) NO 3 ( aq )
HCl( aq ) H 2 O( l ) (^) ¡ H 3 O( aq ) Cl( aq )
536 CHAPTER 16 Acids and Bases
Example 16. In a NaOH solution [OH] is 2.9 10 ^4 M. Calculate the pH of the solution. Strategy Solving this problem takes two steps. First, we need to calculate pOH using Equation (16.8). Next, we use Equation (16.9) to calculate the pH of the solution. Solution We use Equation (16.8):
]
Now we use Equation (16.9)
Alternatively, we can use the ion-product constant of water, ] to calculate [H], and then we can calculate the pH from the [H]. Try it. Check The answer shows that the solution is basic (pH 7), which is consistent with a NaOH solution. Practice Exercise The OH^ ion concentration of a blood sample is 2.5 10 ^7 M. What is the pH of the blood?
K w [H][OH
14.00 3.54 10.
pH 14.00 pOH
pH pOH 14.
3.
log (2.9 10 ^4 )
pOH log [OH
Similar problem: 16.17(b).
In reality, no acids are known to ionize completely in water.
Zn reacts more vigorously with a strong acid like HCl (left) than with a weak acid like CH 3 COOH (right) of the same concentration because there are more H^ ions in the former solution.
In this reaction, NH 3 acts as a base by accepting a proton from water to form NH 4 and OH^ ions. It is a weak base because only a small fraction of the molecules undergo this reaction. Table 16.2 lists some important conjugate acid-base pairs, in order of their rela- tive 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.
- H 3 O^ is the strongest acid that can exist in aqueous solution. Acids stronger than H 3 O^ react with water to produce H 3 O^ and their conjugate bases. Thus HCl, which is a stronger acid than H 3 O, reacts with water completely to form H 3 O^ and Cl:
Acids weaker than H 3 O^ react with water to a much smaller extent, producing H 3 O^ and their conjugate bases. For example, the following equilibrium lies largely to the left:
- 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 (O 2 ) is a stronger base than OH, so it reacts with water completely as follows:
For this reason the oxide ion does not exist in aqueous solutions.
O^2 ( aq ) H 2 O( l ) (^) ¡ 2OH( aq )
HF( aq ) H 2 O( l ) (^) Δ H 3 O( aq ) F( aq )
HCl( aq ) H 2 O( l ) (^) ¡ H 3 O( aq ) Cl( aq )
538 CHAPTER 16 Acids and Bases
Acid Conjugate Base
HClO 4 (perchloric acid) ClO 4 (perchlorate ion) HI (hydroiodic acid) I^ (iodide ion) HBr (hydrobromic acid) Br^ (bromide ion) HCl (hydrochloric acid) Cl^ (chloride ion) H 2 SO 4 (sulfuric acid) HSO 4 (hydrogen sulfate ion) HNO 3 (nitric acid) NO 3 (nitrate ion) H 3 O^ (hydronium ion) H 2 O (water) HSO 4 (hydrogen sulfate ion) SO^24 (sulfate ion) HF (hydrofluoric acid) F^ (fluoride ion) HNO 2 (nitrous acid) NO 2 (nitrite ion) HCOOH (formic acid) HCOO^ (formate ion) CH 3 COOH (acetic acid) CH 3 COO^ (acetate ion) NH 4 (ammonium ion) NH 3 (ammonia) HCN (hydrocyanic acid) CN^ (cyanide ion) H 2 O (water) OH^ (hydroxide ion) NH 3 (ammonia) NH 2 (amide ion)
TABLE 16.2 Relative Strengths of Conjugate Acid-Base Pairs
Acid strength increases
7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777
n
Weak acids
Strong acids
m
Base strength increases
⎫⎪^
16.4 Strength of Acids and Bases 539
Calculate the pH of (a) a 1.0 10 ^3 M HCl solution and (b) a 0.020 M Ba(OH) (^2) solution. Strategy Keep in mind that HCl is a strong acid and Ba(OH) 2 is a strong base. Thus, these species are completely ionized and no HCl or Ba(OH) 2 will be left in solution. Solution (a) The ionization of HCl is
The concentrations of all the species (HCl, H, and Cl) before and after ionization can be represented as follows:
HCl( aq ) Initial ( M ): 0.0 0. Change ( M ): Final ( M ): 0.
A positive () change represents an increase and a negative () change indicates a decrease in concentration. Thus,
(b) Ba(OH) 2 is a strong base; each Ba(OH) 2 unit produces two OH^ ions:
The changes in the concentrations of all the species can be represented as follows:
Initial ( M ): 0.020 0.00 0. Change ( M ): 0. Final ( M ): 0.00 0.020 0.
Thus,
Therefore, from Equation (16.9)
Check 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 com- pared with 1.0 10 ^3 M and 0.040 M.
Practice Exercise Calculate the pH of a 1.8 10 ^2 M Ba(OH) 2 solution.
12.
14.00 1.
pH 14.00 pOH
pOH log 0.040 1.
[OH] 0.040 M
0.020 2(0.020)
Ba(OH) 2 ( aq ) ¡ Ba^2 ( aq ) 2OH( aq )
Ba(OH) 2 ( aq ) (^) ¡ Ba^2 ( aq ) 2OH( aq )
3.
pH log (1.0 10 ^3 )
[H] 1.0 10 ^3 M
1.0 10 ^3 1.0 10 ^3
1.0 10 ^3 1.0 10 ^3 1.0 10 ^3
1.0 10 ^3
¡ H( aq ) Cl( aq )
HCl( aq ) ¡ H( aq ) Cl( aq )
Similar problems: 16.17(a), (c).
Example 16.
16.5 Weak Acids and Acid Ionization Constants 541
equilibrium concentrations, is the same one outlined in Chapter 15. However, because acid ionization represents a major category of chemical equilibrium in aqueous solu- tion, we will develop a systematic procedure for solving this type of problem that will also help us to understand the chemistry involved. Suppose we are asked to calculate the pH of a 0.50 M HF solution at 25C. The ionization of HF is given by
From Table 16.3 we write
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 H 2 O, but its very small K w (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
K a
[H][F]
[HF]
7.1 10 ^4
HF( aq ) (^) Δ H( aq ) F( aq )
Name of Acid Formula Structure K a Conjugate Base K b
Hydrofluoric acid HF 7.1 10 ^4 F^ 1.4 10 ^11 Nitrous acid HNO 2 OPNOOOH 4.5 10 ^4 NO 2 2.2 10 ^11 Acetylsalicylic acid C 9 H 8 O 4 3.0 10 ^4 C9H 7 O 4 3.3 10 ^11 (aspirin)
Formic acid HCOOH 1.7 10 ^4 HCOO^ 5.9 10 ^11 Ascorbic acid* C 6 H 8 O 6 8.0 10 ^5 C6H 7 O 6 1.3 10 ^10
Benzoic acid C 6 H 5 COOH 6.5 10 ^5 C6H 5 COO^ 1.5 10 ^10
Acetic acid CH 3 COOH 1.8 10 ^5 CH 3 COO^ 5.6 10 ^10
Hydrocyanic acid HCN HOCqN 4.9 10 ^10 CN^ 2.0 10 ^5 Phenol C 6 H 5 OH 1.3 10 ^10 C6H 5 O^ 7.7 10 ^5 OOOH
O B CH 3 OCOOOH
O B OCOOOH
CHOH A CH 2 OH
C P O O
C
H G D
HOOH C PPPC EOH
O B HOCOOOH
B O
O B OCOOOH OOOCOCH 3
H¬F
TABLE 16.3 Ionization Constants of Some Weak Acids and Their Conjugate Bases at 25 C
*For ascorbic acid it is the upper left hydroxyl group that is associated with this ionization constant.
Interactivity: Calculating pH of Acid Solution—Steps 1– ARIS, Interactives
concerned with the OH^ ions that are also present in solution. The OH^ concentra- tion can be determined from Equation (16.4) after we have calculated [H]. We can summarize the changes in the concentrations of HF, H, and F^ according to the ICE method shown on p. 510 as follows:
Initial ( M ): 0.50 0.00 0. Change (M): Equilibrium ( M ): 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
Rearranging this expression, we write
This is a quadratic equation, which can be solved using the quadratic formula (see Appendix 3). 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
Now the ionization constant expression becomes
Rearranging, we get
Thus, we have solved for x without having to use the quadratic equation. At equilib- rium, we have
and the pH of the solution is
How good is this approximation? Because K a values for weak acids are gener- ally known to an accuracy of only 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 approxima- tion is valid if the following expression is equal to or less than 5%:
Thus, the approximation we made is acceptable.
0.019 M
0.50 M
pH log (0.019) 1.
[F] 0.019 M
[H] 0.019 M
[HF] (0.50 0.019) M 0.48 M
x 2 3.55 10 ^4 0.019 M
x^2 (0.50)(7.1 10 ^4 ) 3.55 10 ^4
x^2 0.50 x
x^2
7.1 10 ^4
0.50 x 0.
x^2 7.1 10 ^4 x 3.6 10 ^4 0
K a
( x )( x ) 0.50 x
7.1 10 ^4
0.50 x
x x x
HF( aq ) (^) Δ H( aq ) F( aq )
542 CHAPTER 16 Acids and Bases
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 appreciably change the overall mass of the load.
544 CHAPTER 16 Acids and Bases
Calculate the pH of a 0.036 M nitrous acid (HNO 2 ) solution:
Strategy Recall that a weak acid only partially ionizes in water. We are given the initial concentration of a weak acid and asked to calculate the pH of the solution at equilibrium. It is helpful to make a sketch to keep track of the pertinent species.
As in Example 16.6, we ignore the ionization of H 2 O so the major source of H^ ions is the acid. The concentration of OH^ ions is very small as we would expect from an acidic solution so it is present as a minor species. Solution We follow the procedure already outlined. Step 1: The species that can affect the pH of the solution are HNO 2 , H, and the conjugate base NO 2. We ignore water’s contribution to [H]. Step 2: Letting x be the equilibrium concentration of H^ and NO 2 ions in mol/L, we summarize:
HNO 2 ( aq ) H( aq ) NO 2 ( aq ) Initial ( M ): 0.036 0.00 0. Change ( M ): Equilibrium ( M ): 0.036 x x x
Step 3: From Table 16.3 we write
Applying the approximation 0.036 x 0.036, we obtain
To test the approximation,
( Continued )
4.0 10 ^3 M 0.036 M
100% 11%
x 4.0 10 ^3 M
x^2 ^ 1.62^ ^10 ^5
4.5 10 ^4 x^2 0.036 x x^2
4.5 10 ^4 x^2 0.036 x
K a
[H][NO 2 ] [HNO 2 ]
x x x
Δ
HNO 2 ( aq ) (^) Δ H( aq ) NO 2 ( aq )
HNO 2
Example 16.
One way to determine K a of an acid is to measure the pH of the acid solution of known concentration at equilibrium. Example 16.9 shows this approach.
16.5 Weak Acids and Acid Ionization Constants 545
Similar problem: 16.45.
HCOOH
Example 16. The pH of a 0.10 M solution of formic acid (HCOOH) is 2.39. What is the K a of the acid? Strategy Formic acid is a weak acid. It only partially ionizes in water. Note that the concentration of formic acid refers to the initial concentration, before ionization has started. The pH of the solution, on the other hand, refers to the equilibrium state. To calculate K a, then, we need to know the concentrations of all three species: [H], [HCOO], and [HCOOH] at equilibrium. As usual, we ignore the ionization of water. The following sketch summarizes the situation.
( Continued )
Because this is greater than 5%, our approximation is not valid and we must solve the quadratic equation, as follows:
The second solution is physically impossible, because 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,
Check Note that the calculated pH indicates that the solution is acidic, which is what we would expect for a weak acid solution. Also, compare the calculated pH with that of a 0.036 M strong acid solution such as HCl to convince yourself of the difference between a strong acid and a weak acid. Practice Exercise What is the pH of a 0.122 M monoprotic acid whose K a is 5.7 10 ^4?
2.
pH log (3.8 10 ^3 )
[H] 3.8 10 ^3 M
3.8 10 ^3 M or 4.3 10 ^3 M
x
4.5 10 ^4 2 (4.5 10 ^4 )^2 4(1)(1.62 10 ^5 ) 2(1)
x^2 4.5 10 ^4 x 1.62 10 ^5 0
Referring to Example 16.8, we see that the percent ionization of a 0.036 M HNO (^2) solution is
Thus, only about one out of every 9 HNO 2 molecules has ionized. This is consistent with the fact that HNO 2 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 16.4). In qualitative terms, when an acid is diluted, the concentration of the “particles” in the solution is reduced. According to Le Châtelier’s principle (see Section 15.4), this reduction in particle concentration (the stress) is counteracted by shifting the reaction to the side with more particles; that is, the equilibrium shifts from the nonionized acid side (one particle) to the side containing the H^ ion and the con- jugate base (two particles). Consequently, the percent ionization of the acid increases.
Diprotic and Polyprotic Acids
Diprotic and polyprotic acids 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 con- centrations of species in the acid solution. For example, for H 2 CO 3 we write
Note that the conjugate base in the first ionization stage becomes the acid in the second ionization stage. Table 16.4 shows the ionization constants of several diprotic acids and a polypro- tic 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 an H^ ion from a neutral molecule than to remove another H^ from a negatively charged ion derived from the molecule.
K a 2
[H][CO^23 ]
[HCO 3 ]
HCO 3 ( aq ) (^) Δ H( aq ) CO^23 ( aq )
K a 1
[H][HCO 3 ]
[H2CO 3 ]
H 2 CO 3 ( aq ) (^) Δ H( aq ) HCO 3 ( aq )
percent ionization
3.8 10 ^3 M
0.036 M
16.5 Weak Acids and Acid Ionization Constants 547
0
% Ionization
100
Initial concentration of acid
Weak acid
Strong acid
Figure 16. Dependence of percent ioniza- tion on initial concentration of acid. Note that at very low concentrations, all acids (weak and strong) are almost com- pletely ionized.
Top to bottom: H 2 CO 3 , HCO 3 , and CO^23 .
Oxalic acid (H 2 C 2 O 4 ) is a poisonous substance used chiefly as a bleaching and cleans- ing agent (for example, to remove bathtub rings). Calculate the concentrations of all the species present at equilibrium in a 0.10 M solution. Strategy Determining the equilibrium concentrations of the species of a diprotic acid in aqueous solution is more involved than for a monoprotic acid. We follow the same procedure as that used for a monoprotic acid for each stage, as in Example 16.8. Note that the conjugate base from the first stage of ionization becomes the acid for the second stage ionization. ( Continued )
Example 16.
H 2 C 2 O 4
548 CHAPTER 16 Acids and Bases
Name of Acid Formula Structure K a Conjugate Base K b
Sulfuric acid H 2 SO 4 Very large HSO 4 Very small
Hydrogen sulfate ion HSO 4 1.3 10 ^2 SO^24 7.7 10 ^13
Oxalic acid H 2 C 2 O 4 6.5 10 ^2 HC 2 O 4 1.5 10 ^13
Hydrogen oxalate ion HC 2 O 4 6.1 10 ^5 C 2 O^24 1.6 10 ^10
Sulfurous acid* H 2 SO 3 1.3 10 ^2 HSO 3 7.7 10 ^13
Hydrogen sulfite ion HSO 3 6.3 10 ^8 SO^23 1.6 10 ^7
Carbonic acid H 2 CO 3 4.2 10 ^7 HCO 3 2.4 10 ^8
Hydrogen carbonate ion HCO 3 4.8 10 ^11 CO 32 ^ 2.1 10 ^4 Hydrosulfuric acid H 2 S H S H 9.5 10 ^8 HS^ 1.1 10 ^7 Hydrogen sulfide ion †^ HS^ H S^1 10 ^19 S^2 ^1 105
Phosphoric acid H 3 PO 4 7.5 10 ^3 H 2 PO 4 1.3 10 ^12
Dihydrogen phosphate ion H 2 PO 4 6.2 10 ^8 HPO 42 ^ 1.6 10 ^7
Hydrogen phosphate ion HPO 42 ^ 4.8 10 ^13 PO 43 ^ 2.1 10 ^2
O B A O
HOOOPOO
O B A O A H HOOOPOO
O B A O A H HOOOPOOOH
¬
¬ ¬
O B HOOOCOO
O B HOOOCOOOH
O B HOOO S OO
O B HOOO S OOOH
O B C
O B HOOO O COO
O B
O B HOOOCOCOOOH
O B B O
HOOOSOO
O B B O
HOOOSOOOH
TABLE 16.
Ionization Constants of Some Diprotic Acids and a Polyprotic Acid and Their Conjugate Bases at 25 C
*H 2 SO 3 has never been isolated and exists in only minute concentration in aqueous solution of SO 2. The K a value here refers to the process
†. The ionization constant of HS^ is very low and difficult to measure. The value listed here is only an estimate.
SO 2 ( g ) H 2 O( l ) Δ H( aq ) HSO 3 ( aq )
