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Biochemistry (CHEM 262) Dr. Abdellatif Ibdah Department of chemistry Office: N4L0 Office Hour: will be ready.

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Presentation on theme: "Biochemistry (CHEM 262) Dr. Abdellatif Ibdah Department of chemistry Office: N4L0 Office Hour: will be ready."— Presentation transcript:

1 Biochemistry (CHEM 262) Dr. Abdellatif Ibdah Department of chemistry Office: N4L0 Email: aaibdah@just.edu.joaaibdah@just.edu.jo Office Hour: will be ready soon

2 Water: the solvent for Biochemical reactions water polarity Hydrogen bonds Acids, Bases, and Buffers Titration curves Buffers

3 Chemical Bonds Ionic Bond: Ionic Bond: electrons transfer completely to form oppositely charged ions Covalent bond Covalent bond: electrons are shared between two atoms. -Nonpolar covalent bonds -Nonpolar covalent bonds when the electrons are shared equally between the two atoms - Polar covalent bonds - Polar covalent bonds when the electrons are not shared equally slightly positive,negative. One end is slightly positive, the other is slightly negative. Cl + Cl Cl Cl Na Cl +Na + Cl  + H - F ++++ ---- Cl Cl –

4 Definition Covalent Bonding: is formed when one atom shares its electron with another atom

5 Bond Polarity shifted The density of electron cloud is shifted towards one of the two bonded atoms -Polarity H - F ++++ ----

6 Electronegativity is the ability or tendency of an atom to attract electrons in a bond (become more negative) Used to Determine Bond Polarity  EN <.45 Nonpolar Bond 2 >  EN >.45 Polar Bond  EN > 2 Ionic Bond Atoms Electronegativity

7 Electronegativity: The Pauling Scale

8 The bond is non-polar if the atoms share electrons equally O-O F-F C-H ( very small electronigativity difference)

9 Geometry and Polarity  +  - H – F H-F is polar bond Polar Molecule -The C-O bond is a polar bond - The CO 2 is not polar Molecules CO 2 H2OH2O -The H-O bond is a polar bond - The H 2 O is polar Molecules

10 Properties of water Water is a bent molecule Uneven sharing of electrons in the two bond Oxygen, has a partial negative charge Hydrogen have a partial positive charge. Water is a polar The partial charge that develops across the water molecule helps make it an excellent solvent. moleculesolvent

11 Solubility - Solubility can be predicted based on “like dissolves like ” -Polar Solutes like to dissolve in polar Solvents - water and methanol are both polar and dissolve in each other Ions readily dissolve in polar solvent

12 Water: Solvent Properties Ionic compound such as KCl and polar compounds such as ethyl alcohol or acetone tend to dissolve in water This is because of the electrostatic attraction between different charges which makes the system more stable by lowering the energy When salt is added to water the partial charges on the water molecule are attracted to the K + and Cl - ions.saltmoleculeions

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14 Intermolecular Forces 1- Intermolecular forces are the attractive forces holding particles together in the condensed (liquid and solid) phases of matter Dipole-dipole interactions – attractive forces between polar molecules + - + - + - + - + - + - + - + - + - + -

15 Intermolecular Forces 2- Hydrogen Bonds Hydrogen bond hydrogen atomHydrogen bond is an intermolecular force in which the hydrogen atom bound to highly electronegative atom (Nitrogen, Oxygen, Fluoride) Hydrogen BondHydrogen Bond is very strong type of dipole- dipole interaction  + H-F  - ---  + H-F  -

16 Hydrogen Bonds Between HF Molecules

17 Water molecules Hydrogen bonding is also the reason for the unusually high boiling point of water

18 Hydrogen Bonds The hydrogen of one water molecule are attracted to the oxygen from other water molecules. Hydrogen bond: is a non covalent bond between hydrogen which is covalently boned to a very electronegative atom such as oxygen or nitrogen and unshared pair of electron on another electronegative atom

19 Potential of water to form Hydrogen bonding affect its Solvent properties Melting point: Boiling point: although hydrogen bonding is week and required less Energy to break it down as compared to covalent bond, its enough to raise water melting and boiling point Ice: 4 H-bonds per water molecule liquid: 2.3 H-bonds per water molecule.

20 Stabilizing the three dimensional structure of Proteins, DNA, RNA Other biologically important hydrogen bonds

21 Intermolecular Forces 3- Van der Waals (london) forces Non-polar molecules also exert forces on each other induced dipole- induced dipole interaction As a result, induced dipole- induced dipole interaction would take place.

22 Polar Vs Non-Polar ionic compounds and polar compounds are hydrophilic (water loving). ionic compounds Non polar molecules such as hydrocarbons are hydrophobic (water hating) and do not dissolve in water but tend to sequester themselves from water. Oil is a non-polar molecule. molecule It is not attracted to water molecules and, therefore, it does not dissolve in water.molecules Molecules that have both polar (hydrophilic) and non polar hydrophobic portions are amphipathic

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24 This compound tend to form micelles in which the polar head are in contact with water and the nonpolar tails sequestered from water

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26 2.3 Acids, Bases and pH Acid is a proton (hydrogen ion, H + ) donor molecule A base is a proton acceptor molecule The strength of acids depends on the degree of dissociation in water Strength range from complete dissociation in water for strong acid to no dissociation for very week acids Any intermediate value is possible

27 strength of an acid is amount of hydrogen ion released when a given amount of acid is dissolved in water). Describe by K a : Acid dissociation constant Written correctly, The greater the Ka, the stronger the acid. Acid Strength

28 Conjugate Acid-Base Pairs

29 Water (H 2 O) can be dissociated into a hydrogen ion (H + ) and a hydroxide ion (OH - ). Notice that if you put the hydrogen and hydroxide ions back together, you will restore H 2 O

30 Ionization of H 2 O and pH Lets quantitatively examine the dissociation of water: Molar concentration of water is constant K w =K a × [H 2 O] = [H + ][OH - ] K w is called the ion product constant for water. Must define a quantity to express hydrogen ion concentrations…pH

31 what I mean by "water balance" is that the concentration of hydrogen ions is the same as the concentration of hydroxide ions ([H+] = [OH-]). A solution with a pH from 0 to 6.9 is an acid, while a solution with a pH from 7.1 to 14 is a base (can also be called an "alkaline" solution). Acids and bases do not have an even balance of hydrogen ions with hydroxide ions. Acids have more hydrogen ions, while bases have more hydroxide ions.

32 pH A more practical quantity is known as pH. pH = -log [H + ] Note that a difference of one pH unit implies a ten fold difference in [H+]. pH of Pure water is 7 which is neutral, whereas basic solutions have pH > 7 and acidic solution have pH < 7. In biochemistry most acids are weak acids. These have K a well bellow one. Therefore, pK a has been defined by analogy to pH definition: pK a = -log K a

33 Calculation Calculate the pH for the following solutions: 0.01 M HCl = 10 -2 = [H+] pH= -log [H+] = -log [10 -2 ] = 2 2) 0.01 M NaOH =10 -2 = [OH - ] [H + ] [OH - ] = 10 -14 [H + ] = 10 -12 pH= -log [H + ] = -log [10 -12 ] = 12

34 Henderson-Hasselbalch Equation to connect K a of any weak acid with the pH solution containing both that acid and its conjugate base.

35 Henderson-Hasselbalch (Cont’d) From this equation, we see that when the concentrations of weak acid and its conjugate base are equal, the pH of the solution equals the pK a of the weak acid pH = pK a when pH < pK a, the weak acid predominates (Protonated) when pH > pK a, the conjugate base predominates ( deprotonated)

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37 The Henderson-Hasselbalch equation can be used to calculate the pH of the solution Assume 0.1 and 0.5 mole (NaOH) base has been added to 1 mole solution of acetic acid what is the pH?. Assuming the volume is 1L. CH 3 COOH + NaOH → CH 3 COO - Na + + H 2 O When 0.1 mol of NaOH is added, 0.1 mol of acetic acid react with it to form 0.1 mol of acetate ion, leaving 0.9 mol of acetic acid pH = 4.76 + (-0.95) = 3.81 With 0.5 Mole OH¯ added: pH = pKa + log [0.5 ] [0.5] pH = 4.76 + 0 pH = 4.76 = pKa

38 Titration Curves an experiment in which measured mounts of acid (or base) are added to measured amounts of base (or acid) You can follow the course of reaction with pH meter Monoprotic acid releases one H + and has 1 Pka Diprotic acid releases two H + and has 2 pka Triprotic acid releases three H + and has 3 pKa

39 Acetic Acid Acetate Pka of acetic acid is 4.76 Inflection point % %

40 Polyprotic acids 2.14 7.2 12.4 the flatness of the curve near its starting and end points in comparison with the titration curves in the previous figure. This indicats that phosphoric acid is close to a strong acid and phosphate a strong base.

41 Titration Curves Equivalence point: The point in the titration at which the acid is exactly neutralized. The slope of each titration curve is much less near its midpoint than it is near its wing. When [HA] = [A - ], the pH of the solution is relatively insensitive to the addition of strong base or strong acid. This solution known as acid-base buffer.

42 Buffers Buffer: tends to resists change in pH when small to moderate amounts of a strong acid or a strong base is added. consists of a weak acid and its conjugate base Buffers can only be used effectively within one pH unit of their pKa. Examples of acid-base buffers are solutions containing CH 3 COOH and CH 3 COONa H 2 CO 3 and NaHCO 3 NaH 2 PO 4 and Na 2 HPO 4

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44 Q1) Calculate the pH when 1 ml of 0.1 M HCl is added to a) 99 ml water b) 99 ml buffer Q2) Calculate the pH when 0.1 M NaOH is added to a) 99 ml water b) 99 ml buffer

45 1) 1 ml of 0.1 M HCl is added to 99 ml water, and 2) 1 ml of 0.1 M NaOH is added to 99 ml water. HCl + H 2 O → H 3 O + + Cl - We assume that 0.1 M HCl dissociate completely to give 0.1M H 3 O + because HCl is a strong acid. M 2 V 2 = M 1 V 1 → M 2 = (0.1M× 1ml) / 100ml M 2 = 0.001 M We have 100 ml of 0.001 M HCl pH= -log [H 3 O + ] = -log 10 -3, pH = 3 We have also 100 ml of 0.001 M NaOH [H + ][OH - ]= 10 -14 then [H + ]= 10 -11 ; therefore, pH = -log(10 -11 ), pH = 11 The pH calculation

46 Calculate the pH when 1)1 ml of 0.1 M HCl is added to 99 ml buffer (H 2 PO 4 - /HPO 4 -2 ) buffer solution 2)1 ml of 0.1 M NaOH is added to 99 ml Buffer. Know that pK a = 7.2, [H 2 PO 4 - ]= 0.1M, and [HPO 4 2- ]= 0.063M. HPO 4 2- + H +  H 2 PO 4 - Before addition of acid 0.063 M 0.1 M Acid added 0.0001 M After acid react with HPO 4 2- 0.062 M (-001) 0.101 (+0.001) pH = 7.2 + log 0.062/0.101 = 6.99. Calculation

47 2) When OH added. H 2 PO 4 - + OH -  HPO 4 2- + H 2 O Before addition of base 0.1 M 0.063 M Base added 0.001M (-0.001) (+ 0.001) After base react with HPO 4 2- 0.099 M 0.064M PH = 7.2 + log 0.064/0.099 = 7.01. Calculation

48 Buffer Range A buffer is effective in a range of about +/- 1 pH unit of the pK a of the weak acid When pH increased by 1 unit from Pka the ratio of Base to acid increased by 10 units When pH increased by 2 unit from Pka the ratio of Base to acid increased by 100 units

49 Buffer Capacity Buffer capacity is related to the ratio of the weak acid and its conjugate base the greater the concentration of the weak acid and its conjugate base, the greater the buffer capacity

50 Naturally Occurring Buffers H 2 PO 4 - /HPO 4 -2 pair is the principal buffer in cells Carbonic acid (H 2 CO 3 /HCO 3 - ) is an important (but not the only) buffer in blood H 2 CO 3 pK a is 6.37 (the pH of human blood is 7.4 which is near the end of the buffering range) CO 2 can dissolve in water (Blood) The dissolved CO 2 can form Carbonic acid Carbonic acid in turn react to produce bicarbonate ion

51 Naturally Occurring Buffers

52 pH of some body fluid Extracellular fluid = 7.4 Intracelluar fluid = 7.0 Gastric juice = 1.5-3.0 Pancreatic juice = 7.8-8.0 Saliva = 6.4-7.0 Urine = 5.0-8.0

53 Acid-base balance Our bodies are extremely sensitive to blood pH. Any blood pH more acidic than 6.8 or more basic than 8.0 causes death. Maintaining pH within narrow limits Altered acid-base balance affects Osmolarity/fluid volume enzyme action transport process membrane potential nerve and muscle action. Because acids are produced in the course of normal metabolism, the body must have buffers to maintain the pH.

54 Selecting a Buffer The following are typical criteria suitable pK a no interference with the reaction or detection of the assay suitable ionic strength suitable solubility

55 Laboratory Buffers

56 Zwitterions: compounds that have both a positive and negative charge Appropriate for in vitro use


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