22. Calculation of the pH of a Solution of a Polyprotic Acid Histidine has ionizable groups with pk, values of 1.8, 6.0,

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22 Calculation Of The Ph Of A Solution Of A Polyprotic Acid Histidine Has Ionizable Groups With Pk Values Of 1 8 6 0 1 (28.7 KiB) Viewed 82 times

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22 Calculation Of The Ph Of A Solution Of A Polyprotic Acid Histidine Has Ionizable Groups With Pk Values Of 1 8 6 0 3 (72 KiB) Viewed 82 times

22. Calculation of the pH of a Solution of a Polyprotic Acid Histidine has ionizable groups with pk, values of 1.8, 6.0, and 9.2, as shown below (His imidazole group). A biochemist makes up 100 mL of a 0.100 M solution of histidine at a pH of 5.40. She then adds 40 mL of 0.10 M HCI. What is the pH of the resulting solution? COOH L H₂N-CH 1 CH₂ H N H lonizable group CH COO™ I H₂N-CH 1.8 pk₁ CH₂ H -N -COOH-COO™ CH -HisH+ COO™ H₂N-CH 6.0 pK₂ CH₂ H -N H -His CH -NH3 COO™ H₂N-CH 9.2 pk3 CH₂ H N H -NH₂ CH
Consequently you have Three consumable H in the molecule (symbolize as H3A) with 3 pka values. At a pH of 5.40 you are accurate is presumptuous the pka close to 6 is relevant and we are looking at the eqm H2A- <--eqm--> H+ + HA^2- You use the HH equation properly to discover the ratio of HA^2-/H2A- = 0.25 except you cannot use an ICE table as you did. The HCI almost certainly proceeds totally through the HA2- consequently we require to believe the number of moles of every species. mol histidine = 0.100 M x 0.100 L = 0.0100 mol so mol HA^2- + mol H2A- = 0.0100 mol and mol HA^2-/mol H2A- = 0.25 solving gives mol HA^2- = 0.002 mol HA^2- mol HCI added = 0.10 M x 0.040 L = 0.004 mol HCI in additional terms the HCI will respond with every the HA^2- and moreover exchange 0.002 mol of H2A-to H3A. Consequently we require to believe the next equilibrium H3A <--eqm--> H+ + H2A-pka = 1.8 where mol H3A is 0.002 mol and mol H2A- is 0.0100 -0.002 = 0.00800 mol using HH once more pH = 1.8+ log(0.00800/0.00200) = 2.4