Water & pH Lecture 2

1. Water & pHLecture 2 Ahmad Razali Ishak Department of Environmental Health Faculty of Health Sciences UiTM Puncak Alam

2. 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

3. Properties of water • Polarity • Hydrogen bond • Universal solvent • Hydrophobic interactions • Other non covalent interactions in biomolecules • Nucleophilic nature of water • Ionization of water

4. 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

5. 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

6. 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

7. 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)

8. 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

9. 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.

10. 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.

11. 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

12. 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.

13. p. 48

14. Cont.. • At equilibrium, pH = -log [H+] • - log [1.0 x 10-7 ] • = 7 (neutral) • Low pH represent highest [H+] and lowest [OH]

15. 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.

16. This tendency to ionize can be put in terms of an equation for the equilibrium:

17. Where [ ] = Molar concentration; • K = Ionization constant (acid dissociation constant) • Simplest example is water (H2O):

18. p. 47

19. p. 47

20. Table 2-6, p. 50

21. 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

22. Titration of a Weak Acid illustrating its Ionization and Buffering Property

23. Fig. 2-13b, p. 51

24. 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

25. Fig. 2-16a, p. 58

26. Fig. 2-13b, p. 51

27. 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.

28. 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.

29. Henderson Hasselbach Equation p. 49

30. 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.

31. 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

32. 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 • 0.1 mol NaOH added • 0.3 mol NaOH added • 0.5 mol NaOH added

33. 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

34. 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

35. Summary 2-3, p. 51

36. Summary 2-4, p. 53

37. Summary 2-5, p. 61