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Chapter 18

Chapter 18. Acid-Base Equilibria. Acid-Base Equilibria. 18.1 Acids and Bases in Water. 18.2 Autoionization of Water and the pH Scale. 18.3 Proton Transfer and the Brønsted-Lowry Acid-Base Definition. 18.4 Solving Problems Involving Weak-Acid Equilibria.

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Chapter 18

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  1. Chapter 18 Acid-Base Equilibria

  2. Acid-Base Equilibria 18.1 Acids and Bases in Water 18.2 Autoionization of Water and the pH Scale 18.3 Proton Transfer and the Brønsted-Lowry Acid-Base Definition 18.4 Solving Problems Involving Weak-Acid Equilibria 18.5 Weak Bases and Their Relations to Weak Acids 18.6 Molecular Properties and Acid Strength 18.7 Acid-Base Properties of Salt Solutions 18.8 Generalizing the Brønsted-Lowry Concept: The Leveling Effect 18.9 Electron-Pair Donation and the Lewis Acid-Base Definition

  3. The Nature of Acids and Bases: • Acids: Acids taste sour. React with metals and produce H gas. turns blue litmus red pH < 7 • * Bases: Bases taste bitter. They are slippery. turns red litmus blue. pH >7

  4. Arrhenius concept • Acids produce H+ ions. Bases produce OH- ions. • This concept is limited because it applies to only aqueous solution and defines only OH containing bases.

  5. Neutralization: • acid + base _______salt + water.

  6. Acid Dissociation Constant (Ka) • Write Ka expression for strong and weak acids.

  7. Strong acid: HA(g or l) + H2O(l) H2O+(aq) + A-(aq) The extent of dissociation for strong acids. Figure 18.1

  8. Weak acid: HA(aq) + H2O(l) H2O+(aq) + A-(aq) The extent of dissociation for weak acids. Figure 18.2

  9. 1M HCl(aq) 1M CH3COOH(aq) Reaction of zinc with a strong and a weak acid. Figure 18.3

  10. 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

  11. Classifying the Relative Strengths of acids and Bases: • Strong acids: • HCl, HBr, HI • Oxo acids. HNO3, H2SO4, HClO4 • Weak acids: • HF • HCN , H2S (H not bonded to O or halogen) • Oxo acids. HClO, HNO2, H3PO4 • Carboxylic acids. CH3COOH

  12. Classifying the Relative Strengths of acids and Bases: • Strong bases: • M2O, MOH M= (group 1A metal) • MO , M(OH)2 M=group 2A metal • Weak bases: (N atom and lone pair of electrons) • NH3 • Amines.

  13. ACID STRENGTH

  14. 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 18.1: Classifying Acid and Base Strength from the Chemical Formula (a) H2SeO4 (b) (CH3)2CHCOOH (c) KOH (d) (CH3)2CHNH2 SOLUTION:

  15. Autoionization of water and the pH scale • Water dissociates into its ions.

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

  17. 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-]

  18. Divide into Kw [H3O+] > [OH-] [H3O+] = [OH-] [H3O+] < [OH-] The relationship between [H3O+] and [OH-] and the relative acidity of solutions. Figure 18.4 [H3O+] [OH-] ACIDIC SOLUTION BASIC SOLUTION NEUTRAL SOLUTION

  19. 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: 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.

  20. pH scale • pH log[H+] • pH in water ranges from 0 to 14. • Kw = 1.00  1014 = [H+] [OH] • pKw = 14.00 = pH + pOH • As pH rises, pOH falls (sum = 14.00). • pH=7-neutral, pH>7-basic, pH<7-acidic

  21. pH scale • Acidic solns have a higher pOH. • pK=-log K • equation reaches equ, mostly products present, low pK (high K) • Reverse of the above statement is also true. • pH is measure using pH meter, pH paper or acid-base indicator.

  22. pH (indicator) paper pH meter Figure 18.7 Methods for measuring the pH of an aqueous solution

  23. Figure 18.5 The pH values of some familiar aqueous solutions pH = -log [H3O+]

  24. Table 18.3 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

  25. Figure 18.6 The relations among [H3O+], pH, [OH-], and pOH.

  26. Problems • P.3.Calculate pH and pOH at 25 C for: 1.0 M H+ • P.4.pH=6.88, Calculate [ H+] and [ OH-] for this sample. • P.5.Calculate pH of 0.10 M HNO3 • P.6.Calculate pH of 1.0 x 10-10 HCl.

  27. Bronsted-Lowry model: • An acid is a proton (H+ ) donor. A base is a proton acceptor. • * A conjugate acid-base pair only differs by one H. • HCl + H2O _________ H3O+ Cl- Acid base conj acid conj base • HA(aq) + H2O (l)  H3O+ (aq) +A-(aq) CA1 CB2 CA2 CB1

  28. Bronsted-Lowry model: • conjugate base: everything that remains of the acid molecule after a proton is lost. • Has one H less and one more minus charge than the acid. • conjugate acid: formed when the proton is transferred to the base. • Has one more H and one less charge than the base. • Do Follow up problem-18.4-Page.779

  29. 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.

  30. 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 18.4 The Conjugate Pairs in Some Acid-Base Reactions Conjugate Pair Conjugate Pair

  31. 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 18.8 (acid, H+ donor) (base, H+ acceptor) (base, H+ acceptor) (acid, H+ donor)

  32. 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 18.4: 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

  33. 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 18.10 (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 18.5: 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.

  34. Figure 18.9 Strengths of conjugate acid-base pairs

  35. Solving Problems Involving Weak-Acid Equilibria. • Write balanced equation and Ka expression. • Make an ICE table. • Make required assumptions. • Substitute values and solve for x. • Verify assumptions by calculating % error.

  36. Problems • P.7. Calculate the pH of 1.00 M aqueous solution of HF. Ka=7.2 x 10-4 • P.8 Calculate the pH of 0.100 M aq solution of HOCl. Ka=3.5 x 10-8

  37. 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 0.12M 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 18.6: 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

  38. 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 18.6: 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%

  39. 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 18.7: 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

  40. [H3O+][Pr-] (x)(x) 1.3x10-5 = = [HPr] 0.10 SAMPLE PROBLEM 18.7: 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%

  41. Percent dissociation: • For a weak acid % dissociation increases as the acid becomes more dilute. • For solution of any weak acid, [ H+] decreases as [HA]0 decreases, but % dissociation increases as [HA]0 increases.

  42. Problems P.11.Calculate the % dissociation of acetic acid (Ka= 1.8 x 10-5 ) 1.00 M and 0.100 M solutions. P.12.In a 0.100 M HC3H5O3 aqueous solution, lactic acid is 3.7 % dissociated. Calculate ka for the acid.

  43. [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

  44. ACID STRENGTH

  45. Polyprotic Acids • They can furnish more than one proton (H+) to the solution. • Characteristics: 1. Ka values are much smaller than the first value , so only the first dissociation step makes a significant contribution to equilibrium concentration of H+ . • 2. For sulfuric acid, it behaves as a strong acid in the first dissociation step and a weak acid in the second step, where its contribution of H+ ions can be neglected. Use quadratic equation to solve such kind of a problem.

  46. Problems P.13.Calculate the pH of a 5.0 M H3PO4 solution and the equilibrium concentration of the species H3PO4 , H2PO4-, HPO42-, PO43- P.14.Calculate the pH of 1.0 M sulfuric acid solution.

  47. 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 18.8: 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

  48. 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 0 Change - x - + x + x Equilibrium 7.1x10-4 - x - x x x SAMPLE PROBLEM 18.8: 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 = 6x10-8 M

  49. Bases • “Strong” and “weak” are used in the same sense for bases as for acids. • strong = complete dissociation (hydroxide ion supplied to solution) • NaOH(s)  Na+(aq) + OH(aq) • weak = very little dissociation (or reaction with water) • H3CNH2(aq) + H2O(l)  H3CNH3+(aq) + OH(aq)

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