Water Properties: appunti di biochimica

Chapter 2: Water Properties of water Very polar Oxygen is highly electronegative H-bond donor and acceptor High b.p., m.p., heat of vaporization, surface tension Water dissolves polar compounds solvation shell Non-polar substances are insoluble in water Many lipids are amphipathic Hydrogen Bonding of Water One H2O molecule can associate with 4 other H20 molecules Ice: 4 H-bonds per water molecule Water: 2.3 H-bonds per water molecule Crystal lattice of ice Relative Bond Strengths Bond type H3C-CH3 H-H Ionic H-bond Hydrophobic interaction van der Waals kJ/mole 88 104 40 to 200 2 - 20 3 -10 0.4 - 4 Biological Hydrogen Bonds + O H - O C H-Bond distance = 0.2 nm N + H - O C DNA strands held together by H-bonds CG AT Ionization of Water H20 + H20 H20 Keq= [H+] [OH-] [H2O] [H2O] Keq = [H+] [OH-] (1.8 X 10-16M55.5 M ) = [H+] [OH-] 1.0 X 10-14 M2 = [H+] [OH-] = Kw If [HOH-] then [H+] = 1.0 X 10-7 H3O+ + OHH+ + OHKeq=1.8 X 10-16M [H2O] = 55.5 M pH Scale Devised by Sorenson (1902) [H+] can range from 1M and 1 X 10-14M using a log scale simplifies notation pH = -log [H+] Neutral pH = 7.0 Weak Acids and Bases Equilibria Strong acids / bases disassociate completely Weak acids / bases disassociate only partially Enzyme activity sensitive to pH weak acid/bases play important role in protein structure/function Acid/conjugate base pairs HA + H2O HA A- + H3O+ A- + H+ HA = acid ( donates HBronstad Acid) A- = Conjugate base (accepts HBronstad Base) Ka = [HA-] [HA] Ka & pKa value describe tendency to loose H+ large Ka = stronger acid small Ka = weaker acid pKa = - log Ka pKa values determined by titration Phosphate has three ionizable H+ and three pKas Buffers Buffers are aqueous systems that resist changes in pH when small amounts of a strong acid or base are added. A buffered system consist of a weak acid and its conjugate base. The most effective buffering occurs at the region of minimum slope on a titration curve (i.e. around the pKa). Buffers are effective at pHs that are within 1 pH unit of the pKa Henderson-Hasselbach Equation 1) Ka = [HA-] [HA] HA = weak acid A- = Conjugate base 2) [H+] = Ka [HA] [A-] 3) -log[H+] = -log Ka -log [HA] [A-] 4) -log[H+] = -log Ka +log [A-] [HA] 5) pH = pKa +log [A-] [HA] * H-H equation describes the relationship between pH, pKa and buffer concentration Case where 10% acetate ion 90% acetic acid pH = pKa + log10 [0.1 ] pH = 4.76 + 0.95) pH = 3.81 [0.9] Case where 50% acetate ion 50% acetic acid pH = pKa + log10 [0.5 ] pH = 4.76 + 0 pH = 4.76 = pKa [0.5] Case where 90% acetate ion 10% acetic acid pH = pKa + log10 [0.9 ] pH = 4.76 + 0.95 pH = 5.71 [0.1] Cases when buffering fails pH = pKa + log10 [0.99 ] [0.01] pH = 4.76 + 2.00 pH = 6.76 pH = pKa + log10 [0.01 ] [0.99] pH = 4.76 - 2.00 pH = 2.76 Continua »