Version 2012 Updated on 0423 Copyright All rights reserved Dong-Sun Lee, Prof., Ph.D. Chemistry, Seoul Womens University Chapter 13 Titration Curves for Complex Acid/Base Systems A scanning electromicrograph of red blood cells traveling through an artery. The pH of human blood is controlled to be within the range of 7.35 to 7.45 primarily by the carbonic acid / bicarbonate buffer system. CO 2 (g) + H 2 O(l) H 2 CO 3 (aq) H 2 CO 3 + H 2 O H 3 O + + HCO 3 The human body contains a complicated system of buffers within cells and within bodily fluids, such as human blood. Mixtures of Strong and Weak Acids or Strong and Weak Bases The dissociation constant for the weak acid or base is somewhat less than about 1.00 10 4. Ex. pH of M HCl and M weak acid HA (K a = 1.00 10 4 ) [H 3 O + ] = C HCl + [A ] = [A ] If [A ] K a3 K a1 K a2 = [HPO 4 2 ][H 3 O + ] / [H 3 PO 4 ] = 7.11 10 3 6.3210 8 = 4.4910 10 K a1 K a2 K a3 = [PO 4 3 ][H 3 O + ] / [H 3 PO 4 ] = 7.11 10 3 6.3210 8 4.5 10 13 = 2.010 22 The carbon dioxide carbonic acid system CO 2 (aq) + H 2 O H 2 CO 3 K hyd = [H 2 CO 3 ] / [CO 2 (aq)] = 2.8 10 3 H 2 CO 3 + H 2 O HCO 3 + H 3 O + K 1 = [HCO 3 ] [H 3 O + ] / [H 2 CO 3 ] = 1.510 4 HCO 3 + H 2 O CO 3 2 + H 3 O + K 2 = [CO 3 2 ][H 3 O + ] / [HCO 3 ] = 4.69 10 11 CO 2 (aq) + 2H 2 O HCO 3 + H 3 O + K a1 = [HCO 3 ] [H 3 O + ] / [CO 2 (aq)] = 2.8 10 3 1.510 4 = 4.2 10 7 HCO 3 + H 2 O CO 3 2 + H 3 O + K a2 = [CO 3 2 ][H 3 O + ] / [HCO 3 ] = 4.69 10 11 Buffer Solutions Involving Polyprotic Acids Ex. [H 3 O + ] of mixture of 2.00 M phosphoric acid M potassium dihydrogen phosphate H 3 PO 4 + H 2 O H 2 PO 4 + H 3 O + K a1 = [H 2 PO 4 ] [H 3 O + ] / [H 3 PO 4 ] = 7.11 10 3 [H 3 O + ] = K a1 [H 3 PO 4 ] / [H 2 PO 4 ] = 7.11 10 3 (2.00 / 1.50) = 9.48 10 3 M Henderson-Hasselbalch Equation pH = pK a1 + log pH = pK a2 + log [HA - ] [H 2 A] [A 2- ] [HA - ] L 3- + H 2 O HL 2- + OH - K b1 = K w / K a3 HL 2- + H 2 O H 2 L - + OH - K b2 = K w / K a2 H 2 L - + H 2 O H 3 L + OH - K b3 = K w / K a1 Remember ! K a1 K b3 = Kw K a2 K b2 = Kw K a3 K b1 = Kw Diprotic Buffers Polyprotic Acids and Bases H 3 L H 2 L - + H + K a1 = K 1 H 2 L - HL H + K a2 = K 2 HL 2 - L H + K a3 = K 3 Diprotic Acids and Bases L + H 2 O HL + OH K a1 = K w / K 2 HL + H 2 O H 2 L + + OH K b2 = K w / K 1 Remember ! K a1 K b2 = K w K a2 K b1 = K w Diprotic acid Diprotic base Zwitterion H 2 L + HL + H + K a1 = K 1 = 4.6910 3 HL L + H + K a2 = K 2 = 1.7910 10 H 2 L + Leucine HL L NH 2 CHCOOH R N + CHCOO R H3H3 + H + x = 1.31 M [HL] = [H + ]= 1.3110 -2 M K a1 = = = 4.69 [HL][H + ] [H 2 L + ] x x To calculate the pH and composition of individual solution of M H 2 L +, HL and L . The Acid Form, H 2 L H 2 L + HL x x x The concentration of L ? [ L - ][H + ] [HL] K a2 = K 2 = = 1.79 [ L - ] = = 1.79 M = K [HL] [H + ] pH=1.88 [H 2 L + ] = x = 3.6910 -2 M K b2 = K w / K 1 = 2.13 = [ H 2 L + ] = = 2.13 = K b2 ([HL]=[OH - ]) [H 2 L + ][OH - ] [HL] 2.13 [HL] [OH-] The concentration of H 2 L + ? L - + H 2 O HL + OH - K b1 = Kw / K 2 = 5.5910 -5 HL + H 2 O H 2 L + + OH - K b2 = Kw / K 1 = 2.13 K b1 = = = 5.5910 -5 [HL][OH ] [L ] x x 2 The Basic Form, L - L - + H 2 O HL + OH x x x x = 1.6410 -3 M [HL] = 1.6410 -3 M [H + ] = = 6.08 Kw xKw x [L - ] = x = 4.8410 -2 pH=11.22 HL L - + H+ K a2 K 2 = 1.79 HL + H 2 O H 2 L + + OH - K b2 Kw / K 1 = 2.13 [H + ] + [H 2 L + ] = [L ] +[OH ] [H 2 L + ][L ] +[H + ] [OH ] = 0 [HL][H + ] 2 K 1 [HL]K 2 + [H + ] 2 K W = 0 + [H + ] = 0 [HL]K 2 [H + ] K W [H + ] [HL][H + ] K 1 [H + ] 2 ( + 1 ) = K 2 [HL] + K w [HL] K 1 [H + ] 2 = K 2 [HL] + Kw ( + 1 ) [HL] K 1 [H + ] = K 1 K 2 [HL] + K 1 Kw ( K 1 + [HL] ) [H + ] K 1 K 2 F + K 1 Kw ( K 1 + F ) ([HL] = F) The Intermediate Form, HL pH = ( pK 1 + pK 2 ) 1212 [ H 2 L + ] = = 9.34 x M [HL][H + ] K 1 [HL]K 2 [H + ] [L - ] = = 1.02 x M K 2 F > K w K 1 K 2 F + K 1 K w K 1 K 2 F F > K 1 K 1 +F + F [H + ] = (4.69 )(1.79 )( M) + (4.69 )(1.0 ) 4.69 M = 8.76 M pH=6.06 The pH of M Leucine? [H + ] = = = K 1 K 2 K 1 K 2 F + K 1 Kw ( K 1 + F ) K 1 K 2 F F Simplified Calculation for the Intermedate Form Isoelectric and isoionic pH isoelectric pH : The pH value of the dispersion medium of a colloidal suspension at which the colloidal particles do not move in an electric field. At this point, the average charge of the polyprotic acid is zero. Isoionic point : The pH obtained when the pure, neutral polyprotic acid HA(the neutral zwitter ion) is dissolved in water. Ex M Alanine + H 3 NCH(CH) 3 COOH H 3 NCH(CH) 3 COO H 2 NCH(CH) 3 COO Isoionic pH : [H + ] = (K a1 K a2 F + K a1 K w ) / (K a1 +F) pH = 6.12 Isoelectric pH : [H + ] = K a1 K a2 pH = 6.11 The isoelectric pH is not concentration dependent. Acid-base titration of the enzyme ribonuclease. The isoionic point is the pH of the pure protein with no ions present except H + and OH . The isoelectric point is the pH at which the average charge on the protein is 0. Isoelectric focusing of a mixture of proteins: (1) soybean trypsin inhibitor, (2) -lactoglobulin A, (3) -lactoglobulin B, (4) ovotransferrin, (5) horse myoglobin, (6) whale myoglobin, (7) cytochrome C. Principles of electrophoresis Electrophoresis is a separation method based on the fact that charged particles migrate through solution under the influence of an electric field. Their rate of migration depends on the strength of the electric field, their net charge, the size and shape of the molecules, the viscosity and temperature of the medium in which the molecules are moving, and on the ionic strength of the buffer used. If the ion were placed in a vacuum, and if an electric field were imposed, the ion would respond by being accelerated toward the electrode of opposite charge. F = qE F= accelerating force q= the charge on the spherical ion E= electrical field strength (V/cm) F=mA m= mass of spherical ion A= acceleration qE= mA A = qE/m sample Gel tube Upper buffer chamber Lower buffer chamber + Power supply A column gel for polyacrylamide electrophoresis Electrophoresis and densitometry for serum A/G ratio data (44.9: 54.2= 0.83:1) of young Dong-Sun Lee in Normal adult range of A/G ratio is 0.8~2.0. albumin 11 22 22 11 Reactions of weak acids with weak bases HA A + H + K a = [A ][H + ] / [HA] B + H + BH + K = 1/ K a(BH+) = K b / K w HA + B A + BH + K = K a K b / K w 1) K >>1, the reaction will go completion : like the weak + strong Ex. ClCH 2 COOH + CH 3 NH 2 ClCH 2 COO + CH 3 NH M mL M 60.0 mL K = K a K b / K w = 6.0 10 7 pH = pK a + log( [A ] / [HA]) = log (3.60/1.40) = 3.27 2) K is not large : the reaction goes to completion Ex. NH (CH 3 ) 3 N NH 3 + (CH 3 ) 3 NH M M x x x x K = K a K b / K w = 3.60 = x 2 /( x)( x) x = [NH 4 + ] = M [(CH 3 ) 3 N] = M [NH 3 ] = M [(CH 3 ) 3 NH + ] = M Aha ! Two buffers ! pH = pK a + log( [NH 3 ] / [NH 4 + ]) = log ( / ) = 9.81 pH = pK a + log([(CH 3 ) 3 N] / [(CH 3 ) 3 NH + ]) = log ( / ) = 9.81 3) Equimolar mixture of HA and B Ex. 0.10M ammonium bicarbonate NH 4 HCO 3 NH HCO 3 NH 4 + NH 3 + H + HCO 3 + H 2 O CO 3 2 + OH mass balance: [NH 4 + ] + [H + ] = [HCO 3 ] + [OH ] K 1 > K 1 K 1 = [NH 3 ] [H + ] / [NH 4 + ] = 5.7010 10 K 1 = [HCO 3 ] [H + ] / [CO 3 2 ] = 4.45 10 7 K 2 = [CO 3 2 ] [H + ] / [HCO 3 ] = 4.69 10 11 [H + ] = (K 1 K 1 F + K 1 K w ) / (K 1 + F ) = {(4.4510 7 )(5.7010 10 )(0.10) + (4.4510 7 )(K w )} / (4.4510 ) = 7.80 Curves for the titration of polyprotic acid. A) M NaOH solution is used to titrate ml of M H 3 PO 4, B) M oxalic acid, and C) M H 2 SO 4 Titration of ml of M H 2 A with M NaOH. For H 2 A, K a1 = 1.00 10 3 and K a2 = 1.00 10 7. Titration curves for polyfunctional acids Titration curve for ml of M maleic acid H 2 A with M NaOH Molecular model of maleic acid, or (Z)-butenedioic acid and fumaric acid, or (E)-butenedioic acid. Curve for the titration of 5.00 mL of M Na 2 CO 3 with M HCl Titraions in diprotic system 1) Titration reactions : B + H + BH + BH + + H + BH ) Equivalence point : F B NV B mL = F H+ NV e mL 2V e1 = V e2 3) Titration curve : A. Before the addition of any titrant B + H 2 O BH + + OH [OH ] = K b1 F [H + ] = K w / [OH ] Ex. Na 2 CO 3 (mw=105.99) M = N pH = The calculated pH titration curve for the titration of 5.00 mL of M Na 2 CO 3 with M HCl. B. Before the 1st equivalence point (First buffer region) 0V e2 added HCl amount = FV a used HCl amount = FV e Excess HCl amount = F(V a V e ) [H + ] = F HCl (V a V e ) / (V i +V a ) The calculated pH titration curve for the titration of 5.00 mL of M Na 2 CO 3 with M HCl. Calculation of concentration (Finding of HCl concentration) Na 2 CO 3 (mw ) 2 HCl (mw ) 2 Eq wt g 2 1 M 1000 mL g / 2 1 M 1000 mL mg / 2 0.1 M 1 mL a g x M V e mL x M = ( a g 1 M 1000 mL) / ( g V e mL) Principal species The principal form of weak acid or base in aqueous solution under given pH conditions. COOH pK a = 4.20 pH = pK a + log( [A ] / [HA]) pH Major species pH < pK a HA pH= pK a HA and A pH > pK a A COOH pK a1 = 1.25pK a2 = 4.27 pH = pK a1 + log( [HA ] / [H 2 A]) pH = pK a2 + log( [A 2 ] / [HA ]) pH Major species pH < pK a1 H 2 A pH= pK a1 H 2 A and HA pK a1 pK a2 A 2 The predominent molecular form of a triprotic system in the various pH intervals. Amino acid arginine H 3 Arg 2+ H 2 Arg + HArg Arg 2 pK a1 =1.82 pK a2 =8.99 pK a3 =12.48 pH = 1/2(pK a1 + pK a2 )pH = 1/2(pK a2 + pK a3 ) AcidicBasic Fractional composition equation 1) Monoprotic systems Dissociation equilibria : HA A + H + Charge balance : [H + ] = [A ] + [OH ] Mass balance : F = [HA] + [A ] [A ] = F [HA] Dissociation equilibria constants : K a = [A ][H + ] / [HA] K a = [A ][H + ] / [HA] = (F [HA])[H + ] / [HA] [HA] = F [H + ] / ([H + ]+ K a ) 0 = fraction of species in the form HA = fraction of dissociation 1 = fraction of species in the form A 0 + 1 = 1 0 = [HA] / F = [H + ] / ([H + ]+ K a ) 1 = K a / ([H + ]+ K a ) Fractional composition diagram of a monoprotic system with pKa= pH Fraction of species 2) Diprotic system Dissociation equilibria : H 2 A HA + H + HA A 2 + H + Dissociation equilibria constants : K a1 = [HA ][H + ] /[H 2 A] [HA ] = K a1 [H 2 A] / [H + ] K a2 = [A 2 ][H + ] / [HA ] [A 2 ] = K a2 [HA ] / [H + ] = K a1 K a2 [H 2 A] / [H + ] 2 Mass balance : F = [H 2 A] + [HA ] + [A 2 ] = [H 2 A] + (K a1 [H 2 A] / [H + ]) +(K a1 K a2 [H 2 A] / [H + ] 2 ) = [H 2 A](1+K a1 / [H + ] + K a1 K a2 / [H + ] 2 ) 0 + 1 + 2 = 1 0 = [H 2 A] / F = [H + ] 2 / ([H + ] 2 + [H + ]K a1 + K a1 K a2 ) 1 = [HA ] / F= K a1 [H + ] / ([H + ] 2 + [H + ]K a1 + K a1 K a2 ) 2 = [A 2 ] / F= K a1 K a2 / ([H + ] 2 + [H + ]K a1 + K a1 K a2 ) Fractional composition diagram for fumaric acid (trans-butenedioic acid). Composition of H 2 M solutions as a function of pH. Titration of ml of M maleic acid with M NaOH. Logarithmic concentration diagram for 0.100M maleic acid. Summary Polyfunctional acid base diprotic acid system triprotic system Principal species Fractional composition equations Ioselectric point Isoionic pH Isoelectric focusing Electrophoresis pH of the salt of weak acid + weak base