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. Ammonia (NH3) + acid (HA)  ammonium ion (NH4+) + A-
Acid-base reactions
A proton is transferred from one substance (acid) to another molecule
Ammonia (NH3) + acid (HA)  ammonium ion (NH4+) + A-
Ammonia is base
HA is acid
Ammonium ion (NH4+) is conjuagte acid
A- is conjugate base

6. Water: acid or base?
Both
Products: hydronium ion (H3O+) and hydroxide

7. Amphoteric substances
Example: water
NH3 (g) + H2O(l) ↔ NH4+(aq) + OH–(aq)
HCl(g) + H2O(l) → H3O+(aq) + Cl-(aq)

8. Acid-base reactions Acid + base  salt + H2O Exceptions:
Carbonic acid (H2CO3)-Bicarbobate ion (HCO3-)
Ammonia (NH3)-

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)
HC2H3O2 (aq) ↔ H+(aq) + C2H3O2-(aq)
NH3 (aq) + H2O(l) ↔ NH4+(aq) + OH-(aq)

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

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. grams = 58.4 x 5 moles x 0.1 liter = 29.29 g
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 H2SO3 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- - 35.5 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. MacidVacid = MbaseVbase
Another method
MacidVacid = MbaseVbase

24. MacidVacid = 2MbaseVbase
Note
What if one mole of acid produces two moles of H+
MacidVacid = 2MbaseVbase

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

31. pH

32. What is pH?

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

35. pKa

36. What is pKa?

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

47. Conjugate bases Acid Conjugate base CH3COOH CH3COONa (NaCH3COO) H3PO4
NaH2PO4
H2PO4- (or NaH2PO4)
Na2HPO4
H2CO3
NaHCO3

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 CO2 + H20 H2CO3 H+ + HCO3- Blood (instantaneously)
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+ + HCO3-  H2CO3  CO2 + H2O

64. Respiratory Alkalosis
H+ + HCO3-  H2CO3  CO2 + H2O

65. Metabolic Acidosis
H+ + HCO3-  H2CO3  CO2 + H2O

66. Metabolic Alkalosis
H+ + HCO3-  H2CO3  CO2 + H2O