CHAPTER 2 Water, pH, and Ionic Equilibria to accompany Biochemistry, 2/e by Reginald Garrett and Charles Grisham All rights reserved. Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida 32887-6777

Outline 2.1 Properties of Water 2.2 pH 2.3 Buffers 2.4 Water's Unique Role in the Fitness of the Environment

Properties of Water High b.p., m.p., heat of vaporization, surface tension Bent structure makes it polar Non-tetrahedral bond angles H-bond donor and acceptor Potential to form four H-bonds per water

Comparison of Ice and Water Issues: H-bonds and Motion Ice: 4 H-bonds per water molecule Water: 2.3 H-bonds per water molecule Ice: H-bond lifetime - about 10 microsec Water: H-bond lifetime - about 10 psec (10 psec = 0.00000000001 sec) Thats "one times ten to the minus eleven second"!

Solvent Properties of Water Ions are always hydrated in water and carry around a "hydration shell" Water forms H-bonds with polar solutes Hydrophobic interactions - a "secret of life"

Hydrophobic Interactions A nonpolar solute "organizes" water The H-bond network of water reorganizes to accommodate the nonpolar solute This is an increase in "order" of water This is a decrease in ENTROPY

Amphiphilic Molecules Also called "amphipathic" Refers to molecules that contain both polar and nonpolar groups Equivalently - to molecules that are attracted to both polar and nonpolar environments Good examples - fatty acids

Acid-base Equilibria The pH Scale A convenient means of writing small concentrations: pH = -log10 [H+] Sørensen (Denmark) If [H+] = 1 x 10 -7 M Then pH = 7

Dissociation of Weak Electrolytes Consider a weak acid, HA The acid dissociation constant is given by: HA  H+ + A- Ka = [ H + ] [ A - ] ____________________ [HA]

The Henderson-Hasselbalch Equation Know this! You'll use it constantly. For any acid HA, the relationship between the pKa, the concentrations existing at equilibrium and the solution pH is given by: pH = pKa + log10 [A¯ ] ¯¯¯¯¯¯¯¯¯¯ [HA]

Consider the Dissociation of Acetic Acid Assume 0.1 eq base has been added to a fully protonated solution of acetic acid The Henderson-Hasselbalch equation can be used to calculate the pH of the solution: With 0.1 eq OH¯ added: pH = pKa + log10 [0.1 ] ¯¯¯¯¯¯¯¯¯¯ [0.9] pH = 4.76 + (-0.95) pH = 3.81

Consider the Dissociation of Acetic Acid Another case.... What happens if exactly 0.5 eq of base is added to a solution of the fully protonated acetic acid? With 0.5 eq OH¯ added: pH = pKa + log10 [0.5 ] ¯¯¯¯¯¯¯¯¯¯ [0.5] pH = 4.76 + 0 pH = 4.76 = pKa

Consider the Dissociation of Acetic Acid A final case to consider.... What is the pH if 0.9 eq of base is added to a solution of the fully protonated acid? With 0.9 eq OH¯ added: pH = pKa + log10 [0.9 ] ¯¯¯¯¯¯¯¯¯¯ [0.1] pH = 4.76 + 0.95 pH = 5.71

Buffers Buffers are solutions that resist changes in pH as acid and base are added Most buffers consist of a weak acid and its conjugate base Note in Figure 2.15 how the plot of pH versus base added is flat near the pKa Buffers can only be used reliably within a pH unit of their pKa