Download Epjj Lecture 2

Water & pH
Lecture 2
Ahmad Razali Ishak
Department of Environmental Health
Faculty of Health Sciences
UiTM Puncak Alam
Role of water in the life of
organisms
• Mammalian cells 70% water
• Solvent for biological systems & for most
chemical reactions that support life.
• 75% of the earth is covered with water
• Has a very high specific heat-retains heat
better than other materials
Properties of water
1.
2.
3.
4.
5.
Polarity
Hydrogen bond
Universal solvent
Hydrophobic interactions
Other non covalent interactions in
biomolecules
6. Nucleophilic nature of water
7. Ionization of water
1) Polarity
• Water = H2O
• 2 H atoms are linked covalently to
oxygen, each sharing an electron pair.
• Non linear arrangement bond angle104.5
• Oxygen atom more electronegative than
H atom-polar covalent bond
• Creates a permanent dipole in H2O
molecule
2) Hydrogen bonds
• Water molecules attract to each other due to polarity
• H bonds: attraction of one slightly +ve H atom of one
water molecule and one slightly –ve oxygen atom of
another water molecule
• Water molecule can form H- bond with 4 other water
molecules
• H bonds weaker than covalent bonds.
• H bonds gives water a HIGH melting point, specific heat
and heat of vaporization
• The ability to form strong H bond is responsible for the
many unique characteristics of water such as its high
melting point and boilng point
• 3D structures of many important biomolecules including
proteins (Hb) and nucleic acids (DNA) are stabilized by H
bonds
3) Universal Solvent
• Water interact with and dissolve other polar and ionize
(electrolytes) compounds
• Water aligning around electrolytes to form solvation spheres
• Solubility depend on polarity and ability to form H-bonds
• Functional groups on molecules that confer solubility:
carboxylates, protonated amines, amino, hydroxyl and carbonyl
• As the number of polar groups increase in a molecule, so does its
solubility in water.
•
•
•
•
Flavoring and CO2 gas dissolved in water to make soft drinks
Farmers use water to dissolve fertilizers
Medicines in water
Chlorines or fluorides added to water
4)Hydrophobic interaction
• Non polar molecules NOT soluble in water,
hydrophobic
• Amphiphatic molecule: Have both hydrophobic and
hydrophilic portions
• E.g. Detergents, and surfactants
• Form micelles in aqueous solution
• Used to trap grease and oils inside to remove them
• Hydrophilic compounds interact (disslove) with
water eg . Polar cpds (alcohols and ketones)& ionic
cpds (KCl), amino acids
• Hydrophobic compounds do not interact with water
eg. Non polar cpds (hexane, fatty acids,
cholesterol)
5) Other non covalent
interactions in biomolecules
• Four major non covalent forces involved
in structure and function of
biomolecules:
• H-bond
• Hydrophobic interactions
• Ionic bonds
• Van der Waals forces
6) Nucleophilic nature of water
• Nucleophilic: electron rich
• Electrophiles: electron deficient
• Nucleophiles are negatively charged and
have unshared electrons pairs: attack
electrophiles during substitution or
addition reactions.
7)Self-Ionization of Water
• H2O is amphoteric. i.e. as an acid it
gives up H+ to become OH¯, and as a
base it accepts H+ and becomes
H3O+.
• When water reacts with itself
• H2O + H2O
H3O+(aq) + OH¯(aq)
• This reaction of water is called
self-ionization of water.
Kw
• Keq = [H3O+][OH¯]
[H2O]2
• Concentration of pure water is constant
• Keq [H2O]2 = [H3O+][OH¯]
• Kw = [H3O+][OH¯]
• Kw = ion-product constant of water
• At 25 ˚C, both H3O+ and OH¯ ions are
found at concentration of 1.0 x 10-7 M
• Kw = [1.0 x 10-7][1.0 x 10-7]
• Kw = 1.0 x 10-14
• If [H3O+] > 1.0 x 10-7, solution is
acidic
• If [OH¯] > 1.0 x 10-7, solution is
basic
• If [H3O+] = [OH¯] = 1.0 x 10-7,
solution is neutral.
p. 48
Cont..
•
•
•
•
At equilibrium, pH = -log [H+]
- log [1.0 x 10-7 ]
= 7 (neutral)
Low pH represent highest [H+] and
lowest [OH]
Acids, Bases, and Buffers
• Acids are proton donors and base are proton
acceptors (Bronsted-lowry)
• The strength of an acid is measured by its
acid dissociation constant, K
• The larger the K value, the stronger the acid
and more H+ dissociates
• The conc. H+ is expressed as pH, -ve log of H
ion conc.
• This tendency to ionize can be put in
terms of an equation for the
equilibrium:
• Where [ ] = Molar concentration;
• K = Ionization constant (acid
dissociation constant)
• Simplest example is water (H2O):
p. 47
p. 47
Table 2-6, p. 50
Buffers
1. Principle of Ionization of Weak Acids:
• The Fundamental Concept of Buffers is: A Buffer
Resists Change
• pH buffers resist change in pH when either acid (H+)
or base (OH-) is added to it.
• Chemicals which are pH buffers are weak acids or
bases
• Acids = Proton (H+) donors
• Bases = Proton Acceptors
Titration of a Weak Acid illustrating its
Ionization and Buffering Property
Fig. 2-13b, p. 51
• All weak acids have titration curves like this
one. Bases (like ammonium, NH4+) are also
weak acids and have similar titration curves.
• The position where the Buffering zone is on
the pH scale is related to the chemical nature
of the weak acid:
• Acetic acid ionizes in the Acidic portion of
the pH scale
Fig. 2-16a, p. 58
Fig. 2-13b, p. 51
• This relationship is known as the HendersonHasselbalch equation.
• Useful in predicting the properties of buffer
solutions used to control the pH of reaction mixtures.
• The pK of a weak acid is the pH where [A-] = [HA]
• At pH below the pK, [HA] > [A-]
• At pH above the pK, [HA] < [A-]
• Therefore the pK determines the buffering zone for
a weak acid.
• A similar expression pK can be used,
pK=-log K
• The pH of a solution of a weak acid and
its conjugate base is related to the
concentration of the acid and baseHenderson- Hasselbach equation.
Henderson Hasselbach Equation
p. 49
Example 1
• Calculate the pH of a buffer solution
made from 0.20 M HC2H3O2 and 0.50 M
C2H3O2- that has an acid dissociation
constant for HC2H3O2 of 1.8 x 10-5.
Answer
• pH = pKa + log ([A-]/[HA])
• pH = pKa + log ([C2H3O2-] / [HC2H3O2])
• pH = -log (1.8 x 10-5) + log (0.50 M /
0.20 M)
• pH = -log (1.8 x 10-5) + log (2.5)
• pH = 4.7 + 0.40
• pH = 5.1
Example 2
•
Calculate the relative amounts of acetic acid
and acetate ion present at the following
points when 1 mol of acetic acid is titrated
with NaOH. Use HH eqn. to calculate pH
1. 0.1 mol NaOH added
2. 0.3 mol NaOH added
3. 0.5 mol NaOH added
Answer
• Ratio 1:1, when 0. 1 mol of NaOH added,
0.1 mol acetic acid reacts with it to
form 0.1 mol acetate ion, leaving 0.9 mol
acetic acid
pH = pK + log 0.1/0.9
= 4.76 -0.95
= 3.81
Buffer applications
• In humans-in the intracellular and extra
cellular fluid, there is a conjugate acidbase pairs that act as buffer
• Major intracellular buffer is conjugate
base-acid pair H2P04- / HPO42- (pK = 6.8)
• Extracellular fluid- main buffer in blood
and interstitial fluid is the bicarbonate
H2CO3/HCO3- (pK = 6.2). Normal pH for
blood is 7.4
Summary 2-3, p. 51
Summary 2-4, p. 53
Summary 2-5, p. 61