Chapter 15: Acid-Based Equilibria

15.1 Brønsted-Lowry Acids and Bases [Go to section 15.1]

  1. What is the conjugate acid of each of the following? What is the conjugate base of each?
    1. [latex]\ce{OH–}[/latex]
    2. [latex]\ce{H2O}[/latex]
    3. [latex]\ce{HCO3-}[/latex]
    4. [latex]\ce{NH3}[/latex]
    5. [latex]\ce{HSO4-}[/latex]
    6. [latex]\ce{H2O2}[/latex]
    7. [latex]\ce{HS–}[/latex]
    8. [latex]\ce{H5N2+}[/latex]
  2. Write equations that show [latex]\ce{H2PO4^{2-}}[/latex] acting both as an acid and as a base.
  3. What are amphiprotic species? Illustrate with suitable equations.
  4. Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid:
    1. [latex]\ce{HNO3}[/latex]
    2.   [latex]\ce{PH4-}[/latex]
    3. [latex]\ce{H2S}[/latex]
    4. [latex]\ce{CH3CH2COOH}[/latex]
    5.  [latex]\ce{H2PO4-}[/latex]
    6. [latex]\ce{HS–}[/latex]
  5. State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species.
    1. [latex]\ce{NH3}[/latex]
    2.  [latex]\ce{HPO4-}[/latex]
    3. [latex]\ce{Br–}[/latex]
    4.   [latex]\ce{NH4-}[/latex]
    5.   [latex]\ce{ASO4^{3-}}[/latex]
  6. Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base:
    1. [latex]\ce{HS–}[/latex]
    2. [latex]\ce{PO4^{3-}}[/latex]
    3. [latex]\ce{NH2-}[/latex]
    4. [latex]\ce{C2H5OH}[/latex]
    5. [latex]\ce{O^{2–}}[/latex]
    6. [latex]\ce{H2PO4-}[/latex]
  7. Identify the conjugate acids of the species in the previous problem.
  8. Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
    1. [latex]\ce{NO2-} + \ce{H2O} \longrightarrow \ce{HNO2} + \ce{OH-}[/latex]
    2. [latex]\ce{HBr} + \ce{H2O} \longrightarrow \ce{H3O+} + \ce{Br-}[/latex]
    3. [latex]\ce{HS-} + \ce{H2O} \longrightarrow \ce{H2S} + \ce{OH-}[/latex]
    4. [latex]\ce{H2PO4-} + \ce{OH-} \longrightarrow \ce{HPO4^{2-}} + \ce{H2O}[/latex]
    5. [latex]\ce{H2PO4-} + \ce{HCl} \longrightarrow \ce{H3PO4} + \ce{Cl-}[/latex]
    6. [latex]\ce{[Fe(H2O)5(OH)]^{2+}} + \ce{[Al(H2O)6]^{3+}} \longrightarrow \ce{[Fe(H2O)6]^{3+}} + \ce{[Al(H2O)5(OH)]^{2+}}[/latex]
    7. [latex]\ce{CH3OH} + \ce{H-} \longrightarrow \ce{CH3O-} + \ce{H2}[/latex]
  9. Show by suitable net ionic equations that the following species are amphoteric:
    1. [latex]\ce{H2O}[/latex]
    2. [latex]\ce{HSO3-}[/latex]
    3. [latex]\ce{HC2O4-}[/latex]
  10. 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.

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  1. The answers are as follows:
    1. [latex]\ce{H2O}, \hspace{0.25cm} \ce{O^{2–}}[/latex]
    2. [latex]\ce{H3O+}, \hspace{0.25cm} \ce{OH–}[/latex]
    3. [latex]\ce{H2CO3}, \hspace{0.25cm} \ce{CO3^{2-}}[/latex]
    4. [latex]\ce{NH4+}, \hspace{0.25cm} \ce{NH2-}[/latex]
    5. [latex]\ce{H2SO4}, \hspace{0.25cm} \ce{SO4^{2-}}[/latex]
    6. [latex]\ce{H3O2+}, \hspace{0.25cm} \ce{HO2-}[/latex]
    7. [latex]\ce{H2S}, \hspace{0.25cm} \ce{S^{2–}}[/latex]
    8. [latex]\ce{H6N2^{2+}}, \hspace{0.25cm} \ce{H4N2}[/latex]
  2. Amphiprotic species may either gain or lose a proton in a chemical reaction, thus acting as a base or an acid. An example is [latex]\ce{H2O}[/latex]. As an acid: [latex]\ce{H2O}(aq) + \ce{NH3}(aq) \longrightarrow \ce{NH4+}(aq) + \ce{OH-}(aq)[/latex]. As a base: [latex]\ce{H2O}(aq) + \ce{HCl}(aq) \longrightarrow \ce{H3O+}(aq) + \ce{Cl-}(aq)[/latex]
  3.   The answers are as follows:
    1. [latex]\ce{NH3} + \ce{H3O+} \longrightarrow \ce{NH4OH} + \ce{H2O}, \hspace{2cm} \ce{NH3} + \ce{OCH3-} \longrightarrow \ce{NH2-} + \ce{CH3OH}[/latex]
    2. [latex]\ce{HPO4^{2-}} + \ce{OH-} \longrightarrow \ce{PO4^{3-}} + \ce{H2O}, \hspace{2cm} \ce{HPO4^{2-}} + \ce{HClO4} \longrightarrow \ce{H2PO4-} + \ce{ClO4-}[/latex]
      Not Amphiprotic:
    3. [latex]\ce{Br-}[/latex]
    4. [latex]\ce{NH4-}[/latex]
    5. [latex]\ce{AsO4^{3-}}[/latex]
  4. The answers are as follows:
    1. [latex]\ce{H2S}[/latex]
    2. [latex]\ce{HPO4^{3-}}[/latex]
    3. [latex]\ce{NH3}[/latex]
    4. [latex]\ce{C2H5OH2+}[/latex]
    5. [latex]\ce{OH-}[/latex]
    6. [latex]\ce{H3PO4}[/latex]
  5. The answers are as follows:
    1. [latex]\ce{H2O} + \ce{H2O} \leftrightarrows \ce{H3O+} + \ce{OH-}[/latex]
    2. [latex]\ce{HSO3-} + \ce{H2O} \leftrightarrows \ce{H2SO4-} + \ce{OH-}[/latex]; [latex]\ce{HSO3-} + \ce{H2O} \leftrightarrows \ce{SO3^{2-}} + \ce{H3O+}[/latex]
    3. [latex]\ce{HC2O4-} + \ce{H2O} \leftrightarrows \ce{H2C2O4} + \ce{OH-}[/latex]; [latex]\ce{HC2O4} + \ce{H2O} \leftrightarrows \ce{C2O4^{2-}} + \ce{H3O+}[/latex]


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15.2 pH and pOH [Go to section 15.2]

  1. What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52?
  2. The ionization constant for water (Kw) is 2.9 × 10–14 at 40 °C. Calculate , [latex]\ce{[OH–]}[/latex], pH, and pOH for pure water at 40 °C.
  3. Calculate the hydronium ion concentration and the hydroxide ion concentration in lime juice from its pH. See Figure 15.2.1 for useful information.
  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 [latex]\ce{HCl}[/latex]
    2. 0.0143 M [latex]\ce{NaOH}[/latex]
    3. 3.0 M [latex]\ce{HNO3}[/latex]
    4. 0.0031 M [latex]\ce{Ca(OH)2}[/latex]
  5. Calculate the hydronium ion concentrations for the following solutions:
    1. [latex]\ce{HBr}[/latex], pH = 2.50
    2. [latex]\ce{KOH}[/latex], pH = 8.55
    3. [latex]\ce{HClO4}[/latex], pH = 1.02
    4. [latex]\ce{LiOH}[/latex], pH = 11.00
  6. Calculate the hydrogen ion concentration and the hydroxide ion concentration in wine from its pH. See Figure 15.2.1 for useful information.
  7. Calculate the hydroxide ion concentration for the solutions in problem number 15.

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  1. [latex]\ce{[H3O+]}[/latex] = 3.0 × 10–7 M; pOH = 7.48; [latex]\ce{[OH–]}[/latex] = 3.3 × 10–8 M
  2. [latex]\ce{[H3O+]}[/latex] = 1 × 10–2 M; [latex]\ce{[OH–]}[/latex] = 1 × 10–12 M
  3. The answers are as follows:
    1. [latex]\ce{[H3O+]}[/latex] = 3.16 × 10-3 M
    2. [latex]\ce{[H3O+]}[/latex] = 2.82 × 10-9 M
    3. [latex]\ce{[H3O+]}[/latex] = 0.0955 M
    4. [latex]\ce{[H3O+]}[/latex] = 1 × 10-11 M
  4. The answers are as follows:
    1. [latex]\ce{[OH-]}[/latex] = 3.16 × 10-12 M
    2. [latex]\ce{[OH-]}[/latex] = 3.54 × 10-6 M
    3. [latex]\ce{[OH-]}[/latex] = 1.05 × 10-13 M
    4. [latex]\ce{[OH-]}[/latex] = 1 × 10-3 M


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15.3 Relative Strengths of Acids and Bases [Go to section 15.3]

  1. Explain why the neutralization reaction of a strong acid and a weak base gives a weakly acidic solution.
  2. Use this list of important industrial compounds (and Figure 15.3.2) to answer the following questions regarding: [latex]\ce{CaO, Ca(OH)2, CH3CO2H, CO2,HCl, H2CO3, HF, HNO2, HNO3, H3PO4, H2SO4, NH3, NaOH, Na2CO3}[/latex].
    1. Identify the strong Brønsted-Lowry acids and strong Brønsted-Lowry bases.
    2. List those compounds in (a) that can behave as Brønsted-Lowry acids with strengths lying between those of [latex]\ce{H3O+}[/latex] and [latex]\ce{H2O}[/latex].
    3. List those compounds in (a) that can behave as Brønsted-Lowry bases with strengths lying between those of [latex]\ce{H2O}[/latex] and [latex]\ce{OH}[/latex].
  3. What is the ionization constant at 25 °C for the weak acid [latex]\ce{(CH3)2NH2+}[/latex], the conjugate acid of the weak base [latex]\ce{(CH3)2NH}[/latex], Kb = 5.9 × 10–4?

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  1. The answers are as follows:
    1. acids – [latex]\ce{HCl, HNO3, H2SO4}[/latex] bases – [latex]\ce{Ca(OH)2, NaOH}[/latex]
    2. [latex]\ce{H3PO4, HF, HNO2, CH3CO2H, H2CO3}[/latex]
    3. [latex]\ce{Na2CO3}[/latex], [latex]\ce{NH3}[/latex]


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15.4 Hydrolysis of Salts [Go to section 15.4]

  1. Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
    1. [latex]\ce{Al(NO3)3}[/latex]
    2. [latex]\ce{RbI}[/latex]
  2. Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
    1. [latex]\ce{KHCO2}[/latex]
    2. [latex]\ce{CH3NH3Br}[/latex]

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  1. The answers are as follows:
    1. [latex]\ce{Al(NO3)3}[/latex] dissociates into [latex]\ce{Al3+}[/latex] ions (acidic metal cation) and [latex]\ce{NO3-}[/latex] ions (the conjugate base of a strong acid and therefore essentially neutral). The aqueous solution is therefore acidic.
    2. [latex]\ce{RbI}[/latex] dissociates into [latex]\ce{Rb+}[/latex] ions (neutral metal cation) and [latex]\ce{I–}[/latex] ions (the conjugate base of a strong acid and therefore essentially neutral). The aqueous solution is therefore neutral.


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15.5 Polyprotic Acids [Go to section 15.5]

  1. Calculate the concentration of each species present in a 0.050 M solution of [latex]\ce{H2S}[/latex].
  2. Which of the following species will contribute most significantly to the hydronium ion concentration in a solution of [latex]\ce{H3PO4}[/latex]? [latex]\ce{H3PO4, H2PO4-, HPO42-, PO43-}[/latex]

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  1. [latex]\ce{[H2S]}[/latex] = 0.050 M[latex]\ce{[HS–]}[/latex] = 6.7 × 10–5 M [latex]\ce{[H3O+]}[/latex]= 6.7 × 10–5 M[latex]\ce{[S2–]}[/latex] = 1 × 10–19 M[latex]\ce{[OH–]}[/latex] = 1.5 × 10–10 M 


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15.6 Buffers [Go to section 15.6]

  1. Explain why a buffer can be prepared from a mixture of [latex]\ce{NH4Cl}[/latex] and [latex]\ce{NaOH}[/latex] but not from [latex]\ce{NH3}[/latex] and [latex]\ce{NaOH}[/latex].

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  1. [atex]\ce{OH–}[/latex] is a base, and [latex]\ce{NH4+}[/latex] is a weak acid. They react with one another to form [latex]\ce{NH3}[/latex], thereby setting up the equilibrium [latex]\ce{NH4+}(aq) + \ce{OH-}(aq) \longleftrightarrow \ce{NH3}(aq) + \ce{H2O}(l)[/latex]. Because both the base ([latex]\ce{NH3}[/latex]) and the conjugate acid ([latex]\ce{NH4+}[/latex]) are present, a buffer is formed. However, in the second case, NH3 and OH– are both bases, so no buffer is possible.


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15.7 Acid-Base Titrations [Go to section 15.7]

  1. Example

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  1. Example


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