Ap chem unit 15 presentation

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AP Acid Base Equilibria

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Ap chem unit 15 presentation

  1. 1. AP Chem Unit 15
  2. 2.  Solutions of Acids or Bases Containing a Common Ion Buffered Solutions Buffering Capacity Titrations and pH Curves Acid-Base Indicators
  3. 3. Most chemistry in the natural world takes placein an aqueous solution. One of the mostsignificant aqueous reactions is one with acidsand bases. Most living systems are verysensitive to pH and yet are subjected to acidsand bases. The main idea of this unit is to understand the chemistry of a buffered system. Buffered systems contain chemical components that enable a solution to be resistant to change in pH.
  4. 4. 15.1
  5. 5. Solutions that contain a weak acid and thesalt of it conjugate base create a common ionsystem. A common ion system can also be a weak base and the salt of its conjugate acid. ◦ Example: HF and NaF solution or NH3 and NH4Cl solution.
  6. 6.  The soluble salts of conjugate acids and conjugate bases are typically strong electrolytes and therefore dissociate 100%. This dissociation increases the concentration of the conjugate ion and a shift in equilibrium occurs according to Le Chatelier. + - ◦ Example: NH 4Cl(s) ® NH 4(aq) + Cl(aq) + - NH 3(aq) + H 2O(l ) « NH 4(aq) + OH (aq)  Shift Left!
  7. 7. The shift left in the second equation is knownas the common ion effect. NH 4Cl(s) ® NH + + Cl(aq) 4(aq) - NH 3(aq) + H 2O(l ) « NH + + OH (aq) 4(aq) - This effect makes a solution of NH3 and NH4Cl less basic than NH3 alone
  8. 8. The shift left in the second equation is knownas the common ion effect. + - NaF(s) ® Na(aq) + F(aq) + - HF(aq) + H 2O(l ) « H 3O(aq) + F(aq)This effect makes a solution of NaF and HF lessacidic than a solution of HF alone.
  9. 9. Calculations for common ion equilibria issimilar to weak acid calculations except thatthe initial concentrations of the anion are not0. Initial concentrations for the equilibrium calculation must include the concentration of ions from the complete dissociation of the salt.
  10. 10. The equilibrium concentration of H+ in a 1.0 MHF solution is 2.7x10-2M, and the percentdissociation of HF is 2.7%. Calculate [H+] andthe percent dissociation of HF in a solutioncontaining 1.0 M HF (Ka=7.2x10-4) and 1.0 MNaF. Major species? ICE: I C E HF F- H+
  11. 11. The equilibrium concentration of H+ in a 1.0 MHF solution is 2.7x10-2M, and the percentdissociation of HF is 2.7%. Calculate [H+] andthe percent dissociation of HF in a solutioncontaining 1.0 M HF (Ka=7.2x10-4) and 1.0 MNaF. [H + ][F - ] Ka = [HF] ANS: .072%
  12. 12. 15.2
  13. 13. The most important application of a commonion system is buffering. A buffered solution resists a change in its pH when either hydroxide ions or protons are added. ◦ Blood is a practical example of a buffered solution; it is buffered with carbonic acid and the bicarbonate ion (among others).
  14. 14.  A solution can be buffered at any pH by choosing the appropriate components. Smaller amounts of H+ or OH- can be added with little pH effect.
  15. 15. A buffered solution contains 0.50 M acetic acid(Ka= 1.8 x 10-5) and 0.50 M sodium acetate.Calculate the pH of this solution.pH = 4.74
  16. 16.  When OH- ions are added to a solution of a weak acid, the OH- ions react with the best source of H+ (weak acid) to make water: - - OH + HA ® A + H 2O The net result is that OH- ions are replaced by A- ions and water until they are all consumed.
  17. 17.  When H+ ions are added to a solution of a weak acid, the H+ ions react with the strong conjugate base. - + A + H ® HA The net result is that more weak acid is produced, but free H+ do not accumulate to make large changes in the pH.
  18. 18. Buffered solutions are simply solutions of weakacids or bases containing a common ion. The pHcalculations on buffered solutions require exactlythe same procedures introduced in Chapter 14. When a strong acid or base is added to a buffered solution, it is best to deal with the stoichiometry of the resulting reaction first. After the stoichiometric calculations are completed, then consider the equilbrium calculations.
  19. 19. Calculate the change in pH that occurs when0.010 mol solid NaOH is added to the bufferedsolution in #2 (0.50 M acetic and 0.50 M sodiumacetate). Compare this pH change with theoriginal (pH=4.74) and the pH that occurs with0.010 mol of solid NaOH is added to 1.0 L ofwater.
  20. 20.  Major species: OH- +HC2H3O2  C2H3O2- + H2O mole table: HC2H3O2 OH- C 2 H 3 O2 - H2O Before After
  21. 21. [H + ][C2 H 3O2 ] - K a = 1.8x10 -5 = [HC2 H 3O2 ] ICE table: I C E HC2H3O2 OH- C2H3O2- pH= 4.76 (+.02 off original) pH of just OH-=12.00
  22. 22. Buffered solutions work well as long as theconcentration of the salts and weakacids/bases in solution are much larger thanthe amount of OH- and H+ being added. The amount of [H+] being in a buffered solution is often solved using a rearrangement of the equilibrium expression: + [HA] [H ] = K a - [A ]
  23. 23. + [HA] This equation, [H ] = K a - , can be changed into another [A ] useful form by taking the negative log ofboth sides: æ [A - ] ö æ [base]ö pH = pK a + log ç ÷ = pK a + log ç [acid] ÷ è [HA] ø è øThis log form of the expression is called theHenderson-Hasselbalch Equation.
  24. 24. Henderson-Hasselbalch Equation: When usingthis equation it is often assumed that the theequilibrium concentrations of A_ and HA areequal to their initial concentrations due to thevalidity of most approximations. æ [A - ] ö æ [base]ö pH = pK a + log ç ÷ = pK a + log ç [acid] ÷ è [HA] ø è ø Since the initial concentrations of HA and A- are relatively large in a buffered solution, this assumption is generally acceptable.
  25. 25. Calculate the pH of a solution containing 0.75M lactic acid (Ka=1.4x10-4) and 0.25 M sodiumlactate. Lactic acid (HC3H5O3) is a commonconstituent of biologic systems. For example, itis found in milk and is present in humanmuscle tissue during exertion.pH = 3.38
  26. 26. Buffered solutions can be formed from a weakbase and the corresponding conjugate acid. Inthese solutions, the weak base B reacts withany H+ added: + + B + H ® BH The conjugate acid BH reacts with any added + OH-: + - BH + OH ® B + H 2O
  27. 27. A buffered solution contains 0.25 M NH3(Kb=1.8x10-5) and 0.40 M NH4Cl. Calculate thepH of this solution.pH=9.05
  28. 28. Calculate the pH of the solution that resultswhen 0.10 mol gaseous HCl is added to 1.0 Lof the buffered solution from #5 (.25M NH3and .40 NH4Cl).pH= 8.73
  29. 29. 1. Buffered solutions contain relatively large concentrations of a weak acid and the corresponding weak base. They can involve a weak acid HA and the conjugate base A- or a weak base B and the conjugate acid BH+.2. When H+ is added to a buffered solution, it reacts essentially to completion with the weak base: + - + + H + A ® HA or H + B ® BH
  30. 30. 3. When OH- is added to a buffered solution, it reacts essentially to completion with the weak acid present:OH - + HA ® A- + H 2O or OH - + BH + ® B + H 2O3. The pH in the buffered solution is determined by the ratio of the concentrations of the weak acid and weak base. As long as this ratio remains virtually constant, the pH will remain virtually constant. This will be the case as long as the concentration of the buffering materials (HA and A- or B and BH+) are large compared with the amounts of H+ or OH- added.
  31. 31. 15.3
  32. 32. The buffering capacity of a buffered solutionrepresents the amount of protons or hydroxideions the buffer can absorb without a significantchange in pH. The pH of a buffered solution is determined by the ratio [A-] / [HA]. The capacity of a buffered solution is determined by the magnitudes of [HA] and [A-].
  33. 33. Calculate the change in pH that occurs when0.010 mol of gaseous HCl is added to 1.0 L ofeach of the following solutions: ◦ Soln A: 5.00 M HC2H3O2 and 5.00 M NaC2H3O2 ◦ Soln B: 0.050 M HC2H3O2 and 0.050 M NaC2H3O2 Ka=1.8x10-5. Soln A: pH=4.74 Soln B: pH=4.56
  34. 34. The optimal buffering occurs when [HA] isequal to [A-]. This ratio of 1 resists the most amount of pH change. The pKa of the weak acid to be used in the buffer should be as close as possible to the desired pH. ◦ For example, if a buffered solution is needed with a pH of 4.0. The most effect buffering will occur when [HA] is equal to [A-] and the pKa of the acid is close to 4.0 or Ka=1.0x10-4.
  35. 35. A chemist needs a solution buffered at pH=4.30and can choose from the following acids andtheir sodium salts. Calculate the ratio [HA]/[A-]required for each system to yield a pH of 4.30.Which system will work best?a. chloroacetic acid (Ka=1.35x10-3)b. propanoic acid (Ka=1.3x10-5)c. benzoic acid (Ka=6.4x10-5)d. hypochlorous acid (Ka=3.5x10-8) benzoic acid
  36. 36. 15.4
  37. 37. An acid-base titration isoften graphed by plottingthe pH of the solution(y-axis) vs. the amount of titrant added (x-axis). Point (s) is the equivalencepoint/stoichiometric point.This is where [OH-]=[H+].
  38. 38. The net ionic reaction for a strong acid- strongbase titration is: H + + OH - ® H O (aq) (aq) 2 (l )To compute [H+] at any point in a titration, themoles of H+ that remains at a given point mustbe divided by the total volume of the solution. Titrations often include small amounts. The mole is usually very large in comparison. Millimole (mmol) is often used for titration amounts.
  39. 39. mol mmolM= = L ml mmol = ml(M )
  40. 40. 50.0 ml of 0.200 M HNO3 is titrated with 0.100M NaOH. Calculate the pH of the solution atthe following points during the titration.a. NaOH has not been added. Since HNO3 is a strong acid (completedissociation), pH = 0.699
  41. 41. 50.0 ml of 0.200 M HNO3 is titrated with 0.100 MNaOH. Calculate the pH of the solution at thefollowing points during the titration.b. 10.0 ml of 0.100 M NaOH has been added. H+ = 10 mmol – 1.0 mmol of OH- =9.0mmol 9.0 mmol H+/ 60 ml = 0.15 M pH=0.82
  42. 42. 50.0 ml of 0.200 M HNO3 is titrated with 0.100 MNaOH. Calculate the pH of the solution at thefollowing points during the titration.c. 20.0 mL of 0.100 M NaOH has been added.pH=0.942
  43. 43. 50.0 ml of 0.200 M HNO3 is titrated with 0.100 MNaOH. Calculate the pH of the solution at thefollowing points during the titration.d. 50.0 mL of 0.100 M NaOH has been added.pH=1.301
  44. 44. 50.0 ml of 0.200 M HNO3 is titrated with 0.100 MNaOH. Calculate the pH of the solution at thefollowing points during the titration.e. 100.0 mL of 0.100 M NaOH has been added.The stoichiometric point/equivalence point isreached. pH=7
  45. 45. 50.0 ml of 0.200 M HNO3 is titrated with 0.100 MNaOH. Calculate the pH of the solution at thefollowing points during the titration.f. 150.0 mL of 0.100 M NaOH has been added.pH = 12.40
  46. 46. 50.0 ml of 0.200 M HNO3 is titrated with 0.100 MNaOH. Calculate the pH of the solution at thefollowing points during the titration.g. 200.0 mL of 0.100 M NaOH has been added.pH=12.40
  47. 47. The results of these calculations are summarizedby this graph:The pH changes very gradually until the titrationis close to the equivalence point, then dramaticchange occurs. Near the E.P., small changesproduces large changes in the OH-/H+ ratio.
  48. 48. 1. Before the equivalence point, [H+] can be calculated by dividing the number of millimoles of H+ remaining by the total volume of the solution in mL.2. At the equivalence point, pH=7.03. After the equivalence point, [OH-] can be calculated by dividing the number of millimoles of excess OH- by the total volume of the solution. [H+] is calculated from Kw.
  49. 49. When acids don’t completely dissociate,calculations used previously need adjusted. To calculate [H+] after a certain amount of strong base has been added, the weak acid dissociation equilibrium must be used. Remember: that a strong base reacts to completion with a weak acid.
  50. 50. Calculating the pH Curve for a Weak Acid-Strong Base Titration is broken down into twosteps:1. A stoichiometric problem. The reaction of hydroxide ion with the weak acid is assumed to run to completion, and the concentrations of the acid remaining and the conjugate base formed are determined.2. An equilibrium problem: The position of the weak acid equilibrium is determined, and the pH is calculated.
  51. 51. 50.0 mL of 0.10 M acetic acid (HC2H3O2,Ka=1.8x10-5) is titrated with 0.10 M NaOH.a. NaOH has not been added.pH = 2.87
  52. 52. 50.0 mL of 0.10 M acetic acid (HC2H3O2,Ka=1.8x10-5) is titrated with 0.10 M NaOH.b. 10.0 mL of 0.10 M NaOH has been added.pH= 4.14
  53. 53. 50.0 mL of 0.10 M acetic acid (HC2H3O2,Ka=1.8x10-5) is titrated with 0.10 M NaOH.c. 25.0 mL of 0.10 M NaOH has been added.pH=4.74
  54. 54. 50.0 mL of 0.10 M acetic acid (HC2H3O2,Ka=1.8x10-5) is titrated with 0.10 M NaOH.d. 40.0 mL of 0.10 M of NaOH has been added.pH=5.35
  55. 55. 50.0 mL of 0.10 M acetic acid (HC2H3O2,Ka=1.8x10-5) is titrated with 0.10 M NaOH.e. 50.0 mL of NaOH is added.pH=8.72
  56. 56. 50.0 mL of 0.10 M acetic acid (HC2H3O2,Ka=1.8x10-5) is titrated with 0.10 M NaOH.f. 60.0 mL of 0.10 M NaOH is added.pH = 11.96
  57. 57. 50.0 mL of 0.10 M acetic acid (HC2H3O2,Ka=1.8x10-5) is titrated with 0.10 M NaOH.g. 75.0 mL of 0.10 M NaOH is added.pH = 12.30
  58. 58. Some of the differences in the the titrationcurves: The pH increases more rapidly in the beginning of the weak acid titration.
  59. 59. Some of the differences in the the titrationcurves: The titration curve levels off near the halfway point due to buffering effects. ◦ Buffering happens when [HA]=[A-]. This is exactly the halfway point of the titration.
  60. 60. Some of the differences in the the titrationcurves: The value of the pH at the equivalence point is not 7. The value of the e.p. is greater than 7.
  61. 61. Hydrogen cyanide gas (HCN), a powerfulrespiratory inhibitor, is highly toxic. It is a veryweak acid (Ka=6.2x10-10) when dissolved inwater. If a 50.0 mL sample of 0.100 M HCN istitrated with 0.100 M NaOH, calculate the pH ofthe solution.a. after 8.00 mL of 0.100 M NaOH has been added.b. at the halfway point of the titrationc. at the equivalence point of the titration a) pH=8.49 b) pH=9.21 c)pH=10.96
  62. 62. When comparing the example calculation andthe practice problem with weak acids, twomajor conclusions can be made:1. The same amount of 0.10 M NaOH is required to reach the equivalence point even though HCN is a much weaker acid. ◦ It is the amount of acid, not its strength that determines the equivalence point.
  63. 63. When comparing the example calculation andthe practice problem with weak acids, twomajor conclusions can be made:2. The pH value at the equivalence point is affected by the acid stregth. The pH at the e.p. for acetic acid is 8.72. The pH at the e.p. for hydrocyanic acid is 10.96. ◦ The CN- ion is a much stronger base than C2H3O2-. The smaller acid strength and stronger conjugate base produces a higher pH.
  64. 64. The strength of a weak acid has a significanteffect on the shape of its pH curve. The e.p.occurs at the same point, but the shapes of thecurve is dramatically different.
  65. 65. A chemist has synthesized a monoprotic weakacid and wants to determine its Ka value. Todo so, the chemist dissolves 2.00 mmol of thesolid acid in 100.0 mL water and titrates theresulting solution with 0.0500 M NaOH. After20.0 mL NaOH has been added, the pH is 6.00.What is the Ka value for the acid?Ka=1.0x10-6
  66. 66. Find the pH of a solution of 100.0 mL of 0.050M NH3 is titrated with 0.10 M HCl (Kb=1.8x10-5).a. HCl has not been added.pH is found with Kb equilibrium. pH = 10.96
  67. 67. Find the pH of a solution of 100.0 mL of 0.050M NH3 is titrated with 0.10 M HCl (Kb=1.8x10-5).b. 10.0 mL of HCl is added.H+ ions react to completion. pH=9.85
  68. 68. Find the pH of a solution of 100.0 mL of 0.050M NH3 is titrated with 0.10 M HCl (Kb=1.8x10-5).c. 25.0 mL of HCl is added.pH = 9.25
  69. 69. Find the pH of a solution of 100.0 mL of 0.050M NH3 is titrated with 0.10 M HCl (Kb=1.8x10-5).d. 50.0 mL of HCl is added.pH = 5.36
  70. 70. Find the pH of a solution of 100.0 mL of 0.050M NH3 is titrated with 0.10 M HCl (Kb=1.8x10-5).e. 60.0 mL of HCl is added.pH= 2.21
  71. 71. 15.5
  72. 72. There are two common methods for determiningthe equivalence point of a titration:1. Use a pH meter to monitor the pH and then plot the titration curve. The center of the vertical region of the pH curve indicates the equivalence point.2. Use an acid-base indicator, which marks the end point of a titration by changing color.
  73. 73. The equivalence point of a titration is definedby the stoichiometry, but it is not necessarilythe same as the end point where the indicatorchanges color). Selection of the right indicator for the titration is very important.
  74. 74. The most common acid-base indicators arecomplex molecules that are weak acids (HIn). Most exhibit one color when the proton is attached to the molecule and a different color when the proton is absent. ◦ Example: Phenolphthalein is colorless in its HIn form and pink in its In-, or basic form.
  75. 75. + - HIn(aq) « H (aq) + In(aq) (RED) (BLUE) [H + ][In - ] Ka [In - ] Ka = + = [HIn] [H ] [HIn]The color will depend on the ratio of [In-] to[HIn]. For most indicators, approximately1/10th of the initial form must be converted tothe final form before a color is perceived by thehuman eye. the color change will occur at a pH when [In - ] 1 = [HIn] 10
  76. 76. Bromthymol blue, an indicator with Ka=1.0x10-7, is yellow in its HIn form and blue in its In-form. Suppose we put a few drops of thisindicator in a strongly acidic solution. If thesolution is then titrated with NaOH, at what pHwill the indicator color change first be visible?pH=6.0
  77. 77. The H-H equation is very useful in determiningthe pH at which an indicator changes color. æ [In - ] ö [In - ] 1 pH = pK a + log ç ÷ , [HIn] = 10 è [HIn] ø pH = pK a + long( 10 ) = pK a -1 1 Example:Bromthymol blue (Ka=1x10-7, or pKa=7), the pH at the color change is pH=7- 1=6
  78. 78. When a basic solution is titrated, the indicator will initially exist as In- in solution, but as acid is added more HIn is formed. Color change will occur at: [In - ] 10 = [HIn] 1Substituting this reciprocal into the H-Hequation gives us: pH = pKa + log ( 10 ) = pKa +1 1 ◦ Bromthymol blue=pH 7+1=8;The useful range of bromthymol blue is pH(6-8)
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