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

Transcription

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, H2SO4) Base: a substance that produces OH- when dissolved in water (NaOH, KOH) What about ammonia (NH3)?

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, HNO3, CH3COOH Diprotic acid: H2SO4 Triprotic acid: H3PO3 Brønsted-Lowry base: any substance that accepts a proton (H+) from an acid NaOH, NH3, 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 reactions Exceptions: Acid + base salt + H2O Carbonic acid (H 2 CO 3 )-Bicarbobate ion (HCO 3- ) Ammonia (NH 3 )-

9 Acid/base strength

10 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)

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

12

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

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

15 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

16 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

17 Equivalents The amount of molar mass (g) of hydrogen ions that an acid will donate or a base will accept 1M HCl = 1M [H+] = 1 equivalent 1M H2SO4 = 2M [H+] = 2 equivalents

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

19 Example One equivalent of Na+ = 23.1 g One equivalent of Cl g One equivalent of Mg+2 = (24.3)/2 = g Howework: Calculate milligrams of Ca+2 in blood if total concentration of Ca+2 is 5 meq/l.

20 Titration The concentration of acids and bases can be determined by titration

21 Excercise A 25 ml solution of 0.5 M NaOH is titrated until neutralized into a 50 ml sample of HCl. What was 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

22 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

23 Another method M acid V acid = M base V base

24 Note What if one mole of acid produces two moles of H+ M acid V acid = 2M base V base

25 Homework 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, H2SO4, solution, what is the molarity of the sulfuric acid?

26 Equivalence point

27 Ionization of water H3O+ = H+

28 Equilibrium constant Keq = 1.8 x M

29 Kw Kw is called the ion product for water

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31 PH

32 What is ph?

33 Acid dissociation constant Strong acid Strong bases Weak acid Weak bases

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35 pka

36 What is pka?

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38 HENDERSON-HASSELBALCH EQUATION

39 The equation pka is the ph where 50% of acid is dissociated into conjugate base

40 BUFFERS

41 Maintenance of equilibrium

42 What is buffer?

43 Titration

44 Midpoint

45 Buffering capacity

46

47 Conjugate bases CH 3 COOH Acid Conjugate base CH 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

48 How do we choose a buffer?

49 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

50 Exercise What is the ph of a buffer containing 0.1M HF and 0.1M NaF? (Ka = 3.5 x 10-4)

51 Homework What is the ph of a solution containing 0.1M HF and 0.1M NaF, when 0.02M NaOH is added to the solution?

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

53 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.

54 Polyprotic weak acids Example:

55 Hence

56 Excercise 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 NaH2PO3 is mixed with 100 ml of 0.1 M Na2HPO3?

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

58 Blood buffering Blood (instantaneously) CO2 + H20 H2CO3 H+ + HCO3- Lungs (within minutes) Excretion via kidneys (hours to days)

59 Roles of lungs and kidneys Maintaining blood is balanced by the kidneys and the lungs Kidneys control blood HCO3 concentration ([HCO3]) Lungs control the blood CO2 concentration (PCO2)

60 Calculations

61 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)

62 Acid-Base Imbalances ph< 7.35 acidosis ph > 7.45 alkalosis

63 Respiratory Acidosis H + + HCO 3 - H 2 CO 3 CO 2 + H 2 O

64 Respiratory Alkalosis H + + HCO 3 - H 2 CO 3 CO 2 + H 2 O

65 Metabolic Acidosis H + + HCO 3 - H 2 CO 3 CO 2 + H 2 O

66 Metabolic Alkalosis H + + HCO 3 - H 2 CO 3 CO 2 + H 2 O