Buffers-NEW PDF

Weak Acids, Weak Bases and Buffers Prof. Dr. Özlem Dalmızrak Department of Medical Biochemistry Faculty of Medicine Near East University  Ionization/Dissociation of water H2O H+ + OH- H2O + H2O H3O+ + OH-...

Weak Acids, Weak Bases and Buffers Prof. Dr. Özlem Dalmızrak Department of Medical Biochemistry Faculty of Medicine Near East University  Ionization/Dissociation of water H2O H+ + OH- H2O + H2O H3O+ + OH- Hydronium ion Proton hopping results in extremely rapid net movement of a proton over a long distance in a remarkably short time.  H2O H+ + OH- [H+] [OH-] Keq= [H O] Keq=1.8 x 10-16 M 2 In pure water at 25oC, the concentration of water is 55.5 M 55.5 M x Keq = [H+] [OH-] (Kw) (Ion product) [H+] [OH-] = 55.5 M x (1.8 x 10-16 M) [H+] [OH-] = 1 x 10-14 M2 (Kw) (Ion product) [OH-] = [H+] = 1 x 10-7 M  pH indicates H+ and OH- concentrations pH= -log10 [H+] = log 1 + [H ] Log [H+] + log [OH-] = log 10-14 pH + pOH = 14 Indicator dyes Litmus Phenolphthalein Phenol red Accurate determination of pH pH meter  Weak acid and bases have characteristic acid dissociation constants Hydrochloric, sulfuric and nitric acids, commonly called strong acids, are completely ionized in dilute aqueous solutions. The strong bases NaOH and KOH are also completely ionized. Acids may be defined as proton donors and bases as proton acceptors. A proton donor and corresponding proton acceptor make up a conjugate acid-base pair. Acetic acid (CH3COOH), a proton donor and the acetate (CH3COO-) the corresponding proton acceptor. Strong Acids HCl H+ + Cl- 100 % H2SO4 2 H+ + SO42- 100% Strong Bases NaOH Na+ + OH- 100% KOH K+ + OH- 100%  Dissociation of Weak Acids CH3COOH CH3COO- + H+ (HA) (A-) Each acid has a characteristic tendency to lose its proton in an aqueous solution. The stronger the acid, the greater its tendency to lose its proton. [H+] [A-] [HA] Keq = Ka = [H+] = Ka - [HA] [A ] + [HA] -log [H ] = -log Ka - log [A-] [A-] Conjugated base pH = pKa + log [HA] Acid Henderson-Hasselbalch Equation  Calculating the pH of Weak Acid Solutions CH3COOH CH3COO- + H+ Ka = 1.74 x 10-5 [CH3COO-] [H+] Ka = 1.74 x 10-5 = [CH3COOH] [CH3COO-] [H+] = (1.74 x 10-5) [CH3COOH] X X 1-X X2 = 1.74 x 10-5 [1-CH3COOH ] X= 1.74 x 10-5 x 1 pH = - logX = 2.38  BUFFERS Almost every biological process is pH-dependent. A small change in pH produces a large change in the rate of the process. The protonated amino and carboxyl groups of amino acids and the phosphate groups of nucleotides function as weak acids, their ionic state is determined by the pH of the surrounding medium. Buffers are aqueous systems that tend to resist changes in pH when small amounts of acid or base are added. A buffer system consists of a weak acid (proton donor) and its conjugated base (the proton acceptor).  pH of Body Fluids Fluid pH Blood 7.4 Milk 6.6 – 6.9 Urine (normal) 6.0 Gastric juice 0.88 Pancreatic secretions 8.0 Intestinal secretions 7.7 Cerebrospinal fluid 7.4 Saliva 7.2 Tears 7.4 Henderson-Hasselbalch equation fits the calculations for buffer solutions. [A-]  Conjugated base pH = pKa + log [HA]  Acid 1. Calculation of pKa, given pH and molar ratio of proton donor and acceptor 2. Calculation of pH, given pKa and molar ratio of proton donor and acceptor 3. Calculation of molar ratio of proton donor and acceptor, given pH and pKa  The Mechanism of Buffer Action Because of reversible and partial dissociation, weak acids and their conjugated base forms can act as proton donors and proton acceptors: CH3COOH CH3COO- + H+ NaOH Na+ + OH- H2O CH3COOH CH3COO- + H+ HCl Cl- + H* By the same mechanism, weak bases can act as buffers.  Titration Curve CH3COOH CH3COO- + H+  Some weak acids or bases acid Conjugated base pK H-COOH HCOO- + H+ 3.75 CH3-COOH CH3COO- + H+ 4.76 CH3CHCOOH CH3CHCOO- + H+ 3.86 OH OH H3PO4 H++H2PO4- H++HPO4= H+ PO43- 2.34 6.86 12.4 H2CO3 H+ + HCO3- CO32- + H+ 3.8 10.2 C6H5OH C6 H 5 O + H + 9.89 + N H4 NH3 + H+ 9.25  Major Buffers Acid form pKa Cacodylic acid 6.2 BISTRIS 6.5 PIPES 6.8 Imidazole 7.0 HEPES 7.6 Tris 8.3  Physiological Buffer Systems 1. Hemoglobin Non-bicarbonate 2. Proteins buffers 3. Phosphate buffer system 4. Carbonic acid / Bicarbonate system Histidine pKa 6  Phosphate Buffer System H2PO4- H+ + HPO42- pKa = 6.86 HPO42- 7.4 = 6.8 + log H2PO4- HPO42- / H2PO4 = 4/1 * Major intracellular inorganic buffer. *H2PO4 excretion in urine is important for the regulation of blood pH.  Carbonic acid-Bicarbonate Buffer system H2CO3 H+ + HCO3- Ka = 2.7 x 10-4 CO2(d) + H2O H2CO3 Kh = 3 x 10-3 CA CO2(g) CO2(d) CA: Carbonic anhydrase  CO2(d) + H2O H2CO3 Kh = 3 x 10-3 [H2CO3] Kh= [H2CO3] = Kh x [CO2(d)] [CO2(d)] H2CO3 H+ + HCO3- Ka = 2.7 x 10-4 [HCO3-] [H+] [HCO3-] [H+] Ka= Ka= [H2CO3] Kh x [CO2(d)] [HCO3-] [H+] (3 x 10-3) (2.7 x 10-4) = Kcombined = 8.1 x 10-7 Ka x Kh = [CO2(d)] pKcombined = 6.1 [HCO3-] 7.4 = 6.1 + log [H2CO3] HCO3- / H2CO3 ratio is 20/1  QUESTION-1 What is the pH of a mixture of 0.042 M NaH2PO4 and 0.058 M Na2HPO4? (pKa= 6.86) pH = pKa + log ([conjugated base] / [acid]) pH = 6.86 + log (0.058 / 0.042) pH = 7.0  QUESTION-2 If 1 ml of 10 N NaOH is added to a liter of the buffer prepared in Q-1, how much will the pH change? M=n/V 0.042=n/1 liter 0.058=n/1 liter Initially, NaH2PO4 = 0.042 mol , Na2HPO4 = 0.058 mol 1 ml 10 N NaOH = 0.01 mol M=n/V 10=n/0.001 liter H2PO4  H+ + HPO4 NaOH Na+ + OH- 0.042 mol 0.058 mol initially 0.032 mol 0.068 mol after the addition of NaOH pH = pKa + log ([conjugated base] / [acid]) pH = 6.86 + log (0.068 / 0.032) pH = 7.2 Difference 0.2 pH unit  QUESTION-3 If 1 ml of 10 N NaOH is added to a liter of pure water at pH 7.0, What is the final pH? NaOH Na+ + OH- M1 x V1 = M2 x V2 10 x 0.001 = M2 x 1 M2= 0.01 M [Na] = [OH] = 0.01 M pOH = -log [OH] = -log [10-2] = 2 pH + pOH = 14 pH + 2 = 14 pH = 12