Acids and bases

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  • 1. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Acids and Bases18-1
  • 2. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.18-2
  • 3. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Arrhenius Acid-Base Definition An acid is a substance that increase H+ when dissolved in water. HCl(l) + H2O(l) H3O+(aq) + Cl-(aq) The ionization of acids actually produces H3O+ but H3O+  H2O + H+ A base is a substance that increase OH- ions when dissolved in water. NaOH(aq) + H2O(l) Na+(aq) + OH-(aq) Limitation: this theory cannot explain why NH3 is a base18-3
  • 4. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Brønsted-Lowry Acid-Base Definition An acid is a proton donor, any species which donates a H+. A base is a proton acceptor, any species which accepts a H+. HCl(l) + H2O(l) H3O+(aq) + Cl-(aq) An acid reactant will produce a base product and the two will constitute an acid-base conjugate pair.18-4
  • 5. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The Conjugate Acid-Base Pairs An acid and base that differ only in the presence or absence of a proton18-5
  • 6. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Table 18.4 The Conjugate Pairs in Some Acid-Base Reactions Conjugate Pair Acid + Base Base + Acid Conjugate Pair Reaction 1 HF + H2O F- + H3O+ Reaction 2 HCOOH + CN- HCOO- + HCN Reaction 3 NH4+ + CO32- NH3 + HCO3- Reaction 4 H2PO4- + OH- HPO42- + H2O Reaction 5 H2SO4 + N2H5+ HSO4- + N2H62+ Reaction 6 HPO42- + SO32- PO43- + HSO3-18-6
  • 7. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. SAMPLE PROBLEM 18.4: Identifying Conjugate Acid-Base Pairs PROBLEM: The following reactions are important environmental processes. Identify the conjugate acid-base pairs. (a) H2PO4-(aq) + CO32-(aq) HPO42-(aq) + HCO3-(aq) (b) H2O(l) + SO32-(aq) OH-(aq) + HSO3-(aq) PLAN: Identify proton donors (acids) and proton acceptors (bases). conjugate pair conjugate pair2 1 SOLUTION: (a) H2PO4-(aq) + CO32-(aq) HPO42-(aq) + HCO3-(aq) proton proton proton proton donor acceptor acceptor donor conjugate pair1 conjugate pair2 (b) H2O(l) + SO32-(aq) OH-(aq) + HSO3-(aq) proton proton proton proton donor acceptor acceptor donor18-7
  • 8. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Amphoteric substances A substance that can both accept and donate a proton– i.e. act as an acid and as a base Ex. H2O Water acts as a base HCl(l) + H2O(l) H3O+(aq) + Cl-(aq) Water acts as an acid NH3 + H2O(l) NH4+(aq) + OH-(aq)18-8
  • 9. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 18.9 Strengths of conjugate acid- base pairs18-9
  • 10. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Relative Strengths of Acids and Bases In every acid-base reaction, the position of equilibrium favors the weaker acid HCl(l) + H2O(l) H3O+(aq) + Cl-(aq) Stronger acid Weaker acid Since H3O+ is weaker, the forward reaction is favored over the reverse reaction and the equilibrium lies to the right18-10
  • 11. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. SAMPLE PROBLEM 18.5: Predicting the Net Direction of an Acid-Base Reaction PROBLEM: Predict the net direction of equilibrium (a) H2PO4-(aq) + NH3(aq) HPO42-(aq) + NH4+(aq) (b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq) PLAN: Identify the conjugate acid-base pairs and then consult Figure 18.10 (button) to determine the relative strength of each. The stronger the species, the more preponderant its conjugate. SOLUTION: (a) H2PO4-(aq) + NH3(aq) HPO42-(aq) + NH4+(aq) stronger acid weaker acid Net direction is to the right with Kc > 1. (b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq) weaker acid stronger acid Net direction is to the left with Kc < 1.18-11
  • 12. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Autoionization of Water H2O(l) H3O+(aq) + OH-(aq) Kc = [H3O ][OH ] + - The Ion-Product Constant for Water Kw = [H3O+][OH-] = 1.0 x 10-14 at 250C A change in [H3O+] causes an inverse change in [OH-]. In an acidic solution, [H3O+] > [OH-] In a basic solution, [H3O+] < [OH-] In a neutral solution, [H3O+] = [OH-]18-12
  • 13. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The relationship between [H3O+] and [OH-] and the Figure 18.4 relative acidity of solutions. [H3O+] Divide into Kw [OH-] [H3O+] > [OH-] [H3O+] = [OH-] [H3O+] < [OH-] ACIDIC NEUTRAL BASIC SOLUTION SOLUTION SOLUTION18-13
  • 14. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. SAMPLE PROBLEM 18.2: Calculating [H3O+] and [OH-] in an Aqueous Solution PROBLEM: A research chemist adds a measured amount of HCl gas to pure water at 250C and obtains a solution with [H3O+] = 3.0x10-4M. Calculate [OH-]. Is the solution neutral, acidic, or basic? PLAN: Use the Kw at 250C and the [H3O+] to find the corresponding [OH-]. SOLUTION: K = 1.0x10-14 = [H O+] [OH-] so w 3 [OH-] = Kw/ [H3O+] = 1.0x10-14/3.0x10-4 = 3.3x10-11M [H3O+] is > [OH-] and the solution is acidic.18-14
  • 15. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Solve the following: • [ H+] = 1.4 x 10-6 [OH-] = b. [ OH-] = 1.0 x 10-7 [H+] =18-15
  • 16. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The pH Scale -Measures the concentration of hydrogen ions -the power/potential of the hydrogen ion -a way of expressing H+ in a given solution pH = - log [H+] The higher the H+ concentration, the lower the pH pOH = - log [OH-] The higher the OH- concentration, the lower the pOH pOH + pH = 1418-16
  • 17. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 18.5 The pH values of some familiar aqueous solutions pH = -log [H3O+]18-17
  • 18. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Figure 18.6 The relations among [H3O+], pH, [OH-], and pOH.18-18
  • 19. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. SAMPLE PROBLEM 18.3: Calculating [H3O+], pH, [OH-], and pOH PROBLEM: Given [H3O+] concentrations: 2.0M, 0.30M, and 0.0063M, Calculate pH, [OH-], and pOH of the three solutions at 250C. PLAN: HNO3 is a strong acid so [H3O+] = [HNO3]. Use Kw to find the [OH-] and then convert to pH and pOH. SOLUTION: For [H3O+] = 2.0M and -log [H3O+] = -0.30 = pH [OH-] = Kw/ [H3O+] = 1.0x10-14/2.0 = 5.0x10-15M; pOH = 14.30 For [H3O+] = 0.30M and -log [H3O+] = 0.52 = pH [OH-] = Kw/ [H3O+] = 1.0x10-14/0.30 = 3.3x10-14M; pOH = 13.48 For [H3O+] = 0.0063M and -log [H3O+] = 2.20 = pH [OH-] = Kw/ [H3O+] = 1.0x10-14/6.3x10-3 = 1.6x10-12M; pOH = 11.8018-19
  • 20. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. STRONG ACIDS and BASES Strong acids and bases dissociate completely into ions in water. 100% ionization Strong acid: HA(g or l) + H2O(l) H3O+(aq) + A-(aq)18-20
  • 21. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Strong Acids Strong Bases HCl LiOH HBr NaOH HI KOH HNO3 Ca(OH)2 H2 SO4 Sr(OH)2 HClO4 Ba(OH)2 All other acids and bases not listed here are considered weak.18-21
  • 22. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. For STRONG ACIDS, H+ or OH- are equal to the original molarity of the solution Problem: Calculate the pH of 0.5 M HCl and 0.10 M NaOH solution: HCl(g or l) + H2O(l) H3O+(aq) + Cl-(aq) 0.5 M 0.5 M 0.5 M pH = - log (0.5 M) = 0.30 NaOH(aq) Na+(aq) + OH-(aq) 0.10 M 0.10 M 0.10 M pOH = - log (0.10 M) = 1.00 pH = 14 – 1.00 = 13.0018-22
  • 23. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. SAMPLE PROBLEM 18.5: Calculate the pH of a solution prepared by dissolving 5.00 g of KOH in enough water to get 500.0 mL of solution MW of KOH = 56.108 g/mol Answer: 0.178 M of [OH-] pOH = 0.74902 pH = 13.2518-23
  • 24. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. WEAK ACIDS and WEAK BASES Weak acids and bases dissociate very slightly into ions in water. - About 1% ionization, the equilibrium lies far to the left Weak acid: HA(aq) + H2O(l) H2O+(aq) + A-(aq)18-24
  • 25. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The Acid-Dissociation Constant, Ka HA(aq) + H2O(l) H3O+(aq) + A-(aq) [H3O+][A-] Kc = [HA] stronger acid higher [H3O+] larger Ka [H3O+][A-] Ka = [HA] smaller Ka lower [H3O+] weaker acid18-25
  • 26. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. SAMPLE PROBLEM : Write the equilibrium constant expression for the following weak acid: a. NH4 + NH4 + (aq) + H2O(l) H3O+(aq) + NH3 [H3O+][NH3] Ka = [NH4 + ]18-26
  • 27. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Table 18.3 The Relationship Between Ka and pKa pKa = - log Ka Acid Name (Formula) Ka at 250C pKa Hydrogen sulfate ion (HSO4-) 1.02x10-2 1.991 7.1x10-4 3.15 Nitrous acid (HNO2) 1.8x10-5 4.74 Acetic acid (CH3COOH) Hypobromous acid (HBrO) 2.3x10-9 8.64 Phenol (C6H5OH) 1.0x10-10 10.0018-27
  • 28. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The Base-Dissociation Constant, Kb A- (aq) + H2O(l) OH-(aq) + HA (aq) [OH-][HA] Kc = [A-] stronger base higher [OH-] larger Kb [OH-][HA] Kb = [A-]18-28
  • 29. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. SAMPLE PROBLEM : Write the equilibrium constant expression for the following weak base: a. C5H5N C5H5N (aq) + H2O(l) OH-(aq) + C5H5NH+ [C5H5NH+ ][OH-] Kb = [C5H5N]18-29
  • 30. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The Relationship Between Ka and Kb Ka x Kb = Kw Example: Calculate Kb if Ka = 1.5 x 10-5 1.0 x 10-14 Kb = 1.5 x 10-5 = 6.7 x 10-1018-30
  • 31. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Lewis Definition of Acids and Bases An acid is an electron-pair acceptor. A base is an electron-pair donor. F F H H B + N B N F HH F F F HH acid base adduct M(H2O)42+(aq) M2+ H2O(l) adduct18-31
  • 32. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. SAMPLE PROBLEM 18.13: Identifying Lewis Acids and Bases PROBLEM: Identify the Lewis acids and Lewis bases in the following reactions: (a) H+ + OH- H2O (b) Cl- + BCl3 BCl4- (c) K+ + 6H2O K(H2O)6+ PLAN: Look for electron pair acceptors (acids) and donors (bases). SOLUTION: acceptor (a) H+ + OH- H2O donor donor (b) Cl- + BCl3 BCl4- acceptor acceptor (c) K+ + 6H2O K(H2O)6+ donor18-32