ࡱ> Root Entry Fπ{[@Data 1TableWordDocumentF  !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~SummaryInformation( DocumentSummaryInformation8CompObjX0TableRoot Entry Fe\@Data 1TableWordDocumentF  !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~agraph Font!B      N!YF#$%12'(wx !345NOQR     fg13VW!"uate pKa values. The pKa of C6H5CO2H (benzoic acid) is 4.19, and the pKa of H3PO4 (phosphoric aicd) is 2.1. Benzoic acid is a weaker acid than phosphoric acid, so the equilibrium favors benzoic acid and thus lies to the right. f. This molecule is not in Table 2.1 so estimates must be made. The closest match to the ammonium end of glycine is CH3NH3+ (methylammonium ion, pKa 10.64). The closest match to the carboxylic acid end of glycine is CH3CO2H (acetic acid, pKa 4.76). Methylammonium ion is a weaker acid than acetic acid, so the ionic form of glycine is favored and the equilibrium lies to the left. g. To analyze this case, we must use the pKa of CH3CH2OH2+, not CH3CH2OH (ethanol), because CH3CH2OH is serving as a base (proton acceptor), not as an acid in this reaction. The closest match on Table 2.1 to the structure of CH3CH2OH2+ is H3O+ (hydronium ion, pKa -1.74). Because the two acids are very close in pKa, this equilibrium will not significantly favor either side. --link to exercises-- Molecular Structure and Acidity In each case, we remove the circled proton and then consider the ability of the conjugate base to donate electrons to a proton. Because stronger conjugate bases come from weaker acids, the acidity ranking will be reverse of the basicity ranking. a. The conjugate bases are:  There is a difference in electronegativity between the three atoms that would share electrons. Chlorine is most electronegative (3.0), followed by sulfur (2.5) and phosphorus (2.1). Chloride ion will have the greatest resistance to sharing an electron pair and thus be the poorest base. The phosphide ion, (CH3)2P-, will have the least resistance to sharing an electron pair, and thus be the strongest base. Order of basicity: Cl- (weakest base) < CH3S- < (CH3)2P- (strongest base). Order of acidity: HCl (strongest acid) > CH3SH > (CH3)2PH. b. The corresponding conjugate bases are:  The conjugate base of CH3CH2NH2 (ethylamine) has a single resonance structure, whereas the conjugate base of the amide has two resonance structures. The presence of resonance stabilizes the conjugate base, making it a weaker base. Because a weaker conjugate base is derived from a stronger acid, we conclude that the amide is a stronger acid than the amine. c. The corresponding conjugate bases are:  The conjugate base of cyclohexanol (C6H11OH) has a single resonance structure, whereas the conjugate base of phenol (C6H5OH) has four resonance structures. Because the conjugate base of phenol has more resonance forms, phenol is more acidic than cyclohexanol. d. The corresponding conjugate bases are ammonia (NH3) and water. Resonance is not an issue in this case. Ammonia would accept a proton by sharing the nitrogen lone pair. Water would accept a proton by sharing an oxygen lone pair. Nitrogen is less electronegative than oxygen and would thus be more inclined to share the lone pair. Thus ammonia is a stronger base than water, so the ammonium ion (NH4+) is a weaker acid than hydronium ion (H3O+). e. All three protons are removed from the same functional group, a carboxylic acid. The conjugate bases all have two resonance structures. Since each conjugate base would take a proton by sharing electrons from an oxygen atom. The conjugate bases are therefore almost identical. This is because they are derived from exactly the same functional group! The only difference between these three acids is the halogen atom (fluorine, bromine, or chlorine). Because these atoms are not directly involved in sharing electrons with a proton, we consider their inductive effects. (Any time you are considering the acidity of identical functional groups, you can skip immediately to inductive effects for this same reason.) The conjugate base of a carboxylic acid is a carboxylate ion (X = F, Cl, or Br), XCH2CO2-. If the X group is more electronegative than carbon, it pulls electron density from the rest of the molecule toward itself. This decreases the electron density and charge on the oxygen atoms that would share a lone pair with a proton. 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Hardinger.nameless:Temporary Items:Word Work File A 2206Department of Chemistry8Chipmunk:Temporary Items:AutoRecovery save of acids and Department of Chemistry=portable:tutorials:acids and bases :acids and bases solutionsDepartment of Chemistry.Chipmunk:Temporary Items:Word Work File A 2014Department of Chemistry=portable:tutorials:acids and bases :acids and bases solutionsDepartment of Chemistry=portable:tutorials:acids and bases :acids and bases solutionsDepartment of Chemistry=portable:tutorials:acids and bases :acids and bases solutionsDepartment of Chemistry=portable:tutorials:acids and bases :acids and bases solutions@,5A`Zh^ b hj89PTltuwxUu % ' D H!X!^!!!!!!!!!!@B@DD\ @D @"D @>D @ZD @6 6"6$6&6(?@(6462?@66:67788888(9*9.90929::>:N:P:T:V:::::@@: ;; ;$;&;l; <L<&=0=x=z====>>2??@@@@@@\B`B@@@@GTimes New Roman5Symbol3 Arial3Times"h0MASA&  ,!20d"Acids and Bases Exercise SolutionsSteven A. HardingerDepartment of Chemistry FMicrosoft Word DocumentNB6WWord.Document.8 ՜.+,D՜.+,T hp  'UCLAan, b #Acids and Bases Exercise Solutions Title 6> _PID_GUID'AN{1C9DD980-C759-11D3-99AD-000502C397F1} Oh+'0 , H T `lt|'#Acids and Bases Exercise SolutionsscidSteven A. HardingertevNormalADepartment of Chemistry12aMicrosoft Word 8.0s@@%[@P[s Lh jbjb Fkk!]4JJJJf$JO O O ,EO K O O O E gF O 66O  $ {JJY Acids and Bases Exercise Solutions Vocabulary Acid ionization constant: For the equilibrium HA + H20 <===> A- + H3O+, the acid ionization constant defined as is:  Bronsted-Lowry acid: A molecule or ion that donates a proton. Bronsted-Lowry base: A molecule or ion that accepts a proton. Conjugate acid: The structure that results when a Bronsted-Lowry base accepts a proton. Conjugate base: The structure that results when a Bronsted-Lowry acid donates a proton. Inductive effect: The effect of an atom or group of atoms on the electron density in a part of the molecule which is separated by at least two single bonds. Lewis acid: A molecule or ion that accepts a pair of electrons. Lewis base: A molecule or ion that shares a pair of electrons. pKa: The negative logarithm of the Ka. link to exercises Conjugate Acid and Bases The base uses an electron pair to make the new bond with the proton that is transferred. The pair of electrons that used to be the bond to the transferred proton of the acid becomes sole property of the atom to which the proton was bonded. Be sure to consider formal charges as well. Because the atom that used to be attached to the proton gains a pair of electrons that it used to share its charge becomes one unit more negative. The atom of the base that accepts the proton now shares a pair of electrons that formerly were not shared, so its charge becomes one unit more positive.    e. In the first step, water serves as a base to remove a proton from the ammonium end of glycine, forming an amine and a hydronium ion. The carboxylate ion end of glycine then removes a proton from the hydronium ion. The proton that is shuttled in this manner is circled.  --link to exercises-- Use of the pKa Table a. The circled hydrogen is part of an amine functional group. Table 2.1 reveals that NH3 (ammonia; also an amine) has a pKa of 38. Thus we estimate the pKa of H2NCH3 (methylamine) to be approximately 38. Note that while ammonium ions may be somewhat similar to amines, the be bDO`B^ h UnknownDepartment of Chemistry?B$>",KԺGuG"B$ji%zՑ"B$GrB_Ϯ`="@0(   B S  ?!!=K!% m loPU !!!Steven A. Hardinger1nameless:Desktop Folder:acids and bases solutionsSteven A. Hardinger.nameless:Temporary Items:Word Work File A 2206Steven A. Hardinger.nameless:Temporary Items:Word Work File A 2206Steven A. Hardinger.nameless:Temporary Items:Word Work File A 2206Department of Chemistry8Chipmunk:Temporary Items:AutoRecovery save of acids and Department of Chemistry=portable:tutorials:acids and bases :acids and bases solutionsDepartment of Chemistry.Chipmunk:Temporary Items:Word Work File A 2014Department of Chemistry=portable:tutorials:acids and bases :acids and bases solutionsDepartment of Chemistry=portable:tutorials:acids and bases :acids and bases solutionsDepartment of Chemistry=portable:tutorials:acids and bases :acids and bases solutions@,5A`Zh^ b hj89PTltuwxUu % ' D H!X!^!!!!!!!!!!@B@DD\ @D @"D @>D @ZD @6 6"6$6 &6(?@(6462?@66:67788888(9*9.90929::>:N:P:T:V:::::@@: ;; ;$;&;l; <L<&=0=x=z====>>2??@@@@@@\B`B@@@@GTimes New Roman5Symbol3 Arial3Times"h0MApSA& ] ,!0d!"Acids and Bases Exercise SolutionsSteven A. HardingerDepartment of Chemistryst match for methylamine is ammonia, not either of the ammonium ions on Table 2.1 b. The circled hydrogen is part of an alcohol functional group. The only other alcohol on Table 2.1 is CH3CH2OH (ethanol), pKa 15.9. Thus we estimate the pKa of (CH3)3COH (tert-butyl alcohol) as 16. Acetic acid (CH3CO2H) also has an OH group, but the functional group of this carboxylic acid (CO2H) is significantly different from an alcohol functional group (OH). c. The circled hydrogen is part of a phenol (OH attached to a benzene ring) functional group. The only other compound in Table 2.1 with this functional group is phenol, pKa 9.95. Thus we estimate the pKa of our more complex phenol as approximately 10. d. Identify the functional groups, and then estimate their pKa values from Table 2.1:  From this analysis, we conclude that the carboxylic acid proton has the lowest pKa, and so it is the most acidic proton in this molecule. e. Recall that the equilibrium favors the weakest acid/base pair. Use Table 2.1 to evalSummaryInformation( DocumentSummaryInformation8CompObjX0TableR3epartment of Chemistry=portable:tutorials:acids and bases :acids and bases solutionsDepartment of Chemistry=portable:tutorials:acids and bases :acids and bases solutionsDepartment of Chemistry=portable:tutorials:acids and bases :acids and bases solutions@,5A`Zh^ b hj89PTltuwxUu % ' D H!X!^!!!!!!!!!!@B@DD\ @D @"D @>D @ZD @6 6"6$6&6(?@(6462?@66:67788888(9*9.90929::>:N:P:T:V:::::@@: ;; ;$;&;l; <L<&=0=x=z====>>2??@@@@@@\B`B@@@@GTimes New Roman5Symbol3 Arial3Times"h0MApSA& ] ,!0d!"Acids and Bases Exercise SolutionsSteven A. HardingerDepartment of Chemistry [(@(NormalCJmH <A@<Default Par FMicrosoft Word DocumentNB6WWord.Document.8 ՜.+,D՜.+,T hp  'UCLAan, b #Acids and Bases Exercise Solutions Title 6> _PID_GUID'AN{1C9DD980-C759-11D3-99AD-000502C397F1} Oh+'0 , H T `lt|'#Acids and Bases Exercise SolutionsscidSteven A. 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Conjugate acid: The structure that results when a Bronsted-Lowry base accepts a proton. Conjugate base: The structure that results when a Bronsted-Lowry acid donates a proton. Inductive effect: The effect of an atom or group of atoms on the electron density in a part of the molecule which is separated by at least two single bonds. Lewis acid: A molecule or ion that accepts a pair of electrons. Lewis base: A molecule or ion that shares a pair of electrons. pKa: The negative logarithm of the Ka. link to exercises Conjugate Acid and Bases The base uses an electron pair to make the new bond with the proton that is transferred. The pair of electrons that used to be the bond to the transferred proton of the acid becomes sole property of the atom to which the proton was bonded. Be sure to consider formal charges as well. Because the atom that used to be attached to the proton gains a pair of electrons that it used to share its charge becomes one unit more negative. The atom of the base that accepts the proton now shares a pair of electrons that formerly were not shared, so its charge becomes one unit more positive.    e. In the first step, water serves as a base to remove a proton from the ammonium end of glycine, forming an amine and a hydronium ion. The carboxylate ion end of glycine then removes a proton from the hydronium ion. The proton that is shuttled in this manner is circled.  --link to exercises-- Use of the pKa Table a. The circled hydrogen is part of an amine functional group. Table 2.1 reveals that NH3 (ammonia; also an amine) has a pKa of 38. Thus we estimate the pKa of H2NCH3 (methylamine) to be approximately 38. Note that while ammonium ions may be somewhat similar to amines, the best match for methylamine is ammonia, not either of the ammonium ions on Table 2.1 b. The circled hydrogen is part of an alcohol functional group. The only other alcohol on Table 2.1 is CH3CH2OH (ethanol), pKa 15.9. Thus we estimate the pKa of (CH3)3COH (tert-butyl alcohol) as 16. Acetic acid (CH3CO2H) also has an OH group, but the functional group of this carboxylic acid (CO2H) is significantly different from an alcohol functional group (OH). c. The circled hydrogen is part of a phenol (OH attached to a benzene ring) functional group. The only other compound in Table 2.1 with this functional group is phenol, pKa 9.95. Thus we estimate the pKa of our more complex phenol as approximately 10. d. Identify the functional groups, and then estimate their pKa values from Table 2.1:  From this analysis, we conclude that the carboxylic acid proton has the lowest pKa, and so it is the most acidic proton in this molecule. e. Recall that the equilibrium favors the weakest acid/base pair. Use Table 2.1 to evalst match for methylamine is ammonia, not either of the ammonium ions on Table 2.1 b. The circled hydrogen is part of an alcohol functional group. The only other alcohol on Table 2.1 is CH3CH2OH (ethanol), pKa 15.9. Thus we estimate the pKa of (CH3)3COH (tert-butyl alcohol) as 16. Acetic acid (CH3CO2H) also has an OH group, but the functional group of this carboxylic acid (CO2H) is significantly different from an alcohol functional group (OH). c. The circled hydrogen is part of a phenol (OH attached to a benzene ring) functional group. The only other compound in Table 2.1 with this functional group is phenol, pKa 9.95. Thus we estimate the pKa of our more complex phenol as approximately 10. d. Identify the functional groups, and then estimate their pKa values from Table 2.1:  From this analysis, we conclude that the carboxylic acid proton has the lowest pKa, and so it is the most acidic proton in this molecule. e. Recall that the equilibrium favors the weakest acid/base pair. Use Table 2.1 to evalBCDFWX` a b c h "6$688888(9*9.90929N:P:T:V:::::$;&;x=z=@@@BbDjo CJOJQJUmH jU jnUH*jCJOJQJUmHH*?^ _ ` b d e f g h 6"6&6(6^B`B$^ _ ` b d e f g h / =!"#$%  effect The magnitude of this electron density shift depends on the electronegativity of X. The greater the electronegativity, the more electron density is removed from the carboxylate group, and the lower the basicity becomes. Of the three X groups in this case, fluorine is the most electronegative, so FCH2CO2- is the weakest base. Bromine is the least electronegative, so BrCH2CO2- will be the strongest base. The order of acidity is reverse of the order of conjugate base strength, so the relative acidity ranking is: FCH2CO2H (strongest acid) > ClCH2CO2H > BrCH2CO2H (weakest acid).f. The conjugate bases are water (HOH) and CH3OH (methanol). In both cases, the functional group that would share an electron pair with a proton is a HO group. Because these are the same functional group, we need only consider inductive effects. The difference between these structures is a hydrogen atom for water versus a methyl group (CH3) for methanol. Neither is very electronegative, so we expect then to have little if any effect on the electron density at oxygen. (Alkyl groups such as methyl groups are weak electron donors as we will see in Chapter 6). We conclude there is little difference in the electron density of the oxygen atom in water and methanol. Therefore they are of similar basicity, and by extension, the conjugate acids will have similar pKa values./ =!"#$% constant: For the equilibrium --link to exercises--