Biochemistry Falah Shidaifat, DVM, MS, PhD.

Professor of Veterinary Physiology

Department of Basic Biomedical Sciences

Faculty of Veterinary Medicine

Jordan University of Science and Technology

Water: the solvent for Biochemical reactions

• water polarity

• Hydrogen bonds

• Acids, Bases, and Buffers

• Titration curves

Definition

shared electrons can be eitherEqual or unequal

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

Two atoms with unpaired electrons in their outer shells can form bond with each other by sharing electron pair

.

. .

. .

..

. .

• e- are unequally shared• Cl side is slightly -• H side is slightly +

• Unequal sharing of e- in a bond called Polar Covalent Bond or Polar Bond

• Partial Charge indicated by delta +/-

• Molecules with +ve and –ve ends called dipoles.

Polar Covalent Bonds

-+

• Electronegativity is the ability or tendency of an atom to attract

electrons in a bond (become more negative)

The bond is Polar if the two atoms have an unequal share in the bonding electrons (different electronigativity)

The bond is non-polar if the atoms share electrons evenly (same electronigativity)

Example:C-H bond in CH4 (methane)

Geometry of molecule determine its polarity Triatomic with Polar Bonds (CO2)

Non polar Molecule

The bond in a molecule may be polar, but the molecule itself is not polar because of its geometry

O=C=O

The linear geometry of the molecule causes the attraction of the oxygen for the electron in one bond is cancelled out by the equal and opposite attraction for the electron by the oxygen on the other side of the molecule

Solvent Properties of water

Water is a bent molecule

Uneven sharing of electrons in the two bond

one part of the molecule, the Oxygen, has a partial negative charge

and the Hydrogen have a partial positive charge.

Water is a dipole

The partial charge that develops across the water molecule helps make it an excellent solvent. 

Water: Solvent Properties

• Ionic compound such as KCl and polar (dipoles) 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.

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

Potential of water to form Hydrogen bonding affect its

Solvent properties

Melting point:

Boiling point:

Density :

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

Liquid water is less extensively hydrogen bonded and thus is denser than ice thus ice cubes float

Ice: 4 H-bonds per water molecule liquid: 2.3 H-bonds

per water molecule.

http://www.sumanasinc.com/webcontent/animations/content/propertiesofwater/water.html

Stabilizing the three dimensional structure of Proteins, DNA, RNA

Other biologically important hydrogen bonds

Polar Vs Non-Polar

• any substance that carries a net electrical charge such ionic compounds and (dipole) are hydrophilic (water loving).

• 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.  • Because there is no net electrical charge across an oil

molecule• It is not attracted to water molecules and, therefore, it

does not dissolve in water.

Molecules that have both polar (hydrophilic) and non polar hydrophobic portions are amphipathic

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

A similar process is responsible for separation of oil and water

Interaction between nonpolar molecules are very week and depends on a temprorary dipols and dipols they Induced which are caused by clumping of bonding electrons at one end Van der waal interaction

Acids, Bases and pH

Acid is a proton (hydrogen ion) 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 week acids

Any intermediate value is possible

• One can derive a numerical value for the strength of an acid (amount of hydrogen ion released when a given amount of acid is dissolved in water).

• Describe by Ka: Acid dissociation constant

• Written correctly,

• The greater the Ka, the stronger the acid.

Acid Strength

Think of pH as "water balance."

• Water (H2O) 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 H2O

Ionization of H2O and pH

• Lets quantitatively examine the dissociation of water:

• Molar concentration of water (55.5M) 1000 (gm of H20/L) /18 (MW of H20 )

• Kw is called the ion product constant for water.

• Quantity of hydrogen ion concentrations can be expressed as pH

• "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. 

Different enzymes have different optimum pH for their biological activities

pH

• The values of [H+] for most solutions are small and difficult to compare.

• A more practical quantity is known as pH.

• pH = -log10[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 basicsolutions have pH > 7 and acidic solution have pH < 7.

Dr. M. khalifeh

In biochemistry most acids are weak acids. These have Ka well bellow one.Therefore, pKa has been defined by analogy to PH definition:

pKa = -log10 Ka

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 [1012] = 12

Henderson-Hasselbalch

Equation to connect Ka of any weak acid with the pH of

solution containing both that acid and its conjugate base.

pKa has been defined by analogy to PH: pKa = -log10 Ka

Henderson-Hasselbalch (Cont’d)

• Henderson-Hasselbalch equation

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 pKa of the weak acid

PH = PKa

• when pH < pKa, the weak acid predominates (Protonated)

• when pH > pKa, the conjugate base predominates

(deprotonated)

[Weak acid]

[Conjugate base]log=pH pKa +

The Henderson-Hasselbalch equation can be used to calculate the pH of the solution

• Assume 0.1 and 0.5 M (NaOH) base has been added to a fully protonated solution of acetic acid (1 mol), what is the pH?

• 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

• With 0.1 M OH¯ added:

• pH = pKa + log10 [0.1 ]

[0.9]

• pH = 4.76 + (-0.95) = 3.81

With 0.5 M OH¯ added:

pH = pKa + log10 [0.5 ] [0.5]

pH = 4.76 + 0

pH = 4.76 = pKa

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

Acetic Acid

Acetate

Pka of acetic acid is 4.76

Inflection point

%

%

Equivalence point

Titration Curves

• Inflection point: the point in an acid-base titration at which the concentration of the acid equals the concentration of conjugate base. pH = Pka

• 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.

• The form of the curve in figure represents the behavior of any monoprotic weak acid.

Polyprotic acids

Dr. M. khalifeh

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.

Buffers

• buffer: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• CH3COOH and CH3COONa

• H2CO3 and NaHCO3

• NaH2PO4 and Na2HPO4

If 1 ml of 0.1 M HCl is added to 99 ml water or buffer, what will happen?

If 1 ml of 0.1 M NaOH is added to 99 ml water or buffer, what will happen?

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.

We assume that 0.1 M HCl dissociate completely to give 0.1 M H3O because HCl is a strong acid.

Dilution: 0.1/100 = 0.001 M

We have 100 ml of 0.001 M HCl and 100 ml of 0.001 NaOH.

PH= -log [H3O] = 10-3, pH = 3.

Base added, [OH-]= 10-3. [H+][OH-]= 10-14, then [H+]= 10-11; therefore, pH = 11.

Calculate the PH when

At PH 7 and a Pka of 7.2 the ratio of HPO-24 t0 H2PO4

- is 0.63 to 1

pH = Pka + Log A-/HA

7 = 7.2 + log HPO-24 / H2PO4

-

-0.02 = log HPO-24 / H2PO4

-

= antilog -0.2 = 0.63

Calculate the pH when 1)1 ml of 0.1 M HCl is added to 99 ml buffer (H2PO4

-), and

2)1 ml of 0.1 M NaOH is added to 99 ml Buffer.

Calculation

Know that PKa= 7.2, [H2PO4-]= 0.1M, and [HPO4

2-]= 0.063M.

1) When HCl added. All H+ will be used up

HPO42- + H+ H2PO4

-

Before addition of acid 0.063 M 10 -7 M 0.1 M

Acid added- no reaction 0.063 M 10 -3 M(0.1/100) = 0.001) 0.1 M

After acid react with HPO42- 0.062 M (-0.001) ? 0.101 (+0.001)

PH = 7.2 + log 0.062/0.101 = 6.99.

Calculation

2) When OH added.

H2PO4- + OH- HPO4

2- + H2O

Before addition of base 0.1 M 10 -7 M 0.063 M

Base added- no reaction 0.1 M 10 -3 M 0.063 M

(-0.001) (+ 0.001)

After base react with HPO42- 0.099 M ? 0.064M

PH = 7.2 + log 0.064/0.099 = 7.01.

Calculation

Buffer Range

• A buffer is effective in a range of about +/- 1 pH unit of the pKa 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

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

Naturally Occurring Buffers

• H2PO4/HPO4- pair is the principal buffer in cells• Carbonic acid (H2CO3) is an important (but not the

only) buffer in blood• H2CO3 is 6.37 (the pH of human blood is 7.4 which is

near the end of the buffering range)• CO2 can dissolve in water (Blood)• The dissolved CO2 can form Carbonic acid• Carbonic acid in turn react to produce bicarbonate ion

• Respiration decrease acidity of blood• Hyperventilation removes a lot of CO2 and raises pH

of blood (alkalosis)

Naturally Occurring Buffers

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

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 is vital for survival.• 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.

Dr. M. khalifeh

Selecting a Buffer

• The following are typical criteria • suitable pKa

• no interference with the reaction or detection of the assay

• suitable ionic strength • suitable solubility

Laboratory Buffers

• Zwitterions: compounds that have both a positive and negative charge• Less likely to interfere with biochemical reactions• Appropriate for in vitro use