Chapter 15: Acid-Based Equilibria
Chapter 15 Practice
15.1 Brønsted-Lowry Acids and Bases [Go to section 15.1]
- What is the conjugate acid of each of the following? What is the conjugate base of each?
- Write equations that show
acting both as an acid and as a base. - What are amphiprotic species? Illustrate with suitable equations.
- Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid:
- State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species.
- Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base:
- Identify the conjugate acids of the species in the previous problem.
- Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
- Show by suitable net ionic equations that the following species are amphoteric:
- Is the self ionization of water endothermic or exothermic? The ionization constant for water (Kw) is 2.9 × 10–14 at 40 °C and 9.3 × 10–14 at 60 °C.
Show Selected Solutions
- The answers are as follows:
- Amphiprotic species may either gain or lose a proton in a chemical reaction, thus acting as a base or an acid. An example is
. As an acid: . As a base: - The answers are as follows:
Not Amphiprotic:
- The answers are as follows:
- The answers are as follows:
; ;
15.2 pH and pOH [Go to section 15.2]
- What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52?
- The ionization constant for water (Kw) is 2.9 × 10–14 at 40 °C. Calculate ,
, pH, and pOH for pure water at 40 °C. - Calculate the hydronium ion concentration and the hydroxide ion concentration in lime juice from its pH. See Figure 15.2.1 for useful information.
- Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
- 0.200 M
- 0.0143 M
- 3.0 M
- 0.0031 M
- 0.200 M
- Calculate the hydronium ion concentrations for the following solutions:
, pH = 2.50 , pH = 8.55 , pH = 1.02 , pH = 11.00
- Calculate the hydrogen ion concentration and the hydroxide ion concentration in wine from its pH. See Figure 15.2.1 for useful information.
- Calculate the hydroxide ion concentration for the solutions in problem number 15.
15.3 Relative Strengths of Acids and Bases [Go to section 15.3]
- Explain why the neutralization reaction of a strong acid and a weak base gives a weakly acidic solution.
- Use this list of important industrial compounds (and Figure 15.3.2) to answer the following questions regarding:
.- Identify the strong Brønsted-Lowry acids and strong Brønsted-Lowry bases.
- List those compounds in (a) that can behave as Brønsted-Lowry acids with strengths lying between those of
and . - List those compounds in (a) that can behave as Brønsted-Lowry bases with strengths lying between those of
and .
- What is the ionization constant at 25 °C for the weak acid
, the conjugate acid of the weak base , Kb = 5.9 × 10–4?
15.4 Hydrolysis of Salts [Go to section 15.4]
- Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
- Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
Show Selected Solutions
- The answers are as follows:
dissociates into ions (acidic metal cation) and ions (the conjugate base of a strong acid and therefore essentially neutral). The aqueous solution is therefore acidic. dissociates into ions (neutral metal cation) and ions (the conjugate base of a strong acid and therefore essentially neutral). The aqueous solution is therefore neutral.
15.5 Polyprotic Acids [Go to section 15.5]
- Calculate the concentration of each species present in a 0.050 M solution of
. - Which of the following species will contribute most significantly to the hydronium ion concentration in a solution of
?
15.6 Buffers [Go to section 15.6]
- Explain why a buffer can be prepared from a mixture of
and but not from and .
Show Selected Solutions
- [atex]\ce{OH–}[/latex] is a base, and
is a weak acid. They react with one another to form , thereby setting up the equilibrium . Because both the base ( ) and the conjugate acid ( ) are present, a buffer is formed. However, in the second case, NH3 and OH– are both bases, so no buffer is possible.