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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, Sea Harbor Drive, Orlando, Florida
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Outline 2.1 Properties of Water 2.2 pH 2.3 Buffers
2.4 Water's Unique Role in the Fitness of the Environment
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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
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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 = sec) Thats "one times ten to the minus eleven second"!
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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"
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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
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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
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Acid-base Equilibria The pH Scale
A convenient means of writing small concentrations: pH = -log10 [H+] Sørensen (Denmark) If [H+] = 1 x M Then pH = 7
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Dissociation of Weak Electrolytes
Consider a weak acid, HA The acid dissociation constant is given by: HA H+ + A- Ka = [ H + ] [ A - ] ____________________ [HA]
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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]
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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 = (-0.95) pH = 3.81
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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 = pH = 4.76 = pKa
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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 = pH = 5.71
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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
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