Water & pH Lecture 2 Ahmad Razali Ishak Department of Environmental Health Faculty of Health...

Water & pHLecture 2

Ahmad Razali IshakDepartment of Environmental HealthFaculty of Health SciencesUiTM 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. Polarity2. Hydrogen bond3. Universal solvent4. Hydrophobic interactions5. Other non covalent interactions in

biomolecules6. Nucleophilic nature of water7. Ionization of water

1) Polarity

• Water = H2O• 2 H atoms are linked covalently to

oxygen, each sharing an electron pair.• Non linear arrangement bond angle-

104.5• 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

Buffers1. 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 Henderson-Hasselbalch 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 base- Henderson- 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 added2. 0.3 mol NaOH added3. 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 acid-base 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