Library Guides: Chemistry Textbook: Application: Acid-Base Equilibria

Application: Acid-Base Equilibria

Brønsted-Lowry Acids and Bases

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

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

According to the Arrhenius definition, an acid is a compound that dissolves (dissociates or ionizes) in water to yield hydrogen cations (now recognized to be hydronium ions) and a base is a compound that dissolves in water to yield hydroxide anions. Johannes Brønsted and Thomas Lowry proposed a more general definition in 1923 in which acids and bases were defined in terms of the transfer of hydrogen ions, H⁺(Note that these hydrogen ions are often referred to simply as protons). A compound that donates a proton to another compound is called a Brønsted-Lowry acid, and a compound that accepts a proton is called a Brønsted-Lowry base. An acid-base reaction is, thus, the transfer of a proton from a donor (acid) to an acceptor (base).

The concept of conjugate pairs is useful in describing Brønsted-Lowry acid-base reactions. When an acid donates H⁺, the species that remains is called the conjugate base of the acid because it reacts as a proton acceptor in the reverse reaction. Likewise, when a base accepts H⁺, it is converted to its conjugate acid. The reaction between water and ammonia illustrates this idea. In the forward direction, water acts as an acid by donating a proton to ammonia and subsequently becoming a hydroxide ion, OH⁻, the conjugate base of water. The ammonia acts as a base in accepting this proton, becoming an ammonium ion, ${NH}_{4}^{+},$ the conjugate acid of ammonia. In the reverse direction, a hydroxide ion acts as a base in accepting a proton from ammonium ion, which acts as an acid

The reaction between a Brønsted-Lowry acid and water is called acid ionization. Strong acids/bases ionize almost completely in water and weak acids/bases ionize partially in water. Hence, weak acid (or weak base) reactions in water are shown as reversible reactions with undissociated acid (or base) in equilibrium with the products. Consider another example: when hydrogen fluoride which is a weak acid dissolves in water and ionizes, protons are transferred from hydrogen fluoride molecules to water molecules, yielding hydronium ions and fluoride ions:

Base ionization of a species occurs when it accepts protons from water molecules. In the example below, pyridine molecules, C₅NH₅, undergo base ionization when dissolved in water, yielding hydroxide and pyridinium ions

The preceding ionization reactions suggest that water may function as both a base (as in its reaction with hydrogen fluoride) and an acid (as in its reaction with ammonia). Species capable of either donating or accepting protons are called amphiprotric, or more generally, amphoteric, a term that may be used for acids and bases per definitions other than the Brønsted-Lowry one. The equations below show the two possible acid-base reactions for two amphiprotic species, bicarbonate ion and water:

{HCO}_{3}^{-} (aq) + H_{2} O (l) ⇌ {CO}_{3}^{2-} (aq) + H_{3} O^{+} (aq)

{HCO}_{3}^{-} (aq) + H_{2} O (l) ⇌ H_{2} {CO}_{3} (aq) + {OH}^{–} (aq)

The first equation represents the reaction of bicarbonate as an acid with water as a base, whereas the second represents reaction of bicarbonate as a base with water as an acid. When bicarbonate is added to water, both these equilibria are established simultaneously and the composition of the resulting solution may be determined through appropriate equilibrium calculations, as described later in this chapter.

Autoionization of Water

In the liquid state, molecules of an amphiprotic substance can react with one another as illustrated for water in the equations below

The process in which like molecules react to yield ions is called autoionization. Liquid water undergoes autoionization to a very slight extent; at 25 °C, approximately two out of every billion water molecules are ionized. The extent of the water autoionization process is reflected in the value of its equilibrium constant, the ion-product constant for water, K_w:

H_{2} O (l) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + {OH}^{−} (a q) K_{w} = [H_{3} O^{+}] [{OH}^{−}]

The slight ionization of pure water is reflected in the small value of the equilibrium constant; at 25 °C, K_w has a value of 1.0 $\times$ 10⁻¹⁴. The process is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion increase with temperature. For example, at 100 °C, the value for K_w is about 5.6 $\times$ 10⁻¹³, roughly 50 times larger than the value at 25 °C.

EXAMPLE 11.11

Sample Problem: Ion Concentrations in Pure Water

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

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

K_{w} = [H_{3} O^{+}] [{OH}^{−}] = x = 1.0 \times 10^{−14}

So:

x = [H_{3} O^{+}] = [{OH}^{−}] = \sqrt{1.0 \times 10^{−14}} = 1.0 \times 10^{−7} M

The hydronium ion concentration and the hydroxide ion concentration are the same, 1.0 $\times$ 10⁻⁷ M.

Check Your Learning: The ion product of water at 80 °C is 2.4 $\times$ 10⁻¹³. What are the concentrations of hydronium and hydroxide ions in pure water at 80 °C?

Answer:

[H₃O⁺] = [OH⁻] = 4.9 $\times$ 10⁻⁷ M

EXAMPLE 11.12

Sample Problem: The Inverse Relation between [H₃O⁺] and [OH⁻]

A solution of an acid in water has a hydronium ion concentration of 2.0 $\times$ 10⁻⁶ M. What is the concentration of hydroxide ion at 25 °C?

Solution: Use the value of the ion-product constant for water at 25 °C

2H2O(l)⇌H3O+(aq)+OH−(aq)Kw=[H3O+][OH−]=1.0×10−14

to calculate the missing equilibrium concentration.

Rearrangement of the K_w expression shows that [OH⁻] is inversely proportional to [H₃O⁺]:

[{OH}^{−}] = \frac{K_{w}}{[H_{3} O^{+}]} = \frac{1.0 \times 10^{−14}}{2.0 \times 10^{−6}} = 5.0 \times 10^{−9}

Compared with pure water, a solution of acid exhibits a higher concentration of hydronium ions (due to ionization of the acid) and a proportionally lower concentration of hydroxide ions. This may be explained via Le Châtelier’s principle as a left shift in the water autoionization equilibrium resulting from the stress of increased hydronium ion concentration.

Substituting the ion concentrations into the K_w expression confirms this calculation, resulting in the expected value:

K_{w} = [H_{3} O^{+}] [{OH}^{−}] = (2.0 \times 10^{−6}) (5.0 \times 10^{−9}) = 1.0 \times 10^{−14}

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

Answer:

[H₃O⁺] = 1 $\times$ 10⁻¹¹ M

EXAMPLE 11.13

Sample Problem: Representing the Acid-Base Behavior of an Amphoteric Substance

Write separate equations representing the reaction of ${HSO}_{3}^{−}$

(a) as an acid with OH⁻

(b) as a base with HI

Solution: (a) ${HSO}_{3}^{−} (a q) + {OH}^{−} (a q) ⇌ {SO}_{3}^{2−} (a q) + H_{2} O (l)$

(b) ${HSO}_{3}^{−} (a q) + HI (a q) ⇌ H_{2} {SO}_{3} (a q) + I^{−} (a q)$

Check Your Learning: Write separate equations representing the reaction of $H_{2} {PO}_{4}^{−}$

(a) as a base with HBr

(b) as an acid with OH⁻

Answer:

(a) $H_{2} {PO}_{4}^{−} (a q) + HBr (a q) ⇌ H_{3} {PO}_{4} (a q) + {Br}^{−} (a q);$ (b) $H_{2} {PO}_{4}^{−} (a q) + {OH}^{−} (a q) ⇌ {HPO}_{4}^{2−} (a q) + H_{2} O (l)$

pH and pOH

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

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

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

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

pX = −log X

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

pH = −log [H_{3} O^{+}]

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

[H_{3} O^{+}] = 10^{−pH}

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

pOH = −log [{OH}^{−}]

[{OH}^{−}] = 10^{−pOH}

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

K_{w} = [H_{3} O^{+}] [{OH}^{−}]

−log K_{w} = −log ([H_{3} O^{+}] [{OH}^{−}]) = −log [H_{3} O^{+}] + −log [{OH}^{−}]

p K_{w} = pH + pOH

At 25 °C, the value of K_w is 1.0 $\times$ 10⁻¹⁴, and so:

14.00 = pH + pOH

As was shown in Example 11.11, the hydronium ion molarity in pure water (or any neutral solution) is 1.0 $\times$ 10⁻⁷ M at 25 °C. The pH and pOH of a neutral solution at this temperature are therefore:

pH = −log [H_{3} O^{+}] = −log (1.0 \times 1 0^{−7}) = 7.00

pOH = −log [{OH}^{−}] = −log (1.0 \times 1 0^{−7}) = 7.00

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

Since the autoionization constant K_w is temperature dependent, these correlations between pH values and the acidic/neutral/basic adjectives will be different at temperatures other than 25 °C. For example, the “Check Your Learning” exercise accompanying Example 11.11 showed the hydronium molarity of pure water at 80 °C is 4.9 $\times$ 10⁻⁷ M, which corresponds to pH and pOH values of:

pH = −log [H_{3} O^{+}] = −log (4.9 \times 10^{−7}) = 6.31

pOH = −log [{OH}^{−}] = −log (4.9 \times 10^{−7}) = 6.31

At this temperature, then, neutral solutions exhibit pH = pOH = 6.31, acidic solutions exhibit pH less than 6.31 and pOH greater than 6.31, whereas basic solutions exhibit pH greater than 6.31 and pOH less than 6.31. This distinction can be important when studying certain processes that occur at other temperatures, such as enzyme reactions in warm-blooded organisms at a temperature around 36–40 °C. Unless otherwise noted, references to pH values are presumed to be those at 25 °C (Table 11.1).

Summary of Relations for Acidic, Basic and Neutral Solutions
Classification	Relative Ion Concentrations	pH at 25 °C
acidic	[H₃O⁺] > [OH⁻]	pH < 7
neutral	[H₃O⁺] = [OH⁻]	pH = 7
basic	[H₃O⁺] < [OH⁻]	pH > 7

Table 11.1

Figure 11.10 shows the relationships between [H₃O⁺], [OH⁻], pH, and pOH for solutions classified as acidic, basic, and neutral.

A table is provided with 5 columns. The first column is labeled “left bracket H subscript 3 O superscript plus right bracket (M).” Powers of ten are listed in the column beginning at 10 superscript 1, including 10 superscript 0 or 1, 10 superscript negative 1, decreasing by single powers of 10 to 10 superscript negative 15. The second column is labeled “left bracket O H superscript negative right bracket (M).” Powers of ten are listed in the column beginning at 10 superscript negative 15, increasing by single powers of 10 to including 10 superscript 0 or 1, and 10 superscript 1. The third column is labeled “p H.” Values listed in this column are integers beginning at negative 1, increasing by ones up to 14. The fourth column is labeled “p O H.” Values in this column are integers beginning at 15, decreasing by ones up to negative 1. The fifth column is labeled “Sample Solution.” A vertical line at the left of the column has tick marks corresponding to each p H level in the table. Substances are listed next to this line segment with line segments connecting them to the line to show approximate p H and p O H values. 1 M H C l is listed at a p H of 0. Gastric juices are listed at a p H of about 1.5. Lime juice is listed at a p H of about 2, followed by 1 M C H subscript 3 C O subscript 2 H, followed by stomach acid at a p H value of nearly 3. Wine is listed around 3.5. Coffee is listed just past 5. Pure water is listed at a p H of 7. Pure blood is just beyond 7. Milk of Magnesia is listed just past a p H of 10.5. Household ammonia is listed just before a pH of 12. 1 M N a O H is listed at a p H of 0. To the right of this labeled arrow is an arrow that points up and down through the height of the column. A beige strip passes through the table and to this double headed arrow at p H 7. To the left of the double headed arrow in this beige strip is the label “neutral.” A narrow beige strip runs through the arrow. Just above and below this region, the arrow is purple. It gradually turns to a bright red as it extends upward. At the top of the arrow, near the head of the arrow is the label “acidic.” Similarly, the lower region changes color from purple to blue moving to the bottom of the column. The head at this end of the arrow is labeled “basic.” — Figure 11.10 The pH and pOH scales represent concentrations of H₃O⁺ and OH⁻, respectively. The pH and pOH values of some common substances at 25 °C are shown in this chart.

EXAMPLE 11.14

Calculation of pH from [H₃O⁺] What is the pH of stomach acid, a solution of HCl with a hydronium ion concentration of 1.2 $\times$ 10⁻³ M?

Solution

pH = −log [H_{3} O^{+}]

= −log(1.2 \times 10^{−3})

= − (−2.92) = 2.92

(The use of logarithms is explained in Appendix B. When taking the log of a value, keep as many decimal places in the result as there are significant figures in the value.)

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

{CO}_{2} (a q) + H_{2} O(l) ⇌ H_{2} {CO}_{3} (a q)

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

Answer:

5.70

EXAMPLE 11.15

Calculation of Hydronium Ion Concentration from pH Calculate the hydronium ion concentration of blood, the pH of which is 7.3.

Solution

pH = −log [H_{3} O^{+}] = 7.3

log [H_{3} O^{+}] = −7.3

[H_{3} O^{+}] = 10^{−7.3} or [H_{3} O^{+}] = antilog of −7.3

[H_{3} O^{+}] = 5 \times 10^{−8} M

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

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

Answer:

12 M

HOW SCIENCES INTERCONNECT

Environmental Science

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

H_{2} O (l) + {CO}_{2} (g) ⟶ H_{2} {CO}_{3} (a q)

H_{2} {CO}_{3} (a q) ⇌ H^{+} (a q) + {HCO}_{3}^{−} (a q)

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

H_{2} O (l) + {SO}_{3} (g) ⟶ H_{2} {SO}_{4} (a q)

H_{2} {SO}_{4} (a q) ⟶ H^{+} (a q) + {HSO}_{4}^{−} (a q)

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

Acid rain is a particular problem in industrial areas where the products of combustion and smelting are released into the air without being stripped of sulfur and nitrogen oxides. In North America and Europe until the 1980s, it was responsible for the destruction of forests and freshwater lakes, when the acidity of the rain actually killed trees, damaged soil, and made lakes uninhabitable for all but the most acid-tolerant species. Acid rain also corrodes statuary and building facades that are made of marble and limestone (Figure 11.12). Regulations limiting the amount of sulfur and nitrogen oxides that can be released into the atmosphere by industry and automobiles have reduced the severity of acid damage to both natural and manmade environments in North America and Europe. It is now a growing problem in industrial areas of China and India.

For further information on acid rain, visit this website hosted by the US Environmental Protection Agency.

Two photos are shown. Photograph a on the left shows the upper portion of trees against a bright blue sky. The tops of several trees at the center of the photograph have bare branches and appear to be dead. Image b shows a statue of a man that appears to from the revolutionary war era in either marble or limestone. — Figure 11.12 (a) Acid rain makes trees more susceptible to drought and insect infestation, and depletes nutrients in the soil. (b) It also is corrodes statues that are carved from marble or limestone. (credit a: modification of work by Chris M Morris; credit b: modification of work by “Eden, Janine and Jim”/Flickr)

EXAMPLE 11.16

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

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

pOH = −log [{OH}^{−}] = −log 0.0125

= − (- 1.903) = 1.903

The pH can be found from the pOH:

pH + pOH = 14.00

pH = 14.00 - pOH = 14.00 - 1.903 = 12.10

Check Your Learning The hydronium ion concentration of vinegar is approximately 4 $\times$ 10⁻³ M. What are the corresponding values of pOH and pH?

Answer:

pOH = 11.6, pH = 2.4

The acidity of a solution is typically assessed experimentally by measurement of its pH. The pOH of a solution is not usually measured, as it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter (Figure 11.13).

This figure contains two images. The first, image a, is of an analytical digital p H meter on a laboratory counter. The second, image b, is of a portable hand held digital p H meter. — Figure 11.13 (a) A research-grade pH meter used in a laboratory can have a resolution of 0.001 pH units, an accuracy of ± 0.002 pH units, and may cost in excess of $1000. (b) A portable pH meter has lower resolution (0.01 pH units), lower accuracy (± 0.2 pH units), and a far lower price tag. (credit b: modification of work by Jacopo Werther)

The pH of a solution may also be visually estimated using colored indicators (Figure 11.14). The acid-base equilibria that enable use of these indicator dyes for pH measurements are described in a later section of this chapter.

This figure contains two images. The first shows a variety of colors of solutions in labeled beakers. A red solution in a beaker is labeled “0.10 M H C l.” An orange solution is labeled “0.10 M C H subscript 3 C O O H.” A yellow-orange solution is labeled “0.1 M N H subscript 4 C l.” A yellow solution is labeled “deionized water.” A second solution beaker is labeled “0.10 M K C l.” A green solution is labeled “0.10 M aniline.” A blue solution is labeled “0.10 M N H subscript 4 C l (a q).” A final beaker containing a dark blue solution is labeled “0.10 M N a O H.” Image b shows pHydrion paper that is used for measuring pH in the range of p H from 1 to 12. The color scale for identifying p H based on color is shown along with several of the test strips used to evaluate p H. — Figure 11.14 (a) A solution containing a dye mixture, called universal indicator, takes on different colors depending upon its pH. (b) Convenient test strips, called pH paper, contain embedded indicator dyes that yield pH-dependent color changes on contact with aqueous solutions.(credit: modification of work by Sahar Atwa)

Key Concepts and Summary

Concentrations of hydronium and hydroxide ions in aqueous media are often represented as logarithmic pH and pOH values, respectively. At 25 °C, the autoprotolysis equilibrium for water requires the sum of pH and pOH to equal 14 for any aqueous solution. The relative concentrations of hydronium and hydroxide ion in a solution define its status as acidic ([H₃O⁺] > [OH⁻]), basic ([H₃O⁺] < [OH⁻]), or neutral ([H₃O⁺] = [OH⁻]). At 25 °C, a pH < 7 indicates an acidic solution, a pH > 7 a basic solution, and a pH = 7 a neutral solution.

Key Equations

$pH = −log [H_{3} O^{+}]$
pOH = −log[OH⁻]
[H₃O⁺] = 10^−pH
[OH⁻] = 10^−pOH
pH + pOH = pK_w = 14.00 at 25 °C

Relative Strengths of Acids and Bases

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

Assess the relative strengths of acids and bases according to their ionization constants
Carry out equilibrium calculations for weak acid–base systems

Acid and Base Ionization Constants

The relative strength of an acid or base is the extent to which it ionizes when dissolved in water. If the ionization reaction is essentially complete, the acid or base is termed strong; if relatively little ionization occurs, the acid or base is weak. As will be evident throughout the remainder of this chapter, there are many more weak acids and bases than strong ones.

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

HA (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + A^{−} (a q),

the acid ionization constant is written

K_{a} = \frac{[H_{3} O^{+}] [A^{−}]}{[HA]}

where the concentrations are those at equilibrium. Although water is a reactant in the reaction, it is the solvent as well, so we do not include [H₂O] in the equation. The larger the K_a of an acid, the larger the concentration of $H_{3} O^{+}$ and A⁻ relative to the concentration of the nonionized acid, HA, in an equilibrium mixture, and the stronger the acid. An acid is classified as “strong” when it undergoes complete ionization, in which case the concentration of HA is zero and the acid ionization constant is immeasurably large (K_a ≈ ∞). Acids that are partially ionized are called “weak,” and their acid ionization constants may be experimentally measured. A table of ionization constants for weak acids is provided in Appendix H.

To illustrate this idea, three acid ionization equations and K_a values are shown below. The ionization constants increase from first to last of the listed equations, indicating the relative acid strength increases in the order CH₃CO₂H < HNO₂ < ${HSO}_{4}^{−} :$

{CH}_{3} {CO}_{2} H (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + {CH}_{3} {CO}_{2}^{−} (a q) K_{a} = 1.8 \times 10^{−5}

{HNO}_{2} (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + {NO}_{2}^{−} (a q) K_{a} = 4.6 \times 10^{−4}

{HSO}_{4}^{−} (a q) + H_{2} O (a q) ⇌ H_{3} O^{+} (a q) + {SO}_{4}^{2−} (a q) K_{a} = 1.2 \times 10^{−2}

Another measure of the strength of an acid is its percent ionization. The percent ionization of a weak acid is defined in terms of the composition of an equilibrium mixture:

% ionization = \frac{{[H_{3} O^{+}]}_{eq}}{{[HA]}_{0}} \times 100

where the numerator is equivalent to the concentration of the acid's conjugate base (per stoichiometry, [A⁻] = [H₃O⁺]). Unlike the K_a value, the percent ionization of a weak acid varies with the initial concentration of acid, typically decreasing as concentration increases. Equilibrium calculations of the sort described later in this chapter can be used to confirm this behavior.

EXAMPLE 11.17

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

Solution The percent ionization for an acid is:

\frac{{[H_{3} O^{+}]}_{eq}}{{[{HNO}_{2}]}_{0}} \times 100

Converting the provided pH to hydronium ion molarity yields

[H_{3} O^{+}] = 10^{- 2.09} = 0.0081 M

Substituting this value and the provided initial acid concentration into the percent ionization equation gives

\frac{8.1 \times 10^{−3}}{0.125} \times 100 = 6.5 %

(Recall the provided pH value of 2.09 is logarithmic, and so it contains just two significant digits, limiting the certainty of the computed percent ionization.)

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

Answer:

1.3% ionized

LINK TO LEARNING

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

Just as for acids, the relative strength of a base is reflected in the magnitude of its base-ionization constant (K_b) in aqueous solutions. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base has a larger ionization constant than does a weaker base. For the reaction of a base, B:

B (a q) + H_{2} O (l) ⇌ {HB}^{+} (a q) + {OH}^{−} (a q),

the ionization constant is written as

K_{b} = \frac{[{HB}^{+}] [{OH}^{−}]}{[B]}

Inspection of the data for three weak bases presented below shows the base strength increases in the order ${NO}_{2}^{-} < {CH}_{2} {CO}_{2}^{-} < {NH}_{3} .$

\begin{matrix} {NO}_{2}^{−} (a q) + H_{2} O (l) ⇌ {HNO}_{2} (a q) + {OH}^{−} (a q) & K_{b} & = & 2.17 \times 10^{−11} \\ {CH}_{3} {CO}_{2}^{−} (a q) + H_{2} O (l) ⇌ {CH}_{3} {CO}_{2} H (a q) + {OH}^{−} (a q) & K_{b} & = & 5.6 \times 10^{−10} \\ {NH}_{3} (a q) + H_{2} O (l) ⇌ {NH}_{4}^{+} (a q) + {OH}^{−} (a q) & K_{b} & = & 1.8 \times 10^{−5} \end{matrix}

A table of ionization constants for weak bases appears in Appendix I. As for acids, the relative strength of a base is also reflected in its percent ionization, computed as

% ionization = {[{OH}^{−}]}_{e q} / {[B]}_{0} \times 100 %

but will vary depending on the base ionization constant and the initial concentration of the solution.

Relative Strengths of Conjugate Acid-Base Pairs

Brønsted-Lowry acid-base chemistry is the transfer of protons; thus, logic suggests a relation between the relative strengths of conjugate acid-base pairs. The strength of an acid or base is quantified in its ionization constant, K_a or K_b, which represents the extent of the acid or base ionization reaction. For the conjugate acid-base pair HA / A⁻, ionization equilibrium equations and ionization constant expressions are

HA (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + A^{−} (a q) K_{a} = \frac{[H_{3} O^{+}] [A^{−}]}{[HA]}

A^{−} (a q) + H_{2} O (l) ⇌ {OH}^{−} (a q) + HA (a q) K_{b} = \frac{[HA] [OH]}{[A^{−}]}

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

HA(a q) + H_{2} O (l) + A^{−} (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + A^{−} (a q) + {OH}^{−} (a q) + HA(a q)

{2H}_{2} O (l) ⇌ H_{3} O^{+} (a q) + {OH}^{−} (a q)

As discussed in another chapter on equilibrium, the equilibrium constant for a summed reaction is equal to the mathematical product of the equilibrium constants for the added reactions, and so

K_{a} \times K_{b} = \frac{[H_{3} O^{+}] [A^{−}]}{[HA]} \times \frac{[HA] [{OH}^{−}]}{[A^{−}]} = [H_{3} O^{+}] [{OH}^{−}] = K_{w}

This equation states the relation between ionization constants for any conjugate acid-base pair, namely, their mathematical product is equal to the ion product of water, K_w. By rearranging this equation, a reciprocal relation between the strengths of a conjugate acid-base pair becomes evident:

K_{a} = K_{w} / K_{b} or K_{b} = K_{w} / K_{a}

This figure includes a table separated into a left half which is labeled “Acids” and a right half labeled “Bases.” A red arrow points up the left side, which is labeled “Increasing acid strength.” Similarly, a blue arrow points downward along the right side, which is labeled “Increasing base strength.” Names of acids and bases are listed next to each arrow toward the center of the table, followed by chemical formulas. Acids listed top to bottom are sulfuric acid, H subscript 2 S O subscript 4, hydrogen iodide, H I, hydrogen bromide, H B r, hydrogen chloride, H C l, nitric acid, H N O subscript 3, hydronium ion ( in pink text) H subscript 3 O superscript plus, hydrogen sulfate ion, H S O subscript 4 superscript negative, phosphoric acid, H subscript 3 P O subscript 4, hydrogen fluoride, H F, nitrous acid, H N O subscript 2, acetic acid, C H subscript 3 C O subscript 2 H, carbonic acid H subscript 2 C O subscript 3, hydrogen sulfide, H subscript 2 S, ammonium ion, N H subscript 4 superscript +, hydrogen cyanide, H C N, hydrogen carbonate ion, H C O subscript 3 superscript negative, water (shaded in beige) H subscript 2 O, hydrogen sulfide ion, H S superscript negative, ethanol, C subscript 2 H subscript 5 O H, ammonia, N H subscript 3, hydrogen, H subscript 2, methane, and C H subscript 4. The acids at the top of the listing from sulfuric acid through nitric acid are grouped with a bracket to the right labeled “Undergo complete acid ionization in water.” Similarly, the acids at the bottom from hydrogen sulfide ion through methane are grouped with a bracket and labeled, “Do not undergo acid ionization in water.” The right half of the figure lists bases and formulas. From top to bottom the bases listed are hydrogen sulfate ion, H S O subscript 4 superscript negative, iodide ion, I superscript negative, bromide ion, B r superscript negative, chloride ion, C l superscript negative, nitrate ion, N O subscript 3 superscript negative, water (shaded in beige), H subscript 2 O, sulfate ion, S O subscript 4 superscript 2 negative, dihydrogen phosphate ion, H subscript 2 P O subscript 4 superscript negative, fluoride ion, F superscript negative, nitrite ion, N O subscript 2 superscript negative, acetate ion, C H subscript 3 C O subscript 2 superscript negative, hydrogen carbonate ion, H C O subscript 3 superscript negative, hydrogen sulfide ion, H S superscript negative, ammonia, N H subscript 3, cyanide ion, C N superscript negative, carbonate ion, C O subscript 3 superscript 2 negative, hydroxide ion (in blue), O H superscript negative, sulfide ion, S superscript 2 negative, ethoxide ion, C subscript 2 H subscript 5 O superscript negative, amide ion N H subscript 2 superscript negative, hydride ion, H superscript negative, and methide ion C H subscript 3 superscript negative. The bases at the top, from perchlorate ion through nitrate ion are group with a bracket which is labeled “Do not undergo base ionization in water.” Similarly, the lower 5 in the listing, from sulfide ion through methide ion are grouped and labeled “Undergo complete base ionization in water.” — Figure 11.15 This figure shows strengths of conjugate acid-base pairs relative to the strength of water as the reference substance.

The listing of conjugate acid–base pairs shown in Figure 11.15 is arranged to show the relative strength of each species as compared with water, whose entries are highlighted in each of the table’s columns. In the acid column, those species listed below water are weaker acids than water. These species do not undergo acid ionization in water; they are not Bronsted-Lowry acids. All the species listed above water are stronger acids, transferring protons to water to some extent when dissolved in an aqueous solution to generate hydronium ions. Species above water but below hydronium ion are weak acids, undergoing partial acid ionization, wheres those above hydronium ion are strong acids that are completely ionized in aqueous solution.

If all these strong acids are completely ionized in water, why does the column indicate they vary in strength, with nitric acid being the weakest and perchloric acid the strongest? Notice that the sole acid species present in an aqueous solution of any strong acid is H₃O⁺(aq), meaning that hydronium ion is the strongest acid that may exist in water; any stronger acid will react completely with water to generate hydronium ions. This limit on the acid strength of solutes in a solution is called a leveling effect. To measure the differences in acid strength for “strong” acids, the acids must be dissolved in a solvent that is less basic than water. In such solvents, the acids will be “weak,” and so any differences in the extent of their ionization can be determined. For example, the binary hydrogen halides HCl, HBr, and HI are strong acids in water but weak acids in ethanol (strength increasing HCl < HBr < HI).

The right column of Figure 11.15 lists a number of substances in order of increasing base strength from top to bottom. Following the same logic as for the left column, species listed above water are weaker bases and so they don’t undergo base ionization when dissolved in water. Species listed between water and its conjugate base, hydroxide ion, are weak bases that partially ionize. Species listed below hydroxide ion are strong bases that completely ionize in water to yield hydroxide ions (i.e., they are leveled to hydroxide). A comparison of the acid and base columns in this table supports the reciprocal relation between the strengths of conjugate acid-base pairs. For example, the conjugate bases of the strong acids (top of table) are all of negligible strength. A strong acid exhibits an immeasurably large K_a, and so its conjugate base will exhibit a K_b that is essentially zero:

$\begin{array}{l} strong acid : K_{a} \approx \infty \\ conjugate base : K_{b} = K_{w} / K_{a} = K_{w} / \infty \approx 0 \end{array}$

A similar approach can be used to support the observation that conjugate acids of strong bases (K_b ≈ ∞) are of negligible strength (K_a ≈ 0).

EXAMPLE 11.18

Calculating Ionization Constants for Conjugate Acid-Base Pairs Use the K_b for the nitrite ion, ${NO}_{2}^{−},$ to calculate the K_a for its conjugate acid.

SolutionK_b for ${NO}_{2}^{−}$ is given in this section as 2.17 $\times$ 10⁻¹¹. The conjugate acid of ${NO}_{2}^{−}$ is HNO₂; K_a for HNO₂ can be calculated using the relationship:

K_{a} \times K_{b} = 1.0 \times 10^{−14} = K_{w}

Solving for K_a yields

K_{a} = \frac{K_{w}}{K_{b}} = \frac{1.0 \times 10^{−14}}{2.17 \times 10^{−11}} = 4.6 \times 10^{−4}

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

Check Your LearningDetermine the relative acid strengths of ${NH}_{4}^{+}$ and HCN by comparing their ionization constants. The ionization constant of HCN is given in Appendix H as 4.9 $\times$ 10⁻¹⁰. The ionization constant of ${NH}_{4}^{+}$ is not listed, but the ionization constant of its conjugate base, NH₃, is listed as 1.8 $\times$ 10⁻⁵.

Answer:

${NH}_{4}^{+}$ is the slightly stronger acid (K_a for ${NH}_{4}^{+}$ = 5.6 $\times$ 10⁻¹⁰).

Acid-Base Equilibrium Calculations

The chapter on chemical equilibria introduced several types of equilibrium calculations and the various mathematical strategies that are helpful in performing them. These strategies are generally useful for equilibrium systems regardless of chemical reaction class, and so they may be effectively applied to acid-base equilibrium problems. This section presents several example exercises involving equilibrium calculations for acid-base systems.

EXAMPLE 11.19

Determination of K_a from Equilibrium Concentrations Acetic acid is the principal ingredient in vinegar (Figure 11.16) that provides its sour taste. At equilibrium, a solution contains [CH₃CO₂H] = 0.0787 M and $[H_{3} O^{+}] = [{CH}_{3} {CO}_{2}^{−}] = 0.00118 M .$ What is the value of K_a for acetic acid?

An image shows the label of a bottle of distilled white vinegar. The label states that the contents have been reduced with water to 5 percent acidity. — Figure 11.16 Vinegar contains acetic acid, a weak acid. (credit: modification of work by “HomeSpot HQ”/Flickr)

Solution The relevant equilibrium equation and its equilibrium constant expression are shown below. Substitution of the provided equilibrium concentrations permits a straightforward calculation of the K_a for acetic acid.

{CH}_{3} {CO}_{2} H (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + {CH}_{3} {CO}_{2}^{−} (a q)

K_{a} = \frac{[H_{3} O^{+}] [{CH}_{3} {CO}_{2}^{−}]}{[{CH}_{3} {CO}_{2} H]} = \frac{(0.00118) (0.00118)}{0.0787} = 1.77 \times 10^{−5}

Check Your Learning The ${HSO}_{4}^{−}$ ion, weak acid used in some household cleansers:

{HSO}_{4}^{−} (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + {SO}_{4}^{2−} (a q)

What is the acid ionization constant for this weak acid if an equilibrium mixture has the following composition: $[H_{3} O^{+}]$ = 0.027 M; $[{HSO}_{4}^{−}] = 0.29 M;$ and $[{SO}_{4}^{2−}] = 0.13 M ?$

Answer:

K_a for ${HSO}_{4}^{−}$ = 1.2 $\times$ 10⁻²

EXAMPLE 11.20

Determination of K_b from Equilibrium Concentrations Caffeine, C₈H₁₀N₄O₂ is a weak base. What is the value of K_b for caffeine if a solution at equilibrium has [C₈H₁₀N₄O₂] = 0.050 M, $[C_{8} H_{10} N_{4} O_{2} H +]$ = 5.0 $\times$ 10⁻³ M, and [OH⁻] = 2.5 $\times$ 10⁻³ M?

C_{8} H_{10} N_{4} O_{2} (a q) + H_{2} O (l) ⇌ C_{8} H_{10} N_{4} O_{2} H^{+} (a q) + {OH}^{−} (a q)

K_{b} = \frac{[C_{8} H_{10} N_{4} O_{2} H +] [{OH}^{−}]}{[C_{8} H_{10} N_{4} O_{2}]} = \frac{(5.0 \times 10^{−3}) (2.5 \times 10^{−3})}{0.050} = 2.5 \times 10^{−4}

Check Your Learning What is the equilibrium constant for the ionization of the ${HPO}_{4}^{2−}$ ion, a weak base

{HPO}_{4}^{2−} (a q) + H_{2} O (l) ⇌ H_{2} {PO}_{4}^{−} (a q) + {OH}^{−} (a q)

if the composition of an equilibrium mixture is as follows: [OH⁻] = 1.3 $\times$ 10⁻⁶ M; $[H_{2} {PO}_{4}^{−}] = 0.042 M;$ and $[{HPO}_{4}^{2−}] = 0.341 M ?$

Answer:

K_b for ${HPO}_{4}^{2−} = 1.6 \times 10^{−7}$

EXAMPLE 11.21

Determination of K_a or K_b from pH The pH of a 0.0516-M solution of nitrous acid, HNO₂, is 2.34. What is its K_a?

{HNO}_{2} (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + {NO}_{2}^{−} (a q)

Solution The nitrous acid concentration provided is a formal concentration, one that does not account for any chemical equilibria that may be established in solution. Such concentrations are treated as “initial” values for equilibrium calculations using the ICE table approach. Notice the initial value of hydronium ion is listed as approximately zero because a small concentration of H₃O⁺ is present (1 × 10⁻⁷ M) due to the autoprotolysis of water. In many cases, such as all the ones presented in this chapter, this concentration is much less than that generated by ionization of the acid (or base) in question and may be neglected.

The pH provided is a logarithmic measure of the hydronium ion concentration resulting from the acid ionization of the nitrous acid, and so it represents an “equilibrium” value for the ICE table:

[H_{3} O^{+}] = 10^{- 2.34} = 0.0046 M

The ICE table for this system is then

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of “H N O subscript 2 plus sign H subscript 2 O equilibrium sign H subscript 3 O superscript positive sign plus sign N O subscript 2 superscript negative sign.” Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.0516, negative 0.0046, 0.0470. The second column is blank in all three rows. The third column has the following: approximately 0, positive 0.0046, 0.0046. The fourth column has the following: 0, positive 0.0046, 0.0046.

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

K_{a} = \frac{[H_{3} O^{+}] [{NO}_{2}^{−}]}{[{HNO}_{2}]} = \frac{(0.0046) (0.0046)}{(0.0470)} = 4.6 \times 10^{−4}

Check Your Learning. The pH of a solution of household ammonia, a 0.950-M solution of NH_3, is 11.612. What is K_b for NH₃.

Answer:

K_b = 1.8 $\times$ 10⁻⁵

EXAMPLE 11.22

Calculating Equilibrium Concentrations in a Weak Acid Solution Formic acid, HCO₂H, is one irritant that causes the body’s reaction to some ant bites and stings (Figure 11.17). The leather and textile processing industries use formic acid for tanning and dye fixing as well as a neutralizing agent and pH adjuster.

A photograph is shown of a large black ant on the end of a human finger. — Figure 11.17 The pain of some ant bites and stings is caused by formic acid. (credit: John Tann)

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

{HCO}_{2} H (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + {HCO}_{2}^{−} (a q) K_{a} = 1.8 \times 10^{−4}

Solution

The ICE table for this system is

Substituting the equilibrium concentration terms into the K_a expression gives

K_{a} = 1.8 \times 10^{−4} = \frac{[H_{3} O^{+}] [{HCO}_{2}^{−}]}{[{HCO}_{2} H]}

= \frac{(x) (x)}{0.534 - x} = 1.8 \times 10^{−4}

The relatively large initial concentration and small equilibrium constant permits the simplifying assumption that x will be much lesser than 0.534, and so the equation becomes

K_{a} = 1.8 \times 10^{−4} = \frac{x^{2}}{0.534}

Solving the equation for x yields

x^{2} = 0.534 \times (1.8 \times 10^{−4}) = 9.6 \times 10^{−5}

x = \sqrt{9.6 \times 10^{−5}}

= 9.8 \times 10^{−3} M

To check the assumption that x is small compared to 0.534, its relative magnitude can be estimated:

\frac{x}{0.534} = \frac{9.8 \times 10^{−3}}{0.534} = 1.8 \times 10^{−2} (1.8 % of 0.534)

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

As defined in the ICE table, x is equal to the equilibrium concentration of hydronium ion:

x = [H_{3} O^{+}] = 0.0098 M

Finally, the pH is calculated to be

pH = − log [H_{3} O^{+}] = − log (0.0098) = 2.01

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

{CH}_{3} {CO}_{2} H (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + {CH}_{3} {CO}_{2}^{−} (a q) K_{a} = 1.8 \times 10^{−5}

Answer:

percent ionization = 1.3%

EXAMPLE 11.23

Calculating Equilibrium Concentrations in a Weak Base Solution Find the concentration of hydroxide ion, the pOH, and the pH of a 0.25-M solution of trimethylamine, a weak base:

{({CH}_{3})}_{3} N (a q) + H_{2} O (l) ⇌ {({CH}_{3})}_{3} {NH}^{+} (a q) + {OH}^{−} (a q) K_{b} = 6.3 \times 10^{−5}

Solution The ICE table for this system is

Substituting the equilibrium concentration terms into the K_b expression gives

K_{b} = \frac{[{({CH}_{3})}_{3} {NH}^{+}] [{OH}^{−}]}{[{({CH}_{3})}_{3} N]} = \frac{(x) (x)}{0.25 - x} = 6.3 \times 10^{−5}

Assuming x << 0.25 and solving for x yields

x = 4.0 \times 10^{−3} M

This value is less than 5% of the initial concentration (0.25), so the assumption is justified.
As defined in the ICE table, x is equal to the equilibrium concentration of hydroxide ion:

[{OH}^{−}] = ~ 0 + x = x = 4.0 \times 10^{−3} M

= 4.0 \times 10^{−3} M

The pOH is calculated to be

pOH = −log (4.0 \times 10^{−3}) = 2.40

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

pH + pOH = p K_{w} = 14.00

permits the computation of pH:

pH = 14.00 - pOH = 14.00 - 2.40 = 11.60

Check Your LearningCalculate the hydroxide ion concentration and the percent ionization of a 0.0325-M solution of ammonia, a weak base with a K_b of 1.76 $\times$ 10⁻⁵.

Answer:

7.56 $\times$ 10⁻⁴ M, 2.33%

In some cases, the strength of the weak acid or base and its formal (initial) concentration result in an appreciable ionization. Though the ICE strategy remains effective for these systems, the algebra is a bit more involved because the simplifying assumption that x is negligible can not be made. Calculations of this sort are demonstrated in Example 11.24 below.

EXAMPLE 11.24

Calculating Equilibrium Concentrations without Simplifying Assumptions Sodium bisulfate, NaHSO₄, is used in some household cleansers as a source of the ${HSO}_{4}^{−}$ ion, a weak acid. What is the pH of a 0.50-M solution of ${HSO}_{4}^{−} ?$

{HSO}_{4}^{−} (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + {SO}_{4}^{2−} (a q) K_{a} = 1.2 \times 10^{−2}

Solution The ICE table for this system is

Substituting the equilibrium concentration terms into the K_a expression gives

K_{a} = 1.2 \times 10^{−2} = \frac{[H_{3} O^{+}] [{SO}_{4}^{2−}]}{[{HSO}_{4}^{−}]} = \frac{(x) (x)}{0.50 - x}

If the assumption that x << 0.5 is made, simplifying and solving the above equation yields

x = 0.077 M

This value of x is clearly not significantly less than 0.50 M; rather, it is approximately 15% of the initial concentration:
When we check the assumption, we calculate:

\frac{x}{{[{HSO}_{4}^{−}]}_{i}}

\frac{x}{0.50} = \frac{7.7 \times 10^{−2}}{0.50} = 0.15 (15 %)

Because the simplifying assumption is not valid for this system, the equilibrium constant expression is solved as follows:

K_{a} = 1.2 \times 10^{−2} = \frac{(x) (x)}{0.50 - x}

Rearranging this equation yields

6.0 \times 10^{−3} - 1.2 \times 10^{−2} x = x^{2}

Writing the equation in quadratic form gives

x^{2} + 1.2 \times 10^{−2} x - 6.0 \times 10^{−3} = 0

Solving for the two roots of this quadratic equation results in a negative value that may be discarded as physically irrelevant and a positive value equal to x. As defined in the ICE table, x is equal to the hydronium concentration.

\begin{array}{l} x = [H_{3} O^{+}] = 0.072 M \\ pH = - log [H_{3} O^{+}] = - log (0.072) = 1.14 \end{array}

Check Your Learning Calculate the pH in a 0.010-M solution of caffeine, a weak base:

C_{8} H_{10} N_{4} O_{2} (a q) + H_{2} O (l) ⇌ C_{8} H_{10} N_{4} O_{2} H^{+} (a q) + {OH}^{−} (a q) K_{b} = 2.5 \times 10^{−4}

Answer:

pH 11.16

Buffers

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

Describe the composition and function of acid–base buffers
Calculate the pH of a buffer before and after the addition of added acid or base

A solution containing appreciable amounts of a weak conjugate acid-base pair is called a buffer solution, or a buffer. Buffer solutions resist a change in pH when small amounts of a strong acid or a strong base are added (Figure 11.18). A solution of acetic acid and sodium acetate (CH₃COOH + CH₃COONa) is an example of a buffer that consists of a weak acid and its salt. An example of a buffer that consists of a weak base and its salt is a solution of ammonia and ammonium chloride (NH₃(aq) + NH₄Cl(aq)).

Two images are shown. Image a on the left shows two beakers that each contain yellow solutions. The beaker on the left is labeled “Unbuffered” and the beaker on the right is labeled “p H equals 8.0 buffer.” Image b similarly shows 2 beakers. The beaker on the left contains a bright orange solution and is labeled “Unbuffered.” The beaker on the right is labeled “p H equals 8.0 buffer.” — Figure 11.18 (a) The buffered solution on the left and the unbuffered solution on the right have the same pH (pH 8); they are basic, showing the yellow color of the indicator methyl orange at this pH. (b) After the addition of 1 mL of a 0.01-M HCl solution, the buffered solution has not detectably changed its pH but the unbuffered solution has become acidic, as indicated by the change in color of the methyl orange, which turns red at a pH of about 4. (credit: modification of work by Mark Ott)

How Buffers Work

To illustrate the function of a buffer solution, consider a mixture of roughly equal amounts of acetic acid and sodium acetate. The presence of a weak conjugate acid-base pair in the solution imparts the ability to neutralize modest amounts of added strong acid or base. For example, adding strong base to this solution will neutralize hydronium ion and shift the acetic acid ionization equilibrium to the right, partially restoring the decreased H₃O⁺ concentration:

{CH}_{3} {CO}_{2} H (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + {CH}_{3} {CO}_{2}^{−} (a q)

Likewise, adding strong acid to this buffer solution will neutralize acetate ion, shifting the above ionization equilibrium right and returning [H3O+] to near its original value. Figure 11.19 provides a graphical illustration of the changes in conjugate-partner concentration that occur in this buffer solution when strong acid and base are added. The buffering action of the solution is essentially a result of the added strong acid and base being converted to the weak acid and base that make up the buffer's conjugate pair. The weaker acid and base undergo only slight ionization, as compared with the complete ionization of the strong acid and base, and the solution pH, therefore, changes much less drastically than it would in an unbuffered solution.

This figure begins with a chemical reaction at the top: C H subscript 3 C O O H ( a q ) plus H subscript 2 O ( l ) equilibrium arrow H subscript 3 O superscript positive sign ( a q ) plus C H subscript 3 C O O superscript negative sign ( a q ). Below this equation are two arrows: one pointing left and other pointing right. The arrow pointing left has this phrase written above it, “H subscript 3 O superscript positive sign added, equilibrium position shifts to the left.” Below the arrow is the reaction: C H subscript 3 C O O H ( a q ) left-facing arrow C H subscript 3 C O O superscript negative sign ( a q ) plus H subscript 3 O superscript positive sign. The arrow pointing right has this phrase written above it, “O H subscript negative sign added, equilibrium position shifts to the right.” Below the arrow is the reaction: O H superscript negative sign plus C H subscript 3 C O O H ( a q ) right-facing arrow H subscript 2 O ( l ) plus C H subscript 3 C O O superscript negative sign ( a q ). Below all the text is a figure that resembles a bar graph. In the middle are two bars of equal height. One is labeled, “C H subscript 3 C O O H,” and the other is labeled, “C H subscript 3 C O O superscript negative sign.” There is a dotted line at the same height of the bars which extends to the left and right. Above these two bars is the phrase, “Buffer solution equimolar in acid and base.” There is an arrow pointing to the right which is labeled, “Add O H superscript negative sign.” The arrow points to two bars again, but this time the C H subscript 3 C O O H bar is shorter than that C H subscript 3 C O O superscript negative sign bar. Above these two bars is the phrase, “Buffer solution after addition of strong base.” From the middle bars again, there is an arrow that points left. The arrow is labeled, “Add H subscript 3 O superscript positive sign.” This arrow points to two bars again, but this time the C H subscript 3 C O O H bar is taller than the C H subscript 3 C O O superscript negative sign bar. These two bars are labeled, “Buffer solution after addition of strong acid.”

Figure 11.19 Buffering action in a mixture of acetic acid and acetate salt.

EXAMPLE 11.25

pH Changes in Buffered and Unbuffered Solutions Acetate buffers are used in biochemical studies of enzymes and other chemical components of cells to prevent pH changes that might affect the biochemical activity of these compounds.

(a) Calculate the pH of an acetate buffer that is a mixture with 0.10 M acetic acid and 0.10 M sodium acetate.

(b) Calculate the pH after 1.0 mL of 0.10 NaOH is added to 100 mL of this buffer.

(c) For comparison, calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74.

Solution

(a) Following the ICE approach to this equilibrium calculation yields the following:

Substituting the equilibrium concentration terms into the K_a expression, assuming x << 0.10, and solving the simplified equation for x yields

x = 1.8 \times 10^{−5} M

[H_{3} O^{+}] = 0 + x = 1.8 \times 10^{−5} M

pH = −log [H_{3} O^{+}] = −log (1.8 \times 10^{−5})

= 4.74

(b) Calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of this buffer.

Adding strong acid will neutralize some of the acetic acid, yielding the conjugate base acetate ion. Compute the new concentrations of these two buffer components, then repeat the equilibrium calculation of part (a) using these new concentrations.

0.0010 L \times (\frac{0.10 mol NaOH}{1 L}) = 1.0 \times 10^{−4} mol NaOH

The initial molar amount of acetic acid is

0.100 L \times (\frac{0.100 mol {CH}_{3} {CO}_{2} H}{1 L}) = 1.00 \times 10^{−2} mol {CH}_{3} {CO}_{2} H

The amount of acetic acid remaining after some is neutralized by the added base is

(1.0 \times 10^{−2}) - (0.01 \times 10^{−2}) = 0.99 \times 10^{−2} mol {CH}_{3} {CO}_{2} H

The newly formed acetate ion, along with the initially present acetate, gives a final acetate concentration of

(1.0 \times 10^{−2}) + (0.01 \times 10^{−2}) = 1.01 \times 10^{−2} mol {NaCH}_{3} {CO}_{2}

Compute molar concentrations for the two buffer components:

[{CH}_{3} {CO}_{2} H] = \frac{9.9 \times 10^{−3} mol}{0.101 L} = 0.098 M

[{NaCH}_{3} {CO}_{2}] = \frac{1.01 \times 10^{−2} mol}{0.101 L} = 0.100 M

Using these concentrations, the pH of the solution may be computed as in part (a) above, yielding pH = 4.75 (only slightly different from that prior to adding the strong base).

(c) For comparison, calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74.

The amount of hydrogen ion initially present in the solution is

[H_{3} O^{+}] = 10^{- 4.74} = 1.8 \times 10^{- 5} M

mol H_{3} O^{+} = (0.100 L) (1.8 \times 10^{−5} M) = 1.8 \times 10^{−6} mol H_{3} O^{+}

The amount of hydroxide ion added to the solution is

mol {OH}^{-} = (0.0010 L) (0.10 M) = 1.0 \times 10^{−4} mol {OH}^{-}

The added hydroxide will neutralize hydronium ion via the reaction

H_{3} O^{+} (a q) + {OH}^{-} (a q) 2 H_{2} O (l)

The 1:1 stoichiometry of this reaction shows that an excess of hydroxide has been added (greater molar amount than the initially present hydronium ion).

The amount of hydroxide ion remaining is

1.0 \times 10^{−4} mol - 1.8 \times 10^{−6} mol = 9.8 \times 10^{−5} mol {OH}^{-}

corresponding to a hydroxide molarity of

9.8 \times 10^{−5} mol {OH}^{-} / 0.101 L = 9.7 \times 10^{−4} M

The pH of the solution is then calculated to be

pH = 14.00 - pOH = 14.00 - - log (9.7 \times 10^{−4}) = 10.99

In this unbuffered solution, addition of the base results in a significant rise in pH (from 4.74 to 10.99) compared with the very slight increase observed for the buffer solution in part (b) (from 4.74 to 4.75).

Check Your Learning Show that adding 1.0 mL of 0.10 M HCl changes the pH of 100 mL of a 1.8 $\times$ 10⁻⁵ M HCl solution from 4.74 to 3.00.

Answer:

Initial pH of 1.8 $\times$ 10⁻⁵ M HCl; pH = −log[H₃O⁺] = −log[1.8 $\times$ 10⁻⁵] = 4.74
Moles of H₃O⁺ in 100 mL 1.8 $\times$ 10⁻⁵ M HCl; 1.8 $\times$ 10⁻⁵ moles/L $\times$ 0.100 L = 1.8 $\times$ 10⁻⁶
Moles of H₃O⁺ added by addition of 1.0 mL of 0.10 M HCl: 0.10 moles/L $\times$ 0.0010 L = 1.0 $\times$ 10⁻⁴ moles; final pH after addition of 1.0 mL of 0.10 M HCl:

pH = −log [H_{3} O^{+}] = −log (\frac{total moles H_{3} O^{+}}{total volume}) = −log (\frac{1.0 \times 10^{−4} mol + 1.8 \times 10^{−6} mol}{101 mL (\frac{1 L}{1000 mL})}) = 3.00

Buffer Capacity

Buffer solutions do not have an unlimited capacity to keep the pH relatively constant (Figure 11.20). Instead, the ability of a buffer solution to resist changes in pH relies on the presence of appreciable amounts of its conjugate weak acid-base pair. When enough strong acid or base is added to substantially lower the concentration of either member of the buffer pair, the buffering action within the solution is compromised.

No Alt Text — Figure 11.20 The indicator color (methyl orange) shows that a small amount of acid added to a buffered solution of pH 8 (beaker on the left) has little affect on the buffered system (middle beaker). However, a large amount of acid exhausts the buffering capacity of the solution and the pH changes dramatically (beaker on the right). (credit: modification of work by Mark Ott)

The buffer capacity is the amount of acid or base that can be added to a given volume of a buffer solution before the pH changes significantly, usually by one unit. Buffer capacity depends on the amounts of the weak acid and its conjugate base that are in a buffer mixture. For example, 1 L of a solution that is 1.0 M in acetic acid and 1.0 M in sodium acetate has a greater buffer capacity than 1 L of a solution that is 0.10 M in acetic acid and 0.10 M in sodium acetate even though both solutions have the same pH. The first solution has more buffer capacity because it contains more acetic acid and acetate ion.

Selection of Suitable Buffer Mixtures

There are two useful rules of thumb for selecting buffer mixtures:

A good buffer mixture should have about equal concentrations of both of its components. A buffer solution has generally lost its usefulness when one component of the buffer pair is less than about 10% of the other. Figure 11.21 shows how pH changes for an acetic acid-acetate ion buffer as base is added. The initial pH is 4.74. A change of 1 pH unit occurs when the acetic acid concentration is reduced to 11% of the acetate ion concentration.

Figure 11.21 Change in pH as an increasing amount of a 0.10-M NaOH solution is added to 100 mL of a buffer solution in which, initially, [CH₃CO₂H] = 0.10 M and $[{CH}_{3} {CO}_{2}^{−}] = 0.10 M .$ Note the greatly diminished buffering action occurring after the buffer capacity has been reached, resulting in drastic rises in pH on adding more strong base.
Weak acids and their salts are better as buffers for pHs less than 7; weak bases and their salts are better as buffers for pHs greater than 7.

Blood is an important example of a buffered solution, with the principal acid and ion responsible for the buffering action being carbonic acid, H₂CO₃, and the bicarbonate ion, ${HCO}_{3}^{−} .$ When a hydronium ion is introduced to the blood stream, it is removed primarily by the reaction:

H_{3} O^{+} (a q) + {HCO}_{3}^{−} (a q) ⟶ H_{2} {CO}_{3} (a q) + H_{2} O (l)

An added hydroxide ion is removed by the reaction:

{OH}^{−} (a q) + H_{2} {CO}_{3} (a q) ⟶ {HCO}_{3}^{−} (a q) + H_{2} O (l)

The added strong acid or base is thus effectively converted to the much weaker acid or base of the buffer pair (H₃O⁺ is converted to H₂CO₃ and OH^- is converted to HCO₃^-). The pH of human blood thus remains very near the value determined by the buffer pairs pKa, in this case, 7.35. Normal variations in blood pH are usually less than 0.1, and pH changes of 0.4 or greater are likely to be fatal.

The Henderson-Hasselbalch Equation

The ionization-constant expression for a solution of a weak acid can be written as:

K_{a} = \frac{[H_{3} O^{+}] [A^{−}]}{[HA]}

Rearranging to solve for [H₃O⁺] yields:

[H_{3} O^{+}] = K_{a} \times \frac{[HA]}{[A^{−}]}

Taking the negative logarithm of both sides of this equation gives

−log [H_{3} O^{+}] = −log K_{a} − log \frac{[HA]}{[A^{−}]},

which can be written as

pH = p K_{a} + log \frac{[A^{−}]}{[HA]}

where pK_a is the negative of the logarithm of the ionization constant of the weak acid (pK_a = −log K_a). This equation relates the pH, the ionization constant of a weak acid, and the concentrations of the weak conjugate acid-base pair in a buffered solution. Scientists often use this expression, called the Henderson-Hasselbalch equation, to calculate the pH of buffer solutions. It is important to note that the “x is small” assumption must be valid to use this equation.

HOW SCIENCES INTERCONNECT

Medicine: The Buffer System in Blood

The normal pH of human blood is about 7.4. The carbonate buffer system in the blood uses the following equilibrium reaction:

{CO}_{2} (g) + 2 H_{2} O (l) ⇌ H_{2} {CO}_{3} (a q) ⇌ {HCO}_{3}^{−} (a q) + H_{3} O^{+} (a q)

The concentration of carbonic acid, H₂CO₃ is approximately 0.0012 M, and the concentration of the hydrogen carbonate ion, ${HCO}_{3}^{−},$ is around 0.024 M. Using the Henderson-Hasselbalch equation and the pK_a of carbonic acid at body temperature, we can calculate the pH of blood:

pH = p K_{a} + log \frac{[base]}{[acid]} = 6.4 + log \frac{0.024}{0.0012} = 7.7

The fact that the H₂CO₃ concentration is significantly lower than that of the ${HCO}_{3}^{−}$ ion may seem unusual, but this imbalance is due to the fact that most of the by-products of our metabolism that enter our bloodstream are acidic. Therefore, there must be a larger proportion of base than acid, so that the capacity of the buffer will not be exceeded.

Lactic acid is produced in our muscles when we exercise. As the lactic acid enters the bloodstream, it is neutralized by the ${HCO}_{3}^{−}$ ion, producing H₂CO₃. An enzyme then accelerates the breakdown of the excess carbonic acid to carbon dioxide and water, which can be eliminated by breathing. In fact, in addition to the regulating effects of the carbonate buffering system on the pH of blood, the body uses breathing to regulate blood pH. If the pH of the blood decreases too far, an increase in breathing removes CO₂ from the blood through the lungs driving the equilibrium reaction such that [H₃O⁺] is lowered. If the blood is too alkaline, a lower breath rate increases CO₂ concentration in the blood, driving the equilibrium reaction the other way, increasing [H⁺] and restoring an appropriate pH.

LINK TO LEARNING

View information on the buffer system encountered in natural waters.

Key Concepts and Summary

Solutions that contain appreciable amounts of a weak conjugate acid-base pair are called buffers. A buffered solution will experience only slight changes in pH when small amounts of acid or base are added. Addition of large amounts of acid or base can exceed the buffer capacity, consuming most of one conjugate partner and preventing further buffering action.

Key Equations

pK_a = −log K_a
pK_b = −log K_b
$pH = p K_{a} + log \frac{[A^{−}]}{[HA]}$