1. 1 Biomolecules in Water Water, the Biological Solvent Hydrogen Bonding and Solubility Cellular Reactions of Water Ionization, pH and pK The Henderson-Hasselbalch Equation Buffer Systems Water, the Biological Solvent Hydrogen Bonding and Solubility Cellular Reactions of Water Ionization, pH and pK The Henderson-Hasselbalch Equation Buffer Systems

2. 2 Water The solvent of choice for biological systems. Medium for metabolism. Able to absorb large amounts of heat. Solvent for many materials. Used for transport - blood, cerebrospinal fluid, lymph, urine. Serves as a reactant or product of many biochemical reactions.

3. 3 Structure of water Water is a polar molecule Electronegativities of hydrogen (2.1) and oxygen (3.5) result in a polar bond. O HH 104.5 o Water’s ‘bent’ shape results in the molecule having a  + and  - end. ++ -- ++

4. 4 Structure of water

5. 5 Hydrogen bonding The small size of hydrogen along with the shape and polarity of the water molecule all add up to a relatively strong attraction between water molecules. Hydrogen bonding is the strongest of the intermolecular forces.

6. 6 Hydrogen bonding This interaction can occur between water and many biomolecules. It can also occur between two biomolecules. O CH 2 CH 3 H O HH Example Example DNA relies on H bonding to hold the double helix together.

7. 7 Hydrogen bonding and solubility Physical properties of water Due to hydrogen bonding, water is unique for its molecular weight and size. PropertyH 2 ONH 3 CH 4 H 2 S Molecular weight 18 17 16 32 Boiling point ( o C) 100 -33-161-60.7 Melting point( o C) 0 -78-183-85.5 Viscosity a 1.010.250.10 0.15 a Units are centipoise.

8. 8 Water as a solvent hydrophilic Water will dissolve biomolecules that are polar or ionic - hydrophilic. It has a high ion solvating ability. Na + 0.14 M K + 0.004M Cl - 0.10 M Concentrations in blood. Concentrations in blood.

9. 9 Dissolution of NaCl

10. 10 Water as a solvent hydrophobic Nonpolar compounds like fats are not very soluble in water - hydrophobic. amphiphilic Some materials have both polar and nonpolar ends - amphiphilic. One end tends to dissolve in polar solvents and the other in nonpolar ones. C C C C C C C C C C C C C C C C O OH Example - saturated fatty acid

11. 11 How detergents work COO - - OOC COO - - OOC Nonpolar tail dissolves in oil. Polar ‘heads’ are attracted to the water.

12. 12 Micelles Amphiphilic molecules tend to organize into micelle structures. The polar heads will point towards the aqueous environment and the nonpolar tails will be on the inside.

13. 13 Cellular reactions of water Since water is the working solvent for biological systems, it is appropriate to review the acid/base chemistry of this solvent.Autoionization pK, pKw, pKa TitrationsBuffers

14. 14 Autoionization of water amphiprotic Water is an amphiprotic substance that can act either as an acid or a base. HC 2 H 3 O 2(aq) + H 2 O (l) H 3 O + + C 2 H 3 O 2 - (aq) acid base acid base H 2 O (l) + NH 3(aq) NH 4 + (aq) + OH - (aq) acid base acid base

15. 15 Autoionization of water Autoionization When water molecules react with one another to form ions. H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) (10 -7 M) (10 -7 M) K w = [ H 3 O + ] [ OH - ] = 1.0 x 10 -14 at 25 o C ion product of water ion product of water

16. 16 Acid dissociation constant, K a The strength of a weak acid can be expressed as an equilibrium. HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) The strength of a weak acid is related to its equilibrium constant, K a. K a = [A - ][H 3 O + ] [HA] We omit water. It’s already included in the constant. We omit water. It’s already included in the constant.

17. 17 pH and pOH We need to measure and use acids and bases over a very large concentration range. pH and pOH are systems to keep track of these very large ranges. pH = -log[H 3 O + ] pOH = -log[OH - ] pH + pOH = 14

18. 18 pH scale A log based scale used to keep track of the large change important to acids and bases. When you add an acid, the pH gets smaller. When you add a base, the pH gets larger. 14 7 0 10 -14 M 10 -7 M 1 M Very basic Neutral Very acidic H+H+ H+H+ H+H+

19. 19 Titrations Analytical methods based on measurement of volume. If the concentration of an acid is known, the concentration of the base can be found. If we know the concentration of the base, then we can determine the amount of acid. All that is needed is some calibrated glassware and either an indicator or pH meter.

20. 20 Titrations Buret Buret - volumetric glassware used for titrations. It allows you to add a known amount of your titrant to the solution you are testing. If a pH meter is used, the equivalence point can be measured. An indicator will give you the endpoint.

21. 21 Titrations

22. 22 Titrations Note the color change which indicates that the ‘endpoint’ has been reached. StartEnd

23. 23 Indicator examples Acid-base indicators are weak acids that undergo a color change at a known pH. phenolphthalein pH

24. 24 Indicator examples methyl red bromthymol blue

25. 25 Titration of weak acids or bases Titration of a weak acid or base with a strong titrant is a bit more complex than the strong acid/strong base example. We must be concerned with conjugate acid/base pairs and their equilibrium.Example HA + H 2 O H 3 O + + A - acid base

26. 26 Titration of weak acids or bases First, we’ll only be concerned about the titration of a weak acid with a strong base or a weak base with a strong acid. We still have the same four general regions for our titration curve. The calculation will require that you use the appropriate K A or K B relationship. We’ll start by reviewing the type of calculations involved and then work through an example.

27. 27 Titration of weak acids or bases 0% titration If your sample is an acid then use K A = At this point [H 3 O + ] = [A - ]. You can solve for [H 3 O + ] by using either the quadratic or approximation approach. Finally, calculate the pH. [H 3 O + ][A - ] [HA]

28. 28 Titration of weak acids or bases 0% titration If your sample is a base then use K B = At this point [OH - ] = [HA]. You can solve for [OH - ] by using either the quadratic or approximation approach. Then determine pH as 14 - pOH. [OH - ][HA] [A - ]

29. 29 Titration of weak acids or bases 5 - 95% titration In this region, the pH is a function of the K value and the ‘ratio’ of the acid and base forms of our analyte. Henderson-Hasselbalch equation A common format for the equilibrium expression used in the region is the Henderson-Hasselbalch equation. We can present it in two forms depending on the type of material we started with.

30. 30 Titration of weak acids or bases 5 - 95% titration Starting with an acid pH = pK A + log Starting with a base pH = 14 - ( pK B + log ) ( We’re just determining the pOH and then converting it to pH. ) [HA] [A - ] [HA]

31. 31 Titration of weak acids or bases 5 - 95% titration Another approach that can be taken is to simply use the % titration values. For an acidic sample, you would use: pH = pK A + log % titration 100 - % titration

32. 32 Titration of weak acids or bases 5 - 95% titration These equations have their limits and may break down if: K A or K B < 10 -3 or you are working with very dilute solutions. In those cases, you need to consider the equilibrium for water.

33. 33 Titration of weak acids or bases 100% titration - the equivalence point. At the point, we have converted all of our sample to the conjugate form. If your sample was an acid, now solve for the pH using the K B relationship -- do the opposite if you started with a base. Remember that K B + K A = 14 You must also account for dilution of the sample as a result of adding titrant.

34. 34 Titration of weak acids or bases Overtitration (> 100%) These calculations are identical to those covered in our strong acid/strong base example. You simply need to account for the amount of excess titrant added and how much it has been diluted.

35. 35 Titration of weak acids or bases Example A 100 ml solution of 0.10 M benzoic acid is titrated with 0.10 M NaOH. Construct a titration curve. For benzoic acid K A = 6.31 x 10 -5 pK A = 4.20

36. 36 Titration of weak acids or bases 0% titration K A = [H 3 O + ] = [A - ] [HA] + [A - ] = 0.10 M We’ll assume that [A - ] is negligible compared to [HA]. [H 3 O + ][A - ] [HA]

37. 37 Titration of weak acids or bases 0% titration K A = 6.31 x 10 -5 = x = (6.31 x 10 -5 )(0.10) = 0.025 M pH= 2.60 x 2 0.10

38. 38 Titration of weak acids or bases 10% titration Henderson- Hasselbalch Here we can use the Henderson- Hasselbalch equation in % titration format: pH = pK A + log pH = 4.20 + log (10 / 90) = 3.25 We can calculate other points by repeating this process. % titration 100 - % titration

39. 39 Titration of weak acids or bases % titration pH 02.60 103.24 203.60 303.83 404.02 504.20 604.38 704.57 804.80 905.15 Note: At 50% titration, pH = pK A Also, there was only a change of 1.91 pH units as we went from 10 to 90 % titration. Note: At 50% titration, pH = pK A Also, there was only a change of 1.91 pH units as we went from 10 to 90 % titration.

40. 40 Titration of weak acids or bases % titration pH

41. 41 Titration of weak acids or bases 100% titration At this point, virtually all of our acid has been converted to the conjugate base - benzoate. We need to use the K B relationship to readily solve for this point. K B = K B = K W / K A = 1.58 x 10 -10 [OH - ][HA] [A - ]

42. 42 Titration of weak acids or bases 100% titration At the equivalence point: [HA] = [OH - ] We’ve diluted the [HA] + [A - ] = 0.05 Msample and the total volume at this point is 200 ml Finally, we can assume that [benzoic acid] is negligible compared to [benzoate].

43. 43 Titration of weak acids or bases 100% titration K A = 1.58 x 10 -10 = x = (1.58 x 10 -10 )(0.050) = 2.81 x 10 -6 M pOH = 5.55pH = 14 - pOH = 8.45 x 2 0.050

44. 44 Titration of weak acids or bases % titration pH

45. 45 Titration of weak acids or bases Overtitration All we need to do here is to account for the dilution of our titrant. 10 % overtitration (10 ml excess) [OH - ] = 0.10 M = 0.0048 M (when diluted) pOH = 2.32 pH= 14-2.32 = 11.68

46. 46 Titration of weak acids or bases ml titrant total volume [OH - ] pH 1102100.004811.68 1202200.009111.96 1302300.01312.11 1402400.01712.23 1502500.02012.30 This is identical to what we obtained for our strong acid/strong base example.

47. 47 Titration of weak acids or bases % titration pH

48. 48 Buffers Buffer solution A mixture of a conjugate acid - base pair. It corresponds to approximately 10-90% titration - a relative flat region of the curve. This mixture tends to resist changes in pH when an acid or base is added. Buffers are commonly used when pH must be maintained at a relatively constant value and in many biological systems.

49. 49 Buffers Effect of adding an acid or base. We can use the Henderson-Hasselbalch equation to evaluate pH changes of buffered systems. pH = pK A + log The type and degree of buffering is a function of the pK A of our buffering material and the concentration of both forms. [A - ] [HA]

50. 50 Buffers Effect of adding an acid or base. The best way to appreciate the effect of a buffer is to work an example. Example. Example. A total of 100 ml of 1.0 M HCl is added in 10 ml increments to 100 ml of the following solutions: 1) Pure water where pH = 7 2) A solution containing 1.0 M HA, 1.0 M A - where pK A = 7.0 Determine the pH after each addition.

51. 51 Buffers Initially, each solution is at pH 7.00. After adding 10 ml of 1.0 M HCl we have: Pure water [H 3 O + ] = = 0.091 pH = 1.04 This is a pretty big jump! (10 ml)(1.0 M) (110 ml)

52. 52 Buffers Addition of 10 ml 1.0 M HCl to our buffered system. We started with 0.10 moles of both the acid and conjugate base forms. The addition of our first 10 ml can be expected to react with the conjugate base, converting it to the acid form. After addition, we would have 0.09 moles of the base form and 0.11 moles of the acid form.

53. 53 Buffers New concentrations Since all we need to be concerned about is the ratio of A - to HA in the 10-90% region then: pH= pK A + log = 7.00 + log = 6.91 A change of only 0.09 compared to 5.96 with no buffer. [A - ] [HA] 0.09 mmol 1.1 mmol

54. 54 Buffers ml HCl pH added unbuffered buffered 07.007.00 101.046.91 200.786.82 300.646.73 400.546.63 500.486.52 600.436.40 700.396.25 800.356.05 900.325.72

55. 55 Buffers ml HCl added pH buffered unbuffered

56. 56 Buffers For the addition of 100 ml of HCl, we have converted virtually all of A - to HA so the calculation is different. K A = 1.00 x 10 -7 = When we account for dilution, 1.0 M = [HA] + [A - ] where [A - ] is negligible. This is a standard weak acid calculation. [H + ][A - ] [HA]

57. 57 Buffers pH after addition of 100 ml HCl. [H + ]=( K A x [HA] ) 1/2 =(1.00 x 10 -7 x 1.0 M ) 1/2 = 3.16 x 10 -4 pH= 3.5 At this point, we have exceeded the buffer region for our system.

58. 58 Buffers ml HCl added pH buffered unbuffered

59. 59 Buffers Obviously, a buffer only has a limited ability to reduce pH changes. The pK A determines the range where a buffer is useful. The concentration of our buffer system determines how much acid or base it can deal with.

60. 60 Buffers Buffer capacity The number of moles of a strong acid or base that causes 1 liter of a buffer solution to undergo a pH change of 1.00. Since pH = pK A + log then +1 = log So it is the amount of acid or base required to convert ~80% of one form to the other if we start with a 1:1 ratio. [A - ] [HA] [A - ] [HA]

61. 61 Buffers Example Determine the buffer capacity of a 1:1 mixture of a weak acid at the following concentrations: 1.0 M 0.5 M 0.1M 0.05 M 0.01M

62. 62 Buffers Assume that [ HA ] = [ A - ] = C, where C is the initial concentration of either our acid or base. Buffer capacity in general is then: 1 = log [ A - ] = 10 [ HA ] [A - ] = 2 C - [ HA ] [ HA ] = 0.22 C [A - ] [HA] We’ll only worry about addition of a base. We’ll only worry about addition of a base.

63. 63 Buffers Concentration, MBuffer capacity (mol) 1.00.22 0.50.11 0.10.022 0.050.011 0.010.0022 The capacity may be smaller if you don’t start with a 1:1 mixture.

64. 64 Buffers and blood Control of blood pH – Oxygen is transported primarily by hemoglobin in the red blood cells. – CO 2 transported both in plasma and the red blood cells. CO 2 (aq) + 2 H 2 O H 2 CO 3 (aq) H 3 O + (aq) + HCO 3 - (aq)

65. 65 Buffers and blood The amount of CO 2 helps control blood pH. Too much CO 2 Too much CO 2 - Respiratory arrest pH goes down, acid level goes up.acidosis Solution Solution - ventilate and give bicarbonate via IV. Too little CO 2 Too little CO 2 - Hyperventilation, anxiety pH goes up, acid level goes down.alkalosis Solution Solution - rebreath CO 2 in paper bag to raise level.