2. Acids, bases, pH scale, biological buffer systems Saida Almashharawi 2014-2015 Basic Biochemistry CLS 233 1

3. After completing this chapter, students should be able to: - Define buffer, acid, and base. - Define pH and describe how the pH of a solution is measured. - Explain the effect of acids and bases on the pH of a buffered liquid. - Distinguish between acidosis, alkalosis - Explain the role of the buffering systems inside our body in the prevention of acidosis and alkalosis. - To understand how adding a common ion affects the position of an acid–base equilibrium. - Be able to use the Henderson-Hasselbalch equasion to calculate the pH of a buffer. Objetives 2

4. ACIDS Taste sour Turn litmus React with active metals – Fe, Zn React with bases BASES Taste bitter Turn litmus Feel soapy or slippery (react with fats to make soap) React with acids blue to redred to blue General properties 3

5. Acids – produce H + Bases - produce OH - Acids – donate H + Bases – accept H + Acids – accept e - pair Bases – donate e - pair Arrehenius Bronsted-Lowry Lewis only in water any solvent used in organic chemistry, wider range of substances Definitions 4

6. The Bronsted-Lowry Concept Conjugate pairs HCl Cl - CH 3 COOH CH 3 COO - NH 4 + NH 3 HNO 3 NO 3 - How does a conjugate pair differ? H + transfer 5

7. Neutralization In general: Acid + Base  Salt + Water All neutralization reactions are double displacement reactions. HCl + NaOH  NaCl + HOH 6

8. H 2 O  H + + OH - [H + ][OH - ] = 10 -14 For pure water: [H + ] = [OH - ] = 10 -7 M This is neutrality and at 25 o C is a pH = 7. How are (H + ) and (OH - ) related? 7

9. HA Let’s examine the behavior of an acid, HA, in aqueous solution. What happens to the HA molecules in solution? 8

10. HA H+H+ A-A- Strong Acid 100% dissociation of HA Would the solution be conductive? 9

11. HA H+H+ A-A- Weak Acid Partial dissociation of HA 10

12. HA H+H+ A-A- Weak Acid HA  H + + A - At any one time, only a fraction of the molecules are dissociated. 11

13. Strong and Weak Acids/Bases Strong acids/bases – 100% dissociation into ions HClNaOH HNO 3 KOH H 2 SO 4 Weak acids/bases – partial dissociation, both ions and molecules CH 3 COOHNH 3 12

14. pH 23456789101112 neutral @ 25 o C (H + ) = (OH - ) distilled water acidic (H + ) > (OH - ) basic or alkaline (H + ) < (OH - ) natural waters pH = 6.5 - 8.5 0-14 scale for the chemists 13

15. amphoteric (amphiprotic) substances HCO 3 - H 2 CO 3 CO 3 -2 + H + - H + Acting like a base Acting like an acid accepts H + donates H + 14

16. 123456891011 The biological view in the human body gastric juice vaginal fluid urine saliva cerebrospinal fluid blood pancreatic juice bile acidicbasic/alkaline 7 Tortora & Grabowski, Prin. of Anatomy & Physiology, 10 th ed., Wiley (2003) pH 15

17. Show how water can be amphoteric H2OH2O + H + - H + 16

18. Acids, bases, pH scale, biological buffer systems (Part 2) Saida Almashharawi 2014-2015 Basic Biochemistry CLS 233 17

19. Lecture Outlines 1- To Calculate [H 2 O] 2- To Calculate [H + ] and /or [OH - ] using K w 3- To solve problems based on previous point 4- To Calculate pH for some solutions 5- To discus the effect of pH changes on enzyme activity 6- Give some examples for the importance of the pH to the biological system 18

20. K eq & K w For the rxn, A + B C + D Equilibrium Constant, K eq = [C] [D] [A] [B] Note : K eq is fixed & characteristic for any chemical rxn at specific temp. -The ionization of H 2 O is expressed by an equilibrium contestant H2O H + + OH - K eq = [H + ] [OH - ] [H 2 O] 19

21. What is the molar conc. Of H 2 O? In pure water, at 25 C o, [H2O] =?? [H2O] = 55.5 M, HOW? Mass of 1 L of water= 1000 gm, M wt = 18.015 gm/ mol [H2O] = # Moles / L = (1000 gm/L ) / (18.015 gm/mol) [H2O]= 55.5 M K eq = [H + ] [OH - ] (55.5 M)(K eq ) = [H + ] [OH - ]=K w Where K w is the ion product of water at 25 0 C Value of K w of water= 1.8 X 10 -16 So, K w = [H + ] [OH - ] = (55.5 M)(1.8 X 10 -16 M ) K w = 1.0 X 10 -14 M 2 20

22. When [H + ] = [OH - ] in pure water, the solution is neutral (neutral pH) K w = [H + ] [OH - ] = [H + ] 2 [H + ]= = 1X10 -14 M 2 [H + ] = [OH - ] = 1X10 -7 M 21

23. Notes Because the ion product of water is constant, -If [H + ] > 1X10 -7 M [OH - ] < 1X10 -7 M -If [H + ] is very high (e.g HCL [OH - ] will be very low - From the ion product of water, we can calculate [H + ] if we know [OH - ], and vice versa 22

24. Problems 1- what is [H + ] in solution of 0.1 M NaOH? 2- what is [OH - ] in solution in which [H + ] = 0.00013 M? 23

25. Problem Solving 24

26. The pH scale Designates the [H + ]& [OH – ] pH = log 1 = - log [ H + ] [H+] - pH scale is logarithmic - pH of an aqueous solution approximately measured using indicator dyes, e.g. litmus, phenolphthalein, phenol red which undergo color changes Note: pH + pOH = 14 pOH = - Log [OH-] 25

27. Calculating pH pH is the negative log of the hydronium ion concentration. pH = - log [H 3 O + ] Example: For a solution with [H 3 O + ] = 1 x 10 −4 pH =−log [1 x 10 −4 ] pH = - [-4.0] pH = 4.0 26

28. pH of some common fluids 27

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30. Why to study pH??? The most used procedure in biochemistry The pH affects the structure & activity of biological macromolecules e.g enzymes For clinical diagnosis (urine, Blood) pH of blood= 7.4 Acidosis: pH < 7.4 Alkalosis: pH > 7.4 29

31. Effect of pH on enzyme activity When pH  optimum pH  change in enzyme structure  decrease in activity 30

32. Critical Thinking 5.1 The optimum pHs of some enzymes The graph below shows the effect of pH on the activities of three different enzymes. 31

33. Questions 1.Which enzyme works best in alkaline conditions? Ans: Ans: Trypsin 2.Which enzyme works best in acidic conditions? Ans: Ans: Pepsin 3.Which enzyme works best in neutral conditions? Ans: Ans: Salivary amylase 4.What is the range of pH at which each enzyme is active? Ans: Pepsin: pH 1-3; salivary amylase: pH 5.5-8.5; trypsin: pH 7.5-10.5 32

34. Weak Acids & Bases have characteristic dissociation constant Strong Acids & Bases: completely ionized Weak Acids & Bases: not completely ionized when dissolved in water Acids: proton donor Basis: proton acceptor Conjugate acid base pair: a proton donor & its proton acceptor CH3COOH H + + CH3COO - 33

35. Weak Acids & Bases have characteristic dissociation constant 34

36. Titration curves reveal the pKa of Weak Acids Titration is used to determine the amount of an acid in a given solution A measured volume of the acid is titrated with a solution of strong base (NaOH) of known conc. The NaOH is added in small increments until the acid is neutralized (determined by pH meter, or indicator dye) The concentration of the acid in the original solution can be calculated 35

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38. Henderson Hasselbalch equation The shape of the titration curve of any weak acid is described by the Henderson - Hasselbalch equation, which is important for understanding buffer action and acid-base balance in the blood and tissues of vertebrates. For the dissociation of a weak acid HA into H + and A -, the Henderson- Hasselbalch equation can be derived as follows: 37

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40. Henderson Hasselbalch equation 39

41. Henderson Hasselbalch equation This equation also allows us to: (1) calculate pKa, given pH and the molar ratio of proton donor and acceptor (2) calculate pH, given pKa and the molar ratio of proton donor and acceptor (3) calculate the molar ratio of proton donor and acceptor, given pH and pKa. 40

42. Solving Problems Using the Henderson- Hasselbalch Equation 1- Calculate the pKa of lactic acid, given that when the concentration of lactic acid is 0.010 M and the concentration of lactate is 0.087 M, the pH is 4.8 41

43. Solving Problems Using the Henderson- Hasselbalch Equation 1- Calculate the pKa of lactic acid, given that when the concentration of lactic acid is 0.010 M and the concentration of lactate is 0.087 M, the pH is 4.8 42

44. Solving Problems Using the Henderson- Hasselbalch Equation Calculate the pH of a mixture of 0.10 M acetic acid and 0.20 M sodium acetate. The pKa of acetic acid is 4.76 43

45. Solving Problems Using the Henderson- Hasselbalch Equation Calculate the pH of a mixture of 0.10 M acetic acid and 0.20 M sodium acetate. The pKa of acetic acid is 4.76 44

46. Solving Problems Using the Henderson- Hasselbalch Equation Calculate the ratio of the concentrations of acetate and acetic acid required in a buffer system of pH 5.30, The pKa of acetic acid is 4.76 45

47. Solving Problems Using the Henderson- Hasselbalch Equation Calculate the ratio of the concentrations of acetate and acetic acid required in a buffer system of pH 5.30, The pKa of acetic acid is 4.76 46

48. Learning Check Identify each solution as 1) acidic 2) basic3) neutral A. ___ HCl with a pH = 1.5 B. ___ pancreatic fluid [H 3 O + ] = 1 x 10 −8 M C. ___ Sprite ® soft drink pH = 3.0 D. ___ pH = 7.0 E. ___ [OH − ] = 3 x 10 −10 M F. ___ [H 3 O + ] = 5 x 10 −12 47

49. Learning Check A. 1 HCl with a pH = 1.5 B. 2 pancreatic fluid [H 3 O + ] = 1 x 10 −8 M C. 1 Sprite ® soft drink pH = 3.0 D. 3 pH = 7.0 E. 1 [OH − ] = 3 x 10 −10 M F. 2 [H 3 O + ] = 5 x 10 −12 48

50. Learning Check What is the pH of coffee if the [H 3 O + ] is 1 x 10 −5 M? 1) pH = 9.0 2) pH = 7.0 3) pH = 5.0 49

51. Learning Check What is the pH of coffee if the [H 3 O + ] is 1 x 10 −5 M? 3) pH = 5.0 pH = -log [1 x 10 −5 ] = -(-5.0) = 5.0 50

52. Learning Check A. The [H 3 O + ] of tomato juice is 2 x 10 −4 M. What is the pH of the solution? 1) 4.02) 3.73) 10.3 B. The [OH − ] of a solution is 1.0 x 10 −3 M. What is the pH of the solution? 1) 3.00 2) 11.00 3) -11.00 51

53. Learning Check A. 2) 3.7 pH = - log [ 2 x 10 -4 ] = 3.7 2 (EE) 4 (+/-) log (+/-) B. 2) 11.00 Use the K w to obtain [H 3 O + ] = 1.0 x 10 −11 pH = - log [1.0 x 10 −11 ] 1.0 (EE) 11 (+/-) log (+/-) 52

54. Testing the pH of Solutions The pH of solutions can be determined using a) pH meter. b) pH paper. c) indicators that have specific colors at different pH values. 53

55. Acids, bases, pH scale, biological buffer systems (Part 3) Saida Almashharawi 2014-2015 Basic Biochemistry CLS 233 54

56. Biological Buffer systems (3 rd Lect) 55 Buffers : are aqueous systems that tend to resist changes in pH when small amounts of acid (H+) or base (OH-) are added.

57. Weak Acids or Bases Buffer Cells and Tissues against pH Changes 56 The intracellular and extracellular fluids of multicellular organisms have a characteristic and nearly constant pH. The organism’s first line of defense against changes in internal pH is provided by buffer systems. Two important biological buffers are the phosphate and bicarbonate systems. Cytoplasm of most cells contains proteins, which contain many a.a. with functional groups that are weak acids or weak bases. (e.g., proteins containing histidine residues buffer effectively near neutral pH (His pKa=6.0) Nucleotides (e.g., ATP) & many low Mwt metabolites, contain ionizable groups that have buffering power to cytoplasm.

58. Weak Acids or Bases Buffer Cells and Tissues against pH Changes 57

59. Phosphate buffer system 58

60. Bicarbonate Buffer system 59

61. Bicarbonate Buffer system 60

62. Human blood plasma normally has a pH close to 7.4 Should the pH-regulating mechanisms fail or be overwhelmed, as may happen in severe uncontrolled diabetes when an overproduction of metabolic acids causes acidosis, the pH of the blood can fall to 6.8 or below, leading to cell damage and death. In other diseases the pH may rise to lethal levels. Although many aspects of cell structure and function are influenced by pH, it is the catalytic activity of enzymes that is especially sensitive. Enzymes typically show maximal catalytic activity at a characteristic pH, called the pH optimum (Fig. 2–21). On either side of the optimum pH their catalytic activity often declines sharply. Thus, a small change in pH can make a large difference in the rate of some crucial enzyme-catalyzed reactions. Biological control of the pH of cells and body fluids is therefore of central importance in all aspects of metabolism and cellular activities. 61

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65. 64 SUMMARY Buffering against pH Changes in Biological Systems