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UMD BCHM 461 - Solutions to pH and Buffers Problem Set

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1 BCHM461 Spring 2002 Solutions to pH and Buffers Problem Set 1. Here I will provide answers for the pKa = 6.5 example only (a) For this problem, as well as for problems (b) and (c), it is convenient to use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) = pKa =6.5 because when [A-] = [HA], log([A-]/[HA]) = log1 = 0. (b) now log([A-]/[HA]) = log10 = 1,  pH = pKa + 1 = 7.5 (c) now log([A-]/[HA]) = log0.1 = -1,  pH = pKa - 1 = 5.5; (d) here we can use the following equation: [][]apKpHHAA−−=10 pH pH - pKa [A-]/[HA] 4.5 -2 1/100 5.5 -1 1/10 6.5 0 1/1 7.5 1 10/1 8.5 2 100/1 2. It is convenient to use the same equation as in Problem 1(d). (a) when pH = pKa, 10pH-pKa = 100 = 1,  the molar concentrations of acid and conjugate base are equal (b) when the pH is less than the pKa, pH-pKa < 0, hence [A-]/[HA] =10pH-pKa <1  HA is the predominant species (c) when the pH is higher than the pKa, pH-pKa > 0, hence [A-]/[HA] = 10pH-pKa >1  A- is the predominant species (d) When the pH is 1.0 pH unit above the pKa, pH-pKa = 1, [A-]/[HA] = 10,  the ratio of [HA] to [A-] is 1/10; when the pH is 1.0 pH unit below the pKa, pH-pKa = - 1, [A-]/[HA] = 10-1 = 1/10,  the ratio of [HA] to [A-] is 10/1; (e) Likewise, when the pH is 2.0 pH units above the pKa, pH-pKa = 2, [A-]/[HA] = 102=100, and the ratio of [HA] to [A-] is 1/100; when the pH is 2.0 pH units below the pKa, pH-pKa = -2, [A-]/[HA] = 10-2=1/100, and the ratio of [HA] to [A-] is 100/1;2 3. (a) COOH COO- COO- | | | +H3N – C – H +H3N – C – H H2N – C -- H | | | CH3 CH3 CH3 pK1 = 2.34 pK2 = 9.69 +H3N-C2H4-COOH +H3N-C2H4-COO- H2N-C2H4-COO- fully protonated zwitterionic deprotonated (b) The relative concentrations of all three forms of alanine are controlled by the two equilibrium equations: [+H3N-C2H4-COO-]/[+H3N-C2H4-COOH] = 10pH-pK1 [a] [H2N-C2H4-COO-]/[+H3N-C2H4-COO-] = 10pH-pK2 [b] You can use these equations to calculate the ratios of the various forms for the particular values of pH and pKa’s. However, the answers to these problems can be obtained from a simple qualitative analysis, as follows: (1) pH = 1.0 is smaller than both pK1 and pK2, so +H3N-C2H4-COOH is the predominant form. Indeed, because pH < pK1, the concentration of the fully protonated form, +H3N-C2H4-COOH, is greater than that of its conjugate base, +H3N-C2H4-COO-. The concentration of H2N-C2H4-COO- (a conjugate base for the zwitterionic form) is even smaller (practically negligible!) than that of. +H3N-C2H4-COO- because the pH is lower than the pKa (pK2) for the equilibrium between these two forms. (2) Since 6.2 > pK1, the concentration of the zwitterionic form is greater than that of the fully protonated form (its conjugate acid). Since 6.2 < pK2, the concentration of the zwitterionic form is also greater than that for its conjugate base, H2N-C2H4-COO-. (3) Since pK1 < 8.02 < pK2 , the answer is same as in Problem 3(2): the zwitterionic form is predominant. (4) pH 11.9 is higher than both pKa’s. This means that the concentration of the deprotonated form (H2N-C2H4-COO-) is significantly greater than the concentration of the zwitterionic form (11.9 exceeds pK2 by more than 2.0 pH units). The concentration of the fully protonated form is practically negnigible3 (compare pK1 = 2.34 with pH = 11.9!). Therefore, the deprotonated form is predominant at this pH. (c) Let the total concentration of Ala be c and let x denote the fraction of Ala in the zwitterionic form at pH = pI. Recall that pI = (pK1 +pK2)/2. From equation [a] (see above) the concentration of the fully protonated form is then [+H3N-C2H4-COOH] = x/10pH-pK1 = x/103.675 = x/4732. The concentration of the deprotonated form can be determined from equation [b] as [H2N-C2H4-COO-] = x .10pH-pK2 = x .10-3.675 = x/4732. From the ‘matter conservation’ principle, the total concentration of Ala is the sum of the concentrations of all three forms (we neglect the uncharged form): [+H3N-C2H4-COOH] + [+H3N-C2H4-COO-] + [H2N-C2H4-COO-] = c or using the above relationships: x/4732 + x + x/4732 = c which leads to x (1 + 2/4732) = c The fraction of Ala in the zwitterionic form is then x / c = 0.9996, or 99.96%. 4. pH = 2.0 corresponds to [H+] = 10-2 M. When 1M HCl contains 1M H+ and 1M Cl-. This means that in order to make 10-2 M H+, the original 1M HCl has to be diluted 100 fold. Since we are adding HCl to 250 mL of water, a 100-fold dilution will be achieved if we take 250mL/100 = 2.5 mL of 1M HCl. The OH- concentration in pure water is [OH-] = 10-7 M; at pH=2 (i.e. [H+] = 10-2 M) it can be calculated from the ion product of water: [OH-] = 10-14M2/[H+] = 10-12 M. Now assume we added 2.5 mL 1M NaOH instead of HCl. This amounts to 0.01 M concentration of NaOH, which then provides 0.01 M OH-. The H+ concentration can be calculated from the ion product of water: [H+] = 10-14M2/10-2M = 10-12 M, hence the pH = 12.0. 5. (a) [OAc-] = [HOAc] = 0.1M; from the Henderson-Hasselbalch equation: pH = pKa = 4.76 (because log([OAc-]/[HOAc]) = log1 = 0). [H+] = 10-pH M = 1.74 10-5 M (b) 0.01 M HCl will dissociate completely into 0.01M H+ and 0.01M Cl-; 0.01M H+ will essentially completely react with the base OAc- to form an acid, HOAc. As a result, the concentration of OAc- will decrease by 0.01M to 0.09M; the concentration of the acid will increase to 0.11 M. The Henderson-Hasselbalch equation then gives: pH = pKa + log(0.09/0.11) = 4.67, a very small change in pH! [H+] = 10-pH M= 2.12 10-5M. (c ) Initial conditions: [OAc-]/[HOAc] = 0.02/0.18=1/9; the Henderson-Hasselbalch equation gives: pH = pKa + log(1/9) = 3.81, and [H+] = 1.56 10-4M; pH = 3.81. After 0.01 M HCl has been added: [OAc-] will decrease to 0.01M, [HOAc] will increase to 0.19 M, so that pH = pKa + log(1/19) =3.48; and [H+] = 3.3 10-4M. Still small change in the pH.4 (d ) Initial conditions: [OAc-]/[HOAc] = 0.18/0.02=9/1; pH = pKa + log(9/1) = 5.71; [H+] =1.93 10-6M. After 0.01 M HCl has been added: [OAc-] will decrease to 0.17M, [HOAc] will increase to 0.03 M, so that pH = pKa + log(0.17/0.03) =5.51; and [H+] = 3.3 10-4M.  A small change in the pH: there was enough base to resist the addition of this amount of HCl.. (e) The initial concentrations of protons and the pH are same. The added 0.15M HCl is completely ionized, which results in


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UMD BCHM 461 - Solutions to pH and Buffers Problem Set

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