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Chapter 2 Aqueous Solutions PowerPoint Presentation
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Chapter 2 Aqueous Solutions

Chapter 2 Aqueous Solutions

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Chapter 2 Aqueous Solutions

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  1. Chapter 2 Aqueous Solutions View of Earth from Space Read the entire chapter Practice: pH & buffer problems Titration curves WATER Most abundance substance in living systems

  2. Water is an unusual liquid From Lehninger Principles of Biochemistry

  3. Properties of Water The structure of the water molecule • High b.p., m.p., heat of vaporization, surface tension • Bent structure makes it polar • H-bond donor and acceptor • Non-tetrahedral bond angles

  4. Structure of the water molecule More electronegative The electronegativity of oxygen polarizes its covalent bond to H. The lone pair electrons on oxygen make intermolecular hydrogen bonding (H-bonding) possible. Dipolar nature of the water molecule is shown here

  5. A hydrogen bond is most stable when the donor, the hydrogen, and the acceptor are in a straight line. D H………….A D H A ………….

  6. Liquid water forms “flickering clusters”, lasting a nanosecond, with 3.4 bonds per molecule on average Ice forms an ordered hydrogen bonded network with 4 bonds per molecule of water

  7. Weak interactions are crucial to macromolecular structure & function From Lehninger Principles of Biochemistry

  8. Individual hydrogen bonds are relatively weak. • Nevertheless, the interaction of biomolecules with water through hydrogen bonds plays a major role in biomolecular structure & function.

  9. NaCl crystal lattice is disrupted as water molecules cluster around sodium and chloride ions NaCl crystal Hydrated ions DS DG = DH – T DS From Lehninger Principles of Biochemistry

  10. Solvation of ions by oriented water molecules

  11. Compounds that dissolve in water (hydrophilic) The polar nature of the covalent O-H bonds in water make it an excellent solvent for other polar molecules Compounds that dissolve in solvents such as chloroform (hydrophobic) From Lehninger Principles of Biochemistry

  12. 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 Amphiphilic Molecules

  13. Common hydrogen bonds in biological systems Usually oxygen or nitrogen From Lehninger Principles of Biochemistry

  14. Some biologically important hydrogen bonds

  15. The association of amphipathic molecules in aqueous solutions

  16. Amphipathic compounds in aqueous solution Polar head group Hydrophillic Non-polar Hydrophobic tail From Lehninger Principles of Biochemistry

  17. Hydrophobic Interactions • A nonpolar solute "organizes" water • The H-bond network of water reorganizes to accommodate the nonpolar solute • There is an decrease in "order" of water • This is an increase in ENTROPY From Lehninger Principles of Biochemistry

  18. Ionization of water is expressed by an equilibrium constant H2O H+ + OH- Equilibrium constant [H+] [OH-] Keq = G of water in 1 L [H2O] = [H2O] Molecular weight of water [H2O] Keq [H+] [OH-] = At 25 oC [H2O] = 55.5 M = 1.8 x 10-16 M Keq Kw [H+] [OH-] Kw = Ionic product of water at 25 oC

  19. [H+] [OH-] Kw = At neutral pH [H+] = [OH- ] Kw [H+] [H+] = [H+]2 = Kw 1 x 10-14 M2 [H+] = = [H+] = [OH-] = 10-7 M - log [H+] = - log 10-7 M Neutral solution pH is 7 Not an arbitrary number Derived from ionic product of water pH = 7

  20. pH of some aqueous fluids From Lehninger Principles of Biochemistry

  21. Why is pH important? pH optima of some enzymes

  22. Weak acids and bases have characteristic dissociation constants CH3COOH CH3COO- + H+ A- + H+ HA [A-] [H+] = Ka Keq= [HA] Dissociation constant 1 pKa = - log Ka = log Ka Stronger the acid Higher the Ka Lower the pKa

  23. Henderson-Hasselbalch Equation At midpoint of the titration OR when [A-] = [HA] Then pH = pKa

  24. The Henderson-Hasselbach equation is used to • Calculate pKa, given pH and molar ratio • of proton donor and acceptor • Calculate pH, given pKa and molar ratio • of proton donor and acceptor • Calculate the molar ratio of proton donor • and acceptor, given pH and pKa [proton acceptor] pH= pKa+ log [proton donor] When [proton acceptor] = [proton donor], pH= pKa

  25. 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

  26. Conjugate acid-base pairs consist of a proton donor and a proton acceptor From Lehninger Principles of Biochemistry

  27. The acetic acid – acetate pair as a buffer system This system is capable of absorbing either H+ or OH- Buffering occurs due to the simultaneous balancing of water & buffer dissociation reactions as governed by the constants Kw and Ka. From Lehninger Principles of Biochemistry

  28. Titration curve of acetic acid 2 reversible equilibria are involved: H2O H+ + OH- HOAc H+ + OAc- Consider the titration of 0.1 M HOAc with 0.1 M NaOH Weak Acid Buffers work from 1 pH unit below to 1 pH unit above their pKa Strong Acid From Lehninger Principles of Biochemistry

  29. Comparison of the titration curves of 3 weak acids

  30. Titration curve of H3PO4

  31. 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 titration figure 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

  32. Questions to think about…. • Would phosphoric acid or succinic acid be a better buffer at pH 5? • Would ammonia or piperidine be a better buffer at pH 9? • 3) Would HEPES or Tris be a better buffer at pH 7.5? • 4) Indicate the ionic species that predominates at pH 4, 8 and 11 for ammonia