Acids and Bases - Biochemistry - Lecture Slides, Slides of Biochemistry

Download Acids and Bases - Biochemistry - Lecture Slides and more Slides Biochemistry in PDF only on Docsity! Acids and Bases January 27 2003 Docsity.com Ionization of water • Although neutral water has a tendency to ionize H2O <-> H + + OH- • The free proton is associated with a water molecule to form the hydronium ion H3O + • High ionic mobility due to proton jumping Docsity.com Kw Kw = [H +][OH-] • Where Kw is the ionization constant of water • For pure water ionization constant is 10-14 M2 at 25º • For pure water [H+] = [OH-] = (Kw) 1/2 = 10-7 M Docsity.com Acids and bases • For pure water (neutral) [H+] = [OH-] = (Kw) 1/2 = 10-7 M • Acidic if [H+] > 10-7 M • Basic if [H+] < 10-7 M Docsity.com Acids and Bases Lowery definition: • Acid is a substance that can donate a proton. • Base is a substance that can accept a proton. HA + H2O H3O + + A- /OH- Acid Base Conjugate Conjugate Acid Base or HA A- + H+ Acid Conjugate Conjugate Base Acid Docsity.com What to do about the water! The concentration of H2O remains almost unchanged especially in dilute acid solutions. What is the concentration of H2O? Remember the definition: Moles per liter 1 mole of H2O = 18 g = 18 ml 1000 g/liter mlg 11  M molg g 5.55 /18 1000  Docsity.com ][ ]][[ ][ 2 HA AH OHKKa   From now on we will drop the a, in Ka Weak acids (K<1) Strong acids (K>1) Strong acid completely dissociates: Transfers all its protons to H2O to form H3O + HA H+ + A- Docsity.com Weak Acids Weak acids do not completely dissociate: They form an equilibrium:   HAHA If we ADD more H+, the equilibrium shifts to form more HA using up A- that is present. Docsity.com [H+] pH 10-7 = 7 10-3 = 3 10-2 = 2 10-10 = 10 5x10-4 = 3.3 7x10-6 = 5.15 3.3x10-8 = 7.48 pH = -Log[H+] It is easier to think in log of concentrations but it takes practice!! Docsity.com Relationship between pH and [H*] / [OH] concentration | Acidic peule Basic * {, A 14 qo8 lon concentration (MM) Lo-t2 —— Docsity.com Observation If you add .01 ml i.e 1/100 ml of 1M HCl to 1000 ml of water, the pH of the water drops from 7 to 5!! i.e 100 fold increase in H+ concentration: Log = 2 change. Problem: Biological properties change with small changes in pH, usually less than 1 pH unit. How does a system prevent fluctuations in pH? Docsity.com Above and below this range there is insufficient amount of conjugate acid or base to combine with the base or acid to prevent the change in pH. [HA] ][A log pK pH -  1 10 10 1 from variesratio ]HA[ ][A   Docsity.com For weak acids HA A- + H+ This equilibrium depends on concentrations of each component . If [HA] = [A-] or 1/2 dissociated Then 01log ][ ][ log   HA A By definition the pK is the pH where [HA] = [A-] : 50% dissociated : pH = pK Docsity.com The buffer effect can be seen in a titration curve. To a weak acid salt, CH3C00 -, add HC1 while monitoring pH vs. the number of equivalents of acid added. or do the opposite with base. Buffer capacity: the molar amount of acid which the buffer can handle without significant changes in pH. i.e 1 liter of a .01 M buffer can not buffer 1 liter of a 1 M solution of HCl but 1 liter of a 1 M buffer can buffer 1 liter of a .01 M solution of HCl Docsity.com Titration curve for phosphate First Second - Third Starting equivalence equivalence equivalence point point point point Midpoint three | [HPO4 ] =[PO4 ] 4b 12 |- 10)\— Midpoint two | [H2PO; ] = [HPOF] sf | | 6E | = Midpoint one | [H3PO4]=[H2P04] 5 | | | 0.5 1.0 1.5 2.0 2.5 3:0 Equivalents OH . Docsity.com Table 2-3 Dissociation constants and pK’s of Acids & buffers Acid K pK Oxalic 5.37x10-2 1.27 H3PO4 7.08x10 -3 2.15 Succinic Acid 6.17x10-5 4.21 (pK1) Succinate 2.29x10-6 5.65 (pK2) H2PO4 - 1.51x10-7 6.82 NH4 + 5.62x10-10 9.25 Glycine 1.66x10-10 9.78 Docsity.com Table 2-5. Dissociation Constants and pK Values at 25°C of Some Acids Acid K pK Oxalic acid 5.37 x 107? 1.27 (pK) HPO, 7.08 x 107+ 2.15 (pK) Formic acid 1.78 x 10°* 3.75 Succinic acid 6.17 X 107? 4.21 (pK) Oxalate 5.37 x 107° 4.27 (pK) Acetic acid 1.74 x 107° 4.76 Succinate” 2.29 x 10°° 5.64 (pK) 2-(N-Morpholinoethanesulfonic acid (MES) 8.13 x 1077 6.09 HCO; 4.47 x 1077 6.35 (pK,)" Piperazine-N.N’-bis(2-ethanesulfonic acid) (PIPES) 1.74 x 1077 6.76 HPO; L5lx 107? 6.82 (pK) 3-(N-Morpholino)propanesullonic acid (MOPS) 7.08 x 10°" 7.15 N-2-Hydroxyethylpiperazine-N’-2-ethanesulfonic 3.39 x 10° TAF acid (HEPES) Tris(hydroxymethy|aminomethane (Tris) 8.32 x 10°? 8.08 NH} 562X107 = 9.25 Glycine 166x107" 9.78 Hco; 4.68 x 107'' 10.33 (pK») Piperidine RIM LD HPO}” 417X107" — 12.38 (pK) Smee: Dawson, R-M.C., Elliett, D.C., Eliott, WH, and Jones, K.M., Date for Biochemical Research (3rd ed.), pp. 424-425, Oxford Science Publications (1986) and Good, N.E., Winget, G.D., Winter, W., Connolly, T.N., Izawa, S,, and Singh, R-M.M., Biochemistry 5, 467 (1966) "The pK for the overall reaction CO, + H,0 === H,CO, == H” + HCO,; see Box 2-2. Docsity.com What is the pH of a solution of that contains 0.1M CH3C00 - and 0.9 M CH3C00H? 1) pH = pK + Log [A-] [HA] 2) CH3C00H CH3C00 - + H+ 3) Find pH 4) pK = 4.76 A- = 0.1 M HA = 0.9 M 5) Already at equilibrium 6) X = 4.76 + Log 0.1 0.9 Log 0.111 = -.95 X = 4.76 + (-.95) X = 3.81 Docsity.com What would the concentration of CH3C00 - be at pH 5.3 if 0.1M CH3C00H was adjusted to that pH. 1) pH = pK + Log [A-] [HA] 2) CH3C00H CH3C00 - + H+ 3) Find equilibrium value of [A-] i.e [CH3C00 -] 4) pH = 5.3; pK = 4.76 5) Let X = amount of CH3C00H dissociated at equilibrium [A-] = [X] [HA] = [0.1 - X] 6) 5.3 = 4.76 + Log [X] [0.1 - X] Now solve. Docsity.com Blood Buffering System • Bicarbonate most significant buffer • Formed from gaseous CO2 CO2 + H2O <-> H2CO3 H2CO3 <-> H + + HCO3 - • Normal value blood pH 7.4 • Deviations from normal pH value lead to acidosis Docsity.com