Chapter 15: Acid-Based Equilibria
Learning Outcomes
- Explain the characterization of aqueous solutions as acidic, basic, or neutral
- Express hydronium and hydroxide ion concentrations on the pH and pOH scales
- Perform calculations relating pH and pOH
As discussed earlier, hydronium and hydroxide ions are present both in pure water and in all aqueous solutions, and their concentrations are inversely proportional as determined by the ion product of water (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:
[latex]\text{pX} = −\text{log X}[/latex]
The pH of a solution is therefore defined as shown here, where [latex]\ce{H2O+}[/latex] is the molar concentration of hydronium ion in the solution:
[latex]\text{pH} = −\text{log}[\ce{H3O+}][/latex]
Rearranging this equation to isolate the hydronium ion molarity yields the equivalent expression:
[latex][\ce{H3O+}] = 10^{-\text{pH}}[/latex]
Likewise, the hydroxide ion molarity may be expressed as a p-function, or pOH:
[latex]\text{pOH} = −\text{log}[\ce{OH-}][/latex]
or
[latex][\ce{OH-}] = 10^{-\text{pOH}}[/latex]
Finally, the relation between these two ion concentration expressed as p-functions is easily derived from the [latex]K_{\text{w}}[/latex] expression:
[latex]K_{\text{w}}= [\ce{H3O+}][\ce{OH-}][/latex]
[latex]\text{-log}K_{\text{w}}=\text{-log}(\ce{[H3O+]}\ce{[OH-]})=\text{-log}\ce{[H3O+]}+\text{-log}\ce{[OH-]}[/latex]
[latex]\text{p}K_{\text{w}}=\text{pH}+\text{pOH}[/latex]
At 25 °C, the value of [latex]K_{\text{w}}[/latex] is [latex]1.0 \times {10}^{−14}[/latex], and so:
[latex]14.00 = \text{pH} +\text{pOH}[/latex]
As we learned earlier, the hydronium ion molarity in pure water (or any neutral solution) is [latex]1.0 \times {10}^{-7} M[/latex] at 25 °C. The pH and pOH of a neutral solution at this temperature are therefore:
[latex]\text{pH}=−\text{log}\ce{[H3O+]}=−\text{was log}(1.0 \times{10}^{-7}) = 7.00[/latex]
[latex]\text{pOH} = −\text{log}[\ce{OH-}] = −\text{log}(1.0 \times{10}^{-7}) = 7.00[/latex]
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 hydronium molarity of pure water is 80 °C is 4.9 × 10−7 M, which corresponds to pH and pOH values of:
[latex]\text{pH}=−\text{log}[\ce{H3O+}]=−\text{log}(4.9 \times{10}^{-7}) = 6.31[/latex]
[latex]\text{pOH} = −\text{log}[\ce{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 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 15.2.1).
Figure 15.2.1 shows the relationships between [latex]\ce{[H3O+]}[/latex], [latex]\ce{[OH-]}[/latex], pH, and pOH, and gives values for these properties at standard temperatures for some common substances.
Example 15.2.1: 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?
[reveal-answer q=”510167″]Show Solution[/reveal-answer]
[hidden-answer a=”510167″]
pH = −log[H3O+]
= −log(1.2 × 10–3)
= −(−2.92) = 2.92
(The use of logarithms is explained in Essential Mathematics. 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.)
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Check Your Learning
Example 15.2.2: Calculation of Hydronium Ion Concentration from pH
Calculate the hydronium ion concentration of blood, the pH of which is 7.3 (slightly alkaline).
[reveal-answer q=”928286″]Show Solution[/reveal-answer]
[hidden-answer a=”928286″]
pH = −log[latex]\ce{H3O+}[/latex] = 7.3
log[H3O+] = −7.3
[H3O+] = 10–7.3 or [latex]\ce{H3O+}[/latex] = antilog of −7.3
[H3O+] = 5 × 10–8 M
(On a calculator take the antilog, or the “inverse” log, of −7.3, or calculate 10–7.3.)
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Check Your Learning
Environmental Science
Normal rainwater has a pH between 5 and 6 due to the presence of dissolved [latex]\ce{CO2}[/latex] which forms carbonic acid:
[latex]\ce{H2O}(l) + \ce{CO2}(g) \ce{->} \ce{H2CO3}(aq)[/latex]
[latex]\ce{H2CO3}(aq) \ce{<=>} \ce{H+}(aq) + \ce{HCO3-}(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]\ce{H2O}(l) + \ce{SO3}(g) \ce{-> H2SO4}(aq)[/latex]
[latex]\ce{H2SO4}(aq) \ce{-> H+}(aq) + \ce{H2SO2-}(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 15.2.2). 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 US Environmental Protection Agency website.
Example 15.2.3: Calculation of pOH
What are the pOH and the pH of a 0.0125-M solution of potassium hydroxide, [latex]\ce{KOH}[/latex]?
[reveal-answer q=”47438″]Show Solution[/reveal-answer]
[hidden-answer a=”47438″]
Potassium hydroxide is a highly soluble ionic compound and completely dissociates when dissolved in dilute solution, yielding [[latex]\ce{OH-}[/latex] = 0.0125 M:
pOH = −log[latex]\ce{OH-}[/latex] = −log 0.0125
= −(−1.903) = 1.903
The pH can be found from the pOH:
pH + pOH = 14.00
pH = 14.00 − pOH = 14.00 − 1.903 = 12.10
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Check Your Learning
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 15.2.3).
The pH of a solution may also be visually estimated using colored indicators (Figure 15.2.4).
Key Concepts and Summary
Concentrations of hydronium and hydroxide ions in aqueous media are often represented as logarithmic pH and pOH values, respectively. At 25 °C, the autoprotolysis equilibrium for water requires the sum of pH and pOH to equal 14 for any aqueous solution. The relative concentrations of hydronium and hydroxide ion in a solution define its status as acidic ([latex]\ce{[H3O+] > [OH-]}[/latex]), basic ([latex]\ce{[H3O+] < [OH-]}[/latex]), or neutral ([latex]\ce{[H3O+] = [OH-]}[/latex]). At 25 °C, a pH < 7 indicates an acidic solution, a pH > 7 a basic solution, and a pH = 7 a neutral solution.
Key Equations
- [latex]\text{pH} = −\text{log}[\ce{H3O+}][/latex]
- [latex]\text{pOH} = −\text{log}[\ce{OH-}][/latex]
- [latex][\ce{H3O+}] = 10^{-\text{pH}}[/latex]
- [latex][\ce{OH-}] = 10^{-\text{pOH}}[/latex]
- [latex]\text{p}K_{\text{w}}=\text{pH}+\text{pOH}=14.00[/latex] at 25 °C
Try It
- Explain why a sample of pure water at 40 °C is neutral even though [latex]\ce{[H3O+]}[/latex] = 1.7 × 10−7 M. Kw is 2.9 × 10−14 at 40 °C.
- The ionization constant for water (Kw) is 9.614 × 10−14 at 60 °C. Calculate [latex]\ce{[H3O+]}[/latex], [latex]\ce{[OH-]}[/latex], pH, and pOH for pure water at 60 °C.
- Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
- 0.000259 M [latex]\ce{HClO4}[/latex]
- 0.21 M [latex]\ce{NaOH}[/latex]
- 0.000071 M [latex]\ce{Ba(OH)2}[/latex]
- 2.5 M [latex]\ce{KOH}[/latex]
- The hydroxide ion concentration in household ammonia is 3.2 × 10−3 M at 25 °C. What is the concentration of hydronium ions in the solution?
[reveal-answer q=”188973″]Show Selected Solutions[/reveal-answer]
[hidden-answer a=”188973″]
- In a neutral solution [latex]\ce{[H3O+] = [OH-]}[/latex]. At 40 °C, [latex]\ce{[H3O+] = [OH-]}[/latex] = (2.910−14)1/2 = 1.7 × 10−7.
- x = 3.101 × 10−7 M = [latex]\ce{[H3O+] = [OH-]}[/latex]
pH = −log3.101 × 10−7 = −(−6.5085) = 6.5085
pOH = pH = 6.5085 - The pH and the pOH are as follows:
- pH = 3.587; pOH = 10.413;
- pH = 0.68; pOH = 13.32;
- pOH = 3.85; pH = 10.15;
- pH = −0.40; pOH = 14.4
- [latex]\ce{[OH-]}[/latex] = 3.1 × 10−12 M
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Glossary
acidic: describes a solution in which [latex]\ce{[H3O+] > [OH-]}[/latex]
basic: describes a solution in which [latex]\ce{[H3O+] < [OH-]}[/latex] neutral: describes a solution in which [latex]\ce{[H3O+] = [OH-]}[/latex]
pH: logarithmic measure of the concentration of hydronium ions in a solution
pOH: logarithmic measure of the concentration of hydroxide ions in a solution
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- Chemistry 2e. Provided by: OpenStax. Located at: https://openstax.org/. License: CC BY: Attribution. License Terms: Access for free at
https://openstax.org/books/chemistry-2e/pages/1-introduction
describes a solution in which [latex]\ce{[H3O+] = [OH-]}[/latex]
describes a solution in which [latex]\ce{[H3O+] > [OH-]}[/latex]
describes a solution in which [latex]\ce{[H3O+] < [OH-]}[/latex]
logarithmic measure of the concentration of hydronium ions in a solution
logarithmic measure of the concentration of hydroxide ions in a solution