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Acids and bases, pH and buffers Dr. Mamoun Ahram Lecture 2.

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Presentation on theme: "Acids and bases, pH and buffers Dr. Mamoun Ahram Lecture 2."— Presentation transcript:

1 Acids and bases, pH and buffers Dr. Mamoun Ahram Lecture 2

2 ACIDS AND BASES

3 Acids versus bases Acid: a substance that produces H+ when dissolved in water (e.g., HCl, H 2 SO 4 ) Base: a substance that produces OH - when dissolved in water (NaOH, KOH) What about ammonia (NH 3 )?

4 Brønsted-Lowry acids and bases The Brønsted-Lowry acid: any substance able to give a hydrogen ion (H+-a proton) to another molecule – Monoprotic acid: HCl, HNO 3, CH3COOH – Diprotic acid: H 2 SO 4 – Triprotic acid: H 3 PO 3 Brønsted-Lowry base: any substance that accepts a proton (H+) from an acid – NaOH, NH 3, KOH

5 Acid-base reactions A proton is transferred from one substance (acid) to another molecule Ammonia (NH 3 ) + acid (HA)  ammonium ion (NH 4 + ) + A- – Ammonia is base – HA is acid – Ammonium ion (NH 4 + ) is conjuagte acid – A - is conjugate base

6 Water: acid or base? Both Products: hydronium ion (H 3 O + ) and hydroxide

7 Amphoteric substances Example: water NH 3 (g) + H 2 O (l) ↔ NH 4 + (aq) + OH – (aq) HCl (g) + H 2 O (l) → H 3 O + (aq) + Cl - (aq)

8 Acid/base strength

9 Rule The stronger the acid, the weaker the conjugate base HCl (aq) → H + (aq) + Cl - (aq) NaOH (aq) → Na + (aq) + OH - (aq) HC 2 H 3 O 2 (aq) ↔ H + (aq) + C 2 H 3 O 2 - (aq) NH 3 (aq) + H 2 O (l) ↔ NH 4 + (aq) + OH - (aq)

10 Equilibrium constant HA H+ + A- K a : >1 vs. <1

11

12 Expression Molarity (M) Normality (N) Equivalence (N)

13 Molarity of solutions moles = grams / MW M = moles / volume (L) grams = M x vol (L) x MW

14 Exercise How many grams do you need to make 5M NaCl solution in 100 ml (MW 58.4)? grams = 58.4 x 5 moles x 0.1 liter = g

15 Normal solutions N= n x M (where n is an integer) n =the number of donated H + Remember! The normality of a solution is NEVER less than the molarity

16 Exercise What is the normality of H 2 SO 3 solution made by dissolving 6.5 g into 200 mL? (MW = 98)?

17 But… Molarity (and normality) is not useful for understanding neutralization reactions. 1M HCL neutralizes 1M NaOH But… 1M HCl does not neutralize 1M H 2 SO 3 Why?

18 Equivalents The amount of molar mass (g) of hydrogen ions that an acid will donate – or a base will accept 1 mole HCl = 1 mole [H + ] = 1 equivalent 1 mole H 2 SO 4 = 2 mole [H + ] = 2 equivalents

19 Examples One equivalent of Na + = 23.1 g One equivalent of Cl g One equivalent of Mg +2 = (24.3)/2 = g

20 Exercise calculate the number of equivalents of: 40g of Mg +2 16g of Al +3 Mg ++ : 40g x (1mol/24g) x (2eq/1mol) = 3.3 eq Al 3+ : 16g x (27g/1mol) x (3eq/1mol) = 1.8 eq

21 Exercises Calculate milligrams of Ca +2 in blood if total concentration of Ca +2 is 5 mEq/L. (MW = 40.1) (20.1 g/1000 mEq) x (1000 mg/g) x (5 mEq /L) = 100 mg/L What is the normality of H 2 SO 3 made by dissolving 6.5g in 200 ml? equivalents? (MW = 98)

22 Examples (calculate grams) Physiological grammEq/L1 EqMajor electrolytes ? = 23.1 g Na + ? gCl- ?3(24.3)/2 = g Mg +2 ? (40.1/2) = ~20.05 g Ca 2+ ? g K+K+ ? g HCO3 - ?2 SO and HPO 4 3-

23 Titration and equivalence point The concentration of acids and bases can be determined by titration

24 Excercise A 25 ml solution of 0.5 M NaOH is titrated until neutralized into a 50 ml sample of HCl. What is the concentration of the HCl? Step 1 - Determine [OH - ] Step 2 - Determine the number of moles of OH - Step 3 - Determine the number of moles of H + Step 4 - Determine concentration of HCl

25 A 25 ml solution of 0.5 M NaOH is titrated until neutralized into a 50 ml sample of HCl Moles of base = Molarity x Volume Moles base = moles of acid Molarity of acid= moles/volume

26 Another method M acid V acid = M base V base

27 Note What if one mole of acid produces two moles of H+ Consider the charges (or normality)

28 Modified equation N a x V a = N b x V b N a = normality of acid V a = volume of acid N b = normality of base V b = volume of base NaOH=1 H 2 SO 3 =2 H 3 PO 4 =3

29 Exercises If 19.1 mL of M HCl is required to neutralize mL of a sodium hydroxide solution, what is the molarity of the sodium hydroxide? If 12.0 mL of 1.34 M NaOH is required to neutralize mL of a sulfuric acid, H 2 SO 4, solution, what is the molarity of the sulfuric acid?

30 Ionization of water H 3 O + = H +

31 Equilibrium constant Keq = 1.8 x M

32 Kw Kw is called the ion product for water

33

34 PH

35 What is pH?

36 Exercise What is the pH of 0.01 M HCl? What is the pH of 0.01 N H 2 SO 3 ? What is the pH of a solution of 1 x HCl?

37 Acid dissociation constant Strong acid Strong bases Weak acid Weak bases

38

39 pKa

40 What is pKa?

41

42 HENDERSON-HASSELBALCH EQUATION

43 The equation pKa is the pH where 50% of acid is dissociated into conjugate base

44 BUFFERS

45 Maintenance of equilibrium Le Châtelier’s principle

46 What is buffer?

47 Titration

48 Midpoint

49 Buffering capacity

50

51 Conjugate bases AcidConjugate base CH 3 COOHCH 3 COONa (NaCH 3 COO) H 3 PO 4 NaH 2 PO 4 H 2 PO 4 - (or NaH 2 PO 4 )Na 2 HPO 4 H 2 CO 3 NaHCO 3

52 How do we choose a buffer?

53 Problems and solutions A solution of 0.1 M acetic acid and 0.2 M acetate ion. The pKa of acetic acid is 4.8. Hence, the pH of the solution is given by Similarly, the pKa of an acid can be calculated

54 Exercise What is the pH of a buffer containing 0.1M HF and 0.1M NaF? (Ka = 3.5 x 10-4) What is the pH of a solution containing 0.1M HF and 0.1M NaF, when 0.02M NaOH is added to the solution?

55 At the end point of the buffering capacity of a buffer, it is the moles of H + and OH - that are equal Equivalence point

56 Exercise What is the concentration of 5 ml of acetic acid knowing that 44.5 ml of 0.1 N of NaOH are needed to reach the end of the titration of acetic acid? Also, calculate the normality of acetic acid.

57 Polyprotic weak acids Example:

58 Hence

59 Excercises What is the pH of a lactate buffer that contain 75% lactic acid and 25% lactate? (pKa = 3.86) What is the pKa of a dihydrogen phosphae buffer when pH of 7.2 is obtained when 100 ml of 0.1 M NaH 2 PO 3 is mixed with 100 ml of 0.1 M Na 2 HPO 3 ?

60 Buffers in human body Carbonic acid-bicarbonate system (blood) Dihydrogen phosphate-monohydrogen phosphate system (intracellular) Proteins

61 Bicarbonate buffer CO 2 + H 2 0H 2 CO 3 H + + HCO 3 -

62 Blood buffering CO 2 + H 2 0H 2 CO 3 H + + HCO 3 - Blood (instantaneously) Lungs (within minutes) Excretion via kidneys (hours to days)

63 Arterial blood gases (ABG)

64 Calculations… The ratio of bicarbonate to carbonic acid determines the pH of the blood – Normally the ratio is about 20:1 bicarbonate to carbonic acid Blood pH can be calculated from this equation: – pH = pK + log (HCO 3 - /H 2 CO 2 ) – pK is the dissociation constant of the buffer, 6.10 – H 2 CO 3 =0.03 x pCO 2

65 Titration curve of bicarbonate buffer Note pKa

66 Why is this buffer effective? Even though the normal blood pH of 7.4 is outside the optimal buffering range of the bicarbonate buffer, which is 6.1, this buffer pair is important due to two properties: – bicarbonate is present in a relatively high concentration in the ECF (24mmol/L) – the components of the buffer system are effectively under physiological control: the CO 2 by the lungs, and the bicarbonate by the kidneys – It is an open system (not a closed system like in laboratory)

67 Open system An open system is a system that continuously interacts with its environment.

68 Exercise H + + HCO 3 -  H 2 CO 3  CO 2 + H 2 O Blood plasma contains a total carbonate (HCO 3- and CO 2 ) of 2.52 x M. What is the HCO 3 - /CO 2 ratio and the concentration of each buffer component at pH 7.4?

69 Exercise (continued) H + + HCO 3 -  H 2 CO 3  CO 2 + H 2 O What would the pH be if M H + is added and CO 2 is eliminated (closed system)?

70 Exercise (continued) H + + HCO 3 -  H 2 CO 3  CO 2 + H 2 O What would the pH be if M H + is added under physiological conditions (open system)?

71 Acidosis and alkalosis Can be either metabolic or respiratory Acidosis: – Metabolic: production of ketone bodies (starvation) – Respiratory: pulmonary (asthma; emphysema) Alkalosis: – Metabolic: administration of salts or acids – Respiratory: hyperventilation (anxiety)

72 Acid-Base Imbalances pH< 7.35 acidosis pH > 7.45 alkalosis 72

73 Respiratory Acidosis H + + HCO 3 -  H 2 CO 3  CO 2 + H 2 O

74 Respiratory Alkalosis H + + HCO 3 -  H 2 CO 3  CO 2 + H 2 O

75 Metabolic Acidosis H + + HCO 3 -  H 2 CO 3  CO 2 + H 2 O

76 Causes of respiratory acid-base disorders

77 Causes of metabolic acid-base disorders

78 78 Compensation Compensation: The change in HCO 3 - or pCO 3 that results from the primary event If underlying problem is metabolic, hyperventilation or hypoventilation can help : respiratory compensation. If problem is respiratory, renal mechanisms can bring about metabolic compensation.

79 79 Complete vs. partial compensation May be complete if brought back within normal limits Partial compensation if range is still outside norms.

80 Acid-Base DisorderPrimary Change Compensatory Change Respiratory acidosis pCO 2 up HCO 3 - up Respiratory alkalosis pCO 2 down HCO 3 - down Metabolic acidosis HCO 3 - down PCO 2 down Metabolic alkalosis HCO 3 - up PCO 2 up

81 FULLY COMPENSATED pHpCO 2 HCO 3 - Resp. acidosisNormal But<7.40 Resp. alkalosisNormal but>7.40 Met. AcidosisNormal but<7.40 Met. alkalosisNormal but>7.40

82 Partially compensated pHpCO 2 HCO 3 - Res.Acidosis Res.Alkalosis Met. Acidosis Met.Alkalosis

83 EXAMPLES

84 Example 1 Mr. X is admitted with severe attack of asthma. Her arterial blood gas result is as follows: – pH : 7.22 – PaCO 2 : 55 – HCO 3 - : 25 – pH is low – acidosis – paCO 2 is high – in the opposite direction of the pH. – HCO 3 - is Normal Respiratory Acidosis

85 Example 2 Mr. D is admitted with recurring bowel obstruction has been experiencing intractable vomiting for the last several hours. His ABG is: – pH : 7.5 – PaCO 2 :42 – HCO 3 - : 33 Metabolic alkalosis

86 Example 3 Mrs. H is kidney dialysis patient who has missed his last 2 appointments at the dialysis centre. His ABG results: – pH: 7.32 – PaCO 2 : 32 – HCO 3 - : 18 Partially compensated metabolic Acidosis

87 Example 4 Mr. K with COPD.His ABG is: – pH: 7.35 – PaCO 2 : 48 – HCO 3 - :28 Fully compensated Respiratory Acidosis

88 Example 5 Mr. S is a 53 year old man presented to ED with the following ABG. – pH: 7.51 – PaCO 2 : 50 – HCO 3 - : 40 Metabolic alkalosis

89 Practice ABG’s 1.pH 7.48 PaCO 2 32 HCO pH 7.32 PaCO 2 48 HCO pH 7.30 PaCO 2 40 HCO pH 7.38 PaCO 2 48 HCO pH 7.49 PaCO 2 40 HCO pH 7.35 PaCO 2 48 HCO pH 7.45 PaCO 2 47 HCO pH 7.31 PaCO 2 38 HCO pH 7.30 PaCO 2 50 HCO pH 7.48 PaCO 2 40 HCO

90 Answers to Practice ABG’s 1.Respiratory alkalosis 2.Respiratory acidosis 3.Metabolic acidosis 4.Compensated Respiratory acidosis 5.Metabolic alkalosis 6.Compensated Respiratory acidosis 7.Compensated Metabolic alkalosis 8.Metabolic acidosis 9.Respiratory acidosis 10.Metabolic alkalosis

91 Salivary buffers Salivary pH  6.3 Main buffers: – Bicarbonate – Phosphate – Proteins Below pH 5.5, demineralization usually follows

92 Flow rate [H 2 CO 3 ] = 1.3 mM/L –almost constant, but [HCO 3 - ] is not The greater the salivary flow, the more bicarbonate ions available for combining with free hydrogen ions – Normal salivary flow rates = 0.1 and 0.6 mL/minute

93 Buffering saliva Salivary carbonic anhydrase


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