Bài giảng Biochemistry 2/e - Chapter 2: Water, pH, and Ionic Equilibria to accompany Biochemistry, 2/e

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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CHAPTER 2Water, pH, and Ionic Equilibriato accompanyBiochemistry, 2/ebyReginald Garrett and Charles GrishamAll 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 Outline2.1 Properties of Water2.2 pH2.3 Buffers2.4 Water's Unique Role in the Fitness of the EnvironmentProperties of WaterHigh b.p., m.p., heat of vaporization, surface tensionBent structure makes it polarNon-tetrahedral bond anglesH-bond donor and acceptorPotential to form four H-bonds per waterComparison of Ice and WaterIssues: H-bonds and MotionIce: 4 H-bonds per water moleculeWater: 2.3 H-bonds per water moleculeIce: H-bond lifetime - about 10 microsecWater: H-bond lifetime - about 10 psec(10 psec = 0.00000000001 sec)Thats "one times ten to the minus eleven second"!Solvent Properties of WaterIons are always hydrated in water and carry around a "hydration shell"Water forms H-bonds with polar solutesHydrophobic interactions - a "secret of life"Hydrophobic InteractionsA nonpolar solute "organizes" waterThe H-bond network of water reorganizes to accommodate the nonpolar soluteThis is an increase in "order" of waterThis is a decrease in ENTROPYAmphiphilic MoleculesAlso called "amphipathic"Refers to molecules that contain both polar and nonpolar groupsEquivalently - to molecules that are attracted to both polar and nonpolar environmentsGood examples - fatty acidsAcid-base EquilibriaThe pH ScaleA convenient means of writing small concentrations:pH = -log10 [H+]Sørensen (Denmark)If [H+] = 1 x 10 -7 MThen pH = 7Dissociation of Weak ElectrolytesConsider a weak acid, HAThe acid dissociation constant is given by:HA  H+ + A-Ka = [ H + ] [ A - ] ____________________ [HA] The Henderson-Hasselbalch EquationKnow 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 AcidAssume 0.1 eq base has been added to a fully protonated solution of acetic acidThe 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 = 3.81Consider the Dissociation of Acetic AcidAnother 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 + 0pH = 4.76 = pKaConsider the Dissociation of Acetic AcidA 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 + 0.95pH = 5.71BuffersBuffers are solutions that resist changes in pH as acid and base are addedMost buffers consist of a weak acid and its conjugate baseNote in Figure 2.15 how the plot of pH versus base added is flat near the pKaBuffers can only be used reliably within a pH unit of their pKa
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