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 A d here we can use the following equation 10 pH pK a HA 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 Ais 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 1 3 a COO H3N C H CH3 COOH H3N C H CH3 pK1 2 34 H3N C2H4 COOH fully protonated COO H2N C H CH3 pK2 9 69 H3N C2H4 COO H2N C2H4 COO 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 H3NC2H4 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 negnigible 2 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 3 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 …