14.3 Relative Strengths of Acids and Bases

 

Learning Objectives

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

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

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 most common strong acids and bases are listed in (Figure).

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

This table has seven rows and two columns. The first row is a header row, and it labels each column, “6 Strong Acids,” and, “6 Strong Bases.” Under the “6 Strong Acids” column are the following: H C l O subscript 4 perchloric acid; H C l hydrochloric acid; H B r hydrobromic acid; H I hydroiodic acid; H N O subscript 3 nitric acid; H subscript 2 S O subscript 4 sulfuric acid. Under the “6 Strong Bases” column are the following: L i O H lithium hydroxide; N a O H sodium hydroxide; K O H potassium hydroxide; C a ( O H ) subscript 2 calcium hydroxide; S r ( O H ) subscript 2 strontium hydroxide; B a ( O H ) subscript 2 barium hydroxide.

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, Ka. For the reaction of an acid HA:

HA(aq) + H2O(l) ⇌ H3O+(aq) + A−(aq),

the acid ionization constant is written

 

where the concentrations are those at equilibrium. Although water is a reactant in the reaction, it is the solvent as well, so we do not include [H2O] in the equation. The larger the Ka of an acid, the larger the concentration of H3O+ and A− relative to the concentration of the nonionized acid, HA, 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 (Ka ≈ ∞). 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 Ka 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 < HSO4−:

CH3CO2H(aq) + H2O(l) ⇌ H3O+(aq) + CH3CO2−(aq)          Ka = 1.8 × 10-5
HNO2(aq) + H2O(l) ⇌ H3O+(aq) + NO2−(aq)                        Ka = 4.6 × 10-4
HSO4−(aq) + H2O(aq) ⇌ H3O+(aq) + SO42-(aq)                   Ka = 1.2 × 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:

where the numerator is equivalent to the concentration of the acid’s conjugate base (per stoichiometry, [A−] = [H3O+]). Unlike the Ka 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.

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:

Converting the provided pH to hydronium ion molarity yields

[H3O+] = 10-2.09 = 0.0081 M

The percent ionization for an acid is:

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

(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

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

B(aq) + H2O(l) ⇌ HB+(aq)+ OH−(aq),

the ionization constant is written as

Inspection of the data for three weak bases presented below shows the base strength increases in the order NO2– < CH3CO2– <NH3.

NO2−(aq) + H2O(l) ⇌ HNO2(aq) + OH−(aq)                        Kb = 2.17 × 10-11
CH3CO2–(aq) + H2O(l) ⇌ CH3CO2H(aq) + OH−(aq)          Kb = 5.6 × 10-10
NH3(aq) + H2O(l) ⇌ NH4+(aq)+ OH−(aq)                            Kb=  1.8 × 10-5

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−]eq/[B]0 ×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, Ka or Kb, 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

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

HA(aq) + H2O(l) + A−(aq) + H2O(l) ⇌ H3O+(aq) + A−(aq) + OH−(aq) + HA(aq)
 H2O(l)  + H2O(l) ⇌ H3O+(aq) + OH−(aq)

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

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, Kw. By rearranging this equation, a reciprocal relation between the strengths of a conjugate acid-base pair becomes evident:

Ka = Kw/Kb  or  Kb = Kw/Ka

The inverse proportional relation between Ka and Kb means the stronger the acid or base, the weaker its conjugate partner. (Figure) illustrates this relation for several conjugate acid-base pairs.

 

Relative strengths of several conjugate acid-base pairs are shown.

The diagram shows two horizontal bars. The first, labeled, “Relative acid strength,” at the top is red on the left and gradually changes to purple on the right. The red end at the left is labeled, “Stronger acids.” The purple end at the right is labeled, “Weaker acids.” Just outside the bar to the lower left is the label, “K subscript a.” The bar is marked off in increments with a specific acid listed above each increment. The first mark is at 1.0 with H subscript 3 O superscript positive sign. The second is ten raised to the negative two with H C l O subscript 2. The third is ten raised to the negative 4 with H F. The fourth is ten raised to the negative 6 with H subscript 2 C O subscript 3. The fifth is ten raised to a negative 8 with C H subscript 3 C O O H. The sixth is ten raised to the negative ten with N H subscript 4 superscript positive sign. The seventh is ten raised to a negative 12 with H P O subscript 4 superscript 2 negative sign. The eighth is ten raised to the negative 14 with H subscript 2 O. Similarly the second bar, which is labeled “Relative conjugate base strength,” is purple at the left end and gradually becomes blue at the right end. Outside the bar to the left is the label, “Weaker bases.” Outside the bar to the right is the label, “Stronger bases.” Below and to the left of the bar is the label, “K subscript b.” The bar is similarly marked at increments with bases listed above each increment. The first is at ten raised to the negative 14 with H subscript 2 O above it. The second is ten raised to the negative 12 C l O subscript 2 superscript negative sign. The third is ten raised to the negative ten with F superscript negative sign. The fourth is ten raised to a negative eight with H C O subscript 3 superscript negative sign. The fifth is ten raised to the negative 6 with C H subscript 3 C O O superscript negative sign. The sixth is ten raised to the negative 4 with N H subscript 3. The seventh is ten raised to the negative 2 with P O subscript 4 superscript three negative sign. The eighth is 1.0 with O H superscript negative sign.

 

This figure shows strengths of conjugate acid-base pairs relative to the strength of water as the reference substance.

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.”

The listing of conjugate acid–base pairs shown in (Figure) 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, whereas 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) 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 Ka, and so its conjugate base will exhibit a Kb that is essentially zero:

strong acid:     Ka ≈ ∞

conjugate base:     Kb = Kw/Ka = Kw/∞ ≈ 0

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

Calculating Ionization Constants for Conjugate Acid-Base Pairs

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

Solution:

Kb for NO2− is given in this section as 2.17 × 10−11. The conjugate acid of NO2− is HNO2; Ka for HNO2 can be calculated using the relationship:

Ka × Kb = 1.0 × 10-14 = Kw

Solving for Ka yields

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

Check Your Learning:

Determine the relative acid strengths of NH4+ and HCN by comparing their ionization constants. The ionization constant of HCN is given in Appendix H as 4.9 × 10−10. The ionization constant of NH4+ is not listed, but the ionization constant of its conjugate base, NH3, is listed as 1.8 × 10−5.

Answer:

NH4+ is the slightly stronger acid (Ka for NH4+ = 5.6 × 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.

This image shows two bottles containing clear colorless solutions. Each bottle contains a single p H indicator strip. The strip in the bottle on the left is red, and a similar red strip is placed on a filter paper circle in front of the bottle on surface on which the bottles are resting. Similarly, the second bottle on the right contains and orange strip and an orange strip is placed in front of it on a filter paper circle. Between the two bottles is a pack of p Hydrion papers with a p H color scale on its cover.This photo shows two glass containers filled with a transparent liquid. In between the containers is a p H strip indicator guide. There are p H strips placed in front of each glass container. The liquid in the container on the left appears to have a p H of 10 or 11. The liquid in the container on the right appears to have a p H of about 13 or 14.

Determination of Ka from Equilibrium Concentrations

Acetic acid is the principal ingredient in vinegar ((Figure)) that provides its sour taste. At equilibrium, a solution contains [CH3CO2H] = 0.0787 M and [H3O+] = [CH3CO2−] = 0.00118 M. What is the value of Ka for acetic acid?

 

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

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.

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 Ka for acetic acid.

Check Your Learning:

The HSO4− ion is a weak acid used in some household cleansers:

HSO4−(aq) + H2O(l) ⇌ H3O+(aq) + SO42-(aq)

What is the acid ionization constant for this weak acid if an equilibrium mixture has the following composition: [H3O+] = 0.027 M; [HSO4−] = 0.29 M; and [SO42-] = 0.13 M?

 

Answer:

Ka for HSO4− = 1.2 × 10−2

Determination of Kb from Equilibrium Concentrations

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

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 Kb for caffeine.

Check Your Learning:

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

HPO42-(aq) + H2O(l) ⇌ H2PO4−(aq) + OH−(aq)

if the composition of an equilibrium mixture is as follows: [OH−] = 1.3 × 10−6M; [H2PO4−] = 0.042 M; and [HPO42-] = 0.341 M?

 

Answer:

Kb for HPO42- = 1.6 × 10-7

Determination of Ka or Kb from pH

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

HNO2(aq) + H2O(l) ⇌ H3O+(aq) + NO2−(aq)

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 (1.0 × 10−7M) due to the autoionization 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:

[H3O+] = 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:

Check Your Learning:

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

 

Answer:

Kb = 1.8  × 10−5

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)).

 

The pain of some ant bites and stings is caused by formic acid. (credit: John Tann)

A photograph is shown of a large black ant on the end of a human finger.

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

HCO2H(aq) + H2O(l) ⇌ H3O+(aq) + HCO2−(aq)     Ka = 1.8 × 10-4

Solution:

The ICE table for this system is

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 C O subscript 2 H plus sign H subscript 2 O equilibrium arrow H subscript 3 O superscript positive sign.” Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.534, blank, 0.534 minus x. The second column is blank in all three rows. The third column has the following: approximately 0, positive x, x. The fourth column has the following: 0, positive x, x.

Substituting the equilibrium concentration terms into the Ka expression gives


 The relatively large initial concentration and small equilibrium constant permits the simplifying assumption; ie that x < 5% of 0.534 M, which is 0.0267 M. The equation becomes


 Solving the equation for x yields

x2 = 0.534 × (1.8 × 10-4) = 9.6 × 10-5

 

x = 9.8 × 10-3 M

Now check the 5% assumption:  9.8 × 10-3 M < 0.0267 M   

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 = [H3O+] = 0.0098 M

Finally, the pH is calculated to be

pH = − log[H3O+] = − 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?

CH3CO2H(aq) + H2O(l) ⇌ H3O+(aq) + CH3CO2−(aq)         Ka = 1.8 × 10-5
Answer:

percent ionization = 1.3%

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:

(CH3)3N(aq) + H2O(l) ⇌ (CH3)3NH+(aq) + OH−(aq)     Kb = 6.3 × 10-5

Solution:

The ICE table for this system is

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 “( C H subscript 3 ) subscript 3 N plus sign H subscript 2 O equilibrium arrow ( C H subscript 3 ) subscript 3 N H superscript positive sign plus sign O H superscript positive sign.” Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.25, negative x, 0.25 plus sign negative x. The second column is blank in all three rows. The third column has the following: 0, x, 0 plus x. The fourth column has the following: approximately 0, x, and approximately 0 plus x.


 Substituting the equilibrium concentration terms into the Kb expression gives


 Assuming x < 5% of 0.25 M, which is 0.012 M, and solving for x yields

x = 4.0 × 10-3 M


 This value is less than 0.012 M, so the 5% assumption is valid.
 As defined in the ICE table, x is equal to the equilibrium concentration of hydroxide ion:

[OH−] = x = 4.0 × 10-3 M

The pOH is calculated to be

pOH = −log(4.0 × 10-3) = 2.40


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

pH + pOH = pKw = 14.00


 permits the computation of pH:

pH = 14.00 – pOH = 14.00-2.40 = 11.60

Check Your Learning:

Calculate the hydroxide ion concentration and the percent ionization of a 0.0325-M solution of ammonia, a weak base with a Kb of 1.76 × 10−5.

 

Answer:

7.56 × 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 (Figure) below.

Calculating Equilibrium Concentrations without Simplifying Assumptions

Sodium hydrogen sulfate, NaHSO4, is used in some household cleansers as a source of the HSO4− ion, a weak acid. What is the pH of a 0.50-M solution of HSO4−?

HSO4−(aq) + H2O(l) ⇌ H3O+(aq) + SO42-(aq)     Ka = 1.2 × 10-2

Solution:

The ICE table for this system is

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 ( M ). The second column has the header of “H S O subscript 4 superscript negative sign plus sign H subscript 2 O equilibrium sign H subscript 3 O superscript positive sign plus sign S O subscript 4 superscript 2 superscript negative sign.” Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.50, negative x, 0.50 minus x. The second column is blank for all three rows. The third column has the following: approximately 0, positive x, x. The fourth column has the following: 0, positive x, x.



 If the 5% assumption  (x < 0.05 × 0.50 M, or x < 0.025 M) is made, simplifying and solving the above equation yields

x = 0.077 M
 Because the simplifying assumption is not valid for this system, the equilibrium constant expression is solved as follows:


 Rearranging this equation yields

6.0 × 10-3 – 1.2 × 10-2x = x2


 Writing the equation in quadratic form gives

x2 + 1.2 × 10-2x – 6.0 × 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.

x = [H3O+] = 0.072 M
pH = – log[H3O+] = -log(0.072 M) = 1.14

Check Your Learning:

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

C8H10N4O2(aq) + H2O(l) ⇌ C8H10N4O2H+(aq) + OH−(aq)          Kb = 2.5 × 10-4
Answer:

pH 11.16

Effect of Molecular Structure on Acid-Base Strength

Binary Acids and Bases

In the absence of any leveling effect, the acid strength of binary compounds of hydrogen with nonmetals (A) increases as the H-A bond strength decreases down a group in the periodic table. For group 17, the order of increasing acidity is HF < HCl < HBr < HI. Likewise, for group 16, the order of increasing acid strength is H2O < H2S < H2Se < H2Te.

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

 

The figure shows trends in the strengths of binary acids and bases.

This diagram has two rows and four columns. Red arrows point left across the bottom of the figure and down at the right side and are labeled “Increasing acid strength.” Blue arrows point left across the bottom and up at the right side of the figure and are labeled “Increasing base strength.” The first column is labeled 14 at the top and two white squares are beneath it. The first has the number 6 in the upper left corner and the formula C H subscript 4 in the center along with designation Neither acid nor base. The second square contains the number 14 in the upper left corner, the formula C H subscript 4 at the center and the designation Neither acid nor base. The second column is labeled 15 at the top and two blue squares are beneath it. The first has the number 7 in the upper left corner and the formula N H subscript 3 in the center along with the designation Weak base and K subscript b equals 1.8 times 10 superscript negative 5. The second square contains the number 15 in the upper left corner, the formula P H subscript 3 at the center and the designation Very weak base and K subscript b equals 4 times 10 superscript negative 28. The third column is labeled 16 at the top and two squares are beneath it. The first is shaded tan and has the number 8 in the upper left corner and the formula H subscript 2 O in the center along with the designation neutral. The second square is shaded pink, contains the number 16 in the upper left corner, the formula H subscript 2 S at the center and the designation Weak acid and K subscript a equals 9.5 times 10 superscript negative 8. The fourth column is labeled 17 at the top and two squares are beneath it. The first is shaded pink, has the number 9 in the upper left corner and the formula H F in the center along with the designation Weak acid and K subscript a equals 6.8 times 10 superscript negative 4. The second square is shaded a deeper pink, contains the number 17 in the upper left corner, the formula H C l at the center, and the designation Strong acid.

Ternary Acids and Bases

Ternary compounds composed of hydrogen, oxygen, and some third element (“E”) may be structured as depicted in the image below. In these compounds, the central E atom is bonded to one or more O atoms, and at least one of the O atoms is also bonded to an H atom, corresponding to the general molecular formula OmE(OH)n. These compounds may be acidic, basic, or amphoteric depending on the properties of the central E atom. Examples of such compounds include sulfuric acid, O2S(OH)2, sulfurous acid, OS(OH)2, nitric acid, O2NOH, perchloric acid, O3ClOH, aluminum hydroxide, Al(OH)3, calcium hydroxide, Ca(OH)2, and potassium hydroxide, KOH:

A diagram is shown that includes a central atom designated with the letter E. Single bonds extend above, below, left, and right of the E. An O atom is bonded to the right of the E, and an arrow points to the bond labeling it, “Bond a.” An H atom is single bonded to the right of the O atom. An arrow pointing to this bond connects it to the label, “Bond b.”

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

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

Key Concepts and Summary

The relative strengths of acids and bases are reflected in the magnitudes of their ionization constants; the stronger the acid or base, the larger its ionization constant. A reciprocal relation exists between the strengths of a conjugate acid-base pair: the stronger the acid, the weaker its conjugate base. Water exerts a leveling effect on dissolved acids or bases, reacting completely to generate its characteristic hydronium and hydroxide ions (the strongest acid and base that may exist in water). The strengths of the binary acids increase from left to right across a period of the periodic table (CH4 < NH3 < H2O < HF), and they increase down a group (HF < HCl < HBr < HI).

acid ionization constant (Ka)
equilibrium constant for an acid ionization reaction
base ionization constant (Kb)
equilibrium constant for a base ionization reaction
leveling effect
observation that acid-base strength of solutes in a given solvent is limited to that of the solvent’s characteristic acid and base species (in water, hydronium and hydroxide ions, respectively)
oxyacid
ternary compound with acidic properties, molecules of which contain a central nonmetallic atom bonded to one or more O atoms, at least one of which is bonded to an ionizable H atom
percent ionization
ratio of the concentration of ionized acid to initial acid concentration expressed as a percentage

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