ch2 - Water .s 510mm" 1. Solubility of Ethanol in...

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Unformatted text preview: Water .s 510mm" 1. Solubility of Ethanol in Water Explain why ethanol (CH3CH20H) is more soluble in water than is ethane (CH30H3). Answer Ethanol is polar; ethane is not. The ethanol MOI-l group can hydrogen-bond Mth water. 2. Calculation of pH from Hydrogen Ion Concentration What is the pH of a solutioa that has an H+ concentration of (a) 1.75 x 10”"5 moi/L; (1:06.50 >< 10—i9mol/L; (c) 1.0 x 10"“4 moi/L; (a) 1.50 x 10‘5mol/L? Answer Using pH e -log 01”]: (a) “log (2.75 x 10-5) 2 4.76; (b) “clog (6.50 x 10"”) = 9.19; ((5) —log (1.0 X 10—4) 2 4.0; (d) ~log (1.50 >< 10*?) a 4.82. 3. 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) 11.11? Answer Using [at] = 10%; (a) [EVE W 10-3-32 = 1.51 x 10'4 M; (1)) {W} m 10“6-52 = 3.02 x 10*?M; (c) 01*] m 10”}-11 = 7.76 >< 10""12 M. 4. Acidity of Gastric H0} In a hospital laboratory, a 30.0 mL sample of gastric juice, obtained several hours after a meal, was titrated with 0.1 M NaOH to neutrality; 7.2 mL of NaOH was required. The pa tient’s stomach contained no ingested food or drink, thus assume that no buffers were present. What was the pH of the gastric juice? Answer Multiplying volume (L) by molar concentration (moi/L) gives the mueber of moles in that volume of solution. If :5 is the concentration of gastric H01 (moi/L), (0.010 Lax m (0.0072 L)(0.1 tool/L) x = 0.072 M gastric H01 Given that pH = —log [H +land that H01 is a strong acid, pH m -—10g(7.2 x 10”) = 1.1 5. Calculation of the pH of a Strong Acid or Base (3) Write out the acid dissociation reaction for hydrochloric acid. (b) Calculate the pH of a solution of 5.0 X 10“: M H01. ((2) Write out the acid dissociation reaction for sodium hydroxide. ((1) Calculate the pH of a solution of 7.0 X 10”“5 M NaOH. 3-13 " " 1’! $- 5? 4. . a 1k! :1 3-14 Chapter 2 Water Answer (a) H01 Wi‘ H+ + 01“ (b) H01 is a strong acid and fully dissociates into H4“ and Ci“. Thus, [EV] t [01“1 m [H01]. pH = -log {in} = —log (50 x 10"“; m) = 3.3 (two significant figures) (c) NaOH 7;- Na+ «+~ OH“ ((1) NaOH is a strong base; dissociation in aqueous solution is essentially compiete, so [Na+] = [OHT] = {NaOH}. 1311 + 13011 i 14 pOH = —iog [011"; pH 3 i4 + log {oar} = 14 + log (“320 X 10—5) = 9.8 (two significant figures) 6. Calculation of pH from Concentration of Strong Acid Calculate the pH of a solution prepared by diluting 3.0 mL of 2.5 M H01 to a finai volume of 100 mL with H20. Answer Because H01 is a strong acid, it dissociates completely to W" + Cl“. Therefore, 30 mL X 2.5 M H01 2 7.5 need of H4“. In 100 mL of solution, this is 0.075 M H'“. pH = —log [IF] = —1og(0.07‘5) m WG-11) = 1.1 (two significant figures) 7. Measurement of Acetylcholine Levels by pH Changes The concentration of acetylcholine (a nemotransnutter) 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 quantitatively converted into choline and acetic acid, which dissociates to yield acetate and a hydrogen ion: ii is in H0 CH3MC~O—CH2-CH2W*IEI~CH3 in“? HO‘CflngHg—Jlwafls + CHy-CHI-vO“ vi“ H+ CH3 CH3 0 Acetylcholine Choline Acetate in a typical analysis, 15 mL of an aqueoes solution containing an unknown amount of acetylcholine had a pH of 7.65. When incubated with acetylcholinesterase, the pH of the solution decreased to 6.87. As- suming that there was no buffer in the assay mixture, determine the number of moies of acetylcholine in the 15 mL sample. Answer Given that pH = -iog [if], we can calculate [PF] at the beginning and at the end of the reaction: At pH 7.05, log [n+1 : -—7.65 [an m 104-65 m 2.24 X 1.0“8 M At pH 6.87,1og[H+l m -6.87 111*} m 10""687 m 1.35 x 10"" M The difference in {11+} is (1.35 e 0.22) x 1.0”" M = 1.13 x 10‘7 M For a volume of 15 mL, or 0.015 L, multiplying volume by moiarity gives (0.015 L)(1.13 x 10"? moi/L) m 1.7 x 10"9 mol of acetylcholine 8. Physical Meaning of Mill Which of the foliong aqueous solutions has the lowest pH: 0.1 M H01; 0.1 M acetic acid {33Ka m 4.86); 0.1 M formic acid (pKa = 8.75)? Answer A 0.1 M H01 solution has the lowest pH because H01 is a strong acid and dissociates completely to 11* + Cl”, yielding the highest [H+1. ChapterZ Water 8-15 9. Simulated Vinegar One way to make vinegar (not the preferred way) is to prepare a soiution of acetic acid, the sole acid component of vinegar, at the proper pH (see Fig. 244) and add appropriate flavoring agents. Acetic acid (M, 60) is a liquid at 25 °C, with a density of 1.049 g/mL. Calcuiate the volume that must be added to distilied water to maize l L of simuiated vinegar (see Fig. 2—15). Answer From Figure 2-35, the pKa of acetic acid is 4.76. From Figure 2—14, the pH of vine— gar is ~3; we will caicuiate for a solution of pH 3.0 Using the Hendersonmi—iasselbalch equation {Am} [HA] and the fact that dissociation of HA gives equimolar {H41 and [A‘] (where HA is CHgCOOH, and A“ is CHSCOO‘), we can write 30 = 4.76 + log (gm/{HA3} -1.76 = 10s (MW/{HAD m -los ([HAJ/[A'D [Hm/gm m 101'76 = 58 Thus, [HA] e saw}. At pH 3.0, {at} m IA"] m 10‘3, so [HA] m 58 x 10—3 M = 0.058 moi/L Dividing density (g/mL) by molecular weight (g/moi) for acetic acid gives 1.049 g/mL 60 g/moi Dividing this answer into 0.058 moi/L gives the volume of acetic acid needed to prepare 1.0 L of a 0.058 M solution: pH = pKa + iog = 0.017 mol/mL 0.058 moi/L 0.017 1110th 2 3‘3 mm 10. Identifying the Conjugate Base Which is the conjugate base in each of the pairs beiow‘? II. (a) accustom" (b) RNH3,RNH§ (c) ngog", stripe4 (d) Hgoogncog Answer In each pair, the acid is the species that gives up a proton; the conjugate base is the deprotonated species. By inspection, the conjugate base is the species with fewer hydrogen atoms. (3) R000" (1)) RNH2 (c) HgPO; ((1) H0057 Calculation of the pH of a Mixture of 3 Weak Acid and Its Conjugate Base Calculate the pH of a dilute soiution that contains a molar ratio of potassium acetate to acetic acid (pita : 4.76) of (a) 2:1; (in) 1:8; (c) 5:1; ((1) 1:1; (e) 1:10. Answer Using the HendersonnHasseibalch equation, A- pH m pKa + tog Emlf pH = 4.76 ~i~ log ([acetate1/[acetic acidD, where [acetate]/{acetic acid; is the ratio given for each part of the question. (a) log {2/1) a 0.30; pH = 4.76 + 0.30 = 5.06 (b) iog(1/8) m —0.48; pa m 4.76 + (—0.48) = 4.28 5-16 Chapter 2 Water (c) log (5/1) = 0.70; pH 2 4.76 + 0.70 = 5.46 (a) log (1/1) = 0; pH : 4.76 (e) log (1/10) a *1.00-, pH = 4.76 + (W100) ——~ 3.76 12. Effect of pH on Solubility The strongly polar hydrogewbonding properties of water make it an excellent solvent for ionic (charged) species. By contrast, nonionized, nonpolar organic molecules, such as benzene, are relatively insoluble in water. In principle, the aqueous solubility of an organic acid or base can be increased by converting the molecules to charged species. For example, the solubility of bemoie acid in water is low. The addition of sodium bicarbonate to a mixture of water and benzoic acid raises the pH and deprotonates the benzoic acid to form benzoate ion, which is quite soluble in water. l? l? : C—MOH : C~O_ Benzoic acid Benzoate ion pKa e 5 Are the following compounds more soluble in an aqueous solution of 0.1 M NaOH or 0.1 M H01? (The dis- socia‘ole proton in (c) is that of the —OH group.) O / l a \ I OH ens/N~$—CH2QOH Ni H C I-Ei 0/ \O—CHS Pyridine ion ,B-Naphthol NaAcetyltyrosine methyl ester 9K8 a 5 9K8 a 10 pKa = 10 (a) (b) (:3) Answer (a) Pyridine is ionic in its protonated form and therefore more soluble at the lower pH, in 0.1 M H01. (b) fi-Naphthol is ionic when dcprotonated and thus more soluble at the higher pH, in 0.1 M NaOH. (c) N—Acetyltyrosine methyl ester is ionic when deprotonated and thus more soluble in 0.1 M NaOH. 13. Tremment of Poison Ivy Rash The components of poison ivy and poison oak that produce the characteristic itchy rash are catechols substituted with long—chain alkyl groups. OH OH (Gish—CH3 PKa e 8 If you were exposed to poison ivy, which of the treatments below would you apply to the affected area? Justify your choice. (3.) Wash the area Mtb cold water. (b) Wash the area with diiute vinegar or iemon juice. Chapter2 Water 3-17 (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 emulsin and dissolve the hydrophobic alkyl group of an alkylcatechol. Given that the pKa of an aikylcatechol is about 8, in a mildly alkaline solution of bicarbonate (NaHCO3) its MOH group ionizes, making the compound much more water-soluble. A neutral or acidic soiution, as in (a) or (b), would not be effective. 14. pH and Drug Absorption Aspirin is a weak acid with a pKEL of 8.5. It is absorbed into the blood through the celis lining the stomach and the small intestine. Absorption requires passage through the plasma membrane, the rate of which is determined by the polarity of the molecule: charged and highly polar molecules pass slowly, Whereas neutral hydrophobic ones pass rapidly. The pH of the stomach contents is about 1.5, and the pH of the contents of the smali intestine is about 6. is more aspirin absorbed into the bloodstream from the stomach or from the small intestine? Clearly justify your choice. Answer With a pKa of 3.5, aspirin is in its protonated (neutral) form at pH below 2.5. At higher pH, it becomes increasingly deprotonated (anionic). Thus, aspirin is better absorbed in the more acidic environment of the stomach. 15. Calculation of pH from Molar Concentrations What is the pH of a solution containing 0.12 moi/L of NH4CI and 0.03 Inth of NaOH (pit; of Nilfif/NHQ, is 9.25)? Answer For the equilibrium W NH3 4“ H+ pH = NS +10aCiNHsl/{NHID we know that [NHZ] + [NH3] = 0.12 moi/L, and that NaOH completely dissociates to give iOH‘] *2 0.08 moi/L. Thus, [NH3E 2 0.03 mol/L and [NEE =2 0.09 molfL, and ' pli = 9.25 + log (003/009) = 9.25 — 0.48 2 8.77, which rounds to 9. 16. Calculation of 131-! after Titration of Weak Acid A compound has a pit; of 7.4. To 100 mL of a 1.0 M solution of this compound at pH 8.0 is added 30 mL of 1.0 M hydrochloric acid. What is the pH of the resulting solution? Answer Begin by calculating the ratio of conjugate base to acid in the starting solution, using the Henderson-Hasselbalch equation: pH m pKa +losiiA‘1/iHAi) 8.0 == 7.4 + log ({A‘I/{HAB iog (MW/[HAD = 0.6 [A“]/[HA] = 100-6 = 4 3-}.8 ChapterR Water The solution contains 100 meq of the compound (conjugate base plus acid), so 80 meg are in the conjugate base form and 20 meq are in the acid form, for a {baseygacim ratio of 4. Because H01 is a strong acid and dissociates completely, adding 30 mL of 1.0 M HCl adds 30 meq of 11+ to the solution. These 30 meg titrate 30 Ineq of the conjugate base, so the {base}/ [acid] ratio is 1. Solving the Henderson-Hasselbaich equation for pH: 13H m pKa + 10% UAW/[HAD m 7.4 + log t 2 7.4 1?. Properties of a Buffer The amino acid glycine is often used as the main ingredient of a buffer in ’ biochemical experiments. The amino group of giycine, which has a pKa of 9.6, can exist either in the protonated form (—Niig‘) or as the free base (WNHQD, because of the reversible equilibrium fit—NH; g R—NH2 a» W” (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 giycine has its amino group in the —NH§ form? (c) How much 5 M KOH must be added to 1.0 L of 0.1 M giycine at pH 9.0 to bring its pH to exactly 10.0? (d) When 99% of the glycine is in its ~—NH§” form, what is the numerical relation between the pH of the soiution 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 pH unit 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 [A7 [HA] pH m pita + log we can write _ W [A 1 9.0 ~ 9.6 + log [HA] W1 -0 a *fi— = 1 ‘ fl . which corresponds to a ratio of 1/4. This indicates that the amino group of glycine is about 1/5 deprotonated and 4/5 protonated at pH 9.0. (c) From (1)) 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 is 50% deprotonated), 30%, or 0.3, of the glycine is titrated. We can now calculate from the Henderson-Hasseibalch equation the percentage protonation at pH 10.0: [Ami 10.0 m 9.6 "1* log EAT m_ 5.4- __ [HM—10 waste/2 Chapter 2 Water 8‘19 This ratio indicates that glycine is 5/7, or 71%, deprotonated at pH 10.0, an additional 21%, or 0.2}, deprotonation above that (50%, or 0.5) at the pKa. Thus, the total fractional deprotonation in moving from pH 9.0 to 10.0 is 0.30 + 021 = 0.51, which corresponds to 0.51 X 0.1 mol = 0.05 mol of KOH Thus, the voiume of 5 M KOH solution required is (0.5 rnol)/{5 moi/L) m 0.01 L, or 10 mL. ((1) From the Henderson-Hasselbaich equation, pH 3 13K; + 10s Ci—NHalfimNHED m pix; + log (001/099) m pa; + (“2) a pKa - 2 In general, any group with an ionizable hydrogen is almost completely protonated at a pH at ieast two pH units below its pKa value. 18. Freparation of a Phosphate Bufl’er What molar ratio of HPOE‘ to HEPOL; in solution would produce a pH of 7 .0? Phosphoric acid ($131304), a triprotic acid, has 3 Mg values: 2.14, 6.86, and 12.4. Hint: Only one of the pKa values is relevant here. Answer Only the pKa close to the pH is relevant here, because the concentrations of the other species (H3PO4 and P02“) are insignificant compared with the concentrations of HPOE“ and H2130; . Begin with the Hendersondlasseibalch equation: pH = pit; + log ([conjogate base]/{acid]) log ([HFO4”]/[H2PO;}) = pa - pKa = 7.0 — 6.86 = 0.14 {Heels/{Harm} m loo-It = 1.38 m 1.4 (two significant figures) 19. Preparation of Standard Bufler for Calibration of a pH Meter The glass electrode used in com- merciai pH meters gives an electrical response proportional to the concentration of hydrogen ion. To con- vert these responses to a pH reading, the electrode must be calibrated against standard solutions of lmovm H4“ concentration. Determine the weight in grams of sodium dihydrogen phosphate (NaH2P04 » H20; FW 138} and disodiurn hydrogen phosphate (Na2HP04; FW 142) needed to prepare 1 L of a standard buffer at pH 7.00 with a total phosphate concentration of 0.100 M (see Fig. 2—15). See Problem £8 for the 13K; values of phosphoric acid. Answer In solution, the two salts ionize as indicated beiow. i? E? m... — + mm. - + /P\ O Na /P\ O Na HO OH HO 0‘ Na." Sodium dihydrogen phosphate Disodium hydrogen phosphate {sodium phosphate, monobasic) (sodium phosphate, dibasic} NaH2P04- H20 NaZHPO4 The buffering capacity of the solution is determined by the concentration ratio of proton acw ceptor (A‘) to proton donor (HA), or {HPOE‘szPOg‘}. From Figure 2—15, the pig for the dissociation of the ionizable hydrogen of dihydrogen phosphate HEPO; : H1304“ + H+ is 6.86. Using the Henderson-Hasselbalch equation, __ {1’1"} _ pH m pKa + log ——w~«[HA] 5-20 ChapterZ Water we calculate: . [N] 7.00 — 6.86 m 10g WM} [Ni __ o, u “(HA1 w 10 14 — 1.38 This ratio is approximately 7/5; that is, 7 parts NaZHP04 to 5 parts 01211121304 - H20. Because {HI-’02? ~l~ [HgPOQ] = 0.100 M, [HgPOQ] : 0.100 M - {HPOE‘}, and we can now calculate the amount of each salt required for a 0.100 M solution: [HPOE'E 0.100 M — [upon 5 1'38 Solving for [HPO§‘], 0.138 2.38 and {HZPOfl 2 0.100 M — 0.058 M = 0.042 M = 0.042 moi/L. The amount needed for 1 L of solution 2 FW (moi/L). For NaH2P04 - H20: (138 g/moi) (0.042 moi/L) = 5.8 g/L For NaZHPOé: (142 g/mol) (0.058 moi/L) = 8.2 g/L [0113031 = M = 0.058 M = 0.058 moi/L 20. Calculation of Molar Ratios of Conjugate Base to Weak Acid from pH For a weak acid with a pit; of 6.0, calculate the ratio of conjugate base to acid at a pH of 5.0. Answer Using the Henderson-Hasseibalch equation, DH 3 pKa 4“ 108 ([Aml/[HAD 5.0 m 6.0 4» tog ([A“}/[HA]) log ([A‘MHAJ) = —1.0 [A'1/[HA1 m 10%0 m 0.10 21. i’reparation of Buffer of Known pH and Strength Given 0.10 M solutions of acetic acid (pita m 4.76) and sodium acetate, describe how you would go about preparing 1.0 L of 0.10 M acetate buffer of pH 4.00. Answer Use the Henderson-Hasseibalch equation to calculate the ratio {Ac"]/[HAC] in the final buffer. pH = pKa + log ([Ac"]/{HAcD log ([Ao“)/[HAo]) = 91-1 ~ pKa m 4.00 m 4.76 = ~0.70 [Ao‘i/[HAo] : 10—076 The fraction of the solution that is Ac" z [Ac'fi/{HAC + Ac’] = 10-0-75/{1 + 10”"0'75) = 0.148, which must be rounded to 0.15 (two significant figures). Therefore, to make 1.0 L of acetate buffer, use 150 mL of sodium acetate and 850 mL of acetic acid. 22. Choice of Weak Acid for a Buffer Which of these compounds wouid be the best buffer at pH 5.0: formic acid (pKa m 3.8), acetic acid (pKa m 4.76), or ethyiamine (pigl m 9.0)? Briefly justify your answer. Answer Acetic acid; its pig is closest to the desired pH. Chapter2 Water 3—21 23. Working with Buffers A buffer 0011th 0.010 mol of lactic acid (pKa = 3.86) and 0.050 mol of sodium lactate per liter. (3.) Calcuiate the pH of the buffer. (1}) Calculate the change in pH when 5 mL of 0.5 M H01 is added to l L of the buffer. (c) What pH change would you expect if you added the same quantity of HG: to 1 L of pure water? Answer Using the Henderson-Hasseibalch equation, [AM] [HA] (3.) pH :2 pk}. + log ([lactate§/[1actic acidD : 3.86 + tog (0.050 M/0.010 M) = 3.86 + 0.70 e: 4.56. Thus, the pH is 4.6. (13) Strong acids ionize compietely, so 0.005 L X 0.5 mol/L = 0.002 moi of H+ is added. The added acid will convert some of the salt form to the acid form. Thus, the finai pH is pH = 3.86 + log {(0.050 M 0.0025)/(0.010 — 00025)] W 3.86 + 0.58 = 4.44 The change in pH = 4.56 — 4.44 2 0.12, which rounds to 0.1 pH unit. (c) H01 completely dissociates. So, when 5 mL of 0.5 M H0} is added to l L of water, an] m (0.002 mol)/(1 L} = 0.002 mol/L = 0.002 M pH = ~log 0.002 m 2.7‘ The pH of pure water is 7.0, so the change in pH = 7.0 — 2.7 2 4.3, which rounds to 4 pH units. pH = pKa + log 24. Use of Molar Concentrations to Calculate 1311 What is the pH of a soiution that contains 0.20 M sodium acetate and 0.60 M acetic acid (pKa w 4.76)? Answer pH = pKa + tog ([baSBJ/{acidD = pKa + 10g ({acetatemacetic acidD = 4.76 + iog (020/060) = 4.76 + (—0.48) m 43 (two significant figures, based on precision of concentrations) 25. Preparation of an Acetate Buffer Calculate the concentrations of acetic acid (pKa e 4.7 0) and sodium acetate necessary to prepare a 0.2 M buffer solution at pH 5.0. Answer First, calculate the required ratio of conjugate base to acid. pH = 10Ka + log ([acetate]/[acetic acidD tog ([acetateViacetic acid]) 3 pH - pKa = 5.0 m 4.76 m 0.24 {acetate]/[acetic acid] : 3.0024 = 1.7 iacetate1/[acetate + acetic acid] m 1.7/2.7 = 0.08 (two significant figures) Thus, 63% of the 0.2 M buffer is acetate and 27% is acetic acid. 00 at pH 5.0 the buffer has 0.13 M acetate and 0.07 M acetic acid. 26. pH of Insect Defensive Secretion You have been observing an insect that defends itself from ene- mies by secreting a caustic liquid. Anaiysis of the liquid shows it to have a total concentration of for» mate plus formic acid (K21 2 1.8 X 10“) of 1.45 M; the concentration of formats ion is 0.015 M. What is the pH of the secretion? 8-22 Chapter 2 Water Answer Solve the Henderson-Hasselbaich equation for pH. pH m pita + log (iconjugate base1/[acidD Given the Ka of formic acid (Ka m 1.8 X 10%), you can calculate pKa as wlog K... = 3.7. If the concentration of formate a formic acid = 1.45 M and the concentration of formate is 0.015 M, then the concentration of formic acid is 1.45 M m 0.015 M ~i— 1.435 M. log ([formate]/[formic acidD fi log (0015/1435) = —2.0 pH = 3.7 — 2.0 = 1.7 (two significant figures) 27. Calculation of pit?a An unknown compound, X, is thought to have a carboxyi group with a pKa of 2.0 and another iomzable group with a pKa between 5 and 8. When 75 mL of 0.1 M NaOH is added to 100 ml. of a 0.1 M solution of X at pH 2.0, the pH increases to 6.72. Calculate the pit}. of the second ionizabie group of X. Answer At the first pH (pl-i = 2), 50% of the carboxyi group is dissociated (pKa 2 pH). Then Amount of NaOH added = 0.075 L X 0.1 moi/L : 0.0075 moi Amount ofX present m 0.1 L X 0.1 moi/L = 0.01 mol At the new pit of 6.72, the carboxyl group is completely dissociated (because pH is much greater than the pKa). The amount of NaOH required to titrate this remaining 50% of the carhoxyi group is 0.5 X 0.01 moi 2 0.005 mol. ‘ Thus, 0.0075 moi — 0.005 mol m 0.0025 moi of NaOH is available to titrate the other group, and, using the Henderson-Hassei‘oalch equation, .. fi 7 pH -— pKa ~+~ iog [HA] we can find the pita of the second ioruzabie group of X: 6.72 m pKa + log [0.0025/(003 — 00025)] pKa = 6.72 — (#048) m 7.20, which rounds to 7. 28. Ionic Forms of Alanine Alanine is a diprotic acid that can undergo two dissociation reactions (see Table 8—1 for pit“a values). (a) Given the structure of the partially protonated form (or zudtterion; see Fig. 3—9} below, draw the chemical structures of the other two forms of alanine that predominate in aqueous solution: the fully protonated form and the fully deprotonated form. 000” + l HaN—CEI—H CH3 Alanine Of the three possible forms of alanine, which would be present at the highest concentration in solutions of the foliong pH: (In) 1.0; (c) 6.2; (d) 8.02; (e) 11.9. Explain your answers in terms of 131-1 reiative to the two pi?a values. Answer (:1) COOH COO” Haiti—(Ii—H Hanwcmu 0H3 C‘IHS Fully protonated Fully deprotonated Chapter 2 Water 8-23 (b) At pH 1.0, 1.3 pli units below the 1319 of the carboxyl group, more than 90% of the carboxyl groups are protonated, and protonated amino groups predominate by a factor of more than 107'. (c) At pH 6.2 the zwitterion predominates. This is 4 pH units above the 1311}, of the carboxyl group, so the vast majority of carboxyl groups are deprotonated. It is 3.5 pH units below the pita of the amino group, so the vast majority of amino groups are protonated. ((1) At pH 8.02 the zwitterion still predominates. The carboxyl groups are deprotonated and, with the pH stiil 1.6 units below the pKa of the arcino group, the vast maiority of amino groups are protonated. (e) At pH 11.9, 2.2 pH units above the pKa of the amino group, the vast majority of airline groups are deprotonated; and the carboxyl groups, at 9.6 pH units above their 131%,, remain deprotonated. 29. Control of Blood pH by Respiration Rate (a) The partial pressure of COB in the lungs can be varied rapidly by the rate and depth of breathing. For example, a common remedy to alleviate hiccups is to increase the concentration of (3‘02 in the iungs. This can be achieved by holding one’s breath, by very slow and shailow breathing (hypoventilation), or by breathing in and out of a paper bag. Under such conditions, 1300; in the air space of the lungs rises above normai. Qualitativeiy explain the effect of these procodures on the blood pH. (b) A common practice of competitive short—distance runners is to breathe rapidly and deeply (hyperventilate) for about haif a minute to remove 002 from their lungs just before the race begins. Blood pH may rise to 7.60. Explain why the blood pH increases. (c) During a short-distance run, the muscles produce a large amount of lactic acid (CligCI-I(OH)COOH, Ka = 1.38 X 1W4) from their glucose stores. in view of this fact, why might hyperventilation before a dash be useful? Answer (a) Biood pH is controlled by the carbon dioxide—bicarbonate buffer system, as shown by the net equation 003 + H20 "mm—"b" H+ + During hypoventilatton, the concentration of (302 in the lungs and arterial blood in- creases, dn‘ving the equilibrium to the right and raising the { "*3; that is, the pH is lowered. (1)) During hypeweniilan'on, the concentration of 002 in the lungs and arterial blood fails. This drives the equilibrium to the left, which requires the consumption of hydrogen ions, reducing {HW and increasing pH. ((3) Lactate is a moderateiy strong acid (pKa m 3.86) that completeiy dissociates under phys- iological conditions: CH3CH(OH)COOH : CHgCHCOHDCOO‘ ~+ H+ This lowers the pH of the blood and muscle tissue. Hyperventiiation is useful because it removes hydrogen ions, raising the pH of the blood and tissues in anticipation of the acid buildup. 30. Calculation of Blood pH from (302 and Bicarbonate Levels Calculate the pH of a blood plasma sample With a total 002 concentration of 26.9 mM and bicarbonate concentration of 25.6 mM. Recall from page 63 that the relevant pKa of carbonic acid is 6.1. 3-24 Chapter 2 Water Answer Use the Henderson-Hasselbalch equation: pil 2 pk; + log ({bicarbonatel/[carbonic acidD If total [C102] = 26.9 M and [bicarbonate] : 25.6 M, then the concentration of carbonic acid is 26.9 M -~ 25.6 M m 1.3 M. pH = 6.1 + log (25.65/13) m 7.4 (two significant figures) 31. Effect of Holding Ono’s Breath on Blood pH The pH of the extracellular fluid is buffered by the bicarbonate/carbonic acid system. Holding your breath can increase the concentration of (302 (g) in the blood. What effect might this have on the pH of the extracellular fluid? Expiain by showing the rete- vant equilibrium equationCs) for this buffer system. Answer Dissoiving more 002 in the blood increases [if] in blood and extraceilular fluids, lowering pH: 002(d) + H20 3 32003 : H+ + H003 Data Analysis Problem 32. “Switchable” Surfactants Hydrophobic molecules do not dissolve well in water. Given that water is a very commoniy used solvent, this makes certain processes very difficuit: washing oily food residue off dishes, cieaning up spilled oil, keeping the oil and water phases of salad dressings well mixed, and carrying out chemical reactions that invoive both hydrophobic and hydrophilic components. Surfactants are a class of amphipathic compounds that includes soaps, detergents, and emulsifiers. With the use of surfactants, hydrophobic compounds can be suspended in aqueous solution by forming micelles (see Fig. 2—7). A micelle has a hydrophobic core consisting of the hydrophobic compound and the hydrophobic “tails” of the surfactant; the hydropiuiic “heads” of the surfactant cover the surface of the miceile. A suspension of miceiles is cailed an emulsion. The more hydrophiiic the head group of the surfactant, the more powerful it ismthat is, the greater its capacity to emulsify hydrophobic material. When you use soap to remove grease from dirty dishes, the soap forms an emulsion with the grease that is easily removed by water through interaction with the hydrophilic head of the soap molecules. Likewise, a. detergent can be used to emulsify spilied oil for removal by water. And emuisifiers in com— mercial salad dressings keep the oil suspended evenly throughout the water—based mixture. There are some situations in which it would be very useful to have a “switchable” surfactant: a mol- ecuie that could be reversibly converted between a surfactant and a nonsurfactant. (a) Imagine such a “switchable” surfactant existed. How would you use it to clean up and then recover the oil from an oil spill? Liu et al. describe a prototypical switchabte surfactant in their 2006 article “Switchable Surfactants.” The switching is based on the following reaction: as r13 R C + co + H 0 MW R C ~ \N/i \lf__CH3 2 2 \ZEI?:§1EIMCHB + H003 CH3 H CH3 Amidine form Amidiniurn form (b) Given that the pKa of a typical ion is 12.4, in which direction (left or right) would you expect the equilibrium of the above reaction to lie? (See Fig. 246 for relevant pKa vaiues.) Jus- tify your answer. Hint: Remember the reaction H20 + 002 : H2003. Liu and colleagues produced a switchable surfactant for which R = CmHas. They do not name the molecule in their articie; for brevity, we‘ll call it s—surf. (c) The anudinium form of s~surf is a powerful surfactant; the amidine form is not. Explain this ob- servation. Chapter2 Water 5-25 Liu and colleagues found that they couid switch between the two forms of smut by changing the gas that they bubbled through a solution of the surfactant. They demonstrated this switch by measur- ing the electrical conductivity of the s~surf soiution; aqueous solutions of ionic compounds have higher conductivity than solutions of nonionic compounds. They started with a solution of the anti» dine form of s~surf in water. Their results are shown below; dotted lines indicate the switch from one gas to another. Gas bubbled in: (102 Ar C02 Ar Electrical conductivity 0 100 ‘ 200 Time (min) (a) In which form is the majority of swsurf at point A? At point B? (e) Why does the electricai conductivity rise from time 0 to point A? (f) Why does the electrical conductivity fall from point A to point 13‘? (g) Explain how you would use s—surf to clean up and recover the oil from an oil spill. Answer (3.) Use the substance in its surfactant form to emulsify the spilled oil, coliect the emulsified oil, then switch to the nonsurfactant form. The oil and water will separate and the oil can be collected for further use. (b) The equilibrium lies strongly to the right. The stronger acid (lower pKa), H2003, donates a proton to the conjugate base of the weaker acid (higher pita), amidine. (c) The strength of a surfactant depends on the hydrophilicity of its head groups: the more hydrophilic, the more powerful the surfactant. The amidiniuni form of s-surf is much more hydrophilic than the anudine form, so it is a more powerful surfactant. (6) Point A: anodjnium; the (302 has had plenty of time to react with the amidine to pro— duce the amidinium form. Point 13: anudine; Ar has removed (302 from the solution, leav- ing the amidine form. (e) The conductivity rises as uncharged amidine reacts with {302 to produce the charged amidiniuru form. (f) The conductivity falls as Ar removes 002, shifting the equilibrium to the uncharged ami- dine form. (g) Treat s«surf with C02 to produce the surfactant anudinirun form and use this to emulsify the spill. rl‘reat the emulsion with Ar to remove the 003 and produce the nonsurfactant amidine from. The oil will separate from the water and can be recovered. Reference Lin, Y., Jessop, P.G., Cunningham, NL, Eckert, C.A., & Liotta, C.L. (2006) Science 313, 9584360. ...
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This note was uploaded on 08/02/2011 for the course BCHM 461 taught by Professor Gerratana during the Spring '11 term at University of Maryland Baltimore.

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