Chapter 16 – Review
16.1 Acids and Bases
- Write equations that show NH3 as both a conjugate acid and a conjugate base.
Check answers: [1] - Write equations that show H2PO−4 acting both as an acid and as a base.
Check answers: [2] - Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid:
- H3O+
- HCl
- NH3
- CH3CO2H
- NH+4
- HSO−4
Check answers: [3]
- Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid:
- HNO3
- PH+4
- H2S
- CH3CH2COOH
- H2PO−4
- HS−
Check answers: [4]
- Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base:
- H2O
- OH−
- NH3
- CN−
- S2−
- H2PO−4
Check Answers: [5]
- Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base:
- HS−
- PO3−4
- NH−2
- O2−
- H2PO−4
Check Answers: [6]
- What is the conjugate acid of each of the following? What is the conjugate base of each?
- OH−
- H2O
- HCO−3
- NH3
- HSO−4
- H2O2
- HS−
- H5N+2
Check Answers: For the following the conjugate acid is written first followed by its conjugate base: [7]
- What is the conjugate acid of each of the following? What is the conjugate base of each?
- H2S
- H2PO−4
- PH3
- HS−
- HSO−3
- H3O+2
- H4N2
- CH3OH
Check Answers: [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:
- HNO3+H2O⟶H3O++NO−3
- CN−+H2O⟶HCN+OH−
- H2SO4+Cl−⟶HCl+HSO−4
- HSO−4+OH−⟶SO2−4+H2O
- O2−+H2O⟶2OH−
- [Cu(H2O)3(OH)]++[Al(H2O)6]3+⟶Cu(H2O)4]2++[Al(H2O)5(OH)]2+
- H2S+NH−2⟶HS−+NH3
Check Answer: [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:
- NO−2+H2O⟶HNO2+OH−
- HBr+H2O⟶H3O++Br−
- HS−+H2O⟶H2S+OH−
- H2PO−4+OH−⟶HPO2−4+H2O
- H2PO−4+HCl⟶H3PO4+Cl−
- [Fe(H2O)5(OH)]2++[Al(H2O)6]3+⟶[Fe(H2O)6]3++[Al(H2O)5(OH)]2+
- CH3OH+H−⟶CH3O−+H2
Check Answers: [10]
- What are amphiprotic species? Illustrate with suitable equations.
Check Answers: [11] - State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species.
- NH3
- HPO−4
- Br−
- NH+4
- ASO3−4
Check Answers: [12]
16.2 Reactions of Acids and Bases
- 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.
- NaCl
- MgCl2
- Na2SO4
- NaNO3
- K3PO4
Check Answers: [13]
- When an acid reacts with carbonates what are the characteristic products?
- Predict the products of a reaction of hydrobromic acid with sodium carbonate: Na2CO3(s) + 2HBr(aq) →
Check Answers: [14] - What is the generic formula for the oxidation of metals in acidic solutions?
Check Answers: [15] - Predict the products of a reaction of hydrochloric acids with magnesium metal.
Check Answers: [16] - What is the generic formula for when a metal oxide reacts with an acid?
Check Answers: [17] - Predict the products of a reaction of nitric acid with the metal oxide, copper oxide:
Check Answers: [18]
16.3 Ionization of Water
- Write an equation to show the autoionization of water.
Check Answers: [19] - 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] - 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] - 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] - 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
- 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] - 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] - 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] - Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
- 0.200 M HCl
- 0.0143 M NaOH
- 3.0 M HNO3
- 0.0031 M Ca(OH)2
Check Answers: [27]
- Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
- 0.000259 M HClO4
- 0.21 M NaOH
- 0.000071 M Ba(OH)2
- 2.5 M KOH
Check Answers: [28]
- What are the pH and pOH of a solution of 2.0 M HCl, which ionizes completely?
Check Answers: [29] - What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52?
Check Answers: [30] - Calculate the hydrogen ion concentration and the hydroxide ion concentration in a red wine with a pH of 3.500.
Check Answers: [31] - Calculate the hydronium ion concentration and the hydroxide ion concentration in lime juice with a pH of 2.00.
Check Answers: [32] - 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] - 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
- 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] - 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] - 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] - 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] - What volume of 0.975 M NaOH is needed to neutralize 45.0 mL of 0.225 M sulfuric acid, H2SO4?
Check Answers: [39] - What volume of 0.202 M HNO3 is required to neutralize each of the following solutions?
- 15.5 mL of 0.155 M NaOH
- 25.1 mL of 0.0391 M Ba(OH)2
Check Answers: [40]
16.7 Buffers
- Define buffer. What two related chemical components are required to make a buffer?
Check Answers: [41] - Can a buffer be made by combining a strong acid with a strong base? Why or why not?
Check Answers: [42] - Of the following options (a – d), which combinations of compounds can make a buffer? Assume aqueous solutions.
- HCl and NaCl
- HNO2 and NaNO2
- NH4NO3 and HNO3
- NH4NO3 and NH3
Check Answers: [43]
- 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:
- 16.1 Acids & Bases and 16.3 question #5 are adapted from “14.1 Brønsted-Lowry Acids and Bases” In General Chemistry 1 & 2 by Rice University, a derivative of Chemistry (Open Stax) by Paul Flowers, Klaus Theopold, Richard Langley & William R. Robinson and is licensed under CC BY 4.0. Access for free at Chemistry (OpenStax)
- 16.2 Reactions of Acids and Bases: All questions created by Jackie MacDonald, licensed under CC BY 4.0
- 16.3 Ionization of Water Source: All questions except #5 created by created by Jackie MacDonald, licensed under CC BY 4.0
- 16.5 Neutralization and 16.6 Titrations and Neutralization Calculations – created by created by Jackie MacDonald, licensed under CC BY 4.0
- 16.7 Buffers is adapted from “77. Buffers” In Introductory Chemistry, 1st Canadian Edition by David W. Ball and Jessie A. Key. licensed under CC BY-NC-SA 3.0
- One example for NH3 as a conjugate acid: NH−2+H+⟶NH3; as a conjugate base: NH+4(aq)+OH−(aq)⟶NH3(aq)+H2O(l) ↵
- One example for H2PO−4 behaving as an acid (proton donor): H2PO4-(aq) + OH-(aq) → HPO42-(aq) + H2O(l); One example for H2PO−4 acting as a base (proton acceptor): H2PO4-(aq) + H2O(l) → H3PO4(aq) + OH-(aq) ↵
- (a) H3O+(aq)⟶H+(aq)+H2O(l); (b) HCl(l)⟶H+(aq)+Cl−(aq); (c) NH3(aq)⟶H+(aq)+NH−2(aq); (d) CH3CO2H(aq)⟶H+(aq)+CH3CO−2(aq); (e) NH+4(aq)⟶H+(aq)+NH3(aq); (f) HSO−4(aq)⟶H+(aq)+SO2−4(aq) ↵
- (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) ↵
- (a) H2O(l)+H+(aq)⟶H3O+(aq); (b) OH−(aq)+H+(aq)⟶H2O(l); (c) NH3(aq)+H+(aq)⟶NH+4(aq); (d) CN−(aq)+H+(aq)⟶HCN(aq); (e) S2−(aq)+H+(aq)⟶HS−(aq); (f) H2PO−4(aq)+H+(aq)⟶H3PO4(aq) ↵
- (a) HS−(aq) + H+(aq) → H2S(aq); (b) PO3−4 + H+(aq) → HPO42-(aq); (c) NH−2 + H+(aq) → NH3(aq); (d) O2−(aq) + H+(aq) → OH-(aq); (e) H2PO−4 + H+(aq) → H3PO4(aq) ↵
- (a) H2O, O2−; (b) H3O+, OH−; (c) H2CO3, CO2−3; (d) NH+4, NH−2; (e) H2SO4, SO2−4; (f) H3O+2, HO−2; (g) H2S; S2−; (h) H6N2+2, H4N2 ↵
- (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)) ↵
- 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), NO−3(CB); (b) CN−(BB), H2O(BA), HCN(CA), OH−(CB); (c) H2SO4(BA), Cl−(BB), HCl(CA), HSO−4(CB); (d) HSO−4(BA), OH−(BB), SO2−4(CB), 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), NH−2(BB), HS−(CB), NH3(CA) ↵
- 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) ↵
- 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: H2O(aq)+NH3(aq)⇋NH+4(aq)+OH−(aq) As a base: H2O(aq)+HCl(aq)⇋H3O+(aq)+Cl−(aq) ↵
- amphiprotic: (a) NH3+H3O+⟶NH4OH+H2ONH3+OCH−3⟶NH−2+CH3OH (b) HPO2−4+OH−⟶PO3−4+H2OHPO2−4+HClO4⟶H2PO−4+ClO−4 not amphiprotic: (c) Br− (d) NH+4 (e) AsO3−4 ↵
- (a) HCl, NaOH; (b) HCl, Mg(OH)2; (c) H2SO4, NaOH; (d) HNO3, NaOH; (e) H3PO4, KOH ↵
- Na2CO3(s) + 2HBr(aq) → 2NaBr(aq) + CO2(g) + H2O(l) ↵
- acid + metal → hydrogen + ionic compound ↵
- 2HCl(aq) + Mg(s) → H2(g) + MgCl2(aq) ↵
- acid + metal oxide → salt + H2O(l) ↵
- 2HNO3(aq) + CuO(s) → CuNO3(aq) + 2H2O(l) ↵
- H2O(l)+H2O(l)⇋H3O+(aq)+OH−(aq) ↵
- [H+] = 1.0 x 10-9 M, solution is basic since [OH-] > [H+] ↵
- [OH-] = 5.0 x 10-13 M, solution is acidic since [H+] > [OH-] ↵
- [H+] = 1.0 x 10-7 M, the solution is neutral since [H+] = [OH-] ↵
- 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 ↵
- In a neutral solution [H3O+] = [OH−]. At 40 °C, [H3O ] = [OH−] = (2.9 x 10−14)1/2 = 1.7 × 10−7. ↵
- 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 ↵
- 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 ↵
- (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 ↵
- (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 ↵
- pH = -0.30, pOH = 14.30 ↵
- [H3O+] = 3.0 x 10-7 M, [OH-] = 3.3 x 10-8 M ↵
- [H3O+] = 10-3.500 = 3.16 x 10-4 M, [OH-] = 10-10.5 = 3.16 x 10-11 M ↵
- [H3O+] = 1 × 10−2M; [OH−] = 1 × 10−12M ↵
- [OH-] = 5.9 x 10-9 M ↵
- [H3O+] =3.1 x 10-12 M ↵
- [OH-] = 0.175 M ↵
- [HCl] = 0.690 M ↵
- [Ba(OH)2] = 0.288 M ↵
- [H2SO4] = 0.151 M ↵
- 20.8 mL of NaOH is needed to neutralize the acid. ↵
- (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 ↵
- 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. ↵
- No, buffers cannot be made from a strong acid (or strong base) and its conjugate. This is because they both ionize completely. ↵
- (a) no; (b) yes; (c) no; (d) yes ↵
- 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+ ↵