28 Autoionization and the pH Scale (M15Q4)

Learning Objectives

  • Recognize that water can autoionize and describe the relationship between [H+], [OH–] and Kw.
  • Apply the relationship between the pH scale and [H+] and the relationship between pH and pOH as appropriate.
    | pH Scale |
  • Calculate the pH of strong and weak acid and base solutions.
    | pH Calculations |

| Key Concepts and Summary | Key Equations | Glossary | End of Section Exercises |

In the preceding sections, we 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.

Autoionization of Water

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

HA(aq)  +  H2O(l)   ⇌   H3O+(aq)  +  A–(aq)      Ka = [latex]\frac{[H_{3}O^{+}][A^{-}]}{[HA]}[/latex]
A–(aq)  +  H2O(l)   ⇌   OH–(aq)  +  HA(aq)    Kb = [latex]\frac{[OH^{-}][HA]}{[A^{-}]}[/latex]

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

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

2 H2O(l)   ⇌   H3O+(aq)  +  OH–(aq)

As shown in the previous chapter on equilibrium, the K expression for a chemical equation derived from adding two or more other chemical equations is the product of the other equations’ K expressions. Multiplying the equilibrium expressions together and cancelling common terms, we see that:

Ka × Kb = [latex]  \frac{[H_{3}O^{+}][A^{-}]}{[HA]} \times \frac{[OH^{-}][HA]}{[A^{-}]}[/latex] = [H3O+][OH–] = Kw

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 called the ion-product constant for water (Kw):

Ka × Kb = (1.8 × 10-5) × (5.6 × 10-10) = 1.0 × 10-14 = Kw

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

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

Kw = [H3O+][OH–] = [x]2 = 1.0 × 10-14

So:

[H3O+] = [OH–] =[latex]\sqrt{1.0 \times 10^{-14}}[/latex]=  1.0 × 10-7 M

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

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

Answer:

[H3O+] = [OH−] = 4.9 × 10−7 M

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 a hydronium ion concentration of 2.0 × 10−6 M. What is the concentration of hydroxide ion at 25 °C?

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

2 H2O(l)   ⇌   H3O+(aq)  +  OH–(aq)

Kw = [H3O+][OH–] = 1.0 × 10-14

Thus, we can calculate the missing equilibrium concentration.

Rearrangement of the Kw expression yields that [OH−] is directly proportional to the inverse of [H3O+]:

[OH–] = [latex]\frac{K_{w}}{[H_{3}O^{+}]}=\frac{1.0\times10^{-14}}{2.0\times10^{-6}}[/latex] = 5.0 × 10-9

The hydroxide ion concentration in water is reduced to 5.0 × 10−9 M as the hydrogen 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:

Kw = [H3O+][OH–] = [2.0 × 10-6][5.0 × 10-9]= 1.0 × 10-14

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

Answer:

[H3O+] = 1 × 10−11 M

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

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

pX = -logX

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

pH = -log[H3O+]

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

[H3O+] = 10-pH

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

pOH = -log[OH–]

or

[OH–] = 10-pOH

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

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

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

14.00 = pH + pOH
All of these equations are connected in such a way that if you have the [H3O+], [OH–], pH, or pOH, you can solve for the other three:

This figure is a square with four double-headed arrows pointing between the four corners along the edges. The top left corner is labeled [H 3 O +]. The arrow moving left is labeled "[H 3 O +] [O H -] = 1.0 times 10 ^ -14". The top right corner is labeled [O H -]. The arrow pointing down is labeled "p O H = - log [O H -]." The bottom right corner is labeled p O H. The arrow pointing left is labeled "p H + p O H = 14.00". The bottom left corner is labeled p H. The arrow pointing up is labeled "p H = - log [ H 3 O +]".

As was shown above in Example 1, the hydronium ion molarity in pure water (or any neutral solution) is 1.0 × 10−7M at 25 °C. The pH and pOH of a neutral solution at this temperature are therefore:

pH = -log[H3O+] = -log[1.0 × 10−7] = 7.00
pOH = -log[OH–] = -log[1.0 × 10−7] = 7.00

And so, at this temperature, acidic solutions are those with hydronium ion molarities greater than 1.0 × 10−7 M and hydroxide ion molarities less than 1.0 × 10−7 M (corresponding to pH values less than 7.00 and pOH values greater than 7.00). Basic solutions are those with hydronium ion molarities less than 1.0 × 10−7 M and hydroxide ion molarities greater than 1.0 × 10−7 M (corresponding to pH values greater than 7.00 and pOH values less than 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 “Check Your Learning” exercise accompanying Example 1 in this section showed the hydronium molarity of pure water at 80 °C is 4.9 × 10−7 M, which corresponds to pH and pOH values of:

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

At this temperature, then, neutral solutions exhibit pH = pOH = 6.31, acidic solutions exhibit pH less than 6.31 and pOH greater than 6.31, whereas basic solutions exhibit pH greater than 6.31 and pOH less than 6.31. This distinction can be important when studying certain processes that occur at 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).

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

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 scales represent concentrations of [H3O+] and OH−, respectively. The pH and pOH values of some common substances at standard temperature (25 °C) are shown in this chart.

Example 3

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

Solution

pH = -log[H3O+]
= -log(1.2 × 10-3)
=-(-2.92) = 2.92

(The use of logarithms is explained in Appendix B. Recall that, as we have done here, 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:

CO2(aq)  +  H2O(l)   ⇌   H2CO3(aq)

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 4

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

Solution

pH = -log[H3O+] = 7.3
log[H3O+] = -7.3
[H3O+] = 10-7.3
[H3O+] = 5 × 10-8 M

 

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

Answer:

12 M

Example 5

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

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

pOH = -log[OH–] = -log[0.0125]
= -(-1.903) = 1.903

The pH can be found from the pOH:

pH + pOH = 14.000
pH = 14.000 – pOH = 14.000 – 1.903 = 12.097

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

Answer:

pOH = 11.6, pH = 2.4

The acidity of a solution is typically assessed experimentally by measurement of its pH. The pOH of a solution is not usually measured, as it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter (Figure 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 of around $100. (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)

Chemistry in Real Life: Environmental Science

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

H2O(l)  +  CO2(g)   →   H2CO3(aq)
H2CO3(aq)   ⇌   H+(aq)  +  HCO3–(aq)

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:

H2O(l)  +  SO3(g)   →   H2SO4(aq)
H2SO4(aq)   →   H+(aq)  +  HSO4–(aq)

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 5). 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 5. (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)

Key Concepts and Summary

Although we can conceptually determine which of two molecules is a stronger acid or base, quantitative methods are even more robust. In this section, we introduce the concept of p_, where p stands for the -log of a quantity. The pH is the -log[H3O+], where the [H3O+] is the concentration of hydronium at equilibrium. The same is true for pOH, where pOH = -log[OH–]. If one of these four variables are known (pH, pOH, [H3O+], or [OH–]), the other three can be calculated using the equations below. The lower the pH, the more acidic the solution is and the higher the pH, the less acidic. The lower the pOH, the more basic the solution is and the higher the pOH, the more basic.

Key Equations

  • Ka × Kb = [H3O+][OH–] = Kw
  • 14.00 = pH + pOH
  • pH = -log[H3O+]
  • [H3O+] = 10-pH
  • pOH = -log[OH–]
  • [OH–] = 10-pOH

Glossary

acidic
a solution where [H3O+] > [OH–]

autoionization
a reaction where a two of the same molecule react and a proton transfers between them

basic
a solution where [H3O+] < [OH–]

ion-product constant for water (Kw)
the equilibrium constant for the autoionization of water, 2H2O(l) ⇌ H3O+(aq) + OH–(aq)

neutral
a solution where [H3O+] = [OH–]

Chemistry End of Section Exercises

  1. Explain why a sample of pure water at 40 °C is neutral even though [H3O+] = 1.7 × 10−7 M. Kw is 2.9 × 10−14 at 40 °C.
  2. The ionization constant for water (Kw) is 2.9 × 10−15 at 10 °C. Calculate [H3O+], [OH−], pH, and pOH for pure water at 10 °C.
  3. The ionization constant for water (Kw) is 9.311 × 10−14 at 60 °C. Calculate [H3O+], [OH−], pH, and pOH for pure water at 60 °C.
  4. Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
    1. 0.200 M HCl
    2. 0.0143 M NaOH
    3. 3.0 M HNO3
    4. 0.0031 M Ca(OH)2
  5. What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52?
  6. The hydronium ion concentration in a sample of rainwater is found to be 1.7 × 10−6 M at 25 °C. What is the concentration of hydroxide ions in the rainwater?

Answers to Chemistry End of Section Exercises

  1. In a neutral solution [H3O+] = [OH−]. At 40 °C, [H3O+] = [OH−] = (2.9 × 10−14)1/2 = 1.7 × 10−7 M.
  2. [H3O+] = 5.4 × 10-8 M, [OH−] = 5.4 × 10-8 M, pH = 7.27, and pOH = 7.27
  3. [H3O+] = 3.051 × 10-7, [OH−] = 3.051 × 10-7 M, pH = 6.5156, and pOH = 6.5156
  4. (a) pH = 0.699, pOH = 13.301 (b) pH = 12.155, pOH = 1.845, (c) pH = -0.48, pOH = 14.48, (d) pH = 11.79, pOH = 2.21
  5. [H3O+] = 3.0 × 10-7 M, [OH–] = 3.3 × 10-8 M
  6. [OH–] = 5.9 × 10-9 M

License

Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

CHEM 104: Working Copy Copyright © by Chem 104 Textbook Team is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

Share This Book

Feedback/Errata

Leave a Reply

Your email address will not be published. Required fields are marked *