Additional Reading Materials

Chapter 15: Acid and Base Reactions

Ch15.1 Brønsted-Lowry Acids and Bases

Acids and bases have been known for a long time. When Robert Boyle characterized them in 1680, he noted that acids dissolve many substances, change the color of certain natural dyes (for example, they change litmus from blue to red), and lose these characteristic properties after coming into contact with alkalis (bases). In the eighteenth century, it was recognized that acids have a sour taste, react with limestone to liberate a gaseous substance (now known to be CO2), and interact with alkalis to form neutral substances. In 1815, Humphry Davy contributed greatly to the development of the modern acid-base concept by demonstrating that hydrogen is the essential constituent of acids. Around that same time, Joseph Louis Gay-Lussac concluded that acids are substances that can neutralize bases and that these two classes of substances can be defined only in terms of each other. The significance of hydrogen was reemphasized in 1884 when Carl Axel Arrhenius defined an acid as a compound that dissolves in water to yield hydrogen cations (now recognized to be hydronium ions, H3O+) and a base as a compound that dissolves in water to yield hydroxide anions (OH). This definition is not wrong; it is simply limited.

A more general definition was proposed in 1923 by the Danish chemist Johannes Brønsted and the English chemist Thomas Lowry. Their definition centers on the proton, H+. A proton is what remains when a normal hydrogen atom loses an electron. 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 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, in which acids and bases are identified by their ability to accept or to donate a pair of electrons instead.

Acids may be compounds such as HCl or H2SO4, organic acids like acetic acid (CH3COOH) or ascorbic acid (vitamin C), or H2O. Anions (such as HSO4, H2PO4, HS, and HCO3) and cations (such as H3O+, NH4+, and [Al(H2O)6]3+) may also act as acids. Bases fall into the same three categories. Bases may be neutral molecules (such as H2O, NH3, and CH3NH2), anions (such as OH, HS, HCO3, CO32-, F, and PO43-), or cations (such as [Al(H2O)5OH]2+). The most familiar bases are ionic compounds such as NaOH and Ca(OH)2, which contain the hydroxide ion, OH. The hydroxide ion in these compounds accepts a proton from acids to form water:

[latex]\text{H}^{+}\;+\;\text{OH}^{-}\;{\longrightarrow}\;\text{H}_2\text{O}[/latex]

We call the product that remains after an acid donates a proton the conjugate base of that acid. This species is a base because it can accept a proton, thereby re-forming the acid:

[latex]\begin{array}{rl} \text{acid} &\leftrightharpoons& \text{proton}\;+\;\text{conjugate base} \\[0.5em] \text{HF} &\leftrightharpoons& \text{H}^{+}\;+\;\text{F}^{-} \\[0.5em] \text{H}_2\text{SO}_4 &\leftrightharpoons& \text{H}^{+}\;+\;\text{HSO}_4^{\;\;-} \\[0.5em] \text{H}_2\text{O} &\leftrightharpoons& \text{H}^{+}\;+\;\text{OH}^{-} \\[0.5em] \text{HSO}_4^{\;\;-} &\leftrightharpoons& \text{H}^{+}\;+\;\text{SO}_4^{\;\;2-} \\[0.5em] \text{NH}_4^{\;\;+} &\leftrightharpoons& \text{H}^{+}\;+\;\text{NH}_3 \end{array}[/latex]

We call the product that results when a base accepts a proton that base’s conjugate acid. This species is an acid because it can give up a proton and thus re-form the base:

[latex]\begin{array}{rl} \text{base}\;+\;\text{proton} &\leftrightharpoons& \text{conjugate acid} \\[0.5em] \text{OH}^{-}\;+\;\text{H}^{+} &\leftrightharpoons& \text{H}_2\text{O} \\[0.5em] \text{H}_2\text{O}\;+\;\text{H}^{+} &\leftrightharpoons& \text{H}_3\text{O}^{+} \\[0.5em] \text{NH}_3\;+\;\text{H}^{+} &\leftrightharpoons& \text{NH}_4^{\;\;+} \\[0.5em] \text{S}^{2-}\;+\;\text{H}^{+} &\leftrightharpoons& \text{HS}^{-} \\[0.5em] \text{CO}_3^{\;\;2-}\;+\;\text{H}^{+} &\leftrightharpoons& \text{HCO}_3^{\;\;-} \\[0.5em] \text{F}^{-}\;+\;\text{H}^{+} &\leftrightharpoons& \text{HF} \end{array}[/latex]

In the above reaction equations, the behaviors of acids as proton donors and bases as proton acceptors are represented in isolation. In reality, all acid-base reactions involve the transfer of protons between acids and bases. For example, consider the acid-base reaction that takes place when ammonia is dissolved in water. A water molecule (functioning as an acid) transfers a proton to an ammonia molecule (functioning as a base), 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.

The reaction between a Brønsted-Lowry acid and water is called acid ionization. For example, when hydrogen fluoride dissolves in water and ionizes, protons are transferred from HF molecules to water molecules, yielding hydronium ions and fluoride ions:
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, an H atom is connected to an F atom with a single bond. The F atom has pairs of electron dots at the top, right, and bottom. 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. This is followed by a plus sign and an F atom in purple with pairs of electron dots above, right, below, and to the left. This atom also has a superscript negative sign. The reaction is written in symbolic form below. H F 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 F surrounded by 4 pairs of dots and superscript negative. The label below in purple reads, “Base subscript 1.” To the right of the reactions is the formula, K subscript a equals left bracket H subscript 3 O superscript plus right bracket left bracket F superscript negative right bracket all over left bracket H F right bracket.

When we add a base to water, a base ionization reaction occurs in which protons are transferred from water molecules to base molecules. For example, adding pyridine to water yields hydroxide ions and pyridinium ions:
This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in red, is an O atom which has H atoms singly bonded above and to the right. The O atom has lone pairs of electron dots on its left and lower sides. This is followed by a plus sign. The plus sign is followed, in blue, by an N atom with one lone pair of electron dots. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. An equilibrium arrow follows this structure. To the right, in brackets is a structure where an N atom bonded to an H atom, which is red, appears. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. Outside the brackets, to the right, is a superscript positive sign. This structure is followed by a plus sign. Another structure that appears in brackets also appears. An O atom with three lone pairs of electron dots is bonded to an H atom. There is a superscript negative sign outside the brackets. To the right, is the equation: k equals [ C subscript 5 N H subscript 6 superscript positive sign ] [ O H superscript negative sign] all divided by [ C subscript 5 N H subscript 5 ]. Under the initial equation, is this equation: H subscript 2 plus C subscript 5 N H subscript 5 equilibrium arrow C subscript 5 N H subscript 6 superscript positive sign plus O H superscript negative sign. H subscript 2 O is labeled, “acid,” in red. C subscript 5 N H subscript 5 is labeled, “base,” in blue. C subscript 5 N H subscript 6 superscript positive sign is labeled, “acid” in blue. O H superscript negative sign is labeled, “base,” in red.

Notice that both these ionization reactions are represented as equilibrium processes. The relative extent to which these acid and base ionization reactions proceed is an important topic in a later section.

Ch15.2 Autoionization of Water

In the two above examples, we also saw that water can function as either an acid or a base, depending on the nature of the solute dissolved in it. In fact, in pure water or in any aqueous solution, water acts both as an acid and a base. A very small fraction of water molecules donate protons to other water molecules to form hydronium ions and hydroxide ions:
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 in a sample of pure water are ionized at 25 °C. The equilibrium constant for the ionization of water is called the ionization constant for water (Kw):

[latex]\text{H}_2\text{O}(l)\;+\;\text{H}_2\text{O}(l)\;{\leftrightharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{OH}^{-}(aq)\;\;\;\;\;\;\;K_{\text{w}} = [\text{H}_3\text{O}^{+}][\text{OH}^{-}][/latex]

The slight ionization of pure water is reflected in the small value of the equilibrium constant; at 25 °C, Kw has a value of 1.0 × 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 Kw is about 5.6 × 10−13, 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}[/latex]

So:

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

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

Check Your Learning
At 80 °C is Kw = 2.4 × 10−13. What are the concentrations of hydronium and hydroxide ions in pure water at 80 °C?

Answer:

[H3O+] = [OH] = 4.9 × 10−7M

It is important to realize that the autoionization equilibrium for water is established in all aqueous solutions. Adding an acid or base to water will not change the position of the equilibrium. Example 2 demonstrates the quantitative aspects of this relation between hydronium and hydroxide ion concentrations.

Example 2

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

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

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

Thus, we can calculate the missing equilibrium concentration.

Rearrangement of the Kw expression yields:

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

The hydroxide ion concentration is reduced to 5.0 × 10−9M as the hydronium ion concentration increases to 2.0 × 10−6M. This is expected from Le Châtelier’s principle: the autoionization reaction shifts to the left to reduce the stress of 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})(5.0\;\times\;10^{-9}) = 1.0\;\times\;10^{-14}[/latex]

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 × 10−11M

Ch15.3 Amphiprotic Species

Like water, many molecules and ions may either gain or lose a proton under the appropriate conditions. Such species are said to be amphiprotic. Another term used to describe such species is amphoteric, which is 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 one). Consider for example the bicarbonate ion, which may either donate or accept a proton as shown here:

[latex]\begin{array}{rl} \text{HCO}_3^{\;\;-}(aq)\;+\;\text{H}_2\text{O}(l)&\leftrightharpoons&\text{CO}_3^{\;\;2-}(aq)\;+\;\text{H}_3\text{O}^{+}(aq) \\[0.5em] \text{HCO}_3^{\;\;-}(aq)\;+\;\text{H}_2\text{O}(l)&\leftrightharpoons&\text{H}_2\text{CO}_3(aq)\;+\;\text{OH}^{-}(aq) \end{array}[/latex]

Example 3

Representing the Acid-Base Behavior of an Amphoteric Substance
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^{\;\;-}(aq)\;+\;\text{OH}^{-}(aq)\;{\leftrightharpoons}\;\text{SO}_3^{\;\;2-}(aq)\;+\;\text{H}_2\text{O}(l)[/latex]

(b) [latex]\text{HSO}_3^{\;\;-}(aq)\;+\;\text{HI}(aq)\;{\leftrightharpoons}\;\text{H}_2\text{SO}_3(aq)\;+\;\text{I}^{-}(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^{\;\;-}(aq)\;+\;\text{HBr}(aq)\;{\leftrightharpoons}\;\text{H}_3\text{PO}_4(aq)\;+\;\text{Br}^{-}(aq)[/latex];

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

Ch15.4 pH and pOH

The concentrations of hydronium and hydroxide 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 values 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:

[latex]\text{pX} = -\text{log X}[/latex]

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

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

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

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

Likewise, the hydroxide ion molarity may be expressed as pOH:

[latex]\text{pOH} = -\text{log[OH}^{-}][/latex]

Finally, the relation between pH and pOH is easily derived from the Kw expression:

[latex]K_{\text{w}} = [\text{H}_3\text{O}^{+}][\text{OH}^{-}][/latex]
[latex]-\text{log}\;K_{\text{w}} = -\text{log([H}_3\text{O}^{+}][\text{OH}^{-}]) = -\text{log[H}_3\text{O}^{+}]\;+\;(-\text{log[OH}^{-}])[/latex]
[latex]\text{p}K_{\text{w}} = \text{pH}\;+\;\text{pOH}[/latex]

At 25 °C, the value of Kw is 1.0 × 10−14, and so:

[latex]14.00 = \text{pH}\;+\;\text{pOH}[/latex]

The pH and pOH of a neutral solution at this temperature are therefore:

[latex]\text{pH} = -\text{log[H}_3\text{O}^{+}] = -\text{log}(1.0\;\times\;10^{-7}) = 7.00[/latex]
[latex]\text{pOH} = -\text{log[OH}^{-}] = -\text{log}(1.0\;\times\;10^{-7}) = 7.00[/latex]

Therefore, at this temperature, acidic solutions are those with [H3O+] > 1.0 × 10−7M and [OH] < 1.0 × 10−7M (corresponding to pH < 7.00 and pOH > 7.00). Basic solutions are those with [H3O+] < 1.0 × 10−7M and [OH] > 1.0 × 10−7M (corresponding to pH > 7.00 and pOH < 7.00).

Since the autoionization constant Kw 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 hydronium molarity of pure water at 80 °C is 4.9 × 10−7M, which corresponds to pH and pOH values of:

[latex]\text{pH} = -\text{log[H}_3\text{O}^{+}] = -\text{log}(4.9\;\times\;10^{-7}) = 6.31[/latex]
[latex]\text{pOH} = -\text{log[OH}^{-}] = -\text{log}(4.9\;\times\;10^{-7}) = 6.31[/latex]

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 nonstandard temperatures, such as enzyme reactions in warm-blooded organisms. Unless otherwise noted, references to pH values are presumed to be those at standard temperature (25 °C) (Table 1).

Classification Relative Ion Concentrations pH at 25 °C
acidic [H3O+] > [OH] pH < 7
neutral [H3O+] = [OH] pH = 7
basic [H3O+] < [OH] pH > 7
Table 1. Summary of Relations for Acidic, Basic and Neutral Solutions

Figure 1 shows the relationships between [H3O+], [OH], pH, and pOH, and gives values for these properties at standard temperatures 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 standard temperature (25 °C) are shown in this chart.

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

Air-saturated water has a hydronium ion concentration caused by the dissolved CO2 of 2.0 × 10−6M, 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 of}\;-7.3[/latex]
[latex][\text{H}_3\text{O}^{+}] = 5\;\times\;10^{-8}\;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

Environmental Science

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

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

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:

[latex]\text{H}_2\text{O}(l)\;+\;\text{SO}_3(g)\;{\longrightarrow}\;\text{H}_2\text{SO}_4(aq)[/latex]
[latex]\text{H}_2\text{SO}_4(aq)\;{\longrightarrow}\;\text{H}^{+}(aq)\;+\;\text{HSO}_4^{\;\;-}(aq)[/latex]

Carbon dioxide is naturally present in the atmosphere because we and most other organisms produce it as a waste product of metabolism. Carbon dioxide is also formed when fires release carbon stored in vegetation or when we burn wood or fossil fuels. Sulfur trioxide in the atmosphere is naturally produced by volcanic activity, but it also stems 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 2). 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 2. (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 6

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:

[latex]\text{pOH} = -\text{log[OH}^{-}] = -\text{log}\;0.0125[/latex]
[latex]= -(-1.903) = 1.903[/latex]

The pH can be found from the pOH:

[latex]\text{pH}\;+\;\text{pOH} = 14.00[/latex]
[latex]\text{pH} = 14.00\;-\;\text{pOH} = 14.00\;-\;1.903 = 12.10[/latex]

Check Your Learning
The hydronium ion concentration of vinegar is approximately 4 × 10−3M. 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 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 3. (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 4).

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 4. (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, a neutral substance (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)

Ch15.5 Relative Strength of Acids and Bases

We can rank the strengths of acids by the extent to which they ionize in aqueous solution. The reaction of an acid with water is given by the general expression:

[latex]\text{HA}(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{A}^{-}(aq)[/latex]

Water is the base that reacts with the acid HA, A is the conjugate base of the acid HA, and the hydronium ion is the conjugate acid of water. A strong acid yields 100% (or very nearly so) of H3O+ and A when the acid ionizes in water; Figure 5 lists several strong acids. A weak acid gives small amounts of H3O+ and A.

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.
Figure 5. Some of the common strong acids and bases are listed here.

The relative strengths of acids may be determined 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:

[latex]\text{HA}(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{A}^{-}(aq)[/latex],

we write the equation for the ionization constant as:

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

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. Thus a stronger acid has a larger ionization constant than does a weaker acid.

For example, the following data on acid-ionization constants indicate the order of acid strength:

[latex]\begin{array}{rll} \text{CH}_3\text{CO}_2\text{H}(aq)\;+\;\text{H}_2\text{O}(l) &\rightleftharpoons& \text{H}_3\text{O}^{+}(aq)\;+\;\text{CH}_3\text{CO}_2^{\;\;-}(aq) & K_{\text{a}} = 1.8\;\times\;10^{-5} \\[0.5em] \text{HNO}_2(aq)\;+\;\text{H}_2\text{O}(l) &\rightleftharpoons& \text{H}_3\text{O}^{+}(aq)\;+\;\text{NO}_2^{\;\;-}(aq) & K_{\text{a}} = 4.6\;\times\;10^{-4} \\[0.5em] \text{HSO}_4^{\;\;-}(aq)\;+\;\text{H}_2\text{O}(l) &\rightleftharpoons& \text{H}_3\text{O}^{+}(aq)\;+\;\text{SO}_4^{\;\;2-}(aq) & K_{\text{a}} = 1.2\;\times\;10^{-2} \end{array}[/latex]

We can rank the strengths of bases by their tendency to form hydroxide ions in aqueous solution. The reaction of a Brønsted-Lowry base with water is given by:

[latex]\text{B}(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{HB}^{+}(aq)\;+\;\text{OH}^{-}(aq)[/latex]

Water is the acid that reacts with the base, HB+ is the conjugate acid of the base B, and the hydroxide ion is the conjugate base of water. As we did with acids, we can measure the relative strengths of bases by measuring their base-ionization constant (Kb) in aqueous solutions. We write the equation for the ionization constant as:

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

where the concentrations are those at equilibrium.

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. Figure 5 lists several strong bases; soluble ionic hydroxides such as NaOH are considered strong bases because they dissociate nearly completely when dissolved in water.

For example, the chemical reactions and ionization constants of the three bases are:

[latex]\begin{array}{rll} \text{NO}_2^{\;\;-}(aq)\;+\;\text{H}_2\text{O}(l) &\leftrightharpoons& \text{HNO}_2(aq)\;+\;\text{OH}^{-}(aq) & K_{\text{b}} = 2.22\;\times\;10^{-11} \\[0.5em] \text{CH}_3\text{CO}_2^{\;\;-}(aq)\;+\;\text{H}_2\text{O}(l) &\leftrightharpoons& \text{CH}_3\text{CO}_2\text{H}(aq)\;+\;\text{OH}^{-}(aq) & K_{\text{b}} = 5.6\;\times\;10^{-10} \\[0.5em] \text{NH}_3(aq)\;+\;\text{H}_2\text{O}(l) &\leftrightharpoons& \text{NH}_4^{\;\;+}(aq)\;+\;\text{OH}^{-}(aq) & K_{\text{b}} = 1.8\;\times\;10^{-5} \end{array}[/latex]

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

[latex]\begin{array}{rll} \text{HA}(aq)\;+\;\text{H}_2\text{O}(l) &\leftrightharpoons& \text{H}_3\text{O}^{+}(aq)\;+\;\text{A}^{-}(aq) & K_{\text{a}} = \frac{[\text{H}_3\text{O}^{+}][\text{A}^{-}]}{[\text{HA}]} \\[1.5em] \text{A}^{-}(aq)\;+\;\text{H}_2\text{O}(l) &\leftrightharpoons& \text{OH}^{-}(aq)\;+\;\text{HA}(aq) & K_{\text{b}} = \frac{[\text{HA}][\text{OH}]}{[\text{A}^{-}]} \end{array}[/latex]

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

[latex]\rule[0.75ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{HA}(aq)\;+\;\text{H}_2\text{O}(l)\;+\;\rule[0.75ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{A}^{-}(aq)\;+\;\text{H}_2\text{O}(l)\;{\leftrightharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\rule[0.75ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{A}^{-}(aq)\;+\;\text{OH}^{-}(aq)\;+\;\rule[0.75ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{HA}(aq)[/latex]
[latex]2\text{H}_2\text{O}(l)\;{\leftrightharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{OH}^{-}(aq)[/latex]

The K expression for a chemical equation derived from adding two or more other equations is the mathematical product of the other equations’ K expressions. Multiplying the mass-action expressions together and cancelling common terms, we see that:

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

For example, the acid ionization constant of acetic acid (CH3COOH) is 1.8 × 10−5, and the base ionization constant of its conjugate base, acetate ion (CH3COO), is 5.6 × 10−10. The product of these two constants is indeed equal to Kw:

[latex]K_{\text{a}}\;\times\;K_{\text{b}} = (1.8\;\times\;10^{-5})\;\times\;(5.6\;\times\;10^{-10}) = 1.0\;\times\;10^{-14} = K_{\text{w}}[/latex]

The extent to which an acid, HA, donates protons to water molecules depends on the strength of its conjugate base, A. If A is a strong base, any protons that are donated to water molecules are recaptured by A. Thus there is relatively little A and H3O+ in solution, and the acid, HA, is weak. If A is a weak base, water binds the protons more strongly, and the solution contains primarily A and H3O+—the acid is strong. Strong acids form weak conjugate bases, and weak acids form strong conjugate bases (Figure 6).

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 6. This diagram shows the relative strengths of conjugate acid-base pairs, as indicated by their ionization constants in aqueous solution.

Figure 7 lists a series of acids and bases in order of the decreasing strengths of the acids and the corresponding increasing strengths of the bases. The acid and base in a given row are conjugate to each other.

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 7. The chart shows the relative strengths of conjugate acid-base pairs.

The first six acids in Figure 7 are the most common strong acids. These acids are neary completely dissociated in aqueous solution. The conjugate bases of these acids are weaker bases than water. When these acids dissolves in water, their protons are transferred to water, the stronger base.

The acids that lie between the hydronium ion and water in Figure 7 form conjugate bases that can compete with water for possession of a proton. Both hydronium ions and nonionized acid molecules are present in equilibrium in a solution of one of these acids.

Compounds that are weaker acids than water (those found below water in the column of acids) in Figure 7 exhibit almost no observable acidic behavior when dissolved in water. Their conjugate bases are stronger than the hydroxide ion, and if any conjugate base were formed, it would react with water to re-form the acid.

Similarly, the extent to which a base forms hydroxide ion in aqueous solution depends on the strength of the base relative to that of the hydroxide ion, as shown in the last column in Figure 7. A strong base, such as one lying below hydroxide ion, accepts protons from water to yield the conjugate acid and hydroxide ion. Those bases lying between water and hydroxide ion accept protons from water, but a mixture of the hydroxide ion and the base results. Bases that are weaker than water (those that lie above water in the column of bases) show no observable basic behavior in aqueous solution.

Example 7

The Product Ka × Kb = Kw
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:

[latex]K_{\text{a}}\;\times\;K_{\text{b}} = 1.0\;\times\;10^{-14} = K_{\text{w}}[/latex]

Solving for Ka, we get:

[latex]K_{\text{a}} = \frac{K_{\text{w}}}{K_{\text{b}}} = \frac{1.0\;\times\;10^{-14}}{2.17\;\times\;10^{-11}} = 4.6\;\times\;10^{-4}[/latex]

Check Your Learning
We can determine the relative acid strengths of NH4+ and HCN by comparing their ionization constants. The ionization constant of HCN is 4.9 × 10−10. The ionization constant of the conjugate base of NH4+, NH3, is 1.8 × 10−5. Determine the ionization constant of NH4+, and decide which is the stronger acid, HCN or NH4+.

Answer:

NH4+ is the slightly stronger acid (Ka for NH4+ = 5.6 × 10−10).

Ch15.6 Ionization of Weak Acids and Weak Bases

Many acids and bases are weak; that is, they do not ionize fully in aqueous solution. A solution of a weak acid in water is a mixture of the nonionized acid, hydronium ion, and the conjugate base of the acid, with the nonionized acid present in the greatest concentration. Thus, a weak acid increases the hydronium ion concentration in an aqueous solution (but not as much as the same amount of a strong acid).

Acetic acid, CH3CO2H, is a weak acid. When we add acetic acid to water, it ionizes to a small extent according to the equation:

[latex]\text{CH}_3\text{CO}_2\text{H}(aq)\;+\;\text{H}_2\text{O}(l)\;{\leftrightharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{CH}_3\text{CO}_2^{\;\;-}(aq)[/latex]

and at equilibrium:

[latex]K_{\text{a}} = \frac{[\text{H}_3\text{O}^{+}][\text{CH}_3\text{CO}_2^{\;\;-}]}{[\text{CH}_3\text{CO}_2\text{H}]} = 1.8\;\times\;10^{-5}[/latex]

This equilibrium, like any other equilibria, is dynamic—acetic acid molecules donate protons to water molecules and form hydronium ions and acetate ions at the same rate that hydronium ions donate protons to acetate ions to reform acetic acid molecules and water molecules. We can tell by measuring the pH of an aqueous solution of known concentration that only a fraction of the weak acid is ionized at any moment (Figure 8).

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.
Figure 8. pH paper indicates that a 0.1-M solution of HCl (beaker on left) has a pH of 1. The acid is fully ionized and [H3O+] = 0.1 M. A 0.1-M solution of CH3CO2H (beaker on right) is a pH of 3 ([H3O+] = 0.001 M) because the weak acid CH3CO2H is only partially ionized. In this solution, [H3O+] < [CH3CO2H]. (credit: modification of work by Sahar Atwa)

Table 2 gives the ionization constants for several weak acids.

Ionization Reaction Ka at 25 °C
[latex]\text{HSO}_4^{\;\;-}\;+\;\text{H}_2\text{O}\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}\;+\;\text{SO}_4^{\;\;2-}[/latex] 1.2 × 10−2
[latex]\text{HF}\;+\;\text{H}_2\text{O}\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}\;+\;\text{F}^{-}[/latex] 3.5 × 10−4
[latex]\text{HNO}_2\;+\;\text{H}_2\text{O}\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}\;+\;\text{NO}_2^{\;\;-}[/latex] 4.6 × 10−4
[latex]\text{HCNO}\;+\;\text{H}_2\text{O}\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}\;+\;\text{NCO}^{-}[/latex] 2 × 10−4
[latex]\text{HCO}_2\text{H}\;+\;\text{H}_2\text{O}\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}\;+\;\text{HCO}_2^{\;\;-}[/latex] 1.8 × 10−4
[latex]\text{CH}_3\text{CO}_2\text{H}\;+\;\text{H}_2\text{O}\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}\;+\;\text{CH}_3\text{CO}_2^{\;\;-}[/latex] 1.8 × 10−5
[latex]\text{HCIO}\;+\;\text{H}_2\text{O}\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}\;+\;\text{CIO}^{-}[/latex] 2.9 × 10−8
[latex]\text{HBrO}\;+\;\text{H}_2\text{O}\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}\;+\;\text{BrO}^{-}[/latex] 2.8 × 10−9
[latex]\text{HCN}\;+\;\text{H}_2\text{O}\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}\;+\;\text{CN}^{-}[/latex] 4.9 × 10−10
Table 2. Ionization Constants of Some Weak Acids

At equilibrium, a solution of a weak base in water is a mixture of the nonionized base, the conjugate acid of the weak base, and hydroxide ion, with the nonionized base present in the greatest concentration. For example, a solution of the weak base trimethylamine, (CH3)3N, in water reacts according to the equation:

[latex](\text{CH}_3)_3\text{N}(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;(\text{CH}_3)_3\text{NH}^{+}(aq)\;+\;\text{OH}^{-}(aq)[/latex]
[latex]K_{\text{b}} = \frac{[(\text{CH}_3)_3\text{NH}^{+}][\text{OH}^{-}]}{[(\text{CH}_3)_3\text{N}]}[/latex]

We can confirm by measuring the pH of an aqueous solution of a weak base of known concentration that only a fraction of the base reacts with water (Figure 9). The remaining weak base is present as the unreacted form.

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.
Figure 9. pH paper indicates that a 0.1-M solution of NH3 (left) is weakly basic. The solution has a pOH of 3 ([OH] = 0.001 M) because the weak base NH3 only partially reacts with water. A 0.1-M solution of NaOH (right) has a pOH of 1 because NaOH is a strong base. (credit: modification of work by Sahar Atwa)

The ionization constants of several weak bases are given in Table 3.

Ionization Reaction Kb at 25 °C
[latex](\text{CH}_3)_2\text{NH}\;+\;\text{H}_2\text{O}\;{\rightleftharpoons}\;(\text{CH}_3)_2\text{NH}_2^{\;\;+}\;+\;\text{OH}^{-}[/latex] 5.9 × 10−4
[latex]\text{CH}_3\text{NH}_2\;+\;\text{H}_2\text{O}\;{\rightleftharpoons}\;\text{CH}_3\text{NH}_3^{\;\;+}\;+\;\text{OH}^{-}[/latex] 4.4 × 10−4
[latex](\text{CH}_3)_3\text{N}\;+\;\text{H}_2\text{O}\;{\rightleftharpoons}\;(\text{CH}_3)_3\text{NH}^{+}\;+\;\text{OH}^{-}[/latex] 6.3 × 10−5
[latex]\text{NH}_3\;+\;\text{H}_2\text{O}\;{\rightleftharpoons}\;\text{NH}_4^{\;\;+}\;+\;\text{OH}^{-}[/latex] 1.8 × 10−5
[latex]\text{C}_6\text{H}_5\text{NH}_2\;+\;\text{H}_2\text{O}\;{\rightleftharpoons}\;\text{C}_6\text{N}_5\text{NH}_3^{\;\;+}\;+\;\text{OH}^{-}[/latex] 4.3 × 10−10
Table 3. Ionization Constants of Some Weak Bases

Example 8

Determination of Ka from Equilibrium Concentrations
Acetic acid is the principal ingredient in vinegar (Figure 10); that’s why it tastes sour. At equilibrium, a solution contains [CH3CO2H] = 0.0787 M and [H3O+] = [CH3CO2] = 0.00118 M. What is the value of Ka 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 10. Vinegar is a solution of acetic acid, a weak acid. (credit: modification of work by “HomeSpot HQ”/Flickr)

Solution
We are asked to calculate an equilibrium constant from equilibrium concentrations for the reaction:

[latex]\text{CH}_3\text{CO}_2\text{H}(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{CH}_3\text{CO}_2^{\;\;-}(aq)[/latex]
[latex]K_{\text{a}} = \frac{[\text{H}_3\text{O}^{+}][\text{CH}_3\text{CO}_2^{\;\;-}]}{[\text{CH}_3\text{CO}_2\text{H}]} = \frac{(0.00118)(0.00118)}{0.0787} = 1.77\;\times\;10^{-5}[/latex]

Check Your Learning
What is the equilibrium constant for the ionization of the HSO4 ion, the weak acid used in some household cleansers:

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

In one mixture of NaHSO4 and Na2SO4 at equilibrium, [H3O+] = 0.027 M; [HSO4] = 0.29 M; and [SO42] = 0.13 M.

Answer:

Ka for HSO4 = 1.2 x 10-2

Example 9

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 × 10−3M, and [OH] = 2.5 × 10−3M?

Solution
At equilibrium, for the reaction:

[latex]\text{C}_8\text{H}_{10}\text{N}_4\text{O}_2(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{C}_8\text{H}_{10}\text{N}_4\text{O}_2\text{H}^{+}(aq)\;+\;\text{OH}^{-}(aq)[/latex]
[latex]K_{\text{b}} = \frac{[\text{C}_8\text{H}_{10}\text{N}_4\text{O}_2\text{H}^{+}][\text{OH}^{-}]}{[\text{C}_8\text{H}_{10}\text{N}_4\text{O}_2]} = \frac{(5.0\;\times\;10^{-3})(2.5\;\times\;10^{-3})}{0.050} = 2.5\;\times\;10^{-4}[/latex]

Check Your Learning
What is the equilibrium constant for the ionization of the HPO42- ion, a weak base:

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

In a solution containing a mixture of NaH2PO4 and Na2HPO4 at equilibrium, [OH] = 1.3 × 10−6M; [H2PO4] = 0.042 M; and [HPO42-] = 0.341 M.

Answer:

Kb for HPO42- = 1.6 x 10-7

Example 10

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?

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

Solution
We determine an equilibrium constant starting with the initial concentrations of HNO2, H3O+, and NO2 as well as one of the final concentrations, the concentration of hydronium ion at equilibrium. (Remember that pH is simply another way to express the concentration of hydronium ion.)

We can solve this problem with the following steps in which x is a change in concentration of a species in the reaction:
A diagram is shown that includes six rust-red rectangles. The first rectangle is labeled, “p H,” and there is an arrow that points to the right to the second rectangle. The second rectangle is labeled, “[ H subscript 3 O superscript plus ].” There is an arrow that points right to the third rectangle. The third rectangle is labeled, “Calculate x subscript [ H subscript 3 O superscript plus ].” There is an arrow that points right to the fourth rectangle. The fourth rectangle is labeled, “Calculate x subscript [ H N O subscript 2 ] and x subscript [ N O subscript 2 superscript negative sign ].” There is an arrow that points down to the fifth rectangle. The fifth rectangle is labeled, “Calculate equilibrium concentrations.” There is an arrow that points down to the sixth rectangle. The sixth rectangle is labeled, “Calculate K subscript a.”

To get the various values in the ICE table, we first calculate [H3O+] at equilibrium from the pH:

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

The change in [H3O+], [latex]x_{[\text{H}_3\text{O}^{+}]}[/latex], is the difference between the equilibrium concentration of H3O+, which we determined from the pH, and the initial concentration. The initial concentration of H3O+ is its concentration in pure water, which is so much less than the final concentration that we approximate it as zero (~0).

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

Now we can fill in the ICE table with the concentrations at equilibrium:
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.

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

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

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

Example 11

Equilibrium Concentrations in a Solution of a Weak Acid
Formic acid, HCO2H, is the irritant that causes the body’s reaction to ant bites (Figure 11).

A photograph is shown of a large black ant on the end of a human finger.
Figure 11. The pain of an ant’s bite is caused by formic acid. (credit: John Tann)

What is 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}[/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} = \frac{[\text{H}_3\text{O}^{+}][\text{HCO}_2^{\;\;-}]}{[\text{HCO}_2\text{H}]}[/latex]
    [latex]= \frac{(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 = 0\;+\;9.7\;\times\;10^{-3}\;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]

The following example shows that the concentration of products produced by the ionization of a weak base can be determined by the same series of steps used with a weak acid.

Example 12

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

[latex](\text{CH}_3)_3\text{N}(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;(\text{CH}_3)_3\text{NH}^{+}(aq)\;+\;\text{OH}^{-}(aq)\;\;\;\;\;\;\;K_{\text{b}} = 6.3\;\times\;10^{-5}[/latex]

Solution
This problem requires that we calculate an equilibrium concentration by determining concentration changes as the ionization of a base goes to equilibrium. The solution is approached in the same way as that for the ionization of formic acid. The reactants and products will be different and the numbers will be different, but the logic will be the same:
A diagram is shown with 4 tan rectangles connected with right pointing arrows. The first is labeled “Determine the direction of change.” The second is labeled “Determine x and the equilibrium concentrations.” The third is labeled “Solve for x and the equilibrium concentrations.” The final rectangle is labeled “Check the math.”

  1. Determine x and equilibrium concentrations. The table shows the changes and concentrations
    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.
  2. Solve for x and the equilibrium concentrations. At equilibrium:
    [latex]K_{\text{b}} = \frac{[(\text{CH}_3)_3\text{NH}^{+}][\text{OH}^{-}]}{[(\text{CH}_3)_3\text{N}]} = \frac{(x)(x)}{0.25\;-\;x} = 6.3\;\times\;10^{-5}[/latex]

    If we assume that x is small relative to 0.25, then we can approximate 0.25 − x ≈ 0.25. Solving the simplified equation gives:

    [latex]x = 4.0\;\times\;10^{-3}[/latex]

    To check the validity of the approximation: 0.25 – 0.0040 = 0.25, so the assumption is justified.

    To obtain [OH] and pOH:

    [latex][\text{OH}^{-}] = {\sim}0\;+\;x = x = 4.0\;\times\;10^{-3}\;M[/latex]
    [latex]\text{pOH} = -\text{log}(4.0\;\times\;10^{-3}) = 2.40[/latex]

    Using:

    [latex]\text{pH}\;+\;\text{pOH} = \text{p}K_{\text{w}} = 14.00[/latex]

    we can also compute pH:

    [latex]\text{pH} = 14.00\;-\;\text{pOH} = 14.00\;-\;2.40 = 11.60[/latex]
  3. Check the work. A check of our arithmetic shows that Kb = 6.3 × 10−5.

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.

Answer:

7.56 × 10−4M

Example 13

Equilibrium Concentrations in a Solution of a Weak Acid
Sodium bisulfate, NaHSO4, is used in some household cleansers because it contains the HSO4, a weak acid. What is the pH of a 0.50-M solution of HSO4?

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

Solution
We need to determine the equilibrium concentration of the hydronium ion that results from the ionization of HSO4 so that we can use [H3O+] to determine the pH. As in the previous examples, we can approach the solution by the following steps:
Three rectangles are shown with right pointing arrows between them. The first is labeled “Determine x and the equilibrium concentrations.” The second is labeled “Solve for x and the equilibrium concentrations. The third is labeled “Check the math.”

  1. Determine x and equilibrium concentrations. This table shows the changes and concentrations: 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 plus sign negative x equals 0.50 minus x. The second column is blank for all three rows. The third column has the following: approximately 0, x, 0 plus sign x equals x. The fourth column has the following: 0, x, 0 plus sign x equals x.
  2. Solve for x and the concentrations. At equilibrium:
    [latex]K_{\text{a}} = 1.2\;\times\;10^{-2} = \frac{[\text{H}_3\text{O}^{+}][\text{SO}_4^{\;\;2-}]}{[\text{HSO}_4^{\;\;-}]} = \frac{(x)(x)}{0.50\;-\;x}[/latex]

    which gives:

    [latex]6.0\;\times\;10^{-3}\;-\;1.2\;\times\;10^{-2}x = x^{2}[/latex]

    or

    [latex]x^{2}\;+\;1.2\;\times\;10^{-2}x\;-\;6.0\;\times\;10^{-3} = 0[/latex]

    This equation can be solved using the quadratic formula. For an equation of the form

    [latex]ax^{2}\;+\;bx\;+\;c = 0[/latex],

    x is given by the equation:

    [latex]x = \frac{-b\;{\pm}\;\sqrt{b^{2}\;-\;4\text{ac}}}{2a}[/latex]

    In this problem, a = 1, b = 1.2 × 10−3, and c = −6.0 × 10−3.

    Solving for x gives a negative root (which cannot be correct since concentration cannot be negative) and a positive root:

    [latex]x = 7.2\;\times\;10^{-2}[/latex]

    Now determine the hydronium ion concentration and the pH:

    [latex][\text{H}_3\text{O}^{+}] = {\sim}0\;+\;x = 0\;+\;7.2\;\times\;10^{-2}\;M[/latex]
    [latex]= 7.2\;\times\;10^{-2}\;M[/latex]

    The pH of this solution is:

    [latex]\text{pH} = -\text{log}[\text{H}_3\text{O}^{+}] = -\text{log}\;(7.2\;\times\;10^{-2}) = 1.14[/latex]

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

[latex]\text{C}_8\text{H}_{10}\text{N}_4\text{O}_2(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{C}_8\text{H}_{10}\text{N}_4\text{O}_2\text{H}^{+}(aq)\;+\;\text{OH}^{-}(aq)\;\;\;\;\;\;\;K_{\text{b}} = 2.5\;\times\;10^{-4}[/latex]

(Hint: It will be necessary to convert [OH] to [H3O+] or pOH to pH toward the end of the calculation.)

Answer:

pH 11.16

Ch15.7 Relative Strengths of Strong Acids and Bases

Strong acids, such as HCl, HBr, and HI, all exhibit the same strength in water. The water molecule is such a strong base compared to the conjugate bases Cl, Br, and I that ionization of these strong acids is essentially complete in aqueous solutions. In solvents less basic than water, we find HCl, HBr, and HI differ markedly in their tendency to give up a proton to the solvent. For example, when dissolved in ethanol (a weaker base than water), the extent of ionization increases in the order HCl < HBr < HI, and so HI is demonstrated to be the strongest of these acids. The inability to discern differences in strength among strong acids dissolved in water is known as the leveling effect of water.

Water also exerts a leveling effect on the strengths of strong bases. For example, the oxide ion, O2−, and the amide ion, NH2, are such strong bases that they react completely with water:

[latex]\text{O}^{2-}(aq)\;+\;\text{H}_2\text{O}(l)\;{\longrightarrow}\;\text{OH}^{-}(aq)\;+\;\text{OH}^{-}(aq)[/latex]
[latex]\text{NH}_2^{\;\;-}(aq)\;+\;\text{H}_2\text{O}(l)\;{\longrightarrow}\;\text{NH}_3(aq)\;+\;\text{OH}^{-}(aq)[/latex]

Thus, O2− and [latex]\text{NH}_2^{\;\;-}[/latex] appear to have the same base strength in water; they both give a 100% yield of hydroxide ion.

Ch15.8 Acid Strength and Molecular Structure

If we consider a generic acid:

AH ⇌ A + H+

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

Relative Acid Strength HF HCl HBr HI
H–X Bond Energy (kJ/mol) 570 432 366 298
pKa 3.20 −6.1 −8.9 −9.3

The trend in bond energies is due to a steady decrease in overlap between the 1s orbital of hydrogen and the valence orbital of the halogen atom as the size of the halogen increases. The larger the atom to which H is bonded, the weaker the bond. Thus the bond between H and a large atom in a given family, such as I or Te, is weaker than the bond between H and a smaller atom in the same family, such as F or O. As a result, acid strengths of binary hydrides increase as we go down a column of the periodic table. For example, the order of acidity for the binary hydrides of Group 16 elements is as follows, with pKa values in parentheses:

H2O(14.00=pKw) < H2S(7.05) < H2Se(3.89) < H2Te(2.6)

The stability of the conjugate base also has an effect on the acidity. For the acid HA, the conjugate base (A) contains one more lone pair of electrons than the parent acid. Any factor that stabilizes the lone pair (i.e. the excess negative charge) on the conjugate base favors dissociation of H+ and makes the parent acid a stronger acid. Let’s see how this explains the relative acidity of the binary hydrides of the elements in the second row of the periodic table. The observed order of increasing acidity is the following, with pKa values in parentheses:

CH4(50) ≪ NH3(36) < H2O (14.00) < HF(3.20)

Consider, for example, 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 properties of the conjugate acids. For a series of compounds of the general formula HA, as the electronegativity of “A” 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 conjugate acid, acid strengths of binary hydrides increase as we go from left to right across a row of the periodic table.

The effect of stability of conjugate base is also at play when we consider the acidity of carboxylic acids. The pKa ‘s of some typical carboxylic acids are listed in the following 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 powers of ten.

Compound pKa Compound pKa
HCO2H 3.75 CH3CH2CH2CO2H 4.82
CH3CO2H 4.74 ClCH2CH2CH2CO2H 4.53
FCH2CO2H 2.65 CH3CHClCH2CO2H 4.05
ClCH2CO2H 2.85 CH3CH2CHClCO2H 2.89
BrCH2CO2H 2.90 C6H5CO2H 4.20
ICH2CO2H 3.10 p-O2NC6H4CO2H 3.45
Cl3CCO2H 0.77 p-CH3OC6H4CO2H 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? To answer this question we must return to the nature of acid-base equilibria and the definition of pKa , illustrated by the general equations given below.
acbsKeq.gif
acbspKa.gif
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. In an acid base equilibrium the equilibrium always favors the weaker acid and base (these are the more stable components). Consequently, anything that stabilizes the conjugate base (A) of an acid will necessarily make that acid (H–A) stronger and shift the equilibrium to the right. Both the carboxyl group and the carboxylate anion are stabilized by resonance, but the stabilization of the anion is much greater than that of the neutral function, as shown in the following diagram. In the carboxylate anion the two contributing structures have equal weight in the hybrid, 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, by inducing a change in the distribution of electrons within the molecule. This is called an inductive effect. This can be seen in the table above where electronegative substituents near the carboxyl group act to increase the acidity. 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:

CH3CO2H(4.76) < CH2ClCO2H(2.87) < CHCl2CO2H(1.35) < CCl3CO2H(0.66) < CF3CO2H(0.52)

As you might expect, fluorine, which is more electronegative than chlorine, causes a larger effect than chlorine, 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 large. 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 hypohalous acids (general formula HOX, with X representing a halogen) all have a hydrogen atom bonded to an oxygen atom. In aqueous solution, they all produce the following equilibrium:

HOX(aq) ⇌ H+(aq) + OX(aq)

The acidities of these acids vary by about three orders of magnitude due to the difference in electronegativity of the halogen atoms:

HOX Electronegativity of X pKa
HOCl 3.0 7.40
HOBr 2.8 8.55
HOI 2.5 10.5

As the electronegativity of X increases, the distribution of electron density within the molecule changes: 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. As shown in Figure 12, the Ka values of the oxoacids of chlorine increase by a factor of about 104 to 106 with each oxygen as successive oxygen atoms are added. The increase in acid strength with increasing number of terminal oxygen atoms is due to both an inductive effect and increased stabilization of the conjugate base.

Any inductive effect that withdraws electron density from an O–H bond increases the acidity of the compound. 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. The colors in Figure 12 show how the electrostatic potential, a measure of the strength of the interaction of a point charge at any place on the surface of the molecule, changes as the number of terminal oxygen atoms increases. In Figure 12 and Figure 13, blue corresponds to low electron densities, while red corresponds to high electron densities. The oxygen atom in the O–H unit becomes steadily less red from HClO to HClO4, while the H atom becomes steadily bluer, indicating that the electron density on the O–H unit decreases as the number of terminal oxygen atoms increases. The decrease in electron density in the O–H bond weakens it, making it easier to lose hydrogen as H+ ions, thereby increasing the strength of the acid.

Figure 12: The relationship between the acid strengths of the oxoacids of chlorine and the electron density on the O–H unit. These electrostatic potential maps show how the electron density on the O–H unit decreases 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).

At least as important is also the effect of delocalization of the negative charge in the conjugate base. As shown in Figure 13, the number of important resonance structures that can be written for the oxoanions of chlorine increases as the number of terminal oxygen atoms increases, indicating that the single negative charge is delocalized over successively more oxygen atoms.

Electron delocalization in the conjugate base increases acid strength. The electrostatic potential plots in Figure 13 demonstrate that the electron density on the terminal oxygen atoms decreases steadily as their number increases. The oxygen atom in ClO is red, indicating that it is electron rich, and the color of oxygen progressively changes to green in ClO4, indicating that the oxygen atoms are becoming steadily less electron rich through the series. 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 13: As the number of terminal oxygen atoms increases, the number of important resonance structures that can be written for the oxoanions of chlorine also increases, and the single negative charge 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.

Inductive effects are also 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 from left to right. 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.

Ch15.9 Percent Ionization

Another measure of the strength of an acid is its percent ionization. The percent ionization of a weak acid is the ratio of the concentration of the ionized acid to the initial acid concentration, times 100:

[latex]\%\;\text{ionization} = \frac{[\text{H}_3\text{O}^{+}]_{\text{eq}}}{[\text{HA}]_0}\;\times\;100[/latex]

Because the ratio includes the initial concentration, the percent ionization for a solution of a given weak acid varies depending on the original concentration of the acid, and actually decreases with increasing acid concentration.

Example 14

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]\frac{[\text{H}_3\text{O}^{+}]_{\text{eq}}}{[\text{HNO}_2]_0}\;\times\;100[/latex]

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

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

Since

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

we find that 10−2.09 = 8.1 × 10−3M, so that percent ionization is:

[latex]\frac{8.1\;\times\;10^{-3}\;M}{0.125\;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

Ch15.10 Polyprotic Acids

We can classify acids by the number of protons per molecule that they can give up in an acid-base reaction. Acids such as HCl, HNO3, and HCN that contain one ionizable hydrogen atom in each molecule are called monoprotic acids. Their reactions with water are:

[latex]\begin{array}{rl} \text{HCl}(aq)\;+\;\text{H}_2\text{O}(l) &\longrightarrow& \text{H}_3\text{O}^{+}(aq)\;+\;\text{Cl}^{-}(aq) \\[0.5em] \text{HNO}_3(aq)\;+\;\text{H}_2\text{O}(l) &\longrightarrow& \text{H}_3\text{O}^{+}(aq)\;+\;\text{NO}_3^{\;\;-}(aq) \\[0.5em] \text{HCN}(aq)\;+\;\text{H}_2\text{O}(l) &\longrightarrow& \text{H}_3\text{O}^{+}(aq)\;+\;\text{CN}^{-}(aq) \end{array}[/latex]

Even though it contains four hydrogen atoms, acetic acid, CH3CO2H, is also monoprotic because only the hydrogen atom from the carboxyl group (-COOH) reacts with bases:
This image contains two equilibrium reactions. The first shows a C atom bonded to three H atoms and another C atom. The second C atom is double bonded to an O atom and also forms a single bond to another O atom. The second O atom is bonded to an H atom. There is a plus sign and then the molecular formula H subscript 2 O. An equilibrium arrow follows the H subscript 2 O. To the right of the arrow is H subscript 3 O superscript positive sign. There is a plus sign. The final structure shows a C atom bonded the three H atoms and another C atom. This second C atom is double bonded to an O atom and single bonded to another O atom. The entire structure is in brackets and a superscript negative sign appears outside the brackets. The second reaction shows C H subscript 3 C O O H ( a q ) plus H subscript 2 O ( l ) equilibrium arrow H subscript 3 O ( a q ) plus C H subscript 3 C O O superscript negative sign ( a q ).

Similarly, monoprotic bases are bases that will accept a single proton.

Diprotic acids contain two ionizable hydrogen atoms per molecule; ionization of such acids occurs in two steps. The first ionization always takes place to a greater extent than the second ionization. For example, sulfuric acid, a strong acid, ionizes as follows:

[latex]\begin{array}{lrll} \text{First ionization:} & \text{H}_2\text{SO}_4(aq)\;+\;\text{H}_2\text{O}(l) &\rightleftharpoons& \text{H}_3\text{O}^{+}(aq)\;+\;\text{HSO}_4^{\;\;-}(aq) & K_{\text{a}1} > 10^2 \\[0.5em] \text{Second ionization:} & \text{HSO}_4^{\;\;-}(aq)\;+\;\text{H}_2\text{O}(l) &\rightleftharpoons& \text{H}_3\text{O}^{+}(aq)\;+\;\text{SO}_4^{\;\;2-}(aq) & K_{\text{a}2} = 1.2\;\times\;10^{-2} \end{array}[/latex]

This stepwise ionization process occurs for all polyprotic acids. When we make a solution of a weak diprotic acid, we get a solution that contains a mixture of acids. Carbonic acid, H2CO3, is an example of a weak diprotic acid. The first ionization of carbonic acid yields hydronium ions and bicarbonate ions in small amounts.

[latex]\text{H}_2\text{CO}_3(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{HCO}_3^{\;\;-}(aq)[/latex]
[latex]K_{\text{H}_2\text{CO}_3} = \frac{[\text{H}_3\text{O}^{+}][\text{HCO}_3^{\;\;-}]}{[\text{H}_2\text{CO}_3]} = 4.3\;\times\;10^{-7}[/latex]

The bicarbonate ion can also act as an acid. It ionizes and forms hydronium ions and carbonate ions in even smaller quantities.

[latex]\text{HCO}_3^{\;\;-}(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{CO}_3^{\;\;2-}(aq)[/latex]
[latex]K_{\text{HCO}_3^{\;\;-}} = \frac{[\text{H}_3\text{O}^{+}][\text{CO}_3^{\;\;2-}]}{[\text{HCO}_3^{\;\;-}]} = 5.6\;\times\;10^{-11}[/latex]

[latex]K_{\text{H}_2\text{CO}_3}[/latex] is larger than [latex]K_{\text{HCO}_3^{\;-}}[/latex] by a factor of 104, so H2CO3 is the dominant producer of hydronium ion in the solution. This means that little of the HCO3 formed by the ionization of H2CO3 ionizes to give hydronium ions (and carbonate ions), and the concentrations of H3O+ and HCO3 are practically equal in a pure aqueous solution of H2CO3.

If the first ionization constant of a weak diprotic acid is larger than the second by a factor of at least 20, it is appropriate to treat the first ionization separately and calculate concentrations resulting from it before calculating concentrations of species resulting from subsequent ionization. This can simplify our work considerably because we can determine the concentration of H3O+ and the conjugate base from the first ionization, then determine the concentration of the conjugate base of the second ionization in a solution with concentrations determined by the first ionization.

Example 15

Ionization of a Diprotic Acid
When we buy soda water (carbonated water), we are buying a solution of carbon dioxide in water. The solution is acidic because CO2 reacts with water to form carbonic acid, H2CO3. What are [H3O+], [HCO3], and [CO32-] in a saturated solution of CO2 with an initial [H2CO3] = 0.033 M?

[latex]\begin{array}{rll} \text{H}_2\text{CO}_3(aq)\;+\;\text{H}_2\text{O}(l) &\rightleftharpoons& \text{H}_3\text{O}^{+}(aq)\;+\;\text{HCO}_3^{\;\;-}(aq) & K_{\text{a}1} = 4.3\;\times\;10^{-7} \\[0.5em] \text{HCO}_3^{\;\;-}(aq)\;+\;\text{H}_2\text{O}(l) &\rightleftharpoons& \text{H}_3\text{O}^{+}(aq)\;+\;\text{CO}_3^{\;\;2-}(aq) & K_{\text{a}2} = 5.6\;\times\;10^{-11} \end{array}[/latex]

Solution
As indicated by the ionization constants, H2CO3 is a much stronger acid than HCO3, so H2CO3 is the dominant producer of hydronium ion in solution. Thus there are two parts in the solution of this problem: (1) Using the customary four steps, we determine the concentration of H3O+ and HCO3 produced by ionization of H2CO3. (2) Then we determine the concentration of CO32- in a solution with the concentration of H3O+ and HCO3 determined in (1). To summarize:
Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled “left bracket H subscript 2 C O subscript 3 right bracket.” The second is labeled “left bracket H subscript 3 O superscript plus right bracket and left bracket H C O subscript 3 superscript negative right bracket from H subscript 2 C O subscript 3.” The third is labeled “left bracket C O subscript 3 superscript 2 negative right bracket from H C O subscript 3 superscript negative.”

  1. Determine the concentrations of H3O+ and HCO3.
    [latex]\text{H}_2\text{CO}_3(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{HCO}_3^{\;\;-}(aq)\;\;\;\;\;\;\;K_{\text{a}1} = 4.3\;\times\;10^{-7}[/latex]

    For the ionization of weak acid:

    Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled “Determine the direction of change.” The second is labeled “Determine x and the equilibrium concentrations.” The third is labeled “Solve for x and the equilibrium concentrations.” The fourth is labeled “Check the math.”

    An ICE table shows: 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 constant ( M ). The second column has the header of “H subscript 2 C O subscript 3 plus sign H subscript 2 O equilibrium arrow H subscript 3 O superscript positive sign plus sign H C O subscript 3 superscript negative sign.” Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.033, negative sign x, 0.033 minus sign x. The second column has the following: approximately 0, x, x. The third column has the following: 0, x, x.

    Substituting in the equilibrium concentrations gives us:

    [latex]K_{\text{H}_2\text{CO}_3} = \frac{[\text{H}_3\text{O}^{+}][\text{HCO}_3^{\;\;-}]}{[\text{H}_2\text{CO}_3]} = \frac{(x)(x)}{0.033\;-\;x} = 4.3\;\times\;10^{-7}[/latex]

    Solving for x gives:

    [latex]x = 1.2\;\times\;10^{-4}[/latex]

    Thus:

    [latex][\text{H}_2\text{CO}_3] = 0.033\;M[/latex]
    [latex][\text{H}_3\text{O}^{+}] = [\text{HCO}_3^{\;\;-}] = 1.2\;\times\;10^{-4}\;M[/latex]
  2. Determine the concentration of CO32- in a solution at equilibrium with [H3O+] and [HCO3], both equal to 1.2 ×10−4 M.
    [latex]\text{HCO}_3^{\;\;-}(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{CO}_3^{\;\;2-}(aq)\;\;K_{\text{a}2} = 5.6\;\times\;10^{-11}[/latex]
    [latex]K_{\text{HCO}_3^{\;\;-}} = \frac{[\text{H}_3\text{O}^{+}][\text{CO}_3^{\;\;2-}]}{[\text{HCO}_3^{\;\;-}]} = \frac{(1.2\;\times\;10^{-4})[\text{CO}_3^{\;\;2-}]}{1.2\;\times\;10^{-4}}[/latex]
    [latex][\text{CO}_3^{\;\;2-}] = \frac{(5.6\;\times\;10^{-11})(1.2\;\times\;10^{-4})}{1.2\;\times\;10^{-4}} = 5.6\;\times\;10^{-11}\;M[/latex]

To summarize: In part 1 of this example, we found that the H2CO3 in a 0.033-M solution ionizes slightly and at equilibrium [H2CO3] = 0.033 M; [latex][\text{H}_3\text{O}^{+}] = 1.2\;\times\;10^{-4}\;M[/latex]; and [latex][\text{HCO}_3^{\;\;-}] = 1.2\;\times\;10^{-4}\;M[/latex]. In part 2, we determined that [latex][\text{CO}_3^{\;\;2-}] = 5.6\;\times\;10^{-11}\;M[/latex].

Check Your Learning
The concentration of H2S in a saturated aqueous solution at room temperature is approximately 0.1 M. Calculate [H3O+], [HS], and [S2−] in the solution:

[latex]\begin{array}{rll} \text{H}_2\text{S}(aq)\;+\;\text{H}_2\text{O}(l) &\rightleftharpoons& \text{H}_3\text{O}^{+}(aq)\;+\;\text{HS}^{-}(aq) & K_{\text{a}1} = 8.9\;\times\;10^{-8} \\[0.5em] \text{HS}^{-}(aq)\;+\;\text{H}_2\text{O}(l) &\rightleftharpoons& \text{H}_3\text{O}^{+}(aq)\;+\;\text{S}_2^{\;\;-}(aq) & K_{\text{a}2} = 1.0\;\times\;10^{-19} \end{array}[/latex]

Answer:

[H2S] = 0.1 M; [H3O+] = [HS] = 0.000094 M; [S2−] = 1 × 10−19M

We note that the concentration of the sulfide ion is the same as Ka2. This is due to the fact that each subsequent dissociation occurs to a lesser degree (as acid gets weaker).

A triprotic acid is an acid that has three dissociable protons that undergo stepwise ionization: Phosphoric acid is a typical example:

[latex]\begin{array}{lrll} \text{First ionization:} & \text{H}_3\text{PO}_4(aq)\;+\;\text{H}_2\text{O}(l) &\rightleftharpoons& \text{H}_3\text{O}^{+}(aq)\;+\;\text{H}_2\text{PO}_4^{\;\;-}(aq) & K_{\text{a}1} = 7.5\;\times\;10^{-3} \\[0.5em] \text{Second ionization:} & \text{H}_2\text{PO}_4^{\;\;-}(aq)\;+\;\text{H}_2\text{O}(l) &\rightleftharpoons& \text{H}_3\text{O}^{+}(aq)\;+\;\text{HPO}_4^{\;\;2-}(aq) & K_{\text{a}2} = 6.2\;\times\;10^{-8} \\[0.5em] \text{Third ionization:} & \text{HPO}_4^{\;\;2-}(aq)\;+\;\text{H}_2\text{O}(l) &\rightleftharpoons& \text{H}_3\text{O}^{+}(aq)\;+\;\text{PO}_4^{\;\;3-}(aq) & K_{\text{a}3} = 4.2\;\times\;10^{-13} \end{array}[/latex]

As with the diprotic acids, the differences in the ionization constants of these reactions tell us that in each successive step the degree of ionization is significantly weaker. This is a general characteristic of polyprotic acids.

This set of three dissociation reactions may appear to make calculations of equilibrium concentrations in a solution of H3PO4 complicated. However, because the successive ionization constants differ by a factor of 105 to 106, the calculations can be broken down into a series of parts similar to those for diprotic acids.

Polyprotic bases can accept more than one hydrogen ion in solution. The carbonate ion is an example of a diprotic base, since it can accept up to two protons. Solutions of alkali metal carbonates are quite alkaline, due to the reactions:

[latex]\begin{array}{rl} \text{H}_2\text{O}(l)\;+\;\text{CO}_3^{\;\;2-}(aq) &\rightleftharpoons& \text{HCO}_3^{\;\;-}(aq)\;+\;\text{OH}^{-}(aq) \\[0.5em] \text{H}_2\text{O}(l)\;+\;\text{HCO}_3^{\;\;-}(aq) &\rightleftharpoons& \text{H}_2\text{CO}_3(aq)\;+\;\text{OH}^{-}(aq) \end{array}[/latex]

Ch15.11 Strong Acid Reacting with Strong Base

When we mix solutions of an acid and a base, an acid-base neutralization reaction occurs.

A strong acid and a strong base, such as HCl(aq) and NaOH(aq), will react to form a neutral solution if we mix stoichiometrically equivalent quantities, since the conjugate partners produced are of negligible strength:

[latex]\text{HCl}(aq)\;+\;\text{NaOH}(aq)\;{\rightleftharpoons}\;\text{NaCl}(aq)\;+\;\text{H}_2\text{O}(l)[/latex]

A solution is neutral when it contains equal concentrations of hydronium and hydroxide ions.

Stomach Antacids

Our stomachs contain a solution of roughly 0.03 M HCl, which helps us digest the food we eat. The burning sensation associated with heartburn is a result of the stomach acid leaking through the muscular valve at the top of the stomach into the lower reaches of the esophagus. The lining of the esophagus is not protected from the corrosive effects of stomach acid the way the lining of the stomach is, and the results can be very painful. When we have heartburn, it feels better if we reduce the excess acid in the esophagus by taking an antacid. As you may have guessed, antacids are bases. One of the most common antacids is calcium carbonate, CaCO3. The reaction

[latex]\text{CaCO}_3(s)\;+\;2\text{HCl}(aq)\;{\rightleftharpoons}\;\text{CaCl}_2(aq)\;+\;\text{H}_2\text{O}(l)\;+\;\text{CO}_2(g)[/latex]

not only neutralizes stomach acid, it also produces CO2(g), which may result in a satisfying belch.

Milk of Magnesia is a suspension of the sparingly soluble base magnesium hydroxide, Mg(OH)2. It works according to the reaction:

[latex]\text{Mg(OH)}_2(s)\;{\rightleftharpoons}\;\text{Mg}^{2+}(aq)\;+\;2\text{OH}^{-}(aq)[/latex]

The hydroxide ions generated in this equilibrium then go on to react with the hydronium ions from the stomach acid, so that :

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

This reaction does not produce carbon dioxide, but magnesium-containing antacids can have a laxative effect.

Several antacids have aluminum hydroxide, Al(OH)3, as an active ingredient. The aluminum hydroxide tends to cause constipation, and some antacids use aluminum hydroxide in concert with magnesium hydroxide to balance the side effects of the two substances.

Ch15.12 Strong Acid Reacting with Weak Base

When we mix a weak base with a strong acid, the product is a salt containing the conjugate acid of the weak base. This conjugate acid is a weak acid. For example, ammonium chloride, NH4Cl, is a salt formed by the reaction of the weak base ammonia with the strong acid HCl:

[latex]\text{NH}_3(aq)\;+\;\text{HCl}(aq)\;{\longrightarrow}\;\text{NH}_4\text{Cl}(aq)[/latex]

A solution of this salt contains ammonium ions and chloride ions. The chloride ion has no effect on the acidity of the solution since HCl is a strong acid. Chloride is a very weak base and will not accept a proton to a measurable extent. However, the ammonium ion, the conjugate acid of ammonia, reacts with water and increases the hydronium ion concentration:

[latex]\text{NH}_4^{\;\;+}(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{NH}_3(aq)[/latex]

The equilibrium equation for this reaction is simply the ionization constant, Ka, for the acid NH4+:

[latex]\frac{[\text{H}_3\text{O}^{+}][\text{NH}_3]}{[\text{NH}_4^{\;\;+}]} = K_{\text{a}}[/latex]

Example 16

The pH of a Solution of a Weak Base and a Strong Acid
Aniline is an amine that is used to manufacture dyes. It is isolated as aniline hydrochloride, [C6H5NH3]Cl, a salt prepared by the reaction of the weak base aniline and hydrochloric acid. What is the pH of a 0.233 M solution of aniline hydrochloride? (Kb for aniline, C6H5NH2, is 4.3 × 10−10)

[latex]\text{C}_6\text{H}_5\text{NH}_3^{\;\;+}(aq)\;+\;\text{H}_2\text{O}(l)\;{\leftrightharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{C}_6\text{H}_5\text{NH}_2(aq)[/latex]

Solution
The new step in this example is to determine Ka for the C6H5NH3+ ion, the conjugate acid of a weak base. We can determine value of Ka for this acid from the value of Kb for aniline:

[latex]K_{\text{a}}\;(\text{for\;C}_6\text{H}_5\text{NH}_3^{\;\;+})\;\times\;K_{\text{b}}\;(\text{for\;C}_6\text{H}_5\text{NH}_2) = K_{\text{w}} = 1.0\;\times\;10^{-14}[/latex]
[latex]K_{\text{a}}\;(\text{for\;C}_6\text{H}_5\text{NH}_3^{\;\;+}) = \frac{K_{\text{w}}}{K_{\text{b}}\;(\text{for\;C}_6\text{H}_5\text{NH}_2)} = \frac{1.0\;\times\;10^{-14}}{4.3\;\times\;10^{-10}} = 2.3\;\times\;10^{-5}[/latex]

Now we have the ionization constant and the initial concentration of the weak acid, the information necessary to determine the equilibrium concentration of H3O+, and the pH:
Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled “Determine the direction of change.” The second is labeled “Determine x and the equilibrium concentrations.” The third is labeled “Solve for x and the equilibrium concentrations.” The fourth is labeled “Check the math.”

With these steps we find [H3O+] = 2.3 × 10−3M and pH = 2.64

Check Your Learning
(a) Do the calculations and show that the hydronium ion concentration for a 0.233-M solution of C6H5NH3+ is 2.3 × 10−3 and the pH is 2.64.

(b) What is the hydronium ion concentration in a 0.100-M solution of ammonium nitrate, NH4NO3, a salt composed of the ions NH4+ and NO3. The Kb for ammonia is 1.8 × 10−5. Which is the stronger acid C6H5NH3+ or NH4+?

Answer:

(a) Ka(forNH4+) = 5.6 × 10-10, [H3O+] = 7.5 × 10−6 M; (b) C6H5NH3+ is the stronger acid.

Ch15.13 Weak Acid Reacting with Strong Base

When we mix a weak acid with a strong base, we get a salt that contains the conjugate base of the weak acid. This conjugate base is usually a weak base. For example, sodium acetate, NaCH3CO2, is a salt formed by the reaction of the weak acid acetic acid with the strong base sodium hydroxide:

[latex]\text{CH}_3\text{CO}_2\text{H}(aq)\;+\;\text{NaOH}(aq)\;{\longrightarrow}\;\text{NaCH}_3\text{CO}_2(aq)\;+\;\text{H}_2\text{O}(aq)[/latex]

A solution of this salt contains sodium ions and acetate ions. The sodium ion, as the conjugate acid of a strong base, has no effect on the acidity of the solution. However, the acetate ion, the conjugate base of acetic acid, reacts with water and increases the concentration of hydroxide ion:

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

The equilibrium equation for this reaction is the ionization constant, Kb, for the base CH3CO2.

Some reference tables do not report values of Kb. They only report ionization constants for acids. The value of Kb can be calculated from the value of the ionization constant of water, Kw, and Ka, the ionization constant of the conjugate acid of the base.

Example 17

Equilibrium in a Solution of a Weak Acid and a Strong Base
Determine the acetic acid concentration in a solution with [CH3CO2] = 0.050 M and [OH] = 2.5 × 10−6M at equilibrium. (Ka for acetic acid is 1.8 × 10-5) The reaction is:

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

Solution
We are given two of three equilibrium concentrations and asked to find the missing concentration. If we can find the equilibrium constant for the reaction, the process is straightforward.

The acetate ion behaves as a base in this reaction; hydroxide ions are a product. We determine Kb as follows:

[latex]K_{\text{b}}\;(\text{for\;CH}_3\text{CO}_2^{\;\;-}) = \frac{K_{\text{w}}}{K_{\text{a}}\;(\text{for\;CH}_3\text{CO}_2\text{H})} = \frac{1.0\;\times\;10^{-14}}{1.8\;\times\;10^{-5}} = 5.6\;\times\;10^{-10}[/latex]

Now find the missing concentration:

[latex]K_{\text{b}} = \frac{[\text{CH}_3\text{CO}_2\text{H}][\text{OH}^{-}]}{[\text{CH}_3\text{CO}_2^{\;\;-}]} = 5.6\;\times\;10^{-10}[/latex]
[latex]= \frac{[\text{CH}_3\text{CO}_2\text{H}](2.5\;\times\;10^{-6})}{(0.050)} = 5.6\;\times\;10^{-10}[/latex]

Solving this equation we get [CH3CO2H] = 1.1 × 10−5M.

Check Your Learning
What is the pH of a 0.083-M solution of CN? Use 4.9 × 10−10 as Ka for HCN. Hint: We will probably need to convert pOH to pH or find [H3O+] using [OH] in the final stages of this problem.

Answer:

11.11

Ch15.14 Weak Acid Reacting with Weak Base

In a solution of a salt formed by the reaction of a weak acid and a weak base, to predict the pH, we must know both the Ka of the weak acid and the Kb of the weak base. If Ka > Kb, the solution is acidic, and if Kb > Ka, the solution is basic.

Example 18

Determining the Acidic or Basic Nature of Salts
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:

(a) KBr

(b) NaHCO3

(c) NH4Cl

(d) Na2HPO4

(e) NH4F

Solution
Consider each of the ions separately in terms of its effect on the pH of the solution:

(a) The K+ cation and the Br anion are both spectators, since they are the conjugate acid of a strong base (KOH) and the conjugate base of a strong acid (HBr), respectively. The solution is neutral.

(b) The Na+ cation is a spectator, and will not affect the pH of the solution. The HCO3 anion is amphiprotic, it could either behave as an acid or a base. The Ka of HCO3 is 4.7 × 10−11, the Kb of HCO3 (from Ka of H2CO3) is [latex]\frac{1.0\;\times\;10^{-14}}{4.3\;\times\;10^{-7}} = 2.3\;\times\;10^{-8}[/latex].

Since Kb > Ka, the solution is basic.

(c) The NH4+ ion is acidic and the Cl ion is a spectator. The solution will be acidic.

(d) The Na+ ion is a spectator. while the HPO42- is amphiprotic, with a Ka of 4.2 × 10−13 and a Kb of 1.6 × 10^-7. Because Kb > Ka, the solution is basic.

(e) The NH4+ ion is acidic, and the F ion is a base, so we must directly compare the Ka and the Kb of the two ions. Ka of NH4+ is 5.6 × 10−10, which seems very small, but the Kb of F is 1.4 × 10−11, even smaller. So the solution is acidic, since Ka > Kb.

Check Your Learning
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:

(a) K2CO3

(b) CaCl2

(c) KH2PO4

(d) (NH4)2CO3

(e) AlBr3

Answer:

(a) basic; (b) neutral; (c) acidic; (d) basic; (e) acidic

Ch15.15 Reaction Between Amphiprotic Species

Acid-base reactions can also occur between two amphiprotic species. For example, when you mix hydrogen sulfite (HSO3) and bicarbonate (HCO3) solutions, acid-base reaction will occur. But if both species can act as either an acid or a base, how do you tell which species will act as which? In the example mixture, we can envision two scenarios:

scenario I:     HSO3(aq) + HCO3(aq) ⇌  SO32-(aq) + H2CO3(aq)
scenario II:     HSO3(aq) + HCO3(aq) ⇌  H2SO3(aq) + CO32-(aq)

Quantitatively, we can make use of the ionization constants to determine which set of products is more favored and hence which scenario is the dominant one. In scenario I, we have:

HSO3(aq) + H2O(l) ⇌  SO32-(aq) + H3O+(aq)     [latex]K_1 = K_{\text{a, HSO}_3^-} = 6.3\;\times\;10^{-8}[/latex]
HCO3(aq) + H3O+(aq) ⇌  H2CO3(aq) + H2O(l)     [latex]K_2 = \dfrac{1}{K_{\text{a, H}_2\text{CO}_3}} = \dfrac{1}{4.3\;\times\;10^{-7}}[/latex]

Hence, the reaction HSO3(aq) + HCO3(aq) ⇌  SO32-(aq) + H2CO3(aq) has a total equilibrium constant of:

Ktotal, I = K1 × K2 = (6.3 × 10-8)/(4.3 × 10-7) = 0.15

In scenario II, we have:

HSO3(aq) + H3O+(aq) ⇌ H2SO3(aq) + H2O(l)      [latex]K_1 = \dfrac{1}{K_{\text{a, H}_2\text{SO}_3}} = \dfrac{1}{1.7\;\times\;10^{-2}}[/latex]
HCO3(aq) + H2O(l) ⇌ CO32-(aq) + H3O+(aq)     [latex]K_2 = K_{\text{a, HCO}_3^-} = 4.7\;\times\;10^{-11}[/latex]

Hence, the reaction HSO3(aq) + HCO3(aq) ⇌  H2SO3(aq) + CO32-(aq) has a total equilibrium constant of:

Ktotal, II = K1 × K2 = (4.7 × 10-11)/(1.7 × 10-2) = 2.8 × 10-9

Scenario I is significantly more product-favored than scenario II, Ktotal, I >> Ktotal, II. Therefore, we would expect HSO3 to act as the acid and to HCO3 act as the base.

Qualitatively, H2CO3 (with Ka = 4.3 × 10-7) is less acidic than H2SO3 (with Ka = 1.7 × 10-2); SO32- (with Kb = 1.6 × 10-7) is less basic than CO32- (with Kb = 2.1 × 10-4). Hence, the reaction leading to the formation of H2CO3 and SO32- (scenario I) is thermodynamically more favorable than the reaction leading to the formation of H2SO3 and CO32- (scenario II).

Ch15.16 Amino Acids

Another important group of amphiprotic species is the amino acids. Each amino acid molecule contains an carboxyl group and an amino group. We have seen that carboxylic acids are moderately acidic. Most of them have pKa in the range of 3 to 5. We have also seen that organic amines are somewhat basic, and are easily protonated if protons are available.

Because the carboxylic acid is a pretty good source of protons and because protons bind to amines pretty well, it seems reasonable that a proton transfer may occur from one site to the other.
realalanine.gif
Does one of these forms dominate the equilibrium? Compare the pKa values. The pKa of the acid is near 5, and the pKa of the ammonium is near 9. The ammonium holds the proton more tightly than does the acid. The proton stays on the nitrogen.

This is analogous to the acid-base reaction between methylamine (Kb = 4.4 × 10-4) and acetic acid (Ka = 1.8 × 10-5):

CH3NH2(aq) + CH3COOH(aq) ⇌  CH3NH3+(aq) + CH3COO(aq)

where the equilibrium would favor the product-side because:

[latex]K_{\text{total}} = \dfrac{1.8\;\times\;10^{-5}}{2.3\;\times\;10^{-11}} = 7.8\;\times\;10^{5}[/latex]

In fact, amino acids usually exist in zwitterion (German for “double ion”) form, where the proton has transferred from the carboxyl to the amino group. A zwitterion is a compound with no overall electrical charge, but which contains separate parts which are positively and negatively charged.

If you increase the pH of a solution of an amino acid by adding hydroxide ions, the hydrogen ion is removed from the -NH3+ group.

zwittbase.gif

If you decrease the pH by adding an acid to a solution of an amino acid, the -COO part of the zwitterion picks up a hydrogen ion.
zwittacid.gif

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