Exp_1_Key_S11

Exp_1_Key_S11 - Exp 1 Key Spring 2011 MCB 120L Part A....

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Exp 1 Key Spring 2011 MCB 120L 1 Part A. Phosphate buffer preparation. Typically, the Henderson-Hasslebalch equation is used with concentrations (i) or (ii) moles: (i) pH = pKa + log ([A] / [HA]) (ii) pH = pKa + log (mole A / mole HA) Expressions (i) and (ii) are essentially identical remembering that Volume x Concentration = Moles (iii) pH = pKa + log (vol A x [A] / vol HA x [HA]) For the Tris and acetate buffers, the A and HA species of the buffer are made to the same final concentration, hence the volumes cancel in (iii) giving (i); however, for the phosphate buffer, equal concentrations of the A and HA species are being mixed at different volumes, so the concentrations cancel giving a ratio of the volumes: (iv) pH = pKa + log (vol A / vol HA ) 1. Staring with 30 ml monobasic phosphate (H 2 PO 4 ): 7.4 = 7.22 + log (vol A / 30 ml) 1.51 = vol A/ 30 ml vol A (dibasic phosphate: HPO 4 2– ) = 45.3 ml The volume of dibasic phosphate added to reach pH 7.4 is much more (Theory does not necessarily correspond to Reality!). The activity coefficients of the dibasic and monobasic phosphate species at 0.1 M concentrations are not ~ 1.0 (see question 2) 2. Using activity coefficients: The activity of z, a Z = [z] x γz And, for ideal dilute solutions, γz ~1.0, thus a Z ~ [z]. However, the activity coefficients of the dibasic and monobasic phosphate species at 0.1 M concentrations are < 1.0: for 0.1 M solutions: γA = 0.445 and γHA = 0.744 pH = pKa + log (vol A x γA / vol HA x γHA) 7.4 = 7.22 + log (0.445 x vol A / 0.744 x 30 ml) 1.51 = 0.445 x Vol A / 0.744 x 30 ml vol A = 75.7 ml still does not account for all the of the dibasic phosphate needed. Other contributions include the effects of ionic strength (of all ions present) and temperature; and variance introduced in weighing solids and measuring liquids. An important point here is that when making a buffer, the buffer should be titrated to the desired pH and the actual pH recorded , not merely “hoping” the pH is correct from theoretical calculations. Part B. Buffer Capacity The practical buffer capacity definitions and equations used in the calculations below are found on pg 50 of Segel’s Biochem Calc’s Buffer capacity in the acid direction: Buffer capacity in the base direction: BCa = [H + ] = 9[HA][A] BCb = [OH ] = 9[HA][A] . 10[HA] + [A] 10[A] + [HA] BCa: the concentration of strong acid required to lower the pH of the solution by one-pH unit BCb: the concentration of strong base required to raise the pH of the solution by one-pH unit
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Exp 1 Key Spring 2011 MCB 120L 2 1. 0.1 M acetate pH 4.2 BCa = (9)(0.0784 M)(0.0216 M) = 0.0189 M 4.2 = 4.76 + log (A / HA) (10)(0.0784 M) + (0.0216 M) A/HA = 0.275 [A] = (0.1 M)(0.275 / 1.275) = 0.0216 M BCb = (9)(0.0784 M)(0.0216 M) = 0.0518 M [HA] = (0.1 M)( 1.0 / 1.275) = 0.0784 M (10)(0.0216 M) + (0.0784 M) The pH of the solution is below the pKa of the buffer (on the „acid side‟), thus it takes less added strong
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This note was uploaded on 11/10/2011 for the course MCB 120L taught by Professor Fairclough during the Spring '08 term at UC Davis.

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Exp_1_Key_S11 - Exp 1 Key Spring 2011 MCB 120L Part A....

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