Unit Four

Day 34: Acids and Bases

As you work through this section, if you find that you need a bit more background material to help you understand the topics at hand, you can consult “Chemistry: The Molecular Science” (5th ed. Moore and Stanitski) Chapter 3-4 and 14-1 through 14-5, and/or Chapter 15.1-15.7 in the Additional Reading Materials section.

In the last unit, we will use the knowledge we learned so far, e.g. thermodynamics, kinetics, molecular structure, etc., to explore two prevalent types of chemical reactions: acid-base reactions and redox/electrochemical reactions.

D34.1 Definition of Acids and Bases

Around 1815, Humphry Davy demonstrated that hydrogen is the essential constituent of acids, and Joseph Louis Gay-Lussac concluded that acids are substances that can neutralize bases and these two classes of substances can be defined only in terms of each other. The significance of hydrogen was reemphasized in 1884 when Svante Arrhenius defined an acid as a compound that dissolves in water to yield hydrogen cations (or protons, H+) and a base as a compound that dissolves in water to yield hydroxide anions (OH). This Arrhenius acid-base definition is useful in some contexts but limited in scope.

A more general definition, centered on the proton, was proposed in 1923 by the Danish chemist Johannes Brønsted and the English chemist Thomas Lowry: According to the Brønsted-Lowry acid-base definition, a chemical species (ion or molecule) that donates a proton to another chemical species is called an acid and a chemical species that accepts a proton is a base. An acid-base reaction is the transfer of a proton from a proton donor (acid) to a proton acceptor (base).

Although we will not discuss it much in this course, there is an even more general model of acid-base behavior introduced by the American chemist G. N. Lewis. The Lewis acid-base definition says that an acid is a chemical species that can accept a pair of electrons and form a bond; a base is a chemical species that can donate a pair of electrons.

Here we will use the Brønsted-Lowry definition. The chemical species that remains after an acid has donated a proton is called the conjugate base of that acid. Consider the following examples of the generic reaction acid + H2O ⇌ conjugate base + H3O+:

HF + H2O ⇌ F + H3O+
H2SO4 + H2O ⇌ HSO4 + H3O+
H2O + H2O ⇌ OH + H3O+
HSO4 + H2O ⇌ SO42- + H3O+
NH4+ + H2O ⇌ NH3 + H3O+

(Note that in each of these reactions there is a second acid-base pair: H2O is the conjugate base of the acid H3O+.)

Similarly, the chemical species that forms after a base accepts a proton is called the conjugate acid of that base. Consider the following examples of the generic reaction base + H2O ⇌ conjugate acid + OH:

OH + H2O ⇌ H2O + OH
H2O + H2O ⇌ H3O+ + OH
NH3 + H2O ⇌ NH4+ + OH
S2- + H2O ⇌ HS + OH
CO32- + H2O ⇌ HCO3 + OH
F + H2O ⇌ HF + OH

(Note that in these reactions there is also a second acid-base pair: H2O is the conjugate acid of the base OH)

According to the Brønsted-Lowry definition, all acid-base reactions involve the transfer of a protons between an acid and a base. For example, consider the acid-base reaction that takes place when ammonia is dissolved in (and reacts with) water. A water molecule (functioning as an acid here) transfers a proton to an ammonia molecule, yielding the conjugate base of water, OH, and the conjugate acid of ammonia, NH4+:

This figure has three parts in two rows. In the first row, two diagrams of acid-base pairs are shown. On the left, a space filling model of H subscript 2 O is shown with a red O atom at the center and two smaller white H atoms attached in a bent shape. Above this model is the label “H subscript 2 O (acid)” in purple. An arrow points right, which is labeled “Remove H superscript plus.” To the right is another space filling model with a single red O atom to which a single smaller white H atom is attached. The label in purple above this model reads, “O H superscript negative (conjugate base).” Above both of these red and white models is an upward pointing bracket that is labeled “Conjugate acid-base pair.” To the right is a space filling model with a central blue N atom to which three smaller white H atoms are attached in a triangular pyramid arrangement. A label in green above reads “N H subscript 3 (base).” An arrow labeled “Add H superscript plus” points right. To the right of the arrow is another space filling model with a blue central N atom and four smaller white H atoms in a tetrahedral arrangement. The green label above reads “N H subscript 3 superscript plus (conjugate acid).” Above both of these blue and white models is an upward pointing bracket that is labeled “Conjugate acid-base pair.” The second row of the figure shows the chemical reaction, H subscript 2 O ( l ) is shown in purple, and is labeled below in purple as “acid,” plus N H subscript 3 (a q) in green, labeled below in green as “base,” followed by a double sided arrow arrow and O H superscript negative (a q) in purple, labeled in purple as “conjugate base,” plus N H subscript 4 superscript plus (a q)” in green, which is labeled in green as “conjugate acid.” The acid on the left side of the equation is connected to the conjugate base on the right with a purple line. Similarly, the base on the left is connected to the conjugate acid on the right side.

Note that acid-base reactions are often represented as equilibrium processes. The relative extent to which these reactions proceed is an important aspect when we consider acid and base strengths later.

Exercise 1: Brønsted-Lowry Acids and Bases

Identify whether each molecule is a Brønsted-Lowry acid or a  base (relative to water), and write the formula for its conjugate partner (i.e. if it’s an acid, write its conjugate base, if it’s a base, write its conjugate acid).

[Enter the formula for each without subscripts and superscripts, and put any charge in parentheses.  For example,  enter HSO4 as HSO4(-)]

D34.2 Autoionization of Water

In previous discussion, it was clear that water can function as an acid or a base depending on the nature of the solute dissolved in it. In pure water, water acts both as an acid and a base—a very small fraction of water molecules donate protons to other water molecules:

This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in purple, O atom which has H atoms singly bonded above and to the right. The O atom has pairs of electron dots on its left and lower sides. This is followed by a plus sign, which is followed in green by an O atom which has H atoms singly bonded above and to the right. The O atom has pairs of electron dots on its left and lower sides. A double arrow follows. To the right, in brackets is a structure with a central O atom in green, with green H atoms singly bonded above and to the right. A pair of green electron dots is on the lower side of the O atom. To the left of the green O atom, a purple H atom is singly bonded. Outside the brackets to the right is a superscript plus. This is followed by a plus sign and an O atom in purple with pairs of electron dots above, left, and below. An H atom is singly bonded to the right. This atom has a superscript negative sign. The reaction is written in symbolic form below. H subscript 2 O is labeled in purple below as “Acid subscript 1.” This is followed by plus H subscript 2 O, which is labeled in green below as “Base subscript 2.” A double sided arrow follows. To the right is H subscript 3 O superscript plus, which is labeled in green as below in as “Acid subscript 2.” This is followed by plus and O with pairs of dots above, below, and to the left with a singly bonded H on the right with a superscript negative. The label below in purple reads, “ Base subscript 1.”

This type of reaction, in which a substance ionizes when one molecule of the substance reacts with another molecule of the same substance, is referred to as autoionization.

Pure water undergoes autoionization to a very slight extent: only about two out of every 109 molecules are ionized at 25 °C. The very small concentrations of H3O+ and H3O+ give an autoionization constant for water, Kw = 1.0 × 10−14 M2 at 25 °C. Because it is the mathmatical product of concentrations of two ions, it is also called the ion-product constant for water.

H2O(l) + H2O(l) ⇌ H3O+(aq) + OH(aq)     Kw = [H3O+][OH] = 1.0 × 10−14 M2     at 25 °C

The process is endothermic, and so the extent of ionization increases with temperature. For example, at 100 °C, Kw is about 5.6 × 10−13 M2, roughly 50 times larger than the value at 25 °C.

Example 1

Ion Concentrations in Pure Water
What are the H3O+ concentration and the OH 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]. At 25 °C,

[latex]K_{\text{w}} = [\text{H}_3\text{O}^{+}][\text{OH}^{-}] = [\text{H}_3\text{O}^{+}]^2 = [\text{OH}^{-}]^2 = 1.0\;\times\;10^{-14}\;\text{M}^2[/latex]

so

[latex][\text{H}_3\text{O}^{+}] = [\text{OH}^{-}] = \sqrt{1.0\;\times\;10^{-14}\;\text{M}^2} = 1.0\;\times\;10^{-7}\;\text{M}[/latex]

The hydronium ion concentration and the hydroxide ion concentration are the same, and we find that both equal 1.0 × 10−7 M at 25 °C.

Check Your Learning
At 80 °C, Kw = 2.4 × 10−13 M2. Determine the concentrations of hydronium and hydroxide ions in pure water at 80 °C.

Answer:

[H3O+] = [OH] = 4.9 × 10−7M

Example 2

The Inverse Proportionality of [H3O+] and [OH]
A solution of carbon dioxide in water has [H3O+] = 2.0 × 10−6 M. Calculate the concentration of OH at 25 °C.

Solution
The ion-product constant for water at 25 °C is

[latex]2\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{OH}^{-}(aq)\;\;\;\;\;\;\;K_{\text{w}} = [\text{H}_3\text{O}^{+}][\text{OH}^{-}] = 1.0\;\times\;10^{-14}\;\text{M}^2[/latex]

Thus, we can calculate the missing equilibrium concentration.

Rearrangement of the Kw expression yields:

[latex][\text{OH}^{-}] = \dfrac{K_{\text{w}}}{[\text{H}_3\text{O}^{+}]} = \dfrac{1.0\;\times\;10^{-14}\;\text{M}^2}{2.0\;\times\;10^{-6}\;\text{M}} = 5.0\;\times\;10^{-9}\;\text{M}[/latex]

The hydroxide ion concentration is reduced to 5.0 × 10−9 M as the hydronium ion concentration increases to 2.0 × 10−6 M. This is expected from Le Châtelier’s principle: the autoionization reaction shifts to the left to partly counteract the increased hydronium ion concentration and the [OH] is reduced relative to that in pure water.

A check of these concentrations confirms that our arithmetic is correct:

[latex]K_{\text{w}} = [\text{H}_3\text{O}^{+}][\text{OH}^{-}] = (2.0\;\times\;10^{-6}\;\text{M})(5.0\;\times\;10^{-9}\;\text{M}) = 1.0\;\times\;10^{-14}\;\text{M}^2[/latex]

Check Your Learning
Calculate the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 M at 25 °C.

Answer:

[H3O+] = 1 × 10−11 M

Water is an example of a molecule or ion that can gain or lose a proton in acid-base reactions, an amphiprotic chemical species. Amphiprotic species are also amphoteric, a more general term for a species that may act either as an acid or a base by any definition (not just the Brønsted-Lowry definition). Consider for example the bicarbonate ion:

HCO3(aq) + H2O(l) ⇌ CO32-(aq) + H3O+(aq)
HCO3(aq) + H2O(l) ⇌ H2CO3(aq) + OH(aq)

Example 3

Representing the Acid-Base Behavior of an Amphiprotic Species
Write separate equations representing the reaction of HSO3

(a) as an acid with OH

(b) as a base with HI

Solution
(a) [latex]\text{HSO}_3^-(\text{aq})\;+\;\text{OH}^{-}(\text{aq})\;{\rightleftharpoons}\;\text{SO}_3^{2-}(\text{aq})\;+\;\text{H}_2\text{O}(l)[/latex]

(b) [latex]\text{HSO}_3^{-}(\text{aq})\;+\;\text{HI}(\text{aq})\;{\rightleftharpoons}\;\text{H}_2\text{SO}_3(\text{aq})\;+\;\text{I}^{-}(\text{aq})[/latex]

Check Your Learning
Write separate equations representing the reaction of H2PO4

(a) as a base with HBr

(b) as an acid with OH

Answer:

(a) [latex]\text{H}_2\text{PO}_4^{-}(\text{aq})\;+\;\text{HBr}(\text{aq})\;{\rightleftharpoons}\;\text{H}_3\text{PO}_4(\text{aq})\;+\;\text{Br}^{-}(\text{aq})[/latex];

(b) [latex]\text{H}_2\text{PO}_4^{-}(\text{aq})\;+\;\text{OH}^{-}(\text{aq})\;{\rightleftharpoons}\;\text{HPO}_4^{2-}(\text{aq})\;+\;\text{H}_2\text{O}(l)[/latex]

D34.3 pH and pOH

The concentrations of hydronium cations and hydroxide anions in a solution are important for the solution’s acid-base properties and often affect the chemical behaviors of its other solutes. A solution is neutral if it contains equal concentrations of hydronium and hydroxide ions; acidic if [H3O+] > [OH]; and basic if [H3O+] < [OH].

A common means of expressing values that may span many orders of magnitude is to use a logarithmic scale. One such scale is based on the p-function:

pX = -logX

where “X” is the quantity of interest and “log” is the base-10 logarithm. The pH of a solution is therefore defined as:

pH = -log{[H3O+]/(mol/L)}

The reason for dividing by the units mol/L is that [H3O+] has units mol/L and taking the logarithm of a unit makes no sense. From here on we will assume that you are aware that only the numeric value of a concentration (or other quantity) needs to be used as the argument of a logarithm and we will not explicitly divide by the units.

If a pH value is known, the concentration of hydronium ions can be calculated

[H3O+] = 10-pH

Here we assume that you know that units are required for the concentration obtained from this equation and the units are mol/L. Similarly, the hydroxide ion concentration may be expressed as pOH:

pOH = -log[OH]     and     [OH] = 10−pOH

Finally, the relation between pH and pOH can be derived from the Kw expression:

Kw = [H3O+][OH]
-log(Kw) = -log([H3O+][OH]) = -log([H3O+]) + (-log([OH]))
pKw = pH + pOH

At 25 °C:

pKw = 14 = pH + pOH

Therefore, at this temperature:

Classification Relative Ion Concentrations pH at 25 °C
acidic [H3O+] > [OH] pH < 7
neutral [H3O+] = [OH] pH = 7
basic [H3O+] < [OH] pH > 7

Because Kw is temperature dependent, these correlations between pH values and the acidic/neutral/basic adjectives are different at different temperatures. For example, [H3O+] in pure water at 80 °C is 4.9 × 10−7 M, which corresponds to pH and pOH values of:

pH = -log[H3O+] = -log(4.9 × 10−7) = 6.31
pOH = -log[OH] = -log(4.9 × 10−7) = 6.31

At this temperature, then, neutral solutions exhibit pH = pOH = 6.31, acidic solutions exhibit pH < 6.31 and pOH > 6.31, whereas basic solutions exhibit pH > 6.31 and pOH < 6.31. This distinction can be important when studying certain processes that occur at temperatures other than 25 °C.

Unless otherwise noted, references to pH values are presumed to be those at 25 °C. Figure 1 shows the relationships between [H3O+], [OH], pH, and pOH, and gives values for these properties for some common substances.

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 1. The pH and pOH values of some common substances at 25 °C are shown in this chart.

Exercise 2: pH and Relative Strengths of Acids

Experiments show that the pH of a 0.010 M HCN aqueous solution is higher than the pH of a 0.010 M CH3COOH aqueous solution. Based on this observation, what can you deduce about the relative acidity of HCN (prussic acid) and CH3COOH (acetic acid)? Explain your answer.

Example 4

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

Solution

[latex]\begin{array}{rcl} \text{pH} & = & -\text{log[H}_3\text{O}^{+}] \\[0.5em] & = & -\text{log}(1.2\;\times\;10^{-3})\\[0.5em] & = & -(-2.92) =2.92 \end{array}[/latex]

(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:

[latex]\text{CO}_2(\text{aq})\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_2\text{CO}_3(\text{aq})[/latex]

Air-saturated water has a hydronium ion concentration caused by the dissolved CO2 of 2.0 × 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 5

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

Solution

[latex]\text{pH} = -\text{log[H}_3\text{O}^{+}] = 7.3[/latex]
[latex]\text{log[H}_3\text{O}^{+}] = -7.3[/latex]
[latex][\text{H}_3\text{O}^{+}] = 10^{-7.3}\;\;\;\;\text{or}\;\;\;\;[\text{H}_3\text{O}^{+}] = \text{antilog}\;\text{of}\;-7.3[/latex]
[latex][\text{H}_3\text{O}^{+}] = 5\;\times\;10^{-8}\;\text{M}[/latex]

(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

The acidity of a solution is typically determined by measuring its pH. The pOH of a solution is not usually measured, but it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter (Figure 2), or visually estimated using colored indicators (Figure 3).

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 2. (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)
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 3. (a) A universal indicator assumes a different color in solutions of different pH values. Thus, it can be added to a solution to determine the pH of the solution. The eight vials each contain a universal indicator and 0.1-M solutions of progressively weaker acids: HCl (pH = l), CH3CO2H (pH = 3), and NH4Cl (pH = 5), deionized water, which is neutral (pH = 7); and 0.1-M solutions of the progressively stronger bases: KCl (pH = 7), aniline, C6H5NH2 (pH = 9), NH3 (pH = 11), and NaOH (pH = 13). (b) pH paper contains a mixture of indicators that give different colors in solutions of differing pH values. (credit: modification of work by Sahar Atwa)

D34.4 Acid Constant Ka and Base Constant Kb

The relative strengths of acids and bases may be determined by comparing the equilibrium constants for their ionization reactions. For the reaction of a generic acid, HA, in water:

HA(aq) + H2O(l) ⇌ A(aq) + H3O+(aq)

we write the acid ionization constant (Ka) expression as:

[latex]K_{\text{a}} = \dfrac{[\text{H}_3\text{O}^{+}][\text{A}^{-}]}{[\text{HA}]}[/latex]

(Although water is a reactant in the reaction, it is the solvent as well, so we do not include [H2O] in the expression.) The larger the Ka of an acid, the larger the concentration of H3O+ and A relative to the concentration of the nonionized acid, HA. Thus a stronger acid, which ionizes to a greater extent, has a larger ionization constant than a weaker acid.

For example, the following data on acid ionization constants indicate the order of acid strength CH3COOH < HNO2 < HSO4:

CH3COOH(aq) + H2O(l) ⇌ CH3COO(aq) + H3O+(aq)     Ka = 1.8 × 10-5
HNO2(aq)    +    H2O(l)   ⇌   O2(aq) + H3O+(aq)               Ka = 7.4 × 10-4
HSO4(aq)    +    H2O(l)  ⇌   SO42-(aq) + H3O+(aq)            Ka = 1.1 × 10-2

We can consider the strength of a base (B) similarly by its tendency to form hydroxide ions in aqueous solution:

B(aq) + H2O(l) ⇌ HB+(aq) + OH(aq)

where the base ionization constant (Kb) expression is:

[latex]K_{\text{b}} = \dfrac{[\text{HB}^{+}][\text{OH}^{-}]}{[\text{B}]}[/latex]

In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion and HB+ concentrations than do weaker bases. Therefore, a stronger base has a larger Kb than a weaker base.

Notice that Ka and Kb provide a quantitative measure of acid and base strengths—significantly more accurate than just saying “strong acid” or “weak acid”. For example, some weak acids are stronger than other weak acids, as indicated by their larger Ka values.

Consider the ionization reactions for a conjugate acid base pair, HA and A:

HA(aq) + H2O(l) ⇌ A(aq) + H3O+(aq)     [latex]K_{\text{a}} = \dfrac{[\text{H}_3\text{O}^{+}][\text{A}^{-}]}{[\text{HA}]}[/latex]
A(aq) + H2O(l) ⇌ HA(aq) + OH(aq)     [latex]K_{\text{b}} = \dfrac{[\text{HA}][\text{OH}^{-}]}{[\text{A}^{-}]}[/latex]

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

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

When two equilibria sum to a third equilibrium, the product of the first two equilibrium constants equals the third equilibrium constant, so

[latex]K_{\text{a}}\;\times\;K_{\text{b}} = \dfrac{[\text{H}_3\text{O}^{+}][\text{A}^{-}]}{[\text{HA}]}\;\times\;\dfrac{[\text{HA}][\text{OH}^{-}]}{[\text{A}^{-}]} = [\text{H}_3\text{O}^{+}][\text{OH}^{-}] = K_{\text{w}}[/latex]

For example, at 25 °C, Ka of acetic acid (CH3COOH) is 1.8 × 10−5 M, and Kb of its conjugate base, acetate ion (CH3COO), is 5.6 × 10−10 M. The product of these two constants is indeed equal to Kw:

Ka × Kb = (1.8 × 10−5 M) × (5.6 × 10−10 M) = 1.0 × 10−14 M2 = Kw

This relationship tells us that stronger acids form weaker conjugate bases, and weaker acids form stronger conjugate bases (Figure 4 and Figure 5).

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.
Figure 4. This diagram shows the relative strengths of conjugate acid-base pairs, as indicated by their ionization constants in aqueous solution.
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 5. The chart shows the relative strengths of conjugate acid-base pairs. The acid and base in a given row are conjugate to each other.

Although “strong” and “weak” are relative terms, in aqueous solutions acids stronger than H3O+ are called strong acids, and bases stronger than OH are strong bases. Because strong acids and strong bases are completely ionized in aqueous solutions, the concentration of nonionized acid or base is essentially zero. For example, in a 0.10-M solution of HCl, [HCl] = 0, [H30+] = 0.10 M, and [Cl] = 0.10 M. A consequence of this complete ionization is that in aqueous solution there is no way to tell whether one strong acid is stronger than another: HCl, HBr, and HI all are completely ionized. This is known as the leveling effect of water. However, when dissolved in ethanol (a weaker base than water), these acids do not ionize completely. The extent of ionization increases in the order HCl < HBr < HI, and so HI is the strongest of these acids. Water exerts a similar leveling effect on strong bases.

Many acids and bases are weak. A solution of a weak acid in water is a equilibrium mixture of the nonionized acid, hydronium ion, and the conjugate base of the acid, with the nonionized acid present in the greatest concentration. Most aqueous acid-base reactions reach equilibrium in less than a millisecond—the reactions are rapid.

Example 6

Determination of Ka or Kb from pH
One way to determine Ka or Kb is to determine [H3O+] by measuring pH for a solution of known initial concentration of acid. For example, calculate Ka for nitrous acid if the pH of a 0.0516-M solution of HNO2 is is measured as 2.34.

[latex]\text{HNO}_2(\text{aq})\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(\text{aq})\;+\;\text{NO}_2^{\;\;-}(\text{aq})[/latex]

Solution
An ICE table is useful in solving this type of problem. Initially, before the rapid acid-base reaction occurs, the concentrations are HNO2, 0.0516 M; H3O+, 1.0 × 10−7 M; and NO2, 0.0 M so the first row of the ICE table can be filled in.

The equilibrium concentration of hydronium ion can be obtained from the pH:

[latex][\text{H}_3\text{O}^{+}] = 10^{-2.34} = 0.0046\;M[/latex]

Let x be the change in [H3O+], that is the difference between the equilibrium concentration of H3O+, 0.0046 M, and the initial concentration, 1.0 × 10−7 M. Thus x = (0.0046 − 1.0 × 10−7) M = 0.0045999 M, which rounds to 0.0046 M to two significant figures. Note that the initial concentration of H3O+ is so small that it is negligible; this is often true, but needs to be checked.

The change in concentration of NO2 equals the change in concentration of H3O+. The equilibrium concentration of HNO2 is equal to its initial concentration plus the change in its concentration, −x.

Now complete the ICE table with equilibrium concentrations and calculate Ka:
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 x, 0.0470. The second column is blank in the first row, positive sign, blank for the third row. The third column has the following: approximately 0, x equals 0.0046, 0.0046. The fourth column has the following: 0, negative x, 0.0046.

[latex]K_{\text{a}} = \dfrac{[\text{H}_3\text{O}^{+}][\text{NO}_2^{\;\;-}]}{[\text{HNO}_2]} = \dfrac{(0.0046\;\text{M})(0.0046\;\text{M})}{(0.0470\;\text{M})} = 4.5\;\times\;10^{-4}\;\text{M}[/latex]

Check Your Learning.
Determine Kb for NH3, given that the pH of a 0.950-M solution of NH3, is 11.612.

Answer:

Kb = 1.77 × 10−5 M (There are 3 sig figs in the answer because there are three digits to the right of the decimal point in the pH; digits to the left of the decimal point are the power of 10 and are not significant.)

Given the initial concentration of a weak acid or base, Ka or Kb can be used to calculate the concentrations of ions in solution.

Example 7

Equilibrium Concentrations in a Solution of a Weak Acid
Formic acid, HCO2H, is the irritant that causes the body’s reaction to ant bites. Calculate the concentration of hydronium ion and the pH in a 0.534-M solution of formic acid.

[latex]\text{HCO}_2\text{H}(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{HCO}_2^{\;\;-}(aq)\;\;\;\;\;\;\;K_{\text{a}} = 1.8\;\times\;10^{-4}\;\text{M}[/latex]

Solution

  1. Determine x and equilibrium concentrations. The table shows initial concentrations (concentrations before the acid ionizes), changes in concentration, and equilibrium concentrations (the data given in the problem appear in color): 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 three columns and three rows. The first column has the following: 0.534, negative x, 0.534 plus sign negative x. The second column has the following: approximately 0, x, 0 plus sign x equals x. The third column has the following: 0, x, 0 plus sign x equals x.
  2. Solve for x and the equilibrium concentrations.
    At equilibrium:
[latex]K_{\text{a}} = 1.8\;\times\;10^{-4}\;\text{M} = \dfrac{[\text{H}_3\text{O}^{+}][\text{HCO}_2^{\;\;-}]}{[\text{HCO}_2\text{H}]}[/latex]
[latex]\dfrac{(x)(x)}{0.534\;-\;x} = 1.8\;\times\;10^{-4}[/latex]

solving this quadratic equation gives:

[latex]x = 9.7\;\times\;10^{-3}[/latex]

We find the equilibrium concentration of hydronium ion in this formic acid solution from its initial concentration and the change in that concentration:

[latex][\text{H}_3\text{O}^{+}] = {(\sim}0\;+\;x)\;\text{M} = (0\;+\;9.7\;\times\;10^{-3})\;\text{M}[/latex].
[latex]= 9.7\;\times\;10^{-3}\;M[/latex]

The pH of the solution can be found by taking the negative log of the [H3O+]:

[latex]-\text{log}(9.7\;\times\;10^{-3}) = 2.01[/latex]

Check Your Learning
Find the concentration of hydroxide ion in a 0.0325-M solution of ammonia, a weak base with a Kb of 1.76 × 10−5 M.

Answer:

7.56 × 10−4 M

D34.5 Percent Ionization

The percent ionization of a weak acid is another measure of the strength of an acid, HA:

[latex]\text{percent}\;\text{ionization} = \dfrac{[\text{H}_3\text{O}^{+}]_{\text{eq}}}{(\text{concHA})_{\text{initial}}}\;\times\;100\%[/latex]

Stronger acids, with higher Ka, therefore have higher percent ionization (for a given concentration). The percent ionization for a solution of a given weak acid increases with decreasing acid concentration; this can be seen by applying Le Chatelier’s principle to the ionization equilibrium:

HA(aq) + H2O(l) ⇌ A(aq) + H3O+(aq)

Increasing the solution volume for a given quantity of HA added causes the equilibrium to shift to the product side to partially counteract the decrease in total solute concentration.

Example 8

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:

[latex]\text{percent}\;\text{ionization} = \dfrac{[\text{H}_3\text{O}^{+}]_{\text{eq}}}{[\text{concHNO}_2]_{initial}}\;\times\;100\%[/latex]

The chemical equation for the dissociation of the nitrous acid is:

[latex]\text{HNO}_2(\text{aq})\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{NO}_2^{-}(\text{aq})\;+\;\text{H}_3\text{O}^{+}(\text{aq})[/latex]

Since

[latex]10^{-\text{pH}} = [\text{H}_3\text{O}^{+}][/latex]

we find that 10−2.09 = 8.1 × 10−3, and

[latex]\text{percent}\;\text{ionization} = \dfrac{8.1\;\times\;10^{-3}\;\text{M}}{0.125\;\text{M}}\;\times\;100\% = 6.5\%[/latex]

(The logarithm 2.09 indicates a hydronium ion concentration with only two significant figures.)

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

D34.6 Acid Strength and Molecular Structure

Acid-base reactions, like many other chemical reactions, involve breaking and forming bonds. Hence, we can use our chemical understanding of molecular structure and stability to understand what makes some acids stronger than others.

Let’s consider a generic acid:

HA ⇌ A + H+

In general, the stronger the H–A bond, the less likely the bond is to break to form H+ ions and thus the less acidic the substance. This effect is illustrated by the hydrogen halides:

Relative Acid Strength HF HCl HBr HI
H–X Bond Energy (kJ/mol) 566 431 366 299
pKa 3.2 −6.1 −8.9 −9.3

As you go down the halide group, the overlap between the 1s orbital of hydrogen and the valence orbital of the halogen atom decreases, and the H-X bond becomes weaker. As a result, HX acid strength increases. A similar trend is observed for other groups. For example, for group 16 (with pKa values in parentheses):

H2O(14.0=pKw) < H2S(7.0) < H2Se(3.9) < H2Te(2.6)

We know that an equilibrium favors the thermodynamically more stable side, and that the magnitude of the equilibrium constant reflects the energy difference between the components of each side. Therefore, in an acid-base equilibrium, the equilibrium will always favors the side with weaker acid and base (these are the more stable species). Consequently, anything that stabilizes the conjugate base of an acid will necessarily make that acid stronger.

For instance, factors that stabilize the excess negative charge on the conjugate base (A) would favor the dissociation of H+ and make the parent acid (HA) stronger. For example, the relative acidity of the acids with second row elements is (with pKa values in parentheses):

CH4(50) ≪ NH3(36) < H2O (14.0) < HF(3.2)

Consider the compounds at both ends of this series: methane and hydrogen fluoride. The conjugate base of CH4 is CH3, and the conjugate base of HF is F. Because fluorine is much more electronegative than carbon, fluorine can better stabilize the negative charge in the F ion than carbon can stabilize the negative charge in the CH3 ion. Consequently, HF has a greater tendency to dissociate to form H+ and F than does methane to form H+ and CH3, making HF a much stronger acid than CH4.

The same trend is predicted by analyzing the acids: as the electronegativity of A in HA increases, the A–H bond becomes more polar, favoring dissociation to form A and H+. Due to both the increasing stability of the conjugate base and the increasing polarization of the A–H bond in the acid, acid strengths of binary hydrides increase as we go from left to right across a row of the periodic table.

The stability of conjugate base also affects acidities of carboxylic acids. The pKa values of some typical carboxylic acids are listed in the table. When we compare these values with those of comparable alcohols, such as ethanol (pKa = 16) and 2-methyl-2-propanol (pKa = 19), it is clear that carboxylic acids are stronger acids by over ten orders of magnitude.

Compound pKa Compound pKa
HCOOH 3.75 CH3CH2CH2COOH 4.82
CH3COOH 4.74 ClCH2CH2CH2COOH 4.53
FCH2COOH 2.65 CH3CHClCH2COOH 4.05
ClCH2COOH 2.85 CH3CH2CHClCOOH 2.89
BrCH2COOH 2.90 C6H5COOH 4.20
ICH2COOH 3.10 p-O2NC6H4COOH 3.45
Cl3CCOOH 0.77 p-CH3OC6H4COOH 4.45

Why should the presence of a carbonyl group adjacent to a hydroxyl group have such a profound effect on the acidity of the hydroxyl proton? Both the carboxyl group and the carboxylate anion are stabilized by resonance, but the stabilization of the anion is much greater. In the carboxylate anion, the two contributing structures have equal weight, and the C–O bonds are of equal length (between a double and a single bond). This stabilization leads to a markedly increased acidity.

Atoms (or groups of atoms) in a molecule that are not directly bonded to the acidic H can also influence the molecule’s acidity. They can do so via an inductive effect, that is, they induce a change in the distribution of electrons within the molecule. This can be seen in the table above where electronegative substituents near the carboxyl group act to increase the acidity. For example, fluoroacetic acid, FCH2COOH, is significantly more acidic (lower pKa) than acetic acid, CH3COOH. The magnitude of the inductive effect depends on both the nature and the number of halogen substituents, as shown by the pKa values for several acetic acid derivatives:

CH3COOH(4.76) < CH2ClCOOH(2.87) < CHCl2COOH(1.35) < CCl3COOH(0.66) < CF3COOH(0.52)

Fluorine, which is more electronegative than chlorine, causes a larger effect, and the effect of three halogens is greater than the effect of two or one. Notice from these data that inductive effects can be quite significant. For instance, replacing the –CH3 group of acetic acid by a –CF3 group results in about a 10,000-fold increase in acidity.

In another example, the acidity of hypohalous acids (HOX(aq) ⇌ H+(aq) + OX(aq)) varies by about three orders of magnitude due to the difference in electronegativity of the halogen atoms:

HOX Electronegativity of X pKa
HOCl 2.73 7.2
HOBr 2.64 8.5
HOI 2.11 10.5

As the electronegativity of X increases, the electrons are drawn more strongly toward the halogen atom and, in turn, away from the H in the O–H bond, thus weakening the O–H bond and allowing dissociation of hydrogen as H+.

The acidity of oxoacids, with the general formula HOXOn (with n = 0−3), depends strongly on the number of terminal oxygen atoms attached to the central atom X (Figure 6) due to both an inductive effect and increased stabilization of the conjugate base.

Because oxygen is the second most electronegative element, adding terminal oxygen atoms causes electrons to be drawn away from the O–H bond, making it weaker and thereby increasing the strength of the acid. Figure 6 shows how the H atom becomes steadily more blue from HClO to HClO4, making it easier to lose hydrogen as H+ ion.

Figure 6: The relationship between the acid strengths of the oxoacids of chlorine and the electron density on H in the O–H unit. These electrostatic potential maps show how the electron density on H decreases (darker blue color) as the number of terminal oxygen atoms increases. Blue corresponds to low electron densities, whereas red corresponds to high electron densities. Source: Chlorine oxoacids pKa values from J. R. Bowser, Inorganic Chemistry (Pacific Grove, CA: Brooks-Cole,1993).

Also important is the effect of delocalization of the negative charge in the conjugate base. As shown in Figure 7, the number of important resonance structures for the oxoanions of chlorine increases as the number of terminal oxygen atoms increases, indicating that the excess electron is delocalized over successively more oxygen atoms, stabilizing the conjugate base.

For example, in the perchlorate ion (ClO4), the single negative charge is delocalized over all four oxygen atoms, whereas in the hypochlorite ion (ClO), the negative charge is largely localized on a single oxygen atom. As a result, the perchlorate ion has no localized negative charge to which a proton can bind. Consequently, the perchlorate anion has a much lower affinity for a proton than does the hypochlorite ion, and perchloric acid is one of the strongest acids known.

Figure 7: As the number of terminal oxygen atoms increases, the number of important resonance structures for the oxoanions of chlorine also increases, and the excess electron is delocalized over more oxygen atoms. As these electrostatic potential plots demonstrate, the electron density on the terminal oxygen atoms decreases steadily as their number increases. As the electron density on the oxygen atoms decreases, so does their affinity for a proton, making the anion less basic. As a result, the parent oxoacid is more acidic.

The inductive effect is additionally responsible for the trend in the acidities of oxoacids that have the same number of oxygen atoms as we go across a row of the periodic table. For example, H3PO4 is a weak acid, H2SO4 is a strong acid, and HClO4 is one of the strongest acids known. The number of terminal oxygen atoms increases steadily across the row, consistent with the observed increase in acidity. In addition, the electronegativity of the central atom increases steadily from P to S to Cl, which causes electrons to be drawn from oxygen to the central atom, weakening the O–H bond and increasing the strength of the oxoacid.

Podia Question

Sodium hydrogen sulfate, NaHSO4, is used in some household cleansers because it contains HSO4, a weak acid. Calculate the pH of a 0.50-M solution of NaHSO4.

[latex]\text{HSO}_4^{-}(\text{aq})\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(\text{aq})\;+\;\text{SO}_4^{\;\;2-}(\text{aq})\;\;\;\;\;\;\;K_{\text{a}} = 1.2\;\times\;10^{-2}[/latex]

Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.

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