Chapter 16 – Review

16.1 Acids and Bases

  1. Write equations that show NH3 as both a conjugate acid and a conjugate base.
    Check answers: [1]
  2. Write equations that show [latex]\text{H}_2\text{PO}_4^{\;\;-}[/latex] acting both as an acid and as a base.
    Check answers: [2]
  3. Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid:
    1. [latex]\text{H}_3\text{O}^{+}[/latex]
    2. HCl
    3. NH3
    4. CH3CO2H
    5. [latex]\text{NH}_4^{\;\;+}[/latex]
    6. [latex]\text{HSO}_4^{\;\;-}[/latex]
      Check answers: [3]
  4. Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid:
    1. HNO3
    2. [latex]\text{PH}_4^{\;\;+}[/latex]
    3. H2S
    4. CH3CH2COOH
    5. [latex]\text{H}_2\text{PO}_4^{\;\;-}[/latex]
    6. HS−
      Check answers: [4]
  5. Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base:
    1. H2O
    2. OH−
    3. NH3
    4. CN−
    5. S2−
    6. [latex]\text{H}_2\text{PO}_4^{\;\;-}[/latex]
      Check Answers:  [5]
  6. Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base:
    1. HS−
    2. [latex]\text{PO}_4^{\;\;3-}[/latex]
    3. [latex]\text{NH}_2^{\;\;-}[/latex]
    4. O2−
    5. [latex]\text{H}_2\text{PO}_4^{\;\;-}[/latex]
      Check Answers: [6]
  7. What is the conjugate acid of each of the following? What is the conjugate base of each?
    1. OH−
    2. H2O
    3. [latex]\text{HCO}_3^{\;\;-}[/latex]
    4. NH3
    5. [latex]\text{HSO}_4^{\;\;-}[/latex]
    6. H2O2
    7. HS−
    8. [latex]\text{H}_5\text{N}_2^{\;\;+}[/latex]
      Check Answers: For the following the conjugate acid is written first followed by its conjugate base: [7]
  8. What is the conjugate acid of each of the following? What is the conjugate base of each?
    1. H2S
    2. [latex]\text{H}_2\text{PO}_4^{\;\;-}[/latex]
    3. PH3
    4. HS−
    5. [latex]\text{HSO}_3^{\;\;-}[/latex]
    6. [latex]\text{H}_3\text{O}_2^{\;\;+}[/latex]
    7. H4N2
    8. CH3OH
      Check Answers: [8]
  9. 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]\text{HNO}_3\;+\;\text{H}_2\text{O}\;{\longrightarrow}\;\text{H}_3\text{O}^{+}\;+\;\text{NO}_3^{\;\;-}[/latex]
    2. [latex]\text{CN}^{-}\;+\;\text{H}_2\text{O}\;{\longrightarrow}\;\text{HCN}\;+\;\text{OH}^{-}[/latex]
    3. [latex]\text{H}_2\text{SO}_4\;+\;\text{Cl}^{-}\;{\longrightarrow}\;\text{HCl}\;+\;\text{HSO}_4^{\;\;-}[/latex]
    4. [latex]\text{HSO}_4^{\;\;-}\;+\;\text{OH}^{-}\;{\longrightarrow}\;\text{SO}_4^{\;\;2-}\;+\;\text{H}_2\text{O}[/latex]
    5. [latex]\text{O}^{2-}\;+\;\text{H}_2\text{O}\;{\longrightarrow}\;2\text{OH}^{-}[/latex]
    6. [latex][\text{Cu(H}_2\text{O})_3(\text{OH})]^{+}\;+\;[\text{Al(H}_2\text{O})_6]^{3+}\;{\longrightarrow}\;\text{Cu(H}_2\text{O})_4]^{2+}\;+\;[\text{Al(H}_2\text{O})_5(\text{OH})]^{2+}[/latex]
    7. [latex]\text{H}_2\text{S}\;+\;\text{NH}_2^{\;\;-}\;{\longrightarrow}\;\text{HS}^{-}\;+\;\text{NH}_3[/latex]
      Check Answer: [9]
  10. 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]\text{NO}_2^{\;\;-}\;+\;\text{H}_2\text{O}\;{\longrightarrow}\;\text{HNO}_2\;+\;\text{OH}^{-}[/latex]
    2. [latex]\text{HBr}\;+\;\text{H}_2\text{O}\;{\longrightarrow}\;\text{H}_3\text{O}^{+}\;+\;\text{Br}^{-}[/latex]
    3. [latex]\text{HS}^{-}\;+\;\text{H}_2\text{O}\;{\longrightarrow}\;\text{H}_2\text{S}\;+\;\text{OH}^{-}[/latex]
    4. [latex]\text{H}_2\text{PO}_4^{\;\;-}\;+\;\text{OH}^{-}\;{\longrightarrow}\;\text{HPO}_4^{\;\;2-}\;+\;\text{H}_2\text{O}[/latex]
    5. [latex]\text{H}_2\text{PO}_4^{\;\;-}\;+\;\text{HCl}\;{\longrightarrow}\;\text{H}_3\text{PO}_4\;+\;\text{Cl}^{-}[/latex]
    6. [latex][\text{Fe(H}_2\text{O})_5(\text{OH})]^{2+}\;+\;[\text{Al(H}_2\text{O})_6]^{3+}\;{\longrightarrow}\;[\text{Fe(H}_2\text{O})_6]^{3+}\;+\;[\text{Al(H}_2\text{O})_5(\text{OH})]^{2+}[/latex]
    7. [latex]\text{CH}_3\text{OH}\;+\;\text{H}^{-}\;{\longrightarrow}\;\text{CH}_3\text{O}^{-}\;+\;\text{H}_2[/latex]
      Check Answers: [10]
  11. What are amphiprotic species? Illustrate with suitable equations.
    Check Answers: [11]
  12. State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species.
    1. NH3
    2. [latex]\text{HPO}_4^{\;\;-}[/latex]
    3. Br−
    4. [latex]\text{NH}_4^{\;\;+}[/latex]
    5. [latex]\text{ASO}_4^{\;\;3-}[/latex]
      Check Answers: [12]

16.2 Reactions of Acids and Bases

  1. The following salts were produced in an acid-base neutralization reaction. Write the formulas and names of the acid and base from which each of these salts are formed.
    1. NaCl
    2. MgCl2
    3. Na2SO4
    4. NaNO3
    5. K3PO4
      Check Answers: [13]
  2. When an acid reacts with carbonates what are the characteristic products?
  3. Predict the products of a reaction of hydrobromic acid with sodium carbonate: Na2CO3(s) + 2HBr(aq) →
    Check Answers: [14]
  4. What is the generic formula for the oxidation of metals in acidic solutions?
    Check Answers: [15]
  5. Predict the products of a reaction of hydrochloric acids with magnesium metal.
    Check Answers: [16]
  6. What is the generic formula for when a metal oxide reacts with an acid?
    Check Answers: [17]
  7. Predict the products of a reaction of nitric acid with the metal oxide, copper oxide:
    Check Answers: [18]

16.3 Ionization of Water

  1. Write an equation to show the autoionization of water.
    Check Answers: [19]
  2. Calculate the [H+] for a solution at 25°C that is 1.0 x 10-5 M OH–. Is this solution acidic, neutral or basic?
    Check Answers: [20]
  3. Calculate the [OH–] for a solution at 25°C that is 2.0 x 10-2 M H+. Is this solution acidic, neutral or basic?
    Check Answers: [21]
  4. Calculate the [H+] for a solution at 25°C that is 1.0 x 10-7 M OH–. Is this solution acidic, neutral or basic?
    Check Answers: [22]
  5. 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.
    Check Answer: [23]

16.4 Introduction to pH and pOH

  1. Explain why a sample of pure water at 40 °C is neutral even though [H3O+] = 1.7 × 10−7M. Kw is 2.9 × 10−14 at 40 °C.
    Check Answers: [24]
  2. The ionization constant for water (Kw) is 2.9 × 10−14 at 40 °C. Calculate [H3O+], [OH−], pH, and pOH for pure water at 40 °C.
    Check Answers: [25]
  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.
    Check Answers: [26]
  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
      Check Answers: [27]
  5. Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
    1. 0.000259 M HClO4
    2. 0.21 M NaOH
    3. 0.000071 M Ba(OH)2
    4. 2.5 M KOH
      Check Answers: [28]
  6. What are the pH and pOH of a solution of 2.0 M HCl, which ionizes completely?
    Check Answers: [29]
  7. What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52?
    Check Answers: [30]
  8. Calculate the hydrogen ion concentration and the hydroxide ion concentration in a red wine with a pH of 3.500.
    Check Answers: [31]
  9. Calculate the hydronium ion concentration and the hydroxide ion concentration in lime juice with a pH of 2.00.
    Check Answers: [32]
  10. 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?
    Check Answers: [33]
  11. 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?
    Check Answers: [34]

16.5 Neutralization and 16.6 Titrations and Neutralization Calculations

  1. 35.00 mL of 0.125 M HCl is required to neutralize 25.00 mL of KOH. Determine the concentration of the base solution.
    Check Answers: [35]
  2. Suppose that a titration is performed between a strong acid and strong base: 20.70 mL of 0.500 M NaOH is required to reach the end point when titrated against 15.00 mL of HCl of unknown concentration. Determine the concentration of hydrochloric acid used in this titration.
    Check Answers: [36]
  3. What is the concentration of a Ba(OH)2 solution, if 17.25 mL is required to neutralize 19.10 mL of 0.520 M HBr?
    Check Answers: [37]
  4. In a titration of sulfuric acid with sodium hydroxide, 32.20 mL of 0.250 M NaOH is required to neutralize 26.60 mL of the H2SO4 solution. Calculate the molarity of the sulfuric acid.
    Check Answers: [38]
  5. What volume of 0.975 M NaOH is needed to neutralize 45.0 mL of 0.225 M sulfuric acid, H2SO4?
    Check Answers: [39]
  6. What volume of 0.202 M HNO3 is required to neutralize each of the following solutions?
    1. 15.5 mL of 0.155 M NaOH
    2. 25.1 mL of 0.0391 M Ba(OH)2
      Check Answers: [40]

16.7 Buffers

  1. Define buffer. What two related chemical components are required to make a buffer?
    Check Answers: [41]
  2. Can a buffer be made by combining a strong acid with a strong base? Why or why not?
    Check Answers: [42]
  3. Of the following options (a – d), which combinations of compounds can make a buffer? Assume aqueous solutions.
    1. HCl and NaCl
    2. HNO2 and NaNO2
    3. NH4NO3 and HNO3
    4. NH4NO3 and NH3
      Check Answers: [43]
  4. For each combination in previous question that is a buffer, write the chemical equations for the reactions of the buffer components when a strong acid and a strong base is added.
    Check Answers: [44]

    Attribution & References

    Except where otherwise noted, this page is adapted by Jackie MacDonald from:

    1. One example for NH3 as a conjugate acid: [latex]\text{NH}_2^{\;\;-}\;+\;\text{H}^{+}\;{\longrightarrow}\;\text{NH}_3[/latex]; as a conjugate base: [latex]\text{NH}_4^{\;\;+}(aq)\;+\;\text{OH}^{-}(aq)\;{\longrightarrow}\;\text{NH}_3(aq)\;+\;\text{H}_2\text{O}(l)[/latex] ↵
    2. One example for [latex]\text{H}_2\text{PO}_4^{\;\;-}[/latex] behaving as an acid (proton donor): H2PO4-(aq) + OH-(aq) → HPO42-(aq) + H2O(l); One example for [latex]\text{H}_2\text{PO}_4^{\;\;-}[/latex] acting as a base (proton acceptor): H2PO4-(aq) + H2O(l) → H3PO4(aq) + OH-(aq) ↵
    3. (a) [latex]\text{H}_3\text{O}^{+}(aq)\;{\longrightarrow}\;\text{H}^{+}(aq)\;+\;\text{H}_2\text{O}(l)[/latex]; (b) [latex]\text{HCl}(l)\;{\longrightarrow}\;\text{H}^{+}(aq)\;+\;\text{Cl}^{-}(aq)[/latex]; (c) [latex]\text{NH}_3(aq)\;{\longrightarrow}\;\text{H}^{+}(aq)\;+\;\text{NH}_2^{\;\;-}(aq)[/latex]; (d) [latex]\text{CH}_3\text{CO}_2\text{H}(aq)\;{\longrightarrow}\;\text{H}^{+}(aq)\;+\;\text{CH}_3\text{CO}_2^{\;\;-}(aq)[/latex]; (e) [latex]\text{NH}_4^{\;\;+}(aq)\;{\longrightarrow}\;\text{H}^{+}(aq)\;+\;\text{NH}_3(aq)[/latex]; (f) [latex]\text{HSO}_4^{\;\;-}(aq)\;{\longrightarrow}\;\text{H}^{+}(aq)\;+\;\text{SO}_4^{\;\;2-}(aq)[/latex] ↵
    4. (a) HNO3(aq) → H+(aq) + NO3-(aq); (b) PH4+(aq) → H+(aq) + PH3(aq); (c) H2S(aq) → H+(aq) + HS-(aq); (d) CH3CH2COOH(aq) → H+(aq) + CH3CH2COO-(aq); (e) H2PO4-(aq) → H+(aq) + HPO42-(aq); (f) HS−(aq) → H+(aq) + S2−(aq) ↵
    5. (a) [latex]\text{H}_2\text{O}(l)\;+\;\text{H}^{+}(aq)\;{\longrightarrow}\;\text{H}_3\text{O}^{+}(aq)[/latex]; (b) [latex]\text{OH}^{-}(aq)\;+\;\text{H}^{+}(aq)\;{\longrightarrow}\;\text{H}_2\text{O}(l)[/latex]; (c) [latex]\text{NH}_3(aq)\;+\;\text{H}^{+}(aq)\;{\longrightarrow}\;\text{NH}_4^{\;\;+}(aq)[/latex]; (d) [latex]\text{CN}^{-}(aq)\;+\;\text{H}^{+}(aq)\;{\longrightarrow}\;\text{HCN}(aq)[/latex]; (e) [latex]\text{S}^{2-}(aq)\;+\;\text{H}^{+}(aq)\;{\longrightarrow}\;\text{HS}^{-}(aq)[/latex]; (f) [latex]\text{H}_2\text{PO}_4^{\;\;-}(aq)\;+\;\text{H}^{+}(aq)\;{\longrightarrow}\;\text{H}_3\text{PO}_4(aq)[/latex] ↵
    6. (a) HS−(aq) + H+(aq) → H2S(aq); (b) [latex]\text{PO}_4^{\;\;3-}[/latex] + H+(aq) → HPO42-(aq); (c) [latex]\text{NH}_2^{\;\;-}[/latex] + H+(aq) → NH3(aq); (d) O2−(aq) + H+(aq) → OH-(aq); (e) [latex]\text{H}_2\text{PO}_4^{\;\;-}[/latex] + H+(aq) → H3PO4(aq) ↵
    7. (a) H2O, O2−; (b) H3O+, OH−; (c) H2CO3, [latex]\text{CO}_3^{\;\;2-}[/latex]; (d) [latex]\text{NH}_4^{\;\;+}[/latex], [latex]\text{NH}_2^{\;\;-}[/latex]; (e) H2SO4, [latex]\text{SO}_4^{\;\;2-}[/latex]; (f) [latex]\text{H}_3\text{O}_2^{\;\;+}[/latex], [latex]\text{HO}_2^{\;\;-}[/latex]; (g) H2S; S2−; (h) [latex]\text{H}_6\text{N}_2^{\;\;2+}[/latex], H4N2 ↵
    8. (a) H3S+, HS-; (b) H3PO4, HPO42-; (c) PH4+, PH2-; (d) H2S, S2-; (e) H2SO3, SO32-; (f) H4O22+, H2O2; (g) H5N2+, H3N2- ; (h) CH3OH2+; CH3O- (Watch the video "14.8h | How to find the conjugate acid and conjugate base of CH3OH" for an explanation of the answer for (h)) ↵
    9. The labels are Brønsted-Lowry acid = BA; its conjugate base = CB; Brønsted-Lowry base = BB; its conjugate acid = CA. (a) HNO3(BA), H2O(BB), H3O+(CA), [latex]\text{NO}_3^{\;\;-}(\text{CB})[/latex]; (b) CN−(BB), H2O(BA), HCN(CA), OH−(CB); (c) H2SO4(BA), Cl−(BB), HCl(CA), [latex]\text{HSO}_4^{\;\;-}(\text{CB})[/latex]; (d) [latex]\text{HSO}_4^{\;\;-}(\text{BA})[/latex], OH−(BB), [latex]\text{SO}_4^{\;\;2-}(\text{CB})[/latex], H2O(CA); (e) O2−(BB), H2O(BA) OH−(CB and CA); (f) [Cu(H2O)3(OH)]+(BB), [Al(H2O)6]3+(BA), [Cu(H2O)4]2+(CA), [Al(H2O)5(OH)]2+(CB); (g) H2S(BA), [latex]\text{NH}_2^{\;\;-}(\text{BB})[/latex], HS−(CB), NH3(CA) ↵
    10. The labels are Brønsted-Lowry acid = BA; its conjugate base = CB; Brønsted-Lowry base = BB; its conjugate acid = CA. (a) NO2-(BB), H2O (BA), HNO3 (CA), OH-(CB); (b) HBr (BA), H2O (BB), Br- (CB), H3O+ (CA); (c) HS- (BB), H2O (BA), H2S (CA), OH- (CB); (d) H2PO4- (BA), OH- (BB), HPO42- (CB), H2O (CA); (e) H2PO4- (BB), HCl (BA), H3PO4 (CA), Cl- (CB); (f) [Fe(H20)5(OH)]2+ (BB), [Al(H2O)6]3+ (BA), [Fe(H20)6]3+ (CA), [Al(H2O)5(OH)]2+ (CB); (View the video "14.10f | How to identify the conjugate acid-base pairs in [Fe(H2O)5(OH)]2+ + [Al(H2O)6]3+" for an explanation of the answer for (f): ; (g) CH3OH (BA), H- (BB), CH3O- (CB), H2 (CA) ↵
    11. Amphiprotic species may either gain or lose a proton in a chemical reaction, thus acting as a base or an acid. An example is H2O. As an acid: [latex]H_{2}O\textit{(aq)} + NH_{3}\textit{(aq)} \leftrightharpoons NH_{4}^{+} \textit{(aq)} + OH^{-}\textit{(aq)}[/latex] As a base: [latex]H_{2}O\textit{(aq)} + HCl\textit{(aq)}  \leftrightharpoons  H_{3}O^{+}\textit{(aq)} + Cl^{-}\textit{(aq)}[/latex] ↵
    12. amphiprotic: (a) [latex]\text{NH}_3\;+\;\text{H}_3\text{O}^{+}\;{\longrightarrow}\;\text{NH}_4\text{OH}\;+\;\text{H}_2\text{O}\\[0.5em]\text{NH}_3\;+\;\text{OCH}_3^{\;\;-}\;{\longrightarrow}\;\text{NH}_2^{\;\;-}\;+\;\text{CH}_3\text{OH}[/latex] (b) [latex]\text{HPO}_4^{\;\;2-}\;+\;\text{OH}^{-}\;{\longrightarrow}\;\text{PO}_4^{\;\;3-}\;+\;\text{H}_2\text{O}\\[0.5em]\text{HPO}_4^{\;\;2-}\;+\;\text{HClO}_4\;{\longrightarrow}\;\text{H}_2\text{PO}_4^{\;\;-}\;+\;\text{ClO}_4^{\;\;-}[/latex] not amphiprotic: (c) Br− (d) [latex]\text{NH}_4^{\;\;+}[/latex] (e) [latex]\text{AsO}_4^{\;\;3-}[/latex] ↵
    13. (a) HCl, NaOH; (b) HCl, Mg(OH)2; (c) H2SO4, NaOH; (d) HNO3, NaOH; (e) H3PO4, KOH ↵
    14. Na2CO3(s) + 2HBr(aq) → 2NaBr(aq) + CO2(g) + H2O(l) ↵
    15. acid + metal → hydrogen + ionic compound ↵
    16. 2HCl(aq) + Mg(s) → H2(g) + MgCl2(aq) ↵
    17. acid + metal oxide → salt + H2O(l) ↵
    18. 2HNO3(aq) + CuO(s) → CuNO3(aq) + 2H2O(l) ↵
    19. [latex]\text{H}_2\text{O}(l)\;+\;\text{H}_2\text{O}(l)\;{\leftrightharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{OH}^{-}(aq)\\[0.7em][/latex] ↵
    20. [H+] = 1.0 x 10-9 M, solution is basic since [OH-] > [H+] ↵
    21. [OH-] = 5.0 x 10-13 M, solution is acidic since [H+] > [OH-] ↵
    22. [H+] = 1.0 x 10-7 M, the solution is neutral since [H+] = [OH-] ↵
    23. endothermic, temperature is going up from 40 to 60°C. Water absorbs heat on reactant side to shift to make more product (ionization of water). (View the video "14.14 | Is the self-ionization of water endothermic or exothermic?" for an explanation of this answer ↵
    24. In a neutral solution [H3O+] = [OH−]. At 40 °C, [H3O ] = [OH−] = (2.9 x 10−14)1/2 = 1.7 × 10−7. ↵
    25. In a neutral solution [H3O+] = [OH−]. At 40 °C, [H3O ] = [OH−] = (2.9 x 10−14)1/2 = 1.7 × 10−7. pH = pOH = -log[1.7 × 10−7 M] = 6.77 ↵
    26. In a neutral solution [H3O+] = [OH−]. At 60 °C, [H3O ] = [OH−] = (9.311 × 10−14)1/2 = 3.051 × 10−7. The pH = pOH = -log[3.051 × 10−7M] = 6.5156 ↵
    27. (a) pH = 0.699, pOH = 13.301; (b) pOH = 1.845, pH = 12.155; (c) pH = -0.477; pOH = 15.477; (d) Here 0.0031 M Ca(OH)2 yields 2 [OH-] ions for every one molecule the base. So, [OH-] = 6.2 x 10-3 and pOH = 2.21, pH = 11.79 ↵
    28. (a) pH = 3.587; pOH = 10.413; (b) pOH = 0.68; pH = 13.32; (c) Here 0.000071 M Ba(OH)2 yields 2 [OH-] ions for every one molecule the base. So, [OH-] = 1.42 x 10-4 and pOH = 3.85, pH = 10.15 (d) pH = −0.40; pOH = 14.40 ↵
    29. pH = -0.30, pOH = 14.30 ↵
    30. [H3O+] = 3.0 x 10-7 M, [OH-] = 3.3 x 10-8 M ↵
    31. [H3O+] = 10-3.500 = 3.16 x 10-4 M, [OH-] = 10-10.5 = 3.16 x 10-11 M ↵
    32. [H3O+] = 1 × 10−2M; [OH−] = 1 × 10−12M ↵
    33. [OH-] = 5.9 x 10-9 M  ↵
    34. [H3O+] =3.1 x 10-12 M ↵
    35. [OH-] = 0.175 M ↵
    36. [HCl] = 0.690 M ↵
    37. [Ba(OH)2] = 0.288 M ↵
    38. [H2SO4] = 0.151 M ↵
    39. 20.8 mL of NaOH is needed to neutralize the acid. ↵
    40. (a) 11.9 mL of HNO3 solution is needed to neutralize the base solution; (b) 9.72 mL of Ba(OH)2 is needed to neutralize the acid solution ↵
    41. A buffer is the combination of a weak acid or base and a salt of that weak acid or base and they resist a change in pH upon dilution or upon the addition of small amounts of acid or base. ↵
    42. No, buffers cannot be made from a strong acid (or strong base) and its conjugate. This is because they both ionize completely. ↵
    43. (a) no; (b) yes; (c) no; (d) yes ↵
    44. 3b: when a strong acid is added: NO2− + H+ → HNO2; when a strong base is added: HNO2 + OH− → NO2− + H2O; 3d: strong base added: NH4+ + OH− → NH3 + H2O; strong acid: NH3 + H+ → NH4+ ↵

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