Ap chem unit 14 presentation part 1

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Ap chem unit 14 presentation part 1

  1. 1. UNIT 14: ACIDS AND BASES• The Nature of Acids and Bases• Acid Strength• The pH Scale• Calculating the pH of Strong Acid Solutions• Calculating the pH of Weak Acid Solutions• Bases• Polyprotic Acids• Acid-Base Properties of Salts• The Effect of Structure on Acid-Base Properties• Acid-Base Properties of Oxides• The Lewis Acid-Base Model• Strategy for Solving Acid-Base Problems
  2. 2. ARRHENIUS CONCEPTThe first person to recognize the essential nature of acids and bases was Svante Arrhenius. He stated that acids produce hydrogen ions in aqueous solutions, while bases produce hydroxide ion.• Major step in quantifying acid-base chemistry.• Concept is very limited because it applies to aqueous solutions only and it only allows for only one kind of base – the hydroxide ion.
  3. 3. BRONSTED-LOWRY MODELA more general definition of acids and bases was suggested by the Danish chemist, Bronsted, and the English chemist Lowry. An acid is defined as a proton (H+) donor, and a base is a proton acceptor.• includes reactions of gases and not just aqueous reactions.
  4. 4. WATER AS A BASEWater can act as a base because the water molecule has two unshared electrons pairs that allow a covalent bond with a proton (H+). + - HA(aq) + H 2O(l ) Û H 3O(aq) + A(aq)• When water accepts a proton, H3O+, hydronium is formed.• HA is the acid (proton donor) and H2O is the base (proton acceptor).
  5. 5. CONJUGATE ACID-BASE PAIRA conjugate acid-base pair consists of two substances that are related to each other by the donation and acceptance of a proton.• H3O+is the conjugate acid. The result of the accepted proton.• A-is the conjugate base. The result of the donated proton.• These two components are considered conjugate acid base pairs. + - HA(aq) + H 2O(l ) Û H 3O (aq) +A (aq)
  6. 6. CONJUGATE ACID-BASE PAIRH2O and A-, the base and conjugate base are both competing for a proton.• If H2O is a stronger base, the reaction will lie to the right because water will accept the proton before A- will. + - HA(aq) + H 2O(l ) Þ H3O (aq) +A (aq)
  7. 7. CONJUGATE ACID-BASE PAIRH2O and A-, the base and conjugate base are both competing for a proton.• If A- is a stronger base, the reaction will lie to the left because A- will accept the proton before H2O will. + - HA(aq) + H 2O(l ) Ü H3O (aq) +A (aq)
  8. 8. ACID DISSOCIATION CONSTANTThe equilibrium expression for an acid- base reaction gives Ka, the acid dissociation constant: + - + - [H 3O ][A ] [H ][A ] Ka = = [HA] [HA]**Note that H2O is not in the expression and H+ is commonly substituted for hydronium. + - HA(aq) + H 2O(l ) Û H 3O (aq) +A (aq)
  9. 9. ACID DISSOCIATION CONSTANTStrong acids have large dissociation constants. [H 3O + ][A - ] [H + ][A - ] Ka = = [HA] [HA]Weak acids have small dissociation constants. + - HA(aq) + H 2O(l ) Û H 3O (aq) +A (aq)
  10. 10. PRACTICE PROBLEM #1Write the simple dissociation (ionization) reaction (omitting water) for each of the following acids.a. hydrochloric acidb. acetic acidc. the ammonium iond. the anilinium ion (C6H5NH3+)e. the hydrated aluminum (III) ion [Al(H2O)6]3+
  11. 11. ACID STRENGTHThe strength of an acid is defined by the equilibrium position of its dissociation reaction.• A strong acid has an equilibrium that lies to the right. • Most HA has dissociated.• A strong acid produces a weak conjugate base. • Water is the stronger base + - HA(aq) + H 2O(l ) Þ H3O (aq) +A (aq)
  12. 12. ACID STRENGTHThe strength of an acid is defined by the equilibrium position of its dissociation reaction.• A weak acid has an equilibrium that lies to the left. • Most HA is still present.• A weak acid produces a strong conjugate base. • Water is the weaker base + - HA(aq) + H 2O(l ) Ü H3O (aq) +A (aq)
  13. 13. ACID STRENGTH SUMMARYAcid strength and conjugate base strength are indirectly related.
  14. 14. COMMON STRONG ACIDS1. HCl2. HNO33. HClO44. H2SO4 Sulfuric acid is a diprotic acid, an acid with two acidic protons. H2SO4 is a strong acid and dissociates completely, but HSO4- is a weak acid.
  15. 15. COMMON WEAK OXYACIDS1.H3PO42.HNO23.HOCl• Most acids are oxyacids, in which the acidic proton is attached to an oxygen atom.
  16. 16. ORGANIC ACIDSOrganic acids are those with a carbon atom “backbone”, commonly contain the carboxyl group:Acids of this type are usually weak.1. acetic acid (HC2H3O2 or CH3COOH)2. benzoic acid (C6H5COOH)*note: the rest of the hydrogens are notacidic
  17. 17. COMMON MONOPROTIC ACIDS
  18. 18. PRACTICE PROBLEM #2Using Table 14.2, arrange the following species according to their strengths as bases: H2O, F-, Cl-, NO2-, and CN-.Cl-< H2O < F-< NO2-, < CN-
  19. 19. WATER AS AN ACID AND A BASEA substance that can behave either as an acid or as a base is called amphoteric.• Water is the most common amphoteric substance by autoionization. + - H 2O(l) + H 2O(l) Û H3O (aq) + OH (aq)
  20. 20. WATER AS AN ACID AND A BASEThe equilibrium expression for this reaction is the dissociation constant of water, also called the ion-product constant. + - + - Kw = [H3O ][OH ] = [H ][OH ]• At 25°C, the ion-product constant, Kw, = 1.0 x 10-14• [H+] = [OH-]= 1.0 x 10-7 M + - H 2O(l) + H 2O(l) Û H3O (aq) + OH (aq)
  21. 21. USING THE ION-PRODUCT CONSTANTIt is important to recognize the meaning of Ks. In any aqueous solution at 25°C, no matter what it contains, the product of [H+] and [OH-] must always equal 1.0 x 10-14.• [H+] = [OH-] : neutral solution• [H+] >[OH-] : acidic solution• [H+] <[OH-] : basic solution
  22. 22. PRACTICE PROBLEM #3Calculate [H+] or [OH-] as required for each of the following solutions at 25°C, and state whether the solution is neutral, acidic, or basic.a. 1.0 x 10-5 M OH-b. 1.0 x 10-7 M OH-c. 10.0 M H+a. basic b. neutral c. acidic
  23. 23. PRACTICE PROBLEM #4At 60°C, the value of Kw is 1 x 10-13. + - 2H 2O(l ) Û H 3O(aq) + OH (aq)a. Using Le Chatelier’s principle, predict whether the below is exothermic or endothermic.b.Calculate [H+] and [OH-] in a neutral solution at 60°C.Kw increases with temperature – endothermic[H+] =[OH-] = 3 x 10-7 M
  24. 24. PHThe pH scale represents solution acidity with a log based scale. + pH = -log[H ]• [H+] = 1.0 x 10-7• pH = 7.00*The number of decimal places in the log isequal to the number of sig figs in the originalnumber
  25. 25. PHBecause the pH scale is based on a scale of 10, the pH changes by 1 for every power of 10 change in [H+]• example: a solution of pH = 3 has 10 times the concentration of that a solution of pH 4 and 100 times the concentration of a solution of pH 5.
  26. 26. POHThe pOH scale is similar to the pH scale - pOH = -log[OH ]• [OH-] = 1.0 x 10-7• pOH = 7.00*Because the ion product constant of water,pOH + pH always = 14.
  27. 27. PRACTICE PROBLEM #5Calculate pH and POH for each of the following solutions at 25°C.a. 1.0 x 10-3 M OH-b.1.0 M OH-a. pH=11.00, pOH=3.00 b. pH=0.00,pOH=14.00
  28. 28. PRACTICE PROBLEM #6The pH of a sample of human blood was measured to be 7.41 at 25°C. Calculate pOH, [H+], and [OH-] for the sample.pOH=6.59, [H+]= 3.9x10-8, [OH-]= 2.6x10-7M
  29. 29. SOLVING ACID-BASE PROBLEMSStrategies:1. Think chemistry: Focus on the solution components and their reactions. Choose the reaction that is most important.2. Be systematic: Acid-base problems require a step by step approach.3. Be flexible: Treat each problem as a separate entity. Look for both similarities and the differences to other problems.
  30. 30. SOLVING ACID-BASE PROBLEMSStrategies:4. Be patient: The complete solution to a complicated problem cannot be seen immediately in all its detail. Pick the problem apart into its workable steps.5. Be confident: Do not rely on memorizing solutions to problems. Understand and think, do not memorize.
  31. 31. CALCULATING THE PH OF STRONG ACID SOLUTIONSWhen calculating acid-base equilibria, focus on the main solution components:• example: 1.0 M HCl contains virtually no HCl molecules. Because HCl is a strong acid, it is completely dissociated. The main solution components are H+, Cl-, and H2O.• The first step of solving acid-base problems is the writing of the major species in the solution.
  32. 32. PRACTICE PROBLEM #7, PART 1Calculate the pH of 0.10 M HNO3.major species:H+, NO-3, and H2Oconcentration of species contributing:H+ = 0.10 M • H+ from autoionization of water is negligible compared to H+ from HNO3.pHpH = 1.00
  33. 33. PRACTICE PROBLEM #7 PART 2Calculate the pH of 1.0 x 10-10 M HCl.major species:H+, Cl-, and H2Oconcentration of species contributing: H+from HCl is negligible.H+ from autoionization of water is most important.pHpH = 7.00
  34. 34. CALCULATING THE PH OF A WEAK ACIDWe will walk through a systematic approach to solving pH for a weak acid:• Look for major species.• Narrow down contributing species and its concentration. Write balanced equation• Write equilibrium expression.• ICE• Use approximations when able• Check validity
  35. 35. CALCULATING THE PH OF A WEAK ACIDWhat is the pH of a 1.00 M solution of HF, Ka= 7.2x10-4Major Species:HF and H2O because both have small dissociation constants.Concentration of species contributing:H+ from HF is far more significant than H+ from H2OKa = 7.2x10-4vs Kw= 1.0x10-14
  36. 36. CALCULATING THE PH OF A WEAK ACIDWhat is the pH of a 1.00 M solution of HF, Ka= 7.2x10-4strategy:pH  [H+]  Ka [H + ][F - ] K a = 7.2x10 -4 = [HF]initial[HF] = 1.0 M Equilibrium [H+] = ?ICE!
  37. 37. CALCULATING THE PH OF A WEAK ACIDWhat is the pH of a 1.00 M solution of HF, Ka= 7.2x10-4ICE: Initial Change Equilibriu m H+ F- HF [H + ][F - ] K a = 7.2x10 -4 = [HF]
  38. 38. CALCULATING THE PH OF A WEAK ACIDWhat is the pH of a 1.00 M solution of HF, Ka= 7.2x10-4ICE: Initial Change Equilibriu m H+ 0 x x F- 0 x x HF 1.0 -x 1.0-x [H + ][F - ] K a = 7.2x10 -4 = [HF]
  39. 39. CALCULATING THE PH OF A WEAK ACIDWhat is the pH of a 1.00 M solution of HF, Ka= 7.2x10-4 [x][x] K a = 7.2x10 -4 = [1.0 - x] 1.0 - x @ 1.0• Since Ka for HF is so small, HF will dissociate only slightly, and x is expected to be small. If x is very small compared to concentration (2- 3+ exponents), the denominator can be simplified. + - [H ][F ] -4 K a = 7.2x10 = [HF]
  40. 40. CALCULATING THE PH OF A WEAK ACIDWhat is the pH of a 1.00 M solution of HF, Ka= 7.2x10- 4 [x][x] -4 K a = 7.2x10 = [1.0] -2 x @ 2.7x10The validity of this answer needs to be checked due to the approximation made in the calculations. Typically Ka values are known to an accuracy of ±5%. We will use this same accuracy for answers to be considered valid. + - -4 [H ][F ] K a = 7.2x10 = [HF]
  41. 41. CALCULATING THE PH OF A WEAK ACIDWhat is the pH of a 1.00 M solution of HF, Ka= 7.2x10-4Checking validity: x x100%x=2.7x10-2, [HF] = 1.00 M [HA]o 2.7x10 -2 x100% = 2.7% 1.0Mx is valid. [H+] = 2.7x10-2, pH=1.57 [H + ][F - ] K a = 7.2x10 -4 = [HF]
  42. 42. PRACTICE PROBLEM #8The hypochlorite ion (OCl-) is a strong oxidizing agent often found in household bleaches and disinfectants. It is also the active ingredient that forms when swimming pool water is treated with chlorine. In addition to its oxidizing abilities, the hypochlorite ion has a relatively high affinity for protons (it is much stronger base than Cl-, for example) and forms the weakly acidic hypochlorous acid (HOCl, Ka=3.5x10-8). Calculate the pH of a 0.100M aqueous solution of hypochlorous acid.
  43. 43. PRACTICE PROBLEM #8Major species:HOCl and H2ONarrow down contributing species and itsconcentration. Write balanced equationHOClKa=3.5x10-8 significant ≅ 0.100MH2O Kw=1.0x10-14 negligible + - HOCl(aq) Û H (aq) + OCl (aq)
  44. 44. PRACTICE PROBLEM #8Write equilibrium expression. [H + ][OCl - ] Ka = [HOCl]ICE Initial Change Equilibriu m H+ OCl- HOCl
  45. 45. PRACTICE PROBLEM #8Use approximations when able: (x)(x) Ka = (.100 - x) x2 3.5x10 -8 = (@ .100)x≅5.9x10-5
  46. 46. PRACTICE PROBLEM #8Check validity x 5.7x10 -5 x100% = x100% = 0.059% [HA] 0.100pH:[H+] = 5.7x10-8, pH=4.23
  47. 47. PRACTICE PROBLEM #9Calculate the pH of a solution that contains 1.00M HCN (Ka=6.2 x 10-10) and 5.00 M HNO2 (Ka=4.0 x 10-4). Also calculate the concentration of cyanide ion (CN-) in this solution at equilibrium.
  48. 48. PRACTICE PROBLEM #9Major species:Narrow down contributing species and its concentration. Write balanced equation
  49. 49. PRACTICE PROBLEM #9Write equilibrium expression.ICE Initial Change Equilibriu m H+ NO2- HNO2
  50. 50. PRACTICE PROBLEM #9Use approximations when able:Check validity.pH= 1.35 But wait….what about CN-?
  51. 51. PRACTICE PROBLEM #9Calculate the pH of a solution that contains 1.00M HCN (Ka=6.2 x 10-10) and 5.00 M HNO2 (Ka=4.0 x 10-4). Also calculate the concentration of cyanide ion (CN-) in this solution at equilibrium.We also want to calculate the CN- concentration from the dissociation of HCN HCN(a)<-> H+(aq)+ CN-(aq)Ka=6.2x10-10
  52. 52. PRACTICE PROBLEM #9The molarity of the HCN is known and the [H+]is equal to the prominent contribution of HNO2. The source of the H+ ions is not important in equilibrium. [H + ][CN - ] K a = 6.2x10 -10 = [HCN ][CN-]=1.4x10-8M [4.5x10 -2 ][CN - ] 6.2x10 -10 = [1.00]
  53. 53. PERCENT DISSOCIATIONThe percent dissociation is: [amount dissociated]%dissociation = x100% [initial concentration]For a given weak acid, the percent dissociation increases as the acid becomes more dilute.
  54. 54. PRACTICE PROBLEM #10Calculate the percent dissociation of acetic acid (Ka = 1.8 x 10-5) in each of the following solutions.a. 1.00 M HC2H3O2Major species:HC2H3O2 and H2ONarrow down contributing species and itsconcentration. Write balanced equationHC2H3O2 Ka = 1.8 x 10-5, H2O Kw = 1.0 x 10-14 HC2H3O2(aq) <-> H+(aq) + C2H3O2-(aq) [H + ][C2 H3O2 ] - K a = 1.8x10 -5 = [HC2 H3O2 ]
  55. 55. PRACTICE PROBLEM #10Calculate the percent dissociation of acetic acid (Ka = 1.8 x 10-5) in each of the following solutions.a. 1.00 M HC2H3O2 Initial Change EquilibriuICE m H+ C2H3O2- HC2H3O 2 [H + ][C2 H3O2 ] - K a = 1.8x10 -5 = [HC2 H3O2 ]
  56. 56. PRACTICE PROBLEM #10Calculate the percent dissociation of acetic acid (Ka = 1.8 x 10-5) in each of the following solutions.a. 1.00 M HC2H3O2Use approximations when able: 4.2x10 -3Apply Equation x100% = 0.42% 1.00 [H + ][C2 H3O2 ] - K a = 1.8x10 -5 = [HC2 H3O2 ]
  57. 57. PRACTICE PROBLEM #10Calculate the percent dissociation of acetic acid (Ka = 1.8 x 10-5) in each of the following solutions.b. 0.100 M HC2H3O2Major species:HC2H3O2 and H2ONarrow down contributing species and itsconcentration. Write balanced equation HC2H3O2 Ka = 1.8 x 10-5, H2O Kw = 1.0 x 10-14HC2H3O2(aq) <-> H+(aq) + C2H3O2-(aq)
  58. 58. PRACTICE PROBLEM #10Calculate the percent dissociation of acetic acid (Ka = 1.8 x 10-5) in each of the following solutions.b. 0.100 M HC2H3O2 Initial Change EquilibriuICE m H+ C2H3O2- HC2H3O 2 [H + ][C2 H3O2 ] - K a = 1.8x10 -5 = [HC2 H3O2 ]
  59. 59. PRACTICE PROBLEM #10Calculate the percent dissociation of acetic acid (Ka = 1.8 x 10-5) in each of the following solutions.b. 0.100 M HC2H3O2Use approximations when able: 1.3x10 -3Apply Equation x100% = 1.3% 0.10 [H + ][C2 H3O2 ] - K a = 1.8x10 -5 = [HC2 H3O2 ]
  60. 60. PERCENT DISSOCIATIONThe previous example is a perfect example of the percent dissociation relationship.• For a given weak acid, the percent dissociation increases as the acid becomes more dilute.
  61. 61. PRACTICE PROBLEM #11Lactic acid (HC3H5O3) is a waste product that accumulates in muscle tissue during exertion, leading to pain and a feeling of fatigue. In a 0.100 M aqueous solution, lactic acid is 3.7% dissociated. Calculate the value of Ka for this acid.
  62. 62. PRACTICE PROBLEM #11Major species:HC3H5O3and H2ONarrow down contributing species and itsconcentration. Write balancedequationHC3H5O3(aq)<-> H+(aq) + C3H5O3-(aq) + - [H ][C3H 5O ] Ka = 3 [HC3H 5O3 ]
  63. 63. PRACTICE PROBLEM #11Percent Dissociation x x -33.7% = 100% = 100%, x = 3.7x10 [HC3 H 5O3 ] (.10)Substitute in the expression [H + ][C3H 5O3 ] (3.7x10 -3 )(3.7x10 -3 ) - Ka = = [HC3H 5O3 ] 0.10Ka=1.4x10-4 [H + ][C3H 5O3 ] - Ka = [HC3H 5O3 ]

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