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Chapter 16

Chapter 16. Acid-Base Equilibria. Acid-Base Equilibria. 16.1 Acids and Bases in Water. 16.2 Autoionization of Water and the pH Scale. 16.3 Proton Transfer and the Brønsted-Lowry Acid-Base Definition. 16.4 Solving Problems Involving Weak-Acid Equilibria.

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Chapter 16

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  1. Chapter 16 Acid-Base Equilibria

  2. Acid-Base Equilibria 16.1 Acids and Bases in Water 16.2 Autoionization of Water and the pH Scale 16.3 Proton Transfer and the Brønsted-Lowry Acid-Base Definition 16.4 Solving Problems Involving Weak-Acid Equilibria 16.5 Weak Bases and Their Relations to Weak Acids 16.6 Molecular Properties and Acid Strength 16.7 Acid-Base Properties of Salt Solutions 16.8 Generalizing the Brønsted-Lowry Concept: The Leveling Effect 16.9 Electron-Pair Donation and the Lewis Acid-Base Definition

  3. Table 16.1

  4. Strong acid: HA(g or l) + H2O(l) H2O+(aq) + A-(aq) The extent of dissociation for strong acids. Figure 16.1

  5. Weak acid: HA(aq) + H2O(l) H2O+(aq) + A-(aq) The extent of dissociation for weak acids. Figure 16.2

  6. 1M HCl(aq) 1M CH3COOH(aq) Reaction of zinc with a strong and a weak acid. Figure 16.3

  7. HA(g or l) + H2O(l) H3O+(aq) + A-(aq) HA(aq) + H2O(l) H3O+(aq) + A-(aq) [H3O+][A-] stronger acid higher [H3O+] Kc = [H2O][HA] larger Ka [H3O+][A-] Kc[H2O] = Ka = [HA] smaller Ka lower [H3O+] weaker acid Strong acids dissociate completely into ions in water. Kc >> 1 Weak acids dissociate very slightly into ions in water. Kc << 1 The Acid-Dissociation Constant

  8. Table 16.2 ACID STRENGTH

  9. PROBLEM: Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base. PLAN: Pay attention to the text definitions of acids and bases. Look at O for acids as well as the -COOH group; watch for amine groups and cations in bases. (a)Strong acid - H2SeO4 - the number of O atoms exceeds the number of ionizable protons by 2. (b)Weak acid - (CH3)2CHCOOH is an organic acid having a -COOH group. (c)Strong base - KOH is a Group 1A(1) hydroxide. (d)Weak base - (CH3)2CHNH2 has a lone pair of electrons on the N and is an amine. SAMPLE PROBLEM 16.1: Classifying Acid and Base Strength from the Chemical Formula (a) H2SeO4 (b) (CH3)2CHCOOH (c) KOH (d) (CH3)2CHNH2 SOLUTION:

  10. H2O(l) H2O(l) OH-(aq) H3O+(aq) Autoionization of Water and the pH Scale + +

  11. H2O(l) + H2O(l) H3O+(aq) + OH-(aq) [H3O+][OH-] Kc = [H2O]2 The Ion-Product Constant for Water Kc[H2O]2 = 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-]

  12. Divide into Kw [H3O+] > [OH-] [H3O+] = [OH-] [H3O+] < [OH-] The relationship between [H3O+] and [OH-] and the relative acidity of solutions. Figure 16.4 [H3O+] [OH-] ACIDIC SOLUTION BASIC SOLUTION NEUTRAL SOLUTION

  13. SAMPLE PROBLEM 16.2: 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: Kw = 1.0x10-14 = [H3O+] [OH-] so [OH-] = Kw/ [H3O+] = 1.0x10-14/3.0x10-4 = 3.3x10-11M [H3O+] is > [OH-] and the solution is acidic.

  14. Figure 16.5 The pH values of some familiar aqueous solutions pH = -log [H3O+] pX = -log X

  15. Table 16.3 The Relationship Between Ka and pKa Acid Name (Formula) Ka at 250C pKa 1.02x10-2 Hydrogen sulfate ion (HSO4-) 1.991 3.15 7.1x10-4 Nitrous acid (HNO2) 4.74 1.8x10-5 Acetic acid (CH3COOH) 8.64 2.3x10-9 Hypobromous acid (HBrO) 1.0x10-10 Phenol (C6H5OH) 10.00

  16. Figure 16.6 The relations among [H3O+], pH, [OH-], and pOH.

  17. PLAN: HNO3 is a strong acid so [H3O+] = [HNO3]. Use Kw to find the [OH-] and then convert to pH and pOH. SAMPLE PROBLEM 16.3: Calculating [H3O+], pH, [OH-], and pOH PROBLEM: In an art restoration project, a conservator prepares copper-plate etching solutions by diluting concentrated HNO3 to 2.0M, 0.30M, and 0.0063M HNO3. Calculate [H3O+], pH, [OH-], and pOH of the three solutions at 250C. SOLUTION: For 2.0M HNO3, [H3O+] = 2.0M and -log [H3O+] = -0.30 = pH [OH-] = Kw/ [H3O+] = 1.0x10-14/2.0 = 5.0x10-15M; pOH = 14.30 For 0.3M HNO3, [H3O+] = 0.30M and -log [H3O+] = 0.52 = pH [OH-] = Kw/ [H3O+] = 1.0x10-14/0.30 = 3.3x10-14M; pOH = 13.48 For 0.0063M HNO3, [H3O+] = 0.0063M and -log [H3O+] = 2.20 = pH [OH-] = Kw/ [H3O+] = 1.0x10-14/6.3x10-3 = 1.6x10-12M; pOH = 11.80

  18. pH (indicator) paper pH meter Figure 16.7 Methods for measuring the pH of an aqueous solution

  19. An acid is a proton donor, any species which donates a H+. A base is a proton acceptor, any species which accepts a H+. Brønsted-Lowry Acid-Base Definition An acid-base reaction can now be viewed from the standpoint of the reactants AND the products. An acid reactant will produce a base product and the two will constitute an acid-base conjugate pair.

  20. Lone pair binds H+ + + HCl H2O Cl- H3O+ Lone pair binds H+ + + NH3 H2O NH4+ OH- Proton transfer as the essential feature of a Brønsted-Lowry acid-base reaction. Figure 16.8 (acid, H+ donor) (base, H+ acceptor) (base, H+ acceptor) (acid, H+ donor)

  21. Acid + Base Base + Acid 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- Table 16.4 The Conjugate Pairs in Some Acid-Base Reactions Conjugate Pair Conjugate Pair

  22. PROBLEM: The following reactions are important environmental processes. Identify the conjugate acid-base pairs. (a) H2PO4-(aq) + CO32-(aq) HPO42-(aq) + HCO3-(aq) (a) H2PO4-(aq) + CO32-(aq) HPO42-(aq) + HCO3-(aq) PLAN: Identify proton donors (acids) and proton acceptors (bases). (b) H2O(l) + SO32-(aq) OH-(aq) + HSO3-(aq) (b) H2O(l) + SO32-(aq) OH-(aq) + HSO3-(aq) SAMPLE PROBLEM 16.4: Identifying Conjugate Acid-Base Pairs conjugate pair2 conjugate pair1 SOLUTION: proton donor proton acceptor proton acceptor proton donor conjugate pair2 conjugate pair1 proton donor proton acceptor proton acceptor proton donor

  23. PROBLEM: Predict the net direction and whether Ka is greater or less than 1 for each of the following reactions (assume equal initial concentrations of all species): (a) H2PO4-(aq) + NH3(aq) HPO42-(aq) + NH4+(aq) (a)H2PO4-(aq) + NH3(aq) HPO42-(aq) + NH4+(aq) PLAN: Identify the conjugate acid-base pairs and then consult Figure 16.9 (button) to determine the relative strength of each. The stronger the species, the more preponderant its conjugate. (b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq) (b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq) stronger acid stronger base weaker base weaker acid weaker acid weaker base stronger base stronger acid SAMPLE PROBLEM 16.5: Predicting the Net Direction of an Acid-Base Reaction SOLUTION: Net direction is to the right with Kc > 1. Net direction is to the left with Kc < 1.

  24. Figure 16.9 Strengths of conjugate acid-base pairs

  25. PROBLEM: Phenylacetic acid (C6H5CH2COOH, simplified here as HPAc) builds up in the blood of persons with phenylketonuria, an inherited disorder that, if untreated, causes mental retardation and death. A study of the acid shows that the pH of 0.13M HPAc is 2.62. What is the Ka of phenylacetic acid? PLAN: Write out the dissociation equation. Use pH and solution concentration to find the Ka. SOLUTION: HPAc(aq) + H2O(l) H3O+(aq) + PAc-(aq) [H3O+][PAc-] Ka = [HPAc] SAMPLE PROBLEM 16.6: Finding the Ka of a Weak Acid from the pH of Its Solution With a pH of 2.62, the [H3O+]HPAc >> [H3O+]water. Assumptions: [PAc-] ≈ [H3O+]; since HPAc is weak, [HPAc]initial ≈ [HPAc]initial - [HPAc]dissociation

  26. Initial 0.12 - 1x10-7 0 Change -x - +x +x Equilibrium 0.12-x - x +(<1x10-7) x (2.4x10-3) (2.4x10-3) 0.12 1x10-7M [H3O+]from water; x100 2.4x10-3M 2.4x10-3M [HPAc]dissn; x100 0.12M SAMPLE PROBLEM 16.6: Finding the Ka of a Weak Acid from the pH of Its Solution continued Concentration(M) HPAc(aq) + H2O(l) H3O+(aq) + PAc-(aq) [H3O+] = 10-pH = 2.4x10-3 M which is >> 10-7 (the [H3O+] from water) x ≈ 2.4x10-3 M ≈ [H3O+] ≈ [PAc-] [HPAc]equilibrium = 0.12-x ≈ 0.12 M So Ka = = 4.8 x 10-5 Be sure to check for % error. = 4x10-3 % = 2.0%

  27. PLAN: Write out the dissociation equation and expression; make whatever assumptions about concentration which are necessary; substitute. Assumptions: For HPr(aq) + H2O(l) H3O+(aq) + Pr-(aq) Ka = [H3O+][Pr-] HPr(aq) + H2O(l) H3O+(aq) + Pr-(aq) [HPr] Initial 0.10 - 0 0 Change -x - +x +x Equilibrium 0.10-x - x x SAMPLE PROBLEM 16.7: Determining Concentrations from Ka and Initial [HA] PROBLEM: Propanoic acid (CH3CH2COOH, which we simplify and HPr) is an organic acid whose salts are used to retard mold growth in foods. What is the [H3O+] of 0.10M HPr (Ka = 1.3x10-5)? x = [HPr]diss = [H3O+]from HPr= [Pr-] SOLUTION: Concentration(M) Since Ka is small, we will assume that x << 0.10

  28. [H3O+][Pr-] (x)(x) 1.3x10-5 = = [HPr] 0.10 SAMPLE PROBLEM 16.7: Determining Concentrations from Ka and Initial [HA] continued = 1.1x10-3 M = [H3O+] Check: [HPr]diss = 1.1x10-3M/0.10 M x 100 = 1.1%

  29. [HA]dissociated x 100 Percent HA dissociation = [HA]initial [H3O+][PO43-] [H3O+][HPO42-] [H3O+][H2PO4-] H3PO4(aq) + H2O(l) H2PO4-(aq) + H3O+(aq) Ka2 = Ka3 = Ka1 = [H3PO4] [H2PO4-] [HPO42-] H2PO4-(aq) + H2O(l) HPO42-(aq) + H3O+(aq) HPO42-(aq) + H2O(l) PO43-(aq) + H3O+(aq) Polyprotic acids acids with more than more ionizable proton = 7.2x10-3 = 6.3x10-8 = 4.2x10-13 Ka1 > Ka2 > Ka3

  30. Table 16.5 ACID STRENGTH

  31. PLAN: Write out expressions for both dissociations and make assumptions. [HAsc-][H3O+] H2Asc(aq) + H2O(l) HAsc-(aq) + H3O+(aq) Ka1 = [H2Asc] [Asc2-][H3O+] HAsc-(aq) + H2O(l) Asc2-(aq) + H3O+(aq) Ka2 = [HAsc-] SAMPLE PROBLEM 16.8: Calculating Equilibrium Concentrations for a Polyprotic Acid PROBLEM: Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1 = 1.0x10-5 and Ka2 = 5x10-12) found in citrus fruit. Calculate [H2Asc], [HAsc-], [Asc2-], and the pH of 0.050M H2Asc. Ka1 >> Ka2 so the first dissociation produces virtually all of the H3O+. Ka1 is small so [H2Asc]initial ≈ [H2Asc]diss After finding the concentrations of various species for the first dissociation, we can use them as initial concentrations for the second dissociation. SOLUTION: = 1.0x10-5 = 5x10-12

  32. H2Asc(aq) + H2O(l) HAsc-(aq) + H3O+(aq) Initial 0.050 - 0 0 Change - x - + x + x Equilibrium 0.050 - x - x x x Concentration(M) HAsc-(aq) + H2O(l) Asc2-(aq) + H3O+(aq) Initial 7.1x10-4M - 0 7.1x10-4M Change - x - + x + x Equilibrium 7.1x10-4 - x - x 7.1x10-4M +x SAMPLE PROBLEM 16.8: Calculating Equilibrium Concentrations for a Polyprotic Acid continued Concentration(M) Ka1 = [HAsc-][H3O+]/[H2Asc] = 1.0x10-5 = (x)(x)/0.050 M x = 7.1x10-4 M pH = -log(7.1x10-4) = 3.15 X = Ka2 = 5x10-12 M

  33. Table16.6 [BH+][OH-] [B] BASE STRENGTH Kb =

  34. Lone pair binds H+ + H2O CH3NH2 methylamine + CH3NH3+ OH- methylammonium ion Abstraction of a proton from water by methylamine. Figure 16.11

  35. PROBLEM: Dimethylamine, (CH3)2NH, a key intermediate in detergent manufacture, has a Kb of 5.9x10-4. What is the pH of 1.5M (CH3)2NH? PLAN: Perform this calculation as you did those for acids. Keep in mind that you are working with Kb and a base. (CH3)2NH(aq) + H2O(l) (CH3)2NH2+(aq) + OH-(aq) (CH3)2NH(aq) + H2O(l) (CH3)2NH2+(aq) + OH-(aq) Concentration Initial 1.50M - 0 0 Change - x - + x + x Equilibrium 1.50 - x - x x SAMPLE PROBLEM 16.9: Determining pH from Kb and Initial [B] Assumptions: Kb >> Kw so [OH-]from water is neglible [(CH3)2NH2+] = [OH-] = x ; [(CH3)2NH] - x ≈ [(CH3)2NH]initial SOLUTION:

  36. [(CH3)2NH2+][OH-] Kb = 5.9x10-4 = [(CH3)2NH] (x) (x) 5.9x10-4 = 1.5M SAMPLE PROBLEM 16.9: Determining pH from Kb and Initial [B] continued x = 3.0x10-2M = [OH-] Check assumption: 3.0x10-2M/1.5M x 100 = 2% [H3O+] = Kw/[OH-] = 1.0x10-14/3.0x10-2 = 3.3x10-13M pH = -log 3.3x10-13 = 12.48

  37. PROBLEM: Sodium acetate (CH3COONa, or NaAc for this problem) has applications in photographic development and textile dyeing. What is the pH of 0.25M NaAc? Ka of acetic acid (HAc) is 1.8x10-5. PLAN: Sodium salts are soluble in water so [Ac-] = 0.25M. Concentration Ac-(aq) + H2O(l) HAc(aq) + OH-(aq) Initial 0.25M - 0 0 Change -x - +x +x Equilibrium 0.25M-x - x x [HAc][OH-] Kw 1.0x10-14 Kb = Kb = = [Ac-] Ka 1.8x10-5 SAMPLE PROBLEM 16.10: Determining the pH of a Solution of A- Write the association equation for acetic acid; use the Ka to find the Kb. SOLUTION: = 5.6x10-10M

  38. Kb = Check assumption: 1.2x10-5M/0.25M x 100 = 4.8x10-3 % [H3O+] = Kw/[OH-] = 1.0x10-14/1.2x10-5 = 8.3x10-10M [HAc][OH-] [Ac-] SAMPLE PROBLEM 16.10: Determining the pH of a Solution of A- continued [Ac-] = 0.25M-x ≈ 0.25M 5.6x10-10 = x2/0.25M x= 1.2x10-5M = [OH-] pH = -log 8.3x10-10M = 9.08

  39. 6A(16) 7A(17) H2O HF H2S HCl H2Se HBr H2Te HI Figure 16.12 The effect of atomic and molecular properties on nonmetal hydride acidity. Electronegativity increases, acidity increases Bond strength decreases, acidity increases

  40. H H H O O O Cl Cl Cl H O I > H O Br >       O O     O The relative strengths of oxoacids. Figure 16.13 <<

  41. Fe3+ Fe(H2O)63+(aq) 6 x 10-3 Sn2+ Sn(H2O)62+(aq) 4 x 10-4 Cr3+ Cr(H2O)63+(aq) 1 x 10-4 Al3+ Al(H2O)63+(aq) 1 x 10-5 Cu2+ Cu(H2O)62+(aq) 3 x 10-8 ACID STRENGTH Pb2+ Pb(H2O)62+(aq) 3 x 10-8 Zn2+ Zn(H2O)62+(aq) 1 x 10-9 Co2+ Co(H2O)62+(aq) 2 x 10-10 Ni2+ Ni(H2O)62+(aq) 1 x 10-10 Table 16.7 Ka Values of Some Hydrated Metal Ions at 250C Free Ion Hydrated Ion Ka

  42. Electron density drawn toward Al3+ Nearby H2O acts as base H2O H3O+ Al(H2O)63+ Al(H2O)5OH2+ Figure 16.13 The acidic behavior of the hydrated Al3+ ion.

  43. Table 16.8

  44. PROBLEM: Predict whether aqueous solutions of the following are acidic, basic, or neutral, and write an equation for the reaction of any ion with water: PLAN: Consider the acid-base nature of the anions and cations. Strong acid-strong base combinations produce a neutral solution; strong acid-weak base, acidic; weak acid-strong base, basic. SAMPLE PROBLEM 16.11: Predicting Relative Acidity of Salt Solutions (a) Potassium perchlorate, KClO4 (b) Sodium benzoate, C6H5COONa (c) Chromium trichloride, CrCl3 (d) Sodium hydrogen sulfate, NaHSO4 SOLUTION: (a) The ions are K+ and ClO4- , both of which come from a strong base(KOH) and a strong acid(HClO4). Therefore the solution will be neutral. (b) Na+ comes from the strong base NaOH while C6H5COO- is the anion of a weak organic acid. The salt solution will be basic. (c) Cr3+ is a small cation with a large + charge, so it’s hydrated form will react with water to produce H3O+. Cl- comes from the strong acid HCl. Acidic solution. (d) Na+ comes from a strong base. HSO4- can react with water to form H3O+. So the salt solution will be acidic.

  45. PROBLEM: Determine whether an aqueous solution of zinc formate, Zn(HCOO)2, is acidic, basic, or neutral. PLAN: Both Zn2+ and HCOO- come from weak conjugates. In order to find the relatively acidity, write out the dissociation reactions and use the information in Tables 16.2 and 16.7. Zn(H2O)62+(aq) + H2O(l) Zn(H2O)5OH+(aq) + H3O+(aq) HCOO-(aq) + H2O(l) HCOOH(aq) + OH-(aq) SAMPLE PROBLEM 16.12: Predicting the Relative Acidity of Salt Solutions from Ka and Kb of the Ions SOLUTION: Ka Zn(H2O)62+ = 1x10-9 Ka HCOO- = 1.8x10-4 ; Kb = Kw/Ka = 1.0x10-14/1.8x10-4 = 5.6x10-11 Ka for Zn(H2O)62+ >>> Kb HCOO-, therefore the solution is acidic.

  46. M2+ H2O(l) Molecules as Lewis Acids An acid is an electron-pair acceptor. A base is an electron-pair donor. acid base adduct M(H2O)42+(aq) complex ion

  47. Figure 16.15 The Mg2+ ion as a Lewis acid in the chlorophyll molecule.

  48. PROBLEM: Identify the Lewis acids and Lewis bases in the following reactions: (a) H+ + OH- H2O (a) H+ + OH- H2O (b) Cl- + BCl3 BCl4- (b) Cl- + BCl3 BCl4- PLAN: Look for electron pair acceptors (acids) and donors (bases). (c) K+ + 6H2O K(H2O)6+ (c) K+ + 6H2O K(H2O)6+ SAMPLE PROBLEM 16.13: Identifying Lewis Acids and Bases SOLUTION: acceptor donor donor acceptor acceptor donor

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