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c02Water.qxd4/15/166:56 PMPage S-15Chapter 2 Water5. Calculation of Hydrogen Ion Concentration from pH What is the H concentration of a solutionwith pH of (a) 3.82; (b) 6.52; (c) 11.11?Answer Using [H ] 10 pH:(a) [H ] 10 3.82 1.51 10 4 M; (b) [H ] 10 6.52 3.02 10 7 M;(c) [H ] 10 11.11 7.76 10 12 M.6. Acidity of Gastric HCl In a hospital laboratory, a 10.0 mL sample of gastric juice, obtained severalhours after a meal, was titrated with 0.1 M NaOH to neutrality; 7.2 mL of NaOH was required. The patient’s stomach contained no ingested food or drink; thus assume that no buffers were present. Whatwas the pH of the gastric juice?Answer Multiplying volume (L) by molar concentration (mol/L) gives the number of moles inthat volume of solution. If x is the concentration of gastric HCl (mol/L),(0.010 L)x (0.0072 L)(0.1 mol/L)x 0.072 M gastric HClGiven that pH log [H ] and that HCl is a strong acid,pH log (7.2 10 2) 1.17. Calculation of the pH of a Strong Acid or Base (a) Write out the acid dissociation reaction forhydrochloric acid. (b) Calculate the pH of a solution of 5.0 10 4 M HCl. (c) Write out the aciddissociation reaction for sodium hydroxide. (d) Calculate the pH of a solution of 7.0 10 5 M NaOH.Answerz H Cl (a) HCl y(b) HCl is a strong acid and fully dissociates into H and Cl . Thus, [H ] [Cl ] [HCl].pH log [H ] log (5.0 10 4 M) 3.3 (two significant figures)z Na OH (c) NaOH y(d) NaOH is a strong base; dissociation in aqueous solution is essentially complete, so[Na ] [OH ] [NaOH].pH pOH 14pOH log [OH ]pH 14 log [OH ] 14 log (7.0 10 5) 9.8 (two significant figures)8. Calculation of pH from Concentration of Strong Acid Calculate the pH of a solution preparedby 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 Cl . Therefore,3.0 mL 2.5 M HCl 7.5 meq of H . In 100 mL of solution, this is 0.075 M H .pH log [H ] log (0.075) ( 1.1) 1.1 (two significant figures)9. Measurement of Acetylcholine Levels by pH Changes The concentration of acetylcholine (aneurotransmitter) in a sample can be determined from the pH changes that accompany its hydrolysis.When the sample is incubated with the enzyme acetylcholinesterase, acetylcholine is converted tocholine and acetic acid, which dissociates to yield acetate and a hydrogen ion:OCH3 CCH3OCH2CH2 NCH3H2 OCH3HOCH2 CH2CH3 CH3 CCH3CH3Acetylcholine NCholineOAcetateO H S-15

c02Water.qxdS-164/15/166:56 PMPage S-16Chapter 2 WaterIn a typical analysis, 15 mL of an aqueous solution containing an unknown amount of acetylcholine hada pH of 7.65. When incubated with acetylcholinesterase, the pH of the solution decreased to 6.87. Assuming there was no buffer in the assay mixture, determine the number of moles of acetylcholine inthe 15 mL sample.Answer Given that pH log [H ], we can calculate [H ] at the beginning and at the end ofthe reaction:At pH 7.65, log [H ] 7.65[H ] 10 7.65 2.24 10 8 MAt pH 6.87, log [H ] 6.87[H ] 10 6.87 1.35 10 7 MThe difference in [H ] is(1.35 0.22) 10 7 M 1.13 10 7 MFor a volume of 15 mL, or 0.015 L, multiplying volume by molarity gives(0.015 L)(1.13 10 7 mol/L) 1.7 10 9 mol of acetylcholine10. Physical Meaning of pKa Which of the following aqueous solutions has the lowest pH: 0.1 M HCl; 0.1 Macetic acid (pKa 4.86); 0.1 M formic acid (pKa 3.75)?Answer A 0.1 M HCl solution has the lowest pH because HCl is a strong acid and dissociatescompletely to H Cl , yielding the highest [H ].11. Meanings of Ka and pKa (a) Does a strong acid have a greater or lesser tendency to lose its protonthan a weak acid? (b) Does the strong acid have a higher or lower Ka than the weak acid? (c) Doesthe strong acid have a higher or lower pKa than the weak acid?Answer (a) greater; by definition, the stronger acid has the greater tendency to dissociate aproton; (b) higher; as Ka [H ] [A ]/[HA], whichever acid yields the larger concentration of[H ] has the larger Ka; (c) lower; pKa log Ka, so if Ka is larger, log Ka will be smaller.12. Simulated Vinegar One way to make vinegar (not the preferred way) is to prepare a solution ofacetic acid, the sole acid component of vinegar, at the proper pH (see Fig. 2–15) and add appropriateflavoring agents. Acetic acid (Mr 60) is a liquid at 25 C, with a density of 1.049 g/mL. Calculate thevolume that must be added to distilled water to make 1 L of simulated vinegar (see Fig. 2–16).Answer From Figure 2–16, the pKa of acetic acid is 4.76. From Figure 2–15, the pH of vinegaris 3; we will calculate for a solution of pH 3.0. Using the Henderson-Hasselbalch equation[A ]pH pKa log [HA]and the fact that dissociation of HA gives equimolar [H ] and [A ] (where HA is CH3COOH,and A is CH3COO ), we can write3.0 4.76 log ([A ]/[HA]) 1.76 log ([A ]/[HA]) log ([HA]/[A ])[HA]/[A ] 101.76 58Thus, [HA] 58[A ]. At pH 3.0, [H ] [A ] 10 3, so[HA] 58 10 3 M 0.058 mol/LDividing density (g/mL) by molecular weight (g/mol) for acetic acid gives1.049 g/mL 0.017 mol/mL60 g/molDividing this answer into 0.058 mol/L gives the volume of acetic acid needed to prepare 1.0 Lof a 0.058 M solution:0.058 mol/L 3.3 mL/L0.017 mol/mL

c02Water.qxd4/15/166:56 PMPage S-17Chapter 2 WaterS-1713. Identifying the Conjugate Base Which is the conjugate base in each of the pairs below?(a) RCOOH, RCOO (b) RNH2, RNH 3(c) H2PO ,HPO434(d) H2CO3, HCO 3Answer In each pair, the acid is the species that gives up a proton; the conjugate base is thedeprotonated species. By inspection, the conjugate base is the species with fewer hydrogen atoms. (a) RCOO (b) RNH2 (c) H2PO 4 (d) HCO314. Calculation of the pH of a Mixture of a Weak Acid and Its Conjugate Base Calculate the pHof a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pKa 4.76) of(a) 2:1; (b) 1:3; (c) 5:1; (d) 1:1; (e) 1:10.Answer Using the Henderson-Hasselbalch equation,[A ]pH pKa log [HA]pH 4.76 log ([acetate]/[acetic acid]), where [acetate]/[acetic acid] is the ratio given foreach part of the question.(a) log (2/1) 0.30; pH 4.76 0.30 5.06(b) log (1/3) 0.48; pH 4.76 ( 0.48) 4.28(c) log (5/1) 0.70; pH 4.76 0.70 5.46(d) log (1/1) 0; pH 4.76(e) log (1/10) 1.00; pH 4.76 ( 1.00) 3.7615. Effect of pH on Solubility The strongly polar hydrogen-bonding properties of water make it anexcellent solvent for ionic (charged) species. By contrast, nonionized, nonpolar organic molecules, suchas benzene, are relatively insoluble in water. In principle, the aqueous solubility of any organic acid orbase can be increased by converting the molecules to charged species. For example, the solubility ofbenzoic acid in water is low. The addition of sodium bicarbonate to a mixture of water and benzoic acidraises the pH and deprotonates the benzoic acid to form benzoate ion, which is quite soluble in water.OCOOHBenzoic acidpKa 5CO Benzoate ionAre the following compounds more soluble in an aqueous solution of 0.1 M NaOH or 0.1 M HCl? (The dissociable proton are shown in bold.)OOHN HCHCH3 NHCOCH2OHCOCH3Pyridine ionpKa 5 -NaphtholpKa 10N-Acetyltyrosine methyl esterpKa 10(a)(b)(c)Answer(a) Pyridine is ionic in its protonated form and therefore more soluble at the lower pH, in0.1 M HCl.(b) b-Naphthol is ionic when deprotonated and thus more soluble at the higher pH, in0.1 M NaOH.(c) N-Acetyltyrosine methyl ester is ionic when deprotonated and thus more soluble in0.1 M NaOH.

c02Water.qxdS-184/15/166:56 PMPage S-18Chapter 2 Water16. Treatment of Poison Ivy Rash The components of poison ivy and poison oak that produce thecharacteristic itchy rash are catechols substituted with long-chain alkyl groups.OHOH(CH2)nCH3pKa 8If you were exposed to poison ivy, which of the treatments below would you apply to the affected area?Justify your choice.(a) Wash the area with cold water.(b) Wash the area with dilute vinegar or lemon juice.(c) Wash the area with soap and water.(d) Wash the area with soap, water, and baking soda (sodium bicarbonate).Answer The best choice is (d). Soap helps to emulsify and dissolve the hydrophobic alkylgroup of an alkylcatechol. Given that the pKa of an alkylcatechol is about 8, in a mildly alkalinesolution of bicarbonate (NaHCO3) its OOH group ionizes, making the compound much morewater-soluble. A neutral or acidic solution, as in (a) or (b), would not be effective.17. pH and Drug Absorption Aspirin is a weak acid with a pKa of 3.5 (the ionizable H is shown in bold):OCCH3 OOCOHIt is absorbed into the blood through the cells lining the stomach and the small intestine. Absorptionrequires passage through the plasma membrane, the rate of which is determined by the polarity of themolecule: charged and highly polar molecules pass slowly, whereas neutral hydrophobic ones passrapidly. The pH of the stomach contents is about 1.5, and the pH of the contents of the small intestineis about 6. Is more aspirin absorbed into the bloodstream from the stomach or from the smallintestine? Clearly justify your choice.Answer With a pKa of 3.5, aspirin is in its protonated (neutral) form at pH below 2.5. Athigher pH, it becomes increasingly deprotonated (anionic). Thus, aspirin is better absorbed inthe more acidic environment of the stomach.18. Calculation of pH from Molar Concentrations What is the pH of a solution containing 0.12 mol/Lof NH4Cl and 0.03 mol/L of NaOH (pKa of NH 4 /NH3 is 9.25)?Answer For the equilibriumz NH H yNH 43pH pKa log ([NH3]/[NH 4 ])we know that [NH 4 ] [NH3] 0.12 mol/L, and that NaOH completely dissociates to give[OH–] 0.03 mol/L. Thus, [NH3] 0.03 mol/L and [NH 4 ] 0.09 mol/L, andpH 9.25 log (0.03/0.09) 9.25 0.48 8.77, which rounds to 9.

c02Water.qxd4/15/166:56 PMPage S-19Chapter 2 Water19. Calculation of pH after Titration of Weak Acid A compound has a pKa of 7.4. To 100 mL of a 1.0 Msolution of this compound at pH 8.0 is added 30 mL of 1.0 M hydrochloric acid. What is the pH of theresulting solution?Answer Begin by calculating the ratio of conjugate base to acid in the starting solution, usingthe Henderson-Hasselbalch equation:pH pKa log ([A ]/[HA])8.0 7.4 log ([A ]/[HA])log ([A ]/[HA]) 0.6[A ]/[HA] 100.6 4The solution contains 100 meq of the compound (conjugate base plus acid), so 80 meq are inthe conjugate base form and 20 meq are in the acid form, for a [base]/[acid] ratio of 4.Because HCl is a strong acid and dissociates completely, adding 30 mL of 1.0 M HCl adds30 meq of H to the solution. These 30 meq titrate 30 meq of the conjugate base, so the[base]/[acid] ratio is 1. Solving the Henderson-Hasselbalch equation for pH:pH pKa log ([A ]/[HA]) 7.4 log 1 7.420. Properties of a Buffer The amino acid glycine is often used as the main ingredient of a buffer inbiochemical experiments. The amino group of glycine, which has a pKa of 9.6, can exist either in theprotonated form (ONH 3 ) or as the free base (ONH2), because of the reversible equilibriumz RONH2 H RONH 3 y(a) In what pH range can glycine be used as an effective buffer due to its amino group?(b) In a 0.1 M solution of glycine at pH 9.0, what fraction of glycine has its amino group in theONH 3 form?(c) How much 5 M KOH must be added to 1.0 L of 0.1 M glycine at pH 9.0 to bring its pH toexactly 10.0?(d) When 99% of the glycine is in its ONH 3 form, what is the numerical relation between the pH ofthe solution and the pKa of the amino group?Answer(a) In general, a buffer functions best in the zone from about one pH unit below to one pHunit above its pKa. Thus, glycine is a good buffer (through ionization of its amino group)between pH 8.6 and pH 10.6.(b) Using the Henderson-Hasselbalch equation[A ]pH pKa log [HA]we can write[A ]9.0 9.6 log [HA][A ] 10 0.6 0.25[HA]which corresponds to a ratio of 1/4. This indicates that the amino group of glycine isabout 1/5 deprotonated and 4/5 protonated at pH 9.0.S-19

c02Water.qxdS-204/15/166:56 PMPage S-20Chapter 2 Water(c) From (b) we know that the amino group is about 1/5, or 20%, deprotonated at pH 9.0.Thus, in moving from pH 9.0 to pH 9.6 (at which, by definition, the amino group is50% deprotonated), 30%, or 0.3, of the glycine is titrated. We can now calculate fromthe Henderson-Hasselbalch equation the percentage protonation at pH 10.0:[A ]10.0 9.6 log [HA] [A ] 100.4 2.5 5/2[HA]This ratio indicates that glycine is 5/7, or 71%, deprotonated at pH 10.0, an additional21%, or 0.21, deprotonation above that (50%, or 0.5) at the pKa. Thus, the totalfractional deprotonation in moving from pH 9.0 to 10.0 is 0.30 0.21 0.51, whichcorresponds to0.51 0.1 mol 0.05 mol of KOHThus, the volume of 5 M KOH solution required is (0.5 mol)/(5 mol/L) 0.01 L, or 10 mL.(d) From the Henderson-Hasselbalch equation,pH pKa log ([—NH2]/[ONH 3 ]) pKa log (0.01/0.99) pKa ( 2) pKa 2In general, any group with an ionizable hydrogen is almost completely protonated at apH at least two pH units below its pKa value.21. Calculation of the pKa of an Ionizable Group by Titration The pKa values of a compound withtwo ionizable groups are pK1 4.10 and pK2 between 7 and 10. A biochemist has 10 mL of a 1.0 Msolution of this compound at a pH of 8.00. She adds 10.0 mL of 1.00 M HCl, which changes the PH to3.20. What is pK2?Answer The dibasic acid H2A has two dissociable protons:H2A 8888nHA 8888nA2 pKpK12The initial pH (8.00) is so far above pK1 that we know the first proton is fully dissociated, andsome of the HA has dissociated to A2 . We can calculate how much of the 10 mmol of HCl(10 mL 1.0 mmol/mL) was used to convert HA to H2A, using the Henderson-Hasselbalchequation for the group of pK1:pH pK1 log ([HA ]/[H2A])pH pK1 log ([HA ]/[H2A])pK1 pH log ([H2A]/[HA ])Substituting the final pH of 3.20 and the pK1 of 4.10, we get:4.10 3.20 0.90 log ([H2A]/[HA ])100.90 [H2A]/[HA ]7.94 [H2A]/[HA ]Following titration, we have 7.94 parts H2A per part HA , and can calculate the percentage ofH2A in the final solution.7.94[H2A] [H2A] [HA ] 1 7.94 0.888 88.8%

c02Water.qxd4/15/166:56 PMPage S-21Chapter 2 WaterSince we started with 10 mmol of the compound and an equal amount of HCl, then 88.8% ofthe HCl was used up when 88.8% of the compound was converted to H2A. The remaining11.2%, or 1.12 mmol, of the 10 mmol of HCl must have been used in titrating (protonating)the amount of A2– in the initial solution of pH 8.00. Therefore, the initial solution must havecontained 1.12 mmol of the compound in the form A2– (the conjugate base), and the remaining 8.88 mmol must have been present initially as HA– (the acid). Again using the HendersonHasselbalch equation, we can calculate pK2:pH pK2 log ([A2 ]/[HA ])pK2 pH log ([A2 ]/[HA ])1.12 8.0 log 8.88 8.0 ( 0.90) 8.9 (2 significant figures)22. Calculation of the pH of a Solution of a Polyprotic Acid Histidine has ionizable groups with pKavalues of 1.8, 6.0, and 9.2, as shown below (His imidazole group). A biochemist makes up 100 mL ofa 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 ofthe resulting solution?COO COOH H3N CHCH2CH3NHNCHCHIonizablegroup NHCOOH1.8pK1H3NCHCH2CHN NHCOO 6.0pK2H2NCHCH2CHCHCOO COO CHNCHCH9.2pK3N HisH HisNH3CHCH2 HNCCHCHNNH2Answer The initial pH of 5.40 is so far below pK3 (that of the amino group of histidine), thatwe know the group is completely dissociated at the initial pH, so we only need to consider thegroups of pK1 and pK2 (i.e., 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 hydrogenthat was dissociated (the ratio of the conjugate base His to the acid HisH ):pH pK2 log ([His]/[HisH ])Next, substitute the values for pH and pK2, rearrange, and take the antilog of both sides:5.40 6.00 log ([His]/[HisH ]) 0.60 log ([His]/[HisH ])0.60 log ([HisH ]/[His])100.60 ([HisH ]/[His]) 4.0Thus, in the initial solution, the ratio of [HisH ] to [His] is 4 to 1; 4 out of 5 of the imidazolegroups were initially protonated.S-21

c02Water.qxdS-224/15/166:56 PMPage S-22Chapter 2 WaterThe initial solution contains 10.0 mmol of histidine (10.0 mL 0.100 mmol/mL), 1/5 of which(2.0 mmol) had unprotonated imidazole groups. The amount of HCl added was 4.0 mmol, ofwhich 2.0 mmol was consumed in protonating the remaining imidazole groups. The other2.0 mmol of HCl protonated a fraction (2.0 of the 10.0 mmol) of the carboxylate groups (withpK1), leaving 8.0 mmol in the deprotonated form. From these ratios of acid and base after thetitration, we can calculate the final pH:pH pK1 log ([ COO ]/[ COOH])8.0 1.82 log 1.82 0.60 2.422.0 2.4 (2 significant figures)23. Calculation of the Orginal pH from the Final pH after Titration A biochemist has 100 mL ofa .10 M solution of a weak acid with a pKa of 6.3. Sh

(a) log (1.75 10 5) 4.76; (b) log (6.50 10 10) 9.19; (c) log (1.0 10 4) 4.0; (d) log (1.50 10 5) 4.82. Q 1Q 2 er2 Water chapter 2 Solutions Manual for Lehninger Principles of Biochemistry 7th Edition by Nelson IBSN 9781464126116c02Water.qxd 4/15/16 6:56 PM Page S-14