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CHEM160 General Chemistry II Lecture Presentation Aqueous Equilibria: Acids and Bases

CHEM160 General Chemistry II Lecture Presentation Aqueous Equilibria: Acids and Bases. Chapter 16. Dr. Daniel Autrey. Why study acids and bases? . Acids and bases are common in the everyday world as well as in the lab. Some common acidic products

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CHEM160 General Chemistry II Lecture Presentation Aqueous Equilibria: Acids and Bases

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  1. CHEM160 General Chemistry IILecture PresentationAqueous Equilibria: Acids and Bases Chapter 16 Dr. Daniel Autrey Chapter 16

  2. Why study acids and bases? • Acids and bases are common in the everyday world as well as in the lab. • Some common acidic products • vinegar (dilute acetic acid), soft drinks (carbonic acid), aspirin (acetylsalicylic acid), vitamin C (ascorbic acid), lemonade (citric acid + ascorbic acid), muriatic acid (hydrochloric acid) • Some common basic products • Antacids such as milk of magnesia (magnesium hydroxide) and tums (calcium carbonate), household ammonia, oven and drain cleaners (sodium hydroxide), some bleaches (sodium hydroxide) Chapter 16

  3. Why study acids and bases? • Many biological and geological processes involve acid-base chemistry. • Gastric juice contains hydrochloric acid • Lactic acid builds up in muscles during strenuous exercise • Basicity of blood must be maintained within a certain narrow range or death can result • Acidity/basicity of soil and water are of importance to animals and plants living there • Cave formation and weathering of rocks is affected by acidity of water Chapter 16

  4. Why study acids and bases? • 15 of the top 50 chemicals produced in the largest quantities annually in the U.S. are acids or bases. (#1 = sulfuric acid) • Used as reactants and catalysts in manufacture of various consumer and industrial products • Plastics, synthetic fibers, detergents, pharmaceuticals, agricultural fertilizers, explosives, etc. Chapter 16

  5. Why study acids and bases? • So why study this acid-base stuff? • Acid-base reactions constitute an important class of chemical reactions • Understanding acid-base chemistry is necessary for chemistry, biology, geology, and other related scientific disciplines Chapter 16

  6. Concepts to Review • Aqueous Reactions and Solution Stoichiometry • Acids and bases, neutralization, electrolytes (section 4.3) • Molarity and molarity calculations (section 4.5) • Solution stoichiometry (section 4.7) • Chemical equilibrium • Equilibrium constants, solving equilibrium problems, LeChatelier’s principle (chapter 14) Chapter 16

  7. Arrhenius Concept • Arrhenius Concept (S. Arrhenius, 1887) • Acids produce hydrogen ions, H+, in water • H+ ions attach to H2O molecules forming hydronium ions, H3O+ • Bases produce hydroxide ions, OH-, in water • Base contains OH group in its formula. • Neutralization reaction: acid + base  salt + water Chapter 16

  8. Arrhenius Concept • Arrhenius Concept (S. Arrhenius, 1887) • Acids produce hydrogen ions, H+, in water • H+ ions attach to H2O molecules forming hydronium ions, H3O+ • Bases produce hydroxide ions, OH-, in water • Base contains OH group in its formula. • Neutralization reaction: acid + base  salt + water • Problems • Some basic substances do not have OH- • Confined to H2O solutions • H+ does not exist free in water • Forms H3O+ Chapter 16

  9. The Hydrated Proton: Hydronium Ion Chapter 16

  10. Bronsted-Lowry Concept • Bronsted-Lowry Concept (J. Bronsted, T. Lowry, 1923) • Acid-base reaction involves proton transfer • HA + B  HB+ + A- • Acid: proton donor • Base: proton acceptor • does not have to have OH- in formula Chapter 16

  11. Bronsted-Lowry Concept Example: acid base Chapter 16

  12. Bronsted-Lowry Concept Example: acid base Chapter 16

  13. Bronsted-Lowry Concept • Bronsted-Lowry Concept (J. Bronsted, T. Lowry, 1923) • Acid-base reaction involves proton transfer • Acid: proton donor • Base: proton acceptor • does not have to have OH- in formula • Water is amphoteric Chapter 16

  14. Bronsted-Lowry Concept acid base base acid Chapter 16

  15. Bronsted-Lowry Concept • Bronsted-Lowry Concept (J. Bronsted, T. Lowry, 1923) • Acid-base reaction involves proton transfer • Acid: proton donor • Base: proton acceptor • does not have to have OH- in formula • Water is amphoteric • Acids & bases can be molecules or ions Chapter 16

  16. Bronsted-Lowry Concept acid base Chapter 16

  17. Bronsted-Lowry Concept • Bronsted-Lowry Concept (J. Bronsted, T. Lowry, 1923) • Acid-base reaction involves proton transfer • Acid: proton donor • Base: proton acceptor • does not have to have OH- in formula • Water is amphoteric • Acids & bases can be molecules or ions • Back reaction is also a proton transfer. Chapter 16

  18. Bronsted-Lowry Concept acid base base acid Chapter 16

  19. Bronsted-Lowry Concept acid base base acid Acid-base conjugate pairs Chapter 16

  20. Bronsted-Lowry Concept • Conjugate Acid-Base Pair (HA/A-, HB+/B) • Two chemical species whose formulas differ by only one proton acid  conj. base + H+ base + H+ conj. acid Chapter 16

  21. Acid-Base Strength • Strong acids • Ionize completely in water HX + H2O  H3O+ + X- • Weak acids • Ionize only partially in water HX + H2O  H3O+ + X- • Common weak acids: Numerous molecules and certain cations (we’ll learn more about these later) Chapter 16

  22. HX H3O+ X- Strong Acids Before Equilibrium At Equilibrium 100 % Chapter 16

  23. Acid-Base Strength • Common Strong Acids (Know these!) HCl HBr HI HNO3 HClO4 H2SO4 Chapter 16

  24. HX Weak Acids Before Equilibrium At Equilibrium HX < 100 % H3O+ X- Chapter 16

  25. Acid-Base Strength • Strong bases • Ionize completely in water NaOH  Na+ + OH- • Common strong bases: MOH and M(OH)2, M2O and MO (M = Grp IA and IIA metals) • Weak bases • Ionize only partially in water B + H2O  BH+ + OH- • Common weak bases: NR3 (R = H or other chemical species) and certain anions Chapter 16

  26. Relative Acid-Base Strength Strongest HClO4 ClO4- Weakest Acids H2SO4 HSO4- Bases HCl Cl- HNO3 NO3- H3O+ H2O HSO4- SO42- HNO2 NO2- HClO ClO- Weakest NH4+ NH3 Strongest Acids H2O OH- Bases OH- O2- H2 H- Strongest acid that can exist in H2O Strongest base that can exist in H2O Chapter 16

  27. More on Neutralization Reactions • Acids react with bases: • acid + base  salt + water or • acid + base  salt Chapter 16

  28. Autoionization of Water • In pure water, the following equilibrium occurs: H2O + H2O  H3O+ + OH- • The equilibrium constant for this reaction is the ion product, Kw: Kw = [H3O+][OH-] • [H3O+] = [OH-] = 1 x 10-7 M (25 °C) so Kw = 1 x 10-14 (25 °C) Chapter 16

  29. Example Problem 1(1a and 1b on Example Problem Handout) • Calculate the [OH-] in a solution with [H3O+] = 0.00028 M (25 °C). (ans.: [OH-] = 3.6 x 10-11 M) • Calculate the [H3O+] in a solution with [OH-] = 0.056 M (25 °C). (ans.: [H3O+] = 1.8 x 10-13 M) Chapter 16

  30. Example Problem 2(5a and 5b on Example Problem Handout) • Calculate the concentration of [H3O+] and [OH-] for the following aqueous solutions at 25 °C: • 0.0035 M HCl (ans.: [H3O+] = 0.0035 M, [OH-] = 2.9 x 10-12 M) (b) 0.22 M NaOH (ans.: [OH-] = 0.22 M, [H3O+] = 4.5 x 10-14 M) Chapter 16

  31. Problem Solving Strategy • Determine whether the acid or base is strong or weak. HCl and NaOH are strong electrolytes. • If autoionization is considered to be negligible (Cacid > 10-6 M, Cbase > 10-6 M), then: • For strong acids, [H3O+] = CAcid • For strong bases, [OH-] = CBase for MOH or [OH-] = 2CBase for M(OH)2 • [OH-] or [H3O+] are then calculated using the expression for Kw: • Kw = 1 x 10-14 = [H3O+] [OH-] Chapter 16

  32. pH and pOH Scales: Convenient ways to express acidity and basicity • We can represent the concentration of [H3O+] as • pH = -log [H3O+] • We can represent the concentration of [OH-] as • pOH = -log [OH-] • The p function simply means -log. For a number X: • pX = -log X Chapter 16

  33. pH and pOH Scales • A relationship between pH and pOH may be derived for all aqueous solutions: • Kw = [H3O+][OH-] • Take the -log of each side: -logKw = -log([H3O+][OH-]) -logKw = -log[H3O+] - log[OH-] pKw = pH + pOH 14.00 = pH + pOH ( at 25 °C) Chapter 16

  34. Example 3(2a on Example Problem Handout) • Calculate the pH and pOH of a solution in which [H3O+] = 0.00028 M at 25 °C. (ans.: pH = 3.55, pOH = 10.45) Chapter 16

  35. 2.5567839 log inv ln yx cos sin tan Exp 8 7 9 x 5 4 6  2 1 3 + +/- 0 · - pH Calculations with Calculators • To calculate pH from [H3O+]: • Enter [H3O+] • Press “log” key • Change sign by pressing “+/-” key • pOH is calculated in an analogous manner. Chapter 16

  36. Example 4(3a on Example Problem Handout) • Calculate the [H3O+] and [OH-] of a solution if the pH is 4.88. (ans.: [H3O+] = 1.3 x 10-5 M, [OH-] = 7.6 x 10-10 M) Chapter 16

  37. 0.00055783198 log inv ln yx cos sin tan Exp 8 7 9 x 5 4 6  2 1 3 + +/- 0 · - Calculating [H3O+] from pH • To calculate [H3O+] from pH: • Enter pH • Change sign by pressing “+/-” key • Take the antilog by pressing the “2nd/inv/shift” key, followed by the “log” key • [OH-] is calculated in an analogous manner. Chapter 16

  38. Classifying Aqueous Solutions [H3O+] [OH-] pH pOH neutral = 10-7 M = 10-7 M = 7.0 = 7.0 acidic > 10-7 M < 10-7 M < 7.0 > 7.0 basic < 10-7 M > 10-7 M > 7.0 < 7.0 Chapter 16

  39. Example 5(6a on Example Problem Handout) • Calculate the [H3O+], [OH-], pH, and pOH for a solution prepared by dissolving 0.100 moles HCl(g) into enough water to give a volume of 0.350 L. (ans.: [H3O+] = 0.29 M, [OH-] = 3.5 x 10-14 M, pH = 0.54, pOH = 13.46) Chapter 16

  40. Problem Solving Strategy • Identify the substance as an acid or base • Determine whether the acid or base is strong or weak • If strong acid, [H3O+] = CAcid • If strong base, [OH-] = CBase for MOH or [OH-] = 2CBase for M(OH)2 • Calculate CHA (moles/L) (= [H3O+]) • Calculate pH = -log[H3O+] Chapter 16

  41. Problem Solving Strategy • Calculate pOH from pOH = 14.00 - pH • Calculate [OH-] from either [OH-] = 10-pOH or [OH-] = Kw/ [H3O+] Chapter 16

  42. Example 6(7a on Example Problem Handout) • Calculate the [H3O+], [OH-], pH, and pOH for a solution prepared by dissolving 3.1 g KOH(s) (FW = 56 g/mol) into enough water to give a volume of 0.500 L. (ans.: [OH-] = 0.11 M, [H3O+] = 9.0 x 10-14 M, pOH = 0.96, pH = 13.04) Chapter 16

  43. Weak Acid and Base Ionizations • How do we calculate the [H3O+] and pH of weak acid or base solutions if these do not ionize completely? • Consider the specific weak acid or base dissociation equilibrium involved. • Perform equilibrium calculations to determine concentrations of species in the reaction mixture at equilibrium. Chapter 16

  44. Weak Acid Ionization • Consider the ionization of a weak acid: HA + H2O <=> H3O+ + A- • The equilibrium constant for this reaction is the acid dissociation constant, Ka: Ka = [H3O+][A-]/[HA] (Note that these are all equilibrium concentrations.) Chapter 16

  45. Ka’s for Some Weak Acids Increasing acid strength H+ ion that ionizes shown in red. Chapter 16

  46. Calculations with Ka • Ka can be used to calculate: • equilibrium concentrations • pH • percent ionization (Any equilibrium constant can be used to calculate equilibrium concentrations for solution species.) Chapter 16

  47. Calculations Using Ka • Basic Steps for Weak Acid Calculations Using Ka • Write balanced chemical equation and the expression for Ka • Look up value for Ka • For each chemical species involved in the equilibrium (except H2O), write: • Initial concentration • Equilibrium concentration • Let the change in the [H3O+] be the variable “x” • Substitute the equilibrium concentrations into Ka and solve for x using either • quadratic approach • simplified approach • Calculate pH, equilibrium concentrations, % ionization, etc., as specified in the problem. Chapter 16

  48. Example 7(8a on the Example Problem Handout) • Calculate the equilibrium concentration of each species, the pH, and the percent ionization for a 0.500 M solution of acetic acid, represented as either HOAc or CH3COOH (Ka = 1.8 x 10-5). (ans.: [HOAc] = 0.497 M, [OAc-] = [H3O+] = 0.0030 M, [OH-] = 3.3 x 10-12 M, 0.60%) Chapter 16

  49. Weak Acid Ionization • Some acids have more than one ionizable proton. • Called polyprotic acids • Ionize in steps • Consider the ionization of a weak diprotic acid: H2A + H2O <=> H3O+ + HA- Ka1 = [H3O+][HA-]/[H2A] HA- + H2O <=> H3O+ + A2- Ka2 = [H3O+][A2-]/[HA-] • The 1st ionization effectively controls solution pH for most polyprotic acids: • Ka1 >> Ka2 > Ka3, etc. • When calculating pH, ignore any ionization after the first ionization if Ka1 >> Ka2. Chapter 16

  50. Example 8(Example 9a on Example Problem Handout) • Calculate the equilibrium concentration of each species and the pH of a 0.250 M solution of H2CO3 (Ka1 = 4.3 x 10-7, Ka2 = 5.6 x 10-11). (ans.: [H2CO3] = 0.250 M, [HCO3-] = [H3O+] = 0.00033 M, [CO32-] = 5.6 x 10-11 M], [OH-] = 3.0 x 10-11 M, pH = 3.48) Chapter 16

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