Chapter 2 Biology Assessment Answers

1. Effect of Local Environment on Ionic Bond Strength If the ATP-binding site of an enzyme is buried in the interior of the enzyme, in a hydrophobic environment, is the ionic interaction between enzyme and substrate stronger or weaker than that same interaction would be on the surface of the enzyme, exposed to water? Why? Answer The strength, or force (F), of ionic interactions in a solution depends on the magni-tude of the charges (Q), the distance between the charged groups (r), and the dielectric con stant (e, which is dimensionless) of the solvent in which the interactions occur, according to the following equation: F-When comparing the strength of the same ionic bond in water and in the hydrophobic envi ronment inside a protein, we know that and rare the same; only e is different. Water has a large dielectric corestant because of the large number of dipoles. A hydrophobic "solvent" such as the inside of a protein has a much smaller dielectric constant. Given that the strength of the ionic interaction (F) varies as the inverse of the dielectric constant (e), the ionic interac tion would be stronger in the protein environment, with the smaller dielectric constant. 5. Calculation of Hydrogen Ion Concentration from pH What is the H concentration of a solution with pH of (a) 3.82; (b) 6.52; (c) 11.119 Answer Using [H+] = 10: (a) [H]10-1.51 × 10 м; (b) [H]10-4803.02 x 10 м (c) [H]1011.117.76 × 10-12 м. 8. Calculation of pH from Concentration of Strong Acid Calculate the pH of a solution prepared by diluting 3.0 mL of 2.5 M HCl to a final volume of 100 mL with H₂O. Answer Because HCl is a strong acid, it dissociates completely to H + CI. Therefore, 3.0 ml. x 2.5 м НСІ 7.5 meq of H. In 100 mL of solution, this is 0.075 м Н. pH-log[H+]-log (0.075)-(-1.1)-1.1 (two significant figures) 11. Meanings of K, and på, (a) Does a strong acid have a greater or lesser tendency to lose its proton than a weak acid (b) Does the strong acid have a higher or lower &, than the weak acid? (e) Does the strong acid have a higher or lower pk, than the weak acid? Answer (a) greater, by definition, the stronger acid has the greater tendency to dissociate a proton; (b) higher, as K [H][A][HA], whichever acid yields the larger concentration of [H] has the larger K; (e) lower, pk-log K, so if K, is larger, -log A, will be smaller. 14. Calculation of the pH of a Mixture of a Weak Acid and Its Conjugate Base Calculate the pH of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (p*K_{a} = 4.76) (a) 2:1; (b) 1:3; (c) 5:1; (d) 1:1; (e) 1:10. Answer Using the Henderson-Hasselbalch equation, DH=pK a +log [A^ - ] [HA] pH =4.76+log Jacetate Macetic acid]), where [acetate)]/[acetic acid] is the ratio given for each part of the question. (a) log(2/1) = 0.3 pH = 4.76 + 0.3 = 5.06 (b) log(1/3) = - 0.48 ; pH = 4.76 + (- 0.48) - 4.28 (e) log(5/1) = 0.7 H = 4.76 + 0.7 - 5.46 (d) log(1/1) = 0 pH = 4.76 (e) log(1/10) = - 1, 1 pH-4.76+(-1.00)-3.76 22. Calculation of the pH of a Solution of a Polyprotic Acid Histidine has ionizable groups with values of 1.6, 6.0, and 9.2, as shown below (His imidazole group). A biochemist makes up 100 ml a 0.100 M solution of histidine at a pH of 5.40. She then adds 40 mL of 0.10 M HCL. What is the pH of the resulting solution? p*K_{a} COOH COO- Coo- HN-CH CH H N HN-CH HN-CH CH H N COO-HAN-CH CH H N CH H N CH 18 6.0 CH CH 92 yd CH H lonizable -COOH group -HisH -His -NH-NH, Answer The initial pH of 5.40 is so far below r*K_{3} (that of the amino group of histidine), that we know the group is completely dissociated at initial pH, so we only need to consider the groups of p*K_{2} and p*K_{2} (Le., the H on the carboxyl group and the H on the imidazole ring). Initially, the pH was 5,40, from which we can calculate the fraction of the imidazole hydrogen that was dissociated (the ratio of the conjugate base His to the acid HisH"): pH=pK 2 +log( [His] /[HisH^ * D Next, substitute the values for pH and rearrange, and take the antilog of both sides: p*K_{2r} 5.4 =6.00+log(lHinV[HisH^ + D) -0.60-log (His)/[HisH*]) 100.40 (HisH [His]) = 4.0 0.60=log([HisH^ + mathcal V [His] ) Thus, in the initial solution, the ratio of [HisH"] to [His) is 4 to 1; 4 out of 5 of the imidazole groups were initially protonated. The initial solution contains 10.0 mmol of histidine (10.0 mL x 0.100 mmol/mL), 1/5 of which (2.0 mmol) had unprotonated imidazole groups. The amount of HCI added was 4.0 mmol, of which 2.0 mmol was consumed in protonating the remaining imidazole groups. The other 2.0 mmol of HCI protonated a fraction (2.0 of the 10.0 mmol) of the carboxylate groups (with PK), leaving 8.0 mmol in the deprotonated form. From these ratios of acid and base after the titration, we can calculate the final pH: pHp+log (1-COO-COOHI) 8.0 1.82 + log 2.0 1.82+0.60-2.42 -2.4 (2 significant figures) 24. Preparation of a Phosphate Buffer What molar ratio of HPO to HPO in solution would produce a pH of 7.0? Phosphoric acid (H₃PO₄), a triprotic acid, has 3 p, values: 2.14, 6.86, and 12.4. Hint: Only one of the př, values is relevant here. Answer Only the ph, close to the pH is relevant here, because the concentrations of the other species (H₂PO, and PO) are insignificant compared with the concentrations of HPO and HPO. Begin with the Henderson-Hasselbalch equation: pHpK+log (conjugate base [acid]) log ([HPO PODPHPK-7.0-6.80 -0.14 [HPOH PO-100441.88 1.4 (two significant figures) 26. Calculation of Molar Ratios of Conjugate Base to Weak Acid from pH For a weak acid with a pk, of 6.0, calculate the ratio of conjugate base to acid at a pH of 5.0. Answer Using the Henderson-Hasselbalch equation, pHpK+log ([A [HA]) 5.0-6.0+log (JA HAI) log (A [HA]) = -1.0 [AHA] 10~1.0-0.10