Fundamental Laboratory Approaches
for Biochemistry and Biotechnology

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by Alex J. Ninfa and David P. Ballou

Notes on Experiments Homework Answers Book Revisions Contact Us Authors Home Links		Individual Sections for Chapter 1 Chapter 1, pp 36-42 Problem 1-1. Problem 1-2. Problem 1-3. Problem 1- 4. Problem 1-5. Chapter 1, pp 43-44. Problem 1-6.
		Problem 1-7. Problem 1-8. Problem 1-9. Solution B Solution C Problem 1-10.     Problem 1-1. 9 mg = 0.009 g The molecular weight of glucose is 180 g/mol 0.009 g / 180 g/mol= 5 X 10-5 mol= 0.05 mmol= 50 mmol   Problem 1-2.   Note, there is an error in the way this question is worded; although concentrations were requested, the units requested were not units of concentration, but were units of mass. The correct wording of this problem is as below: If you dissolve 9 mg of glucose in 5 ml of H2O, what is the concentration of glucose in mM and mM? 9 mg of glucose = 0.05 mmol 0.05 mmol / 5 mL = 0.05 mmol / 0.005 L = 10 mmol / L = 10mM 10 mM = 10,000 mM Problem 1-3. Note, there is an error in the way this question is worded; although concentrations was requested, the units requested were not units of concentration, but were units of mass. The correct wording of this problem is as below: If you take 2 mL of the above solution and dilute it to 50 mL, what is the concentration of glucose in the resulting solution in mM? 2 mL X 10 mM / 50 mL = 0.4 mM = 400 mM (2 mL / 50 mL) 10 mM= 0.4 mM= 400 mM Problem 1-4. (2 mL) (0.4 mM) = 0.8 mmol (2 X 10-3 L) ( 0.4 X 10-3 M) = 0.8 X 10-6 mol = 0.8 mmol Problem 1-5. 70 mg/100 mL = 700 mg / L = 0.7 g /L (0.7 g /L) / (180 g/ mol) = 3.89 X 10-3 M = 3.89 mM (1 g /L) / (180 g /mol) = 5.56 X 10-3 M=5.56 mM   Chapter 1, pp 43-44. Problem 1-6. pH = pKa + log ([acetate]/[acetic acid]) 5.1 = 4.76 + log ([acetate]/[acetic acid]) 0.34 = log([acetate]/[acetic acid]); ([acetate]/[acetic acid])= 2.19 Therefore, 219 mL of acetate are added to each 100 mL of acetic acid. The molarity of the resulting buffer is 0.1 M. Problem 1-7. pKa of Tris = 8.08 8.2 = 8.08 + log ([Tris base]/[HCl acid]) log ([Tris base]/[HCl acid]) = 0.12 [Tris base]/[HCl acid] = 1.32 100 mL of 0.1 MTris base = 0.01 mol (0.01 mol - x)/x = 1.32; x = 4.3 mmol = 43 mL of 0.1 M HCl Therefore, to 100 mL of 0.1 M Tris, add 43 mL of 0.1 M HCl. The total volume = 143 mL and the final Tris concentration is (0.1 M) (100) / 143 = 0.07 M. Problem 1-8. Note: If you are able to do problems 1-8 and 1-9, then you clearly have a good understanding of buffers and pH calculations. pH = pKa + log (base/acid) The appropriate pKa in this case is pK2 = 6.82 pH = 6.82 + log (150 mL X 0.1 M/ 50 ml X 0.1 M)= 6.82 + 0.427 = 7.30 6.30 = 6.82 + log (base/acid) -0.52 = log (base/acid); Therefore base/acid = 0.302 at pH 6.30 Originally we have (150 mL) (0.1 M) base = 15 mmoles and (50 mL ) (0.1 M) acid = 5 mmoles. Therefore, we will convert some of the HPO4 to H2PO4-. We use H3PO4 to titrate the solution. Each time that H3PO4 dissociates to form H+ and H2PO4- (the acidic form in the Henderson-Hasselbach equation), it titrates one HPO42- and adds an equivalent H2PO4- to the solution. Let x= amount of H3PO4 added (or amount of HPO42- converted): 0.302 = 15 -x/5 + 2x 1.51 + 0.604 x=15-x 1.604 x = 13.49 x = 8.41 mmol of H3PO4 Note that since the titration is with 0.1 M H3PO4, the buffer concentration is still 0.1 M.   Problem 1-9. Solution A 4.4 = 4.76 + log (acetate/acetic acid) - 0.36 = log ([base]/[acid]); [base]/[acid] = 0.437; [base] = (0.437) [acid] base] + [acid] = 0.1 M = 0.437 [acid] + [acid]; 1.437 [acid] = 0.1 M 0.1 M / 1.437 = [acid] = 0.0696 M [base] = 1-0.0696 = 0.0304 M   Thus, at pH 4.4 there are (100 mL)(0.0696 M) acid = 6.96 mmol and (100 ml)(0.304 M) base = 3.04 mmol. If you add (10 mL) (0.1 M) NaOH = 1 mmol, you will then have 5.96 mmol acid and 4.04 mmol base. pH= 4.76 + log (4.04/5.96) = 4.59. Thus, a change in pH of 0.19 pH units will be observed. The buffering capacity = (change in OH-) / (change in pH)= 1 mM/0.19 pH units=5.26 mmol/pH unit If we had added 10 mL of 0.1 M HCl instead of NaOH, after the addition, acetic acid = 7.96 mmol and acetate = 2.04 mmol pH = 4.76 + log (2.04/7.96) = 4.17 the change in pH = 0.23, and the buffer capacity is 4.35 mmol/pH unit Solution B. 4.76 = 4.76 + log ([acetate]/[acetic acid]) [acetate] = [acetic acid] = 0.05 M acetate = (0.05 M) (100 mL)= 5 mmol acetic acid = (0.05 M) (100 mL) = 5 mmol After adding 10 mL of 0.1 M NaOH, acid = 6 mmol and acetic acid = 4 mmol pH = 4.76 + log (6/4) = 4.76 + 0.176 = 4.94 Thus, a change in pH of 0.176 is predicted. The buffering capacity is 1 mmol/0.176 pH unit = 5.68mmol/pH unit. If we had added 10 mL of 0.1 M HCl instead of the NaOH, after the addition there would be 6 mmol acetic acid and 4 mmol acetate. pH = 4.76 + log (4/6) = 4.58 The change in pH = -0.176; and buffer capacity = 5.68 mmol/pH unit. Solution C 5.2 = 4.76 + log ([acetate]/[acetic acid]) log ([acetate]/[acetic acid]) =0.44; [acetate]/[acetic acid]=2.75 [acetic acid] + 2.75 [acetic acid] = 0.1 M; [acetic acid] = 0.027 M [acetate]=0.1 M - 0.027 M = 0.073 M acetate = (0.073 M)(100 mL) =7.3 mmol acetic acid = (0.027 M) (100 mL) = 2.7 mmol After addition of 10 ml of 0.1 M NaOH, acetate = 8.3 mmol and acetic acid = 1.7 mmol pH = 4.76 + log (8.3/1.7) = 5.45 The change in pH after addition of the NaOH is thus predicted to be 0.25 pH unit. The buffering capacity = 1 mmol/ 0.25 pH unit= 4.0 mmol/pH unit If we had added 10 mL of 0.1 M HCl instead of the NaOH, after the addition acetic acid = 3.7 mmol and acetate = 6.3 mmol. pH = 4.76 + log (6.3/3.7) = 4.99 Thus, a change in pH of -0.209 unit is predicted and the buffer capacity is 4.79 mmol/pH unit.   Problem 1-10. Glutamic Acid (fully protonated) H(2)OOCCH2CH2CH(NH3+)(3)COOH(1) pK1 = 2.1, pK2 = 4.07, pK3 = 9.47 At pH 5.0 (1) is ionized and (2) is more than 1/2 ionized. (3) is fully protonated. Therefore, -1 > charge > 0 5.0 = 4.07 + log([base]/[acid]); [base]/[acid] = 8.51 8.51[acid] + [acid] = 1 fraction acid = 0.105; fraction base = 0.895 Acidic form is neutral; basic form has charge of -1. Therefore the net charge is -.895 at pH 5.0. The pI is when charge = 0; pH = 3.085   Alanine (fully protonated) CH3CH(NH3+)(2)COOH(1) pK1 = 2.35, pK2 = 9.87 5.0 = 2.35 + log ([base]/[acid]) [base]/[acid] = 447 447[acid] + [acid] = 1 fraction acid = 2.2 x 10-3; fraction base = 0.998 Net charge = 0.002 pI = (pK1 + pK2)/2 = 8.51 We could also calculate the fraction of the -NH3+ group that has ionized and add it to the total net charge, but it is even a smaller contribution than 0.002. Lysine (fully protonated) H3N+(3) (CH2)4CH(NH3+)(2)COOH(1) pK1 = 2.16, pK2 = 9.06, pK3 = 10.54 At pH 5.0 (1) is nearly fully ionized and (2) and (3) are nearly fully protonated. Thereforem, the charge ~ 1. 5.0 = 2.16 + log[base]/[acid]; [base]/[acid] = 692 692[acid] + [acid] = 1 fraction acid = 1.44 x 10-3 fraction base = 0.999 Net charge - 1.001 Again, you could calculate the small contributions of the two -NH3+ groups, but they will be nearly insignificant.