1. Introduction into biochemistry
Dr. Mamoun Ahram
Nursing
Summer semester, 2015

2. What is biochemistry? It is the chemistry of living organisms.
It describes the structure, organization, and functions of living matter in molecular terms.

3. Important terms Electronegativity Covalent bonds
Polar vs. non-polar covalent bonds
Non-covalent interactions
electrostatic interactions
hydrogen bonds (donor and acceptor)
van der Waals interactions
Hydrophobic interactions
Hydrophobic versus hydrophilic molecules
Backbone of a molecule
Nucleophile versus electrophile

4. Chemical elements in living creatures
Living organisms on Earth are composed mainly of 31 elements

5. Why can carbon form complex molecules?

6. Properties of carbon (1)
It can form four bonds, which can be single, double, or triple bonds.
Each bond is very stable.
strength of bonds: triple > double > Single)
They link C atoms together in chains and rings.
These serve as a backbones.

7. Properties of carbon (2)
Carbon bonds have angles giving molecules three-dimensional structure.
In a carbon backbone, some carbon atoms rotate around a single covalent bond producing molecules of different shapes.

8. Properties of carbon (3)
The electronegativity of carbon is between other atoms.
It can form polar and non-polar molecules
Pure carbon is not water soluble, but when carbon forms covalent bonds with other elements like O or N, the molecule that makes carbon compounds to be soluble.
Nonpolar

9. Water

10. Properties of water (1)
Water is a polar molecule as a whole because of
the different electronegativitiy between Hydrogen and oxygen
It is angular
Water is highly cohesive.
Water molecules produce a network.

12. Properties of water (2)
Water is an excellent solvent because It is small and it weakens electrostatic forces and hydrogen bonding between polar molecules.

13. Properties of water (3) It is reactive because it is a nucleophile.
A nucleophile is an electron-rich molecule that is attracted to positively-charged or electron-deficient species (electrophiles)

14. Properties of water (4)
Water molecules are ionized to become a positively-charged hydronium ion (or proton), and a hydroxide ion:

15. Acids and bases

16. Acids versus bases
Acid: a substance that produces H+ when dissolved in water (HCl, HNO3, CH3COOH, H2SO4, H3PO3)
H+ Reacts with water producing hydronium ion (H3O+)
Base: a substance that produces OH- when dissolved in water (NaOH, KOH)

17. Brønsted-Lowry acids and bases
The Brønsted-Lowry acid: any substance (proton donor) 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
NH3

18. According to this… The reaction does not need to occur in water
the Brønsted Lowry base does not need give OH- ions in water.
The products of an acid base reaction can also behave as acids and bases.

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

20. Amphoteric substances
Substances that can act as an acid in one reaction and as a base in another are called amphoteric substances
Example: water
With ammonia (NH3), water acts as an acid because it donates a proton (hydrogen ion) to ammonia
NH3 + H2O ↔ NH4+ + OH–
With hydrochloric acid, water acts as a base water is acting as a bas
HCl+ H2O → H3O+ + Cl-

21. Acid/base strength Acids differ in their ability to release protons
Strong acids dissociate 100%
Bases differ in their ability to accept protons
Strong bases have strong affinity for protons

22. Note
For multi-protic acids (H2SO3, H3PO4), each proton is donated at different strengths

23. Rule The stronger the acid, the weaker the conjugate base
Reactions favors formation of weak acid
Strong acids and bases are one-way reactions
HCl → H+ + Cl-
NaOH → Na+ + OH-
Weak acids and bases do not ionize completely
HC2H3O2 ↔ H+ + C2H3O2-
NH3 + H2O ↔ NH4+ + OH-

24. Equilibrium constant Acid/base solutions are at constant equilibrium
We can write equilibrium constant (Keq) for such reactions   
The value of the Keq indicates direction of reaction
When Keq is greater than 1 the product side is favored
When Keq is less than 1 the reactants are favored
Equilibrium constant for acids is Ka and for bases Kb
Note: H3O+ = H +

26. Expression
Solutions can be expressed in terms of its concentration or molarity
Acids and bases can also be expressed in terms of their normality (N) or equivalence (Eq)

27. Molarity of solutions
We know that moles of a solution are the amount in grams in relation to its molecular weight (MW or a.m.u.).
moles = grams / MW
A molar solution is one in which 1 liter of solution contains the number of grams equal to its molecular weight.
M = moles / volume
 Since (mol = grams / MW), you can calculate the grams of a chemical you need to dissolve in a known volume of water to obtain a certain concentration (M) using the following formula:
 grams = M x vol x MW

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

29. Equivalents
When it comes to acids, bases and ions, it is useful to think of them as equivalents.
An equivalent is the amount of moles or molar mass (g) of hydrogen ions that an acid will donate
or a base will accept
A 1 g-Eq of any ion is defined as the molar mass of the ion divided by the ionic charge.

30. Examples 1 mol HCl = 1 mol [H+] = 1 equivalent
1 mol H2SO4 = 2 mol [H+] = 2 equivalents
One equivalent of Na+ = 23.1 g
One equivalent of Cl g
One equivalent of Mg+2 = (24.3)/2 = g
Remember: One equivalent of any acid neutralizes one equivalent of any base.

31. Exercise
Calculate milligrams of Ca+2 in blood if total concentration of Ca+2 is 5 mEq/L.
1 Eq of Ca2+ = 40.1 g/2 =20.1 g
Grams of Ca2+ in blood =
= (5 mEq/L) x (1 Eq/1000 mEq) x (20.1 g/ 1 Eq)
= 0.1 mg/L
=100 mg/L

32. N= n x M (where n is an integer)
Normal solutions
Normality (N) considers both the molarity of the solution and the equivalent content of the acid or base
N= n x M (where n is an integer)
For an acid solution, n is the number of H+ provided by a formula unit of acid. Similarly, for a base solution is the number of OH- a base can donate

33. The normality of a solution is NEVER less than the molarity
Example
3 M H2SO4 solution is the same as a 6 N H2SO4 solution
1 M Ca(OH)2 solution is the same as a 2N Ca(OH)2 solution
Remember!
The normality of a solution is NEVER less than the molarity

34. Exercise
What is the normality of H2SO3 solution made by dissolving 6.5 g into 200 mL? (MW = 98)?
M = mol / MW or M = grams / (MW x vol)
= 6.5 g / (98 x 0.2 L)
= 6.5 / 19.6
= 0.33
N = M x n = 0.33 x 2 = 0.66 N

35. Normality and equivalents
Based on the equation above, since n eq of an acid is neutralized by the same n eq of a base, then (N x liters) of an acid is neutralized by (N x liters) of a base

36. Problems
If number of H+ (or OH-) is different, convert M to N, find the answer in N, then convert N to M.

37. Note that each one produces 1 mole of H+ or OH-, so 1M of HCl is equal to 1M of NaOH, so 1M HCl produces 1M of H+, which is neutralized by 1M NaOH, which produces 1M OH-
M1 x Vol1 = M2 x Vol2
0.12 x 22.4 = M2 x 12
M2 = (0.12 x 22.4) / 12
M2 = M

38. Note that they 1 mole of HNO3 produces 1 mole of H+, but 1 mole of Ba(OH)2 produces 2 moles of OH-. IN this case, it is better to compare normalities (N), not molarities (M). convert M to N first N of HNO3 = 1 x = N N of Ba(OH)2 = 2 x 0.12 = 0.24 N1 x Vol1 = N2 x Vol x 15 = x Vol2 Vol2 = (0.24 x 15) / Vol2 = mL Apply same method for third question

39. Ionization of water
Water dissociates into hydronium (H3O+) and hydroxyl (OH-) ions
For simplicity, we refer to the hydronium ion as a hydrogen ion (H+) and write the reaction equilibrium as

40. Equilibrium constant
The equilibrium constant Keq of the dissociation of water is
The equilibrium constant for water ionization under standard conditions is 1.8 x M

41. Kw
Since there are 55.6 moles of water in 1 liter, the product of the hydrogen and hydroxide ion concentrations results in a value of 1 x for:
This constant, Kw, is called the ion product for water

42. [H+] and [OH-]
For pure water, there are equal concentrations of [H+] and [OH-], each with a value of 1 x 10-7 M
Since Kw is a fixed value, the concentrations of [H+] and [OH-] are inversely changing
If the concentration of H+ is high, then the concentration of OH- must be low, and vice versa. For example, if [H+] = 10-2 M, then [OH-] = M

43. pH and buffer

44. What is pH?
Changes in [H+] have significant effects on many biochemical processes
A logarithmic quantity, pH, was developed as a convenient scale for working with levels of [H+] The pH of a solution is a measure of its concentration of H+ (acidity)
The pH is defined as

45. Acidic and basic pH On the pH scale, pH of 7.0 is considered neutral
pH values below 7.0 indicate acidic solutions
pH values above 7.0 indicate basic (alkaline) solutions

46. Acids and bases An acid is a proton donor. A base is a proton acceptor
The species formed by the ionization of an acid is proton(s) and conjugate base of the acid. Conversely, protonation of a base yields its conjugate acid
The conjugate base of acetic acid is acetate ion

47. Strong acids
A strong acid dissociates completely, while a strong base completely binds to a proton.
HA  H+ + A-
In a 0.1 M solution of hydrogen chloride (HCl) in water, [H+] = 0.1 M because all the HCl has dissociated into H+ and Cl ions
For this solution pH = log 0.1 = 1.0

48. Strong bases
Similarly, a 0.01 M solution of NaOH will have an OH- concentration of 1 x 10-2
The pH is calculated as shown below:
[H+] = 1 x / 1 x 10-2
= 1 x 10-12
The pH will be 12
Otherwise, you say if [OH-] = 10-2, then [H+] = and the pH = 12

49. Weak acids and bases
Many acids, such as the amino acids and acetic acid, do not dissociate completely in water
Such acids are called weak acids, and similarly, some bases do not completely dissociate and are called weak bases
The ionization equilibrium of a weak acid is given as

50. Acid dissociation constant
Ka, the acid dissociation constant, is the equilibrium constant for the reaction of a weak acid (HA) converting into a proton and the conjugate base (A-).

51. What is pKa?
As with pH, a logarithmic scale is useful for working with Ka values. pKa is defined as:
pKa is the tendency of an acid to dissociate into H+ and the conjugate base A-
Each weak acid and base have their own fixed pKa values
Note: if Ka is very large, pKa is very small

53. The Henderson-Hasselbalch equation
What is the relation between pH, pKa, and the ratio of acid to base?
A useful expression can be derived

54. When pKa = pH
At the point of the dissociation where the concentration of the conjugate base [A-] is equal to that of the acid [HA]:
pH = pKa + log[1]
The log of 1 = 0
pKa = pH
pKa is the pH where 50% of acid is dissociated into conjugate base

55. Maintenance of equilibrium
Because a weak acid (or base) can exist in equilibrium with its conjugate base, the equilibrium can be maintained even when small amounts of [H+] or [OH-] are added to it.

56. Example
If a small amount of [H+] is added to a solution of weak acid, then the [H+] can combine with it to produce [HA], maintaining [H+] level
If a small amount of [OH-] is added, then [OH-] will bind to [H+] producing H2O, [HA] will dissociate to [H+] and [A-] to compensate for the loss of [H+], maintaining [H+] level

57. What is buffer?
The shift in equilibrium will then maintain the pH of the solution
A solution with the ability to resist changes in pH is called a buffer

58. Titration The pKa values of weak acids are determined by titration.
This involves adding small amounts of a strong acid or base to the solution and measuring the resulting changes in pH generating a plot of titration curve.

59. Midpoint
For example, titration of acetic acid gives the curve shown to the right. Note that the pKa value is the midpoint of the curve.
At this point, pH is equal to the pKa since there are equal concentrations of HA and A-.

60. Buffering capacity
The ability of a buffer to minimize changes in pH is known as its buffering capacity.
The buffering capacity of weak acids and bases is one pH unit from their pKa values.
In other words, a buffer is effective at resisting changes in pH at pH +/- 1 of the pKa.
For example, since the pKa of acetic acid is 4.7, then the buffering capacity ranges

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

62. Properties of buffers They all have the same function.
Their pKa’s are different.
Buffering capacities are same.
We choose a buffer based on its pKa, which should be close to the desired pH.

63. Problems and solutions
the Henderson-Hasselbalch (HH) equation is useful for determining the pH of a solution if the molar proportion of A- to HA and the pKa of HA are known.
Example:
Consider 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 if the molar proportion of A- to HA and the pH of the solution are known.

64. Exercise
What is the pH of a buffer containing 0.1M HF and 0.1M NaF? (Ka = 3.5 x 10-4)
[H3O+] = Ka x [HF]/[F-]
= 3.5 x 10-4 x (0.1/ 0.1)
= 3.5 x 10-4
pH = -log 3.5 x 10-4
= 3.46
Can you solve it using the Henderson-Hasselbalch (HH) equation?

65. Polyprotic weak acids
Some weak acids (buffers) can donate more than one proton.
An example is phosphoric acid (H3PO4).
Phosphoric acid can donate up to three protons.

66. Different pKa These protons do not dissociate at the same time.
Each proton has a certain pKa.

67. Hence
Each proton dissociate at certain pH

68. 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?

69. Excellent buffer
This makes phosphoric acid an excellent buffer at different pH ranges.
It also the buffer intracellularly.

70. Buffers in human body Carbonic acid-bicarbonate system (blood)
Dihydrogen phosphate-monohydrogen phosphate system (intracellular)
Proteins (via the amino acids)

71. Blood buffering Blood pH must be maintained at around 7.4.
Any dramatic change (up or down) can be dangerous.
By far the most important buffer for maintaining acid-base balance in the blood is the carbonic-acid-bicarbonate buffer.

72. How?
During metabolism, cells produce CO2. This CO2 is dissolved in blood and is converted to carbonic acid, which dissociates into bicarbonate ion (conjugate base) and H+.
Dissolved CO2 is in equilibrium with pressure of CO2 (pCO2) in lungs.
If blood pH changes, then the body changes the rate of breathing in order to balance blood pH.

73. The Henderson-Hasselbalch equation for blood buffering
The normal ratio of [HCO3-/CO2] in the blood is about 20 to 1 at pH What is this ratio at pH 7.3 (acidosis) and at 7.5 (alkalosis)?

74. The equation
Relationships of the bicarbonate buffer system to the lungs and the kidneys.

75. 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)