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pH and Buffers

pH and Buffers. The Whole Story. Acid & Base and pH. This should be a review but we will go over it to help refresh your memory. 0 7 14. 1 M 10 -7 M 10 -14 M. Very Neutral Very Acidic Basic. Trivia time:.

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pH and Buffers

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  1. pH and Buffers The Whole Story

  2. Acid & Base and pH • This should be a review but we will go over it to help refresh your memory 0 7 14 1 M 10-7 M 10-14 M Very Neutral Very Acidic Basic

  3. Trivia time: • Why was the concept of pH developed?

  4. Trivia time: • Why was the concept of pH developed? • By the Danish Biochemist Sorensen to test the acidity of the beer he was making • Who said biochemistry isn’t cool?

  5. pH - pouvoir hydrogene (the power of hydrogen) • Water undergoes ionization • Water ionizes to form the hydronium (hydroxyl) ion and hydroxide ions • Water can act as both an acid and base • The equilibrium constant for the ionization of water is: • The concentration of pure water • 1 liter = 1000g MW of water is 10.015 • the final concentration of water is 55M and H+ concentration is about 1.8 x 10-9 • Very little water actually dissociates • So Keq is very small – not easily measured or easy to use

  6. _ + 2 H O + H O OH 3 2 pH - pouvoir hydrogene (the power of hydrogen) • Water undergoes ionization • Water ionizes to form the hydronium (hydroxyl) ion and hydroxide ions • Water can act as both an acid and base • The equilibrium constant for the ionization of water is: • The concentration of pure water • 1 liter = 1000g MW of water is 10.015 • the final concentration of water is 55M and H+ concentration is about 1.8 x 10-9 • Very little water actually dissociates • So Keq is very small – not easily measured or easy to use

  7. _ + 2 H O + H O OH 3 2 [products] [H+ ] [OH- ] = K = eq [H2O]2 [reactants] pH - pouvoir hydrogene (the power of hydrogen) • Water undergoes ionization • Water ionizes to form the hydronium (hydroxyl) ion and hydroxide ions • Water can act as both an acid and base • The equilibrium constant for the ionization of water is: • The concentration of pure water • 1 liter = 1000g MW of water is 10.015 • the final concentration of water is 55M and H+ concentration is about 1.8 x 10-9 • Very little water actually dissociates • So Keq is very small – not easily measured or easy to use

  8. _ + 2 H O + H O OH 3 2 [H2O] is effectively constant. pH - pouvoir hydrogene (the power of hydrogen) • Water undergoes ionization • Water ionizes to form the hydronium (hydroxyl) ion and hydroxide ions • Water can act as both an acid and base • The equilibrium constant for the ionization of water is: • The concentration of pure water • 1 liter = 1000g MW of water is 10.015 • the final concentration of water is 55M and H+ concentration is about 1.8 x 10-9 • Very little water actually dissociates • So Keq is very small – not easily measured or easy to use [products] [H+ ] [OH- ] = K = eq [H2O]2 [reactants]

  9. Instead a different constant is used where the denominator is ignored • Kw= 1.0 X 10 -14 • pH is a measure of the acidity and basicity of a solution • when [H+] = [OH-] the solution is neutral and pH is 7 • when [H+] > [OH-] the solution is acidic and pH is less than 7 • when [H+] < [OH-] the solution is basic and pH is more than 7 • a change in 1 pH units is = a ten fold change in hydrogen ion concentration

  10. _ - 7 + [H ] = [OH ] = 1 X 10 • Instead a different constant is used where the denominator is ignored • Kw= 1.0 X 10 -14 • pH is a measure of the acidity and basicity of a solution • when [H+] = [OH-] the solution is neutral and pH is 7 • when [H+] > [OH-] the solution is acidic and pH is less than 7 • when [H+] < [OH-] the solution is basic and pH is more than 7 • a change in 1 pH units is = a ten fold change in hydrogen ion concentration _ + [OH ] K = [H ] Therefore: w And as measured in pure water

  11. _ - 7 + [H ] = [OH ] = 1 X 10 • Instead a different constant is used where the denominator is ignored • Kw= 1.0 X 10 -14 • pH is a measure of the acidity and basicity of a solution • pH = -log[H+] • when [H+] = [OH-] the solution is neutral and pH is 7 • when [H+] > [OH-] the solution is acidic and pH is less than 7 • when [H+] < [OH-] the solution is basic and pH is more than 7 • a change in 1 pH units is = a ten fold change in hydrogen ion concentration _ + [OH ] K = [H ] Therefore: w And as measured in pure water

  12. The extent of ionization of a weak acid is a function of its acid dissociation constant pKa • Bronsted and Lowry acid and bases • acid donates protons • bases accepts protons • Strong acids dissociate nearly fully • Weak acids only partially dissociate • Acids with Ka < 1 are considered weak acids • Ka for acetic acid is 1.76 x 10-5 -> difficult to work with so instead use log scale: • pKa = -log Ka • So the pKa of acetic acid is = -log 1.76 x 10-5 = 4.75 • The pH is a measure of acidity and the pKa is a measure of acid strength

  13. _ + HA + H O H O + A 2 3 acid base conjugate conjugate acid base The extent of ionization of a weak acid is a function of its acid dissociation constant pKa • Bronsted and Lowry acid and bases • acid donates protons • bases accepts protons • Strong acids dissociate nearly fully • Weak acids only partially dissociate • Acids with Ka < 1 are considered weak acids • Ka for acetic acid is 1.76 x 10-5 -> difficult to work with so instead use log scale: • pKa = -log Ka • So the pKa of acetic acid is = -log 1.76 x 10-5 = 4.75 • The pH is a measure of acidity and the pKa is a measure of acid strength

  14. _ + HA + H O H O + A 2 3 acid base conjugate conjugate acid base The extent of ionization of a weak acid is a function of its acid dissociation constant pKa • Bronsted and Lowry acid and bases • acid donates protons • bases accepts protons • Strong acids dissociate nearly fully • [H+] = [ acid] and thus pH = -log [acid] • Weak acids only partially dissociate • Acids with Ka < 1 are considered weak acids • Ka for acetic acid is 1.76 x 10-5 -> difficult to work with so instead use log scale: • pKa = -log Ka • So the pKa of acetic acid is = -log 1.76 x 10-5 = 4.75 • The pH is a measure of acidity and the pKa is a measure of acid strength

  15. pKa and Ka are used for all acids. _ + [H ] [A ] K = a [HA] The extent of ionization of a weak acid is a function of its acid dissociation constant pKa • Bronsted and Lowry acid and bases • acid donates protons • bases accepts protons • Strong acids dissociate nearly fully • [H+] = [ acid] and thus pH = -log [acid] • Weak acids only partially dissociate • Acids with Ka < 1 are considered weak acids • Ka for acetic acid is 1.76 x 10-5 -> difficult to work with so instead use log scale: • pKa = -log Ka • So the pKa of acetic acid is = -log 1.76 x 10-5 = 4.75 • The pH is a measure of acidity and the pKa is a measure of acid strength _ + HA + H O H O + A 2 3 acid base conjugate conjugate acid base

  16. Note relationship between pKA, Ka and acid strength. Dissociation constants and pKa values More Acidic C hemi c al K a ( M ) p K a F o r mic aci d -4 3 . 8 1 . 77 x 10 C a rb oni c a ci d -7 6 . 4 4 . 30 x 10 -11 B ic a r b o na te 10 . 2 5 . 61 x 10 Am m o n ium -10 9 . 2 5 . 62 x 10

  17. pKa of a weak acid is determined experimentally by titration. • pKa is when the concentration of acid and base is equal in a titration (ask yourself if that then equal to pH 7). • There is a point in a titration of a weak acid where the change in pH is very little. This is the buffer action of the acid. Titrations and pKa Many acids have more than one ionizable group (polyprotic)

  18. HA H+ + A- The relationship between pH and pKa FOR A WEAK ACID is described by the Henderson-Hasselbalch equation

  19. HA H+ + A- The relationship between pH and pKa is described by the Henderson-Hasselbalch equation Start with a weak acid

  20. HA H+ + A- [ H+] [A-] K = a [ HA ] The relationship between pH and pKa is described by the Henderson-Hasselbalch equation Start with a weak acid

  21. HA H+ + A- [ H+] [A-] K = a [ HA ] The relationship between pH and pKa is described by the Henderson-Hasselbalch equation Start with a weak acid Arrange to get

  22. HA H+ + A- [ H+] [A-] K = a [ HA ] The relationship between pH and pKa is described by the Henderson-Hasselbalch equation Start with a weak acid Arrange to get [HA] = [H+] K a [A-]

  23. HA H+ + A- [ H+] [A-] K = a [ HA ] The relationship between pH and pKa is described by the Henderson-Hasselbalch equation Start with a weak acid Arrange to get [HA] = [H+] K a [A-] Multiply each side by log

  24. HA H+ + A- [ H+] [A-] K = a [ HA ] The relationship between pH and pKa is described by the Henderson-Hasselbalch equation Start with a weak acid Arrange to get [HA] = [H+] K a [A-] Multiply each side by log [HA] Log[H+] = Log K + Log a [A-]

  25. HA H+ + A- [ H+] [A-] K = a [ HA ] The relationship between pH and pKa is described by the Henderson-Hasselbalch equation Start with a weak acid Arrange to get [HA] = [H+] K a [A-] Multiply each side by log [HA] Log[H+] = Log K + Log a [A-] Multiply by -1

  26. HA H+ + A- [ H+] [A-] K = a [ HA ] The relationship between pH and pKa is described by the Henderson-Hasselbalch equation Start with a weak acid Arrange to get [HA] = [H+] K a [A-] Multiply each side by log [HA] Log[H+] = Log K + Log a [A-] Multiply by -1 [HA] -Log[H+] = -Log K - Log a [A-]

  27. HA H+ + A- [ H+] [A-] K = a [ HA ] The relationship between pH and pKa is described by the Henderson-Hasselbalch equation Start with a weak acid Arrange to get [HA] = [H+] K a [A-] Multiply each side by log [HA] Log[H+] = Log K + Log a [A-] Multiply by -1 [HA] -Log[H+] = -Log K - Log Substitute pH = -Log[H+] pKa = Log Ka a [A-]

  28. HA H+ + A- [ H+] [A-] K = a [ HA ] The relationship between pH and pKa is described by the Henderson-Hasselbalch equation Start with a weak acid Arrange to get [HA] = [H+] K a [A-] Multiply each side by log [HA] Log[H+] = Log K + Log a [A-] Multiply by -1 [HA] -Log[H+] = -Log K - Log Substitute pH = -Log[H+] pKa = Log Ka a [A-] [HA] pH = pKa - Log [A-]

  29. HA H+ + A- [ H+] [A-] K = a [ HA ] The relationship between pH and pKa is described by the Henderson-Hasselbalch equation Start with a weak acid Arrange to get [HA] = [H+] K a [A-] Multiply each side by log [HA] Log[H+] = Log K + Log a [A-] Multiply by -1 [HA] -Log[H+] = -Log K - Log Substitute pH = -Log[H+] pKa = Log Ka a [A-] [HA] pH = pKa - Log [A-] Remove (-) and invert last term

  30. HA H+ + A- [ H+] [A-] K = a [ HA ] The relationship between pH and pKa is described by the Henderson-Hasselbalch equation Start with a weak acid Arrange to get [HA] = [H+] K a [A-] Multiply each side by log [HA] Log[H+] = Log K + Log a [A-] Multiply by -1 [HA] -Log[H+] = -Log K - Log Substitute pH = -Log[H+] pKa = Log Ka a [A-] [HA] pH = pKa - Log [A-] Remove (-) and invert last term [A-] pH = pKa + Log [HA]

  31. HA H+ + A- [ H+] [A-] K = a [ HA ] The relationship between pH and pKa is described by the Henderson-Hasselbalch equation Start with a weak acid Arrange to get [HA] = [H+] K a [A-] Multiply each side by log [HA] Log[H+] = Log K + Log a [A-] Multiply by -1 [HA] -Log[H+] = -Log K - Log Substitute pH = -Log[H+] pKa = Log Ka a [A-] [HA] pH = pKa - Log [A-] Remove (-) and invert last term [A-] pH = pKa + Log [HA] Henderson -Hasselbalch

  32. What is the H-H Equation Used For? • This is used to determine the concentration of acid and base at a given pH. It is Also used to determine the pH of a known solution. These concepts are used to calculate buffer strength and understand the pH of a biological solution. • Remember that buffers are mixtures of weak acids and their conjugate bases that resist pH by shifting the equilibrium between the acid and base in response to the pH of a solution.

  33. Case 1) when the concentration of base equals the acid. • When pH = pKa 50% of the acid is dissociated _ [A ] pH = pKa + Log [HA] [ x ] = pKa + Log [ x ] pH = pKa

  34. Case 2) when the pH is above or below 1 pH unit of the pKa • Then 90% of the buffer is in the conjugate base for • If pH is 2 units different then 99% of buffer is in the conjugate base form • What does this mean about the buffering ability if more acid or base is added? _ [A ] 5.00 = 4.00 + Log [HA] _ [A ] 1.00 = Log [HA] _ [A ] 10.0 = [HA]

  35. Calculate the pH of a mixture of 250 mM acetic acid and 100 mM Na acetate.The pKa of acetic acid is 4.75.

  36. Calculate the pH of a mixture of 250 mM acetic acid and 100 mM Na acetate.The pKa of acetic acid is 4.75. • Start with the HH equation • pH = pKa + Log [A-]/[HA]

  37. Calculate the pH of a mixture of 250 mM acetic acid and 100 mM Na acetate.The pKa of acetic acid is 4.75. • Start with the HH equation • pH = pKa + Log [A-]/[HA] • Recognize that a compound that ends with -ate is the base form of a weak acid and plug in the known terms

  38. Calculate the pH of a mixture of 250 mM acetic acid and 100 mM Na acetate.The pKa of acetic acid is 4.75. • Start with the HH equation • pH = pKa + Log [A-]/[HA] • Recognize that a compound that ends with -ate is the base form of a weak acid and plug in the known terms • pH = 4.75 + Log (0.1/0.25)

  39. Calculate the pH of a mixture of 250 mM acetic acid and 100 mM Na acetate.The pKa of acetic acid is 4.75. • Start with the HH equation • pH = pKa + Log [A-]/[HA] • Recognize that a compound that ends with -ate is the base form of a weak acid and plug in the known terms • pH = 4.75 + Log (0.1/0.25) • Remember [X] indicates molar concentration

  40. Calculate the pH of a mixture of 250 mM acetic acid and 100 mM Na acetate.The pKa of acetic acid is 4.75. • Start with the HH equation • pH = pKa + Log [A-]/[HA] • Recognize that a compound that ends with -ate is the base form of a weak acid and plug in the known terms • pH = 4.75 + Log (0.1/0.25) • Remember [X] indicates molar concentration • pH =4.35

  41. What is the ratio of lactic acid to lactate in a buffer at pH of 5.00.The pKa of lactic acid is 3.86?

  42. What is the ratio of lactic acid to lactate in a buffer at pH of 5.00.The pKa of lactic acid is 3.86? • Again we recognize this is a weak acid and start with the HH equation

  43. What is the ratio of lactic acid to lactate in a buffer at pH of 5.00.The pKa of lactic acid is 3.86? • Again we recognize this is a weak acid and start with the HH equation • pH = pKa + Log [A-]/[HA]

  44. What is the ratio of lactic acid to lactate in a buffer at pH of 5.00.The pKa of lactic acid is 3.86? • Again we recognize this is a weak acid and start with the HH equation • pH = pKa + Log [A-]/[HA] • Plug in the known terms

  45. What is the ratio of lactic acid to lactate in a buffer at pH of 5.00.The pKa of lactic acid is 3.86? • Again we recognize this is a weak acid and start with the HH equation • pH = pKa + Log [A-]/[HA] • Plug in the known terms • 5.00 = 3.86 + Log [A-]/[HA]

  46. What is the ratio of lactic acid to lactate in a buffer at pH of 5.00.The pKa of lactic acid is 3.86? • Again we recognize this is a weak acid and start with the HH equation • pH = pKa + Log [A-]/[HA] • Plug in the known terms • 5.00 = 3.86 + Log [A-]/[HA] • Arrange

  47. What is the ratio of lactic acid to lactate in a buffer at pH of 5.00.The pKa of lactic acid is 3.86? • Again we recognize this is a weak acid and start with the HH equation • pH = pKa + Log [A-]/[HA] • Plug in the known terms • 5.00 = 3.86 + Log [A-]/[HA] • Arrange • 1.14 = Log [A-]/[HA]

  48. What is the ratio of lactic acid to lactate in a buffer at pH of 5.00.The pKa of lactic acid is 3.86? • Again we recognize this is a weak acid and start with the HH equation • pH = pKa + Log [A-]/[HA] • Plug in the known terms • 5.00 = 3.86 + Log [A-]/[HA] • Arrange • 1.14 = Log [A-]/[HA] • Inv Log on both sides

  49. What is the ratio of lactic acid to lactate in a buffer at pH of 5.00.The pKa of lactic acid is 3.86? • Again we recognize this is a weak acid and start with the HH equation • pH = pKa + Log [A-]/[HA] • Plug in the known terms • 5.00 = 3.86 + Log [A-]/[HA] • Arrange • 1.14 = Log [A-]/[HA] • Inv Log on both sides • Ratio = 13.80

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