Answers to Practice Test Questions 12 rganic Acids and Bases 1. (a) (b) pppp aa = llllll(kk aa ) KK aa = 10 pppp aa KK aa = 10 4.20 KK aa = 6.3 10 5 (c) Benzoic acid is a weak acid. We know this because the pka of benzoic acid is larger than 0. Since it is a weak acid, benzoic acid will not fully dissociate in water. 2. The pppp aa value describes the acidity of a species. Stronger acids have lower (or more negative) pppp aa values than weaker acids. Since the pppp aa value for ethanamide is 20 lower than the pppp aa value for ethanamine, ethanamide is 10 20 times more acidic than ethanamine. Since the strength of an acid depends on the stability of its conjugate base, compare the conjugate bases of both species. The conjugate base of ethanamine is: 3 C C N The conjugate base of ethanamide is: 3 C C N 3 C C N As shown by these two resonance structures, the negative charge in the conjugate base of ethanamide is delocalized over (shared by) two atoms the oxygen and the nitrogen. Delocalization of charge over multiple atoms provides stability compared to concentrating the whole charge on one atom (which is what happens in the conjugate base of ethanamine). Furthermore, oxygen is more electronegative than nitrogen so there is slightly more negative charge on oxygen than there is on nitrogen. Shifting charge onto more electronegative atoms is also favourable.
3. The reaction described verbally is: 3 C C C 2 S 4 S 4 3 C (a) (b) The acids (i.e. proton donors) in this reaction are 2 SSSS 4 and the protonated carboxylic acid (example shown is CC 3 CCCC 2 2 ). Because we are told that this reaction is product-favoured, the acid on the reactant side must be stronger. So, 2 SSSS 4 is a stronger acid than the protonated carboxylic acid. base(!) acid conjugate acid of C 3 C 2 conjugate base of 2 S 4 3 C C 3 C C 3 C C These three resonance structures show that the positive charge of this acid is delocalized over (shared by) three atoms two oxygen atoms and one carbon atom. 4. (a) CCCC 3 CCCC 2 2 NN 3 CC 2 NN and 3 CC are stronger bases than (the conjugate base of water), so they cannot exist in aqueous solution. Instead, they react with water to give and either NNNN 3 or CCCC 4 (their *very weak* conjugate acids). 2 SS is a stronger acid than 2 so its conjugate base ( ) is a weaker base than. As such, can exist in water. The same logic explains why CCCC 3 CCCC 2 can exist in water. (b) CCCC 3 CCCC 2 NNNN 4 is a stronger acid than 3 (the conjugate acid of water), so it cannot exist in aqueous solution. Instead, it reacts with water to give a solvated proton ( (aaaa) or 3 (aaaa) ) and BBBB (aaaa). is a weaker acid than 3 (due to the high charge density of FF ) so it only partially dissociates (reacts with water) and therefore can exist in water.
5. The strength of an acid is determined based on the stability of its conjugate base. Acids with weaker conjugate bases (i.e. more stable conjugate bases) give up more easily so they are stronger acids. The conjugate base of is FF which is a much smaller ion than CCCC, BBBB or II. As such, the negative charge is spread over a smaller surface area so FF has a higher charge density than the other halide anions. This makes FF less stable than the other halide anions and therefore makes a weaker acid than the other hydrogen halides. 6. Note that the curves cross at the pppp aa values. The pppp aa of an acid is equal to the pppp at which it exists in a 1 : 1 ratio (or 50/50 mixture) with its conjugate base. At one pppp unit below the pppp aa value, there is 10 times as much acid as conjugate base and, at one pppp unit above the pppp aa value, there is 10 times as much conjugate base as acid.
7. Consider the dominant form of each species (alcohol and sulfate ester) at p ~7. This is best illustrated by looking at distribution curves. In both cases, one species dominates at p ~7. Distribution curve for an alcohol with pppp aa = 1111 Distribution curve for a sulfate ester with pppp aa = 33
The pppp aa of the alcohol is significantly higher than 7, so it exists almost entirely in protonated form at p ~7. As such, it exists as a neutral species at physiological p. The pppp aa of the sulfate ester is significantly lower than 7, so it exists almost entirely in deprotonated form at p ~7. As such, it exists as a charged species at physiological p. Charged species tend to be more soluble in water than neutral species because the strongest intermolecular forces are ion-dipole attractions. These tend to be stronger than dipole-dipole attractions (the strongest possible intermolecular forces between neutral species such as an alcohol and water) and increase the solubility of the drug in polar solvents like water. This trick is primarily used for drugs with large nonpolar regions. Dipole-dipole attractions would tend to be significant enough to dissolve a small alcohol in water. 8. (aaaa) (aaaa) AA (aaaa) I (M) 0.25 10 7 0 C (M) xx xx xx E (M) 0.25 xx 0.25 10 7 xx xx xx Step 1: Write a balanced chemical equation for the reaction see above Step 2: rganize all known information see above ICE tables are useful tools to solve equilibrium problems; they can be constructed using any property that is directly proportional to moles for all species in this case, concentration at constant volume In pure water at 2222, the concentration of is 1111 77 MM. Since the concentration of ascorbic acid is significantly higher than 1111 77 MM, it is reasonable to assume that this initial small amount of will be substantially smaller than the amount of generated by dissociation of the acid. Check this assumption at the end of the calculation! The initial concentration of A is relatively high and it is a weak acid. It is therefore reasonable to assume that the majority of A will remain in protonated form (in other words, that x will be substantially smaller than 0.25). Check this assumption at the end of the calculation! Step 3: Write equilibrium constant expression KK aa = aa (aa AA ) aa Step 4: Calculate KK aa from pppp aa ppkk aa = llllllkk aa KK aa = 10 pppp aa KK aa = 10 4.17 KK aa = 6.8 10 5
Step 5: Calculate x KK aa = aa (aa AA ) aa 6.8 10 5 = (xx)(xx) 0.25 (6.8 10 5 )(0.25) = xx 2 xx = (6.8 10 5 )(0.25) xx = 0.0041 Step 6: Check assumptions. If invalid, repeat calculation without the invalid assumption. Assumption #1: 10 7 xx xx 10 7 0.0041 = 0.0041 assumption is okay Assumption #2: 0.25 xx 0.25 0.25 0.0041 = 0.2459 = 0.25 assumption is okay Step 7: Calculate p from aa aa = xx = 0.0041 pppp = llllll(aa ) pppp = llllll(0.0041) pppp = 2.39 Step 8: Check your work Ascorbic acid is a weak acid, so expect a p below 7. If this had been a strong acid, the p would have been llllll(0.25) = 0.6. Since it is a weak acid, it makes sense that the p is higher than that. 9. BB (aaaa) 2 (ll) BBBB (aaaa) (aaaa) I (M) 0.047 0 10 7 C (M) xx xx xx E (M) 0.047 xx xx 10 7 xx xx Step 1: Write a balanced chemical equation for the reaction see above Step 2: rganize all known information see above ICE tables are useful tools to solve equilibrium problems; they can be constructed using any property that is directly proportional to moles for all species in this case, concentration at constant volume In pure water at 25, the concentration of is 10 7 MM. Since the concentration of piperidine is significantly higher than 10 7 MM, it is reasonable to assume that this initial small amount of will be substantially smaller than the amount of generated by protonation of the base. Check this assumption at the end of the calculation! You might initially want to assume that x will be significantly smaller than 0.047. If you make that assumption, you will find that x is approximately 0.007, so this assumption fails. At that point, you may either use the method of successive approximations or follow the approach below.
Step 3: Write equilibrium constant expression KK bb = aa BBBB (aa ) aa BB Step 4: Calculate KK bb from pppp bb ppkk bb = llllllkk bb KK bb = 10 pppp bb KK bb = 10 2.89 KK bb = 1.3 10 3 Step 5: Calculate x KK bb = aa BBBB (aa ) aa BB 1.3 10 3 = (xx)(xx) 0.047xx (1.3 10 3 )(0.047 xx) = xx 2 xx 2 (1.3 10 3 )xx (6.0 10 5 ) = 0 xx = 0.0072 Step 6: Check assumptions. If invalid, repeat calculation without the invalid assumption. Assumption #1: 10 7 xx xx 10 7 0.0072 = 0.0072 assumption is okay Step 7: Calculate p from aa ne approach is to calculate p from aa then calculate p from p. Alternatively, calculate aa from aa then calculate p from aa. Either approach gives the same p. aa = xx = 0.0072 pppppp = llllll(aa ) pppppp = llllll(0.0072) pppppp = 2.14 pppp = 14 pppppp = 14 2.14 = 11.86 or = (aa )(aa ) aa = = 1.00 1014 = 1.4 10 12 aa 0.0072 pppp = llllll(aa ) pppp = llllll(1.4 10 12 ) pppp = 11.86 Step 8: Check your work Piperidine is a weak base, so expect a p above 7. If this had been a strong base, the p would have been 14 llllll(0.047) = 12.67. Since it is a weak base, it makes sense that the p is lower than that.
10. (a) CC 3 4 3 (b) ketone and carboxylic acid (c) A 10-6 M solution of a strong acid would have a p of approximately 6. A 10-6 M solution of a weak acid would therefore have a p between 6 and 7 (depending on what fraction of the weak acid had been deprotonated). The distribution curve for pyruvic acid (ppkk aa = 2.50) looks like this: (d) This distribution curve shows that, at p 6 (which is 3.5 units higher than the pppp aa of pyruvic acid), pyruvic acid exists almost entirely in deprotonated form. In other words, it exists almost entirely as its conjugate base. Since the pyruvic acid is almost completely dissociated, the equilibrium between water, and is what governs p: 2 (ll) (aaaa) (aaaa) The dissociation of pyruvic acid into and conjugate base disrupts this equilibrium. As a result, some of the produced reacts with already in solution to re-establish equilibrium. In other words, produced reacts with until = 1.00 10 14 again.
(e) Step 1: Calculate the amount of generated when the pyruvic acid dissociates fully See parts (c) and (d) for why pyruvic acid dissociates fully when it is at very low concentrations. This is stoichiometry (not an equilibrium ICE table) since the reaction goes to completion. CC 3 4 3(aaaa) (aaaa) CC 3 3 3(aaaa) Minitial 8.0 10 7 1.0 10 7 0 Mchange 88. 00 1111 77 88. 00 1111 77 88. 00 1111 77 Mfinal 0 99. 00 1111 77 88. 00 1111 77 Step 2: Set up ICE table for the 22 / / equilibrium after the acid fully dissociates 2 (ll) (aaaa) (aaaa) I (M) nn/aa 9.0 10 7 1.0 10 7 C (M) xx xx xx E (M) nn/aa 99. 00 1111 77 xx 11. 00 1111 77 xx You could also write x in the water column and x in the other two columns. If you do that, you ll get a negative value for x. As long as you factor in that negative sign properly, the calculation will work perfectly well. It is very important that you use the correct initial concentrations of and and that you do NT approximate them as 0. When the concentration of acid is very small, the initial amounts of and become significant. Step 3: Write equilibrium constant expression = aa (aa ) = (aa aa )(aa ) since aa 2 = 1 2 Step 4: Calculate x = (aa )(aa ) 1.00 10 14 = (9.0 10 7 xx)(1.0 10 7 xx) 1.00 10 14 = (9.0 10 14 ) (10.0 10 7 )xx xx 2 0 = (8.0 10 14 ) (1.00 10 6 )xx xx 2 or xx 2 (1.00 10 6 )xx 8.0 10 14 = 0 Use quadratic equation (or other tools) to solve, giving: xx = 8.8 10 8 (xx = 9.1 10 7 would give negative concentrations for and which is impossible) Step 5: Calculate p from aa aa = 9.0 10 7 xx = 9.0 10 7 8.8 10 8 = 8.1 10 7 pppp = llllll(aa ) pppp = llllll(8.1 10 7 ) pppp = 6.09 Step 6: Check your work Based on the logic explained in part (c), the expected p was between 6 and 7 closer to 6 because the acid fully dissociates at this low concentration. 6.09 meets this description perfectly, so it is a reasonable answer.
(f) (g) This is another two temperatures; two equilibrium constants question requiring the use of llll KK 2 = rr 1 1. KK 1 RR TT 1 TT 2 Step 1: Write a balanced chemical equation for the reaction CC 3 4 3(aaaa) (aaaa) CC 3 3 3(aaaa) Step 2: Calculate KK aa from pppp aa at 2222 ppkk aa = llllllkk aa KK aa = 10 pppp aa KK aa = 10 2.50 KK aa = 3.2 10 3 Step 3: Match equilibrium constants to the corresponding temperature TT 1 = 25 = 298.15 KK KK 1 = 3.2 10 3 TT 2 = 37 = 310.15 KK KK 2 =??? Step 4: Calculate the standard enthalpy change for the reaction (phase change) rr = ff (pppppppppppppppp) ff (rreeeeeeeeeeeeeeee) rr = ff (aaaa) ff CC 3 3 3(aaaa) ff CC 3 4 3(aaaa) rr = 0 kkkk kkkk kkkk 596.22 607.52 mmmmmm mmmmmm mmmmmm rr = 11.3 kkkk = 1.13 104 JJ mmmmmm mmmmmm Step 5: Crunch the numbers llll KK 2 = rr 1 1 KK 1 RR TT 1 TT 2 KK llll 2 3.2 10 3 = 1.13 104 JJ mmmmmm 1 JJ 1 298.15 KK 310.15 KK llll KK 2 3.2 10 3 = 0.1764 8.314 462 mmmmmm KK KK 2 3.2 103 = ee0.1764 KK 2 = (3.2 10 3 )(ee 0.1764 ) KK 2 = 3.8 10 3 Step 6: Calculate pppp aa from KK aa at 3333 ppkk aa = llllllkk aa ppkk aa = llllll(3.8 10 3 ) ppkk aa = 2.42 Step 8: Check your work The reaction is endothermic so it makes sense that K would be slightly larger at the higher temperature. Since the temperature difference between 25 and 37 is not larger, it makes sense that the two pppp aa values are similar. Since the p is significantly higher than the pppp aa of pyruvic acid, we expect most of the acid to have been deprotonated to give the pyruvate ion at at 37 C and p 7.
11. (a) (b) butanoic acid (higher pka) (weaker acid) pka = 4.9 The reaction is reactant-favoured at 25. The pppp aa values tell us the relative strength of the two acids. Methoxyethanoic acid has a lower pppp aa value, so it is a stronger acid. By definition, the strength of a conjugate base increases as the strength of its conjugate acid decreases. So, the conjugate base of the weaker acid (butanoic acid) is the stronger base. If equal amounts of the four species were combined, we would expect the stronger acid and stronger base to react more than the weaker acid and weaker base. As such, the net direction of reaction would be toward the reactants until equilibrium was reached. At equilibrium, there will be higher concentrations of butanoic acid and methoxyethanoate than of butanoate and methoxyethanoic acid. The pppp aa for butanoic acid can be used to calculate its KK aa value which is the equilibrium constant for dissociation of the acid into and its conjugate base. This equilibrium constant can be used to calculate the standard free energy change for that reaction and therefore the standard free energy of formation of the conjugate base. Step 1: Write a balanced chemical equation for the reaction CCCC 3 CCCC 2 CCCC 2 CCCC 2 (aaaa) (aaaa) CCCC 3 CCCC 2 CCCC 2 CCCC 2(aaaa) Step 2: Convert temperature to Kelvin (if necessary) TT = 25 = 298.15 KK Step 3: Calculate KK aa from pppp aa ppkk aa = llllllkk aa KK aa = 10 pppp aa KK aa = 10 4.9 KK aa = 1 10 5 Step 4: Calculate the standard free energy change from the equilibrium constant rr GG = RRRRRRRRRR JJ rr GG = 8.314 462 (298.15 KK)llll(1 mmmmmm KK 105 ) rr GG = 2.80 10 4 JJ 1 kkkk rr GG = 28.0 kkkk mmmmmm methoxyethanoate (conjugate base of stronger acid) (weaker base) mmmmmm 1000 JJ butanoate (conjugate base of weaker acid) (stronger base) pka = 3.7 methoxyethanoic acid (lower pka) (stronger acid)
Step 5: Calculate the standard free energy of formation for CCCC 33 CCCC 22 CCCC 22 CCCC 22(aaaa) rr GG = ff GG (pppppppppppppppp) ff GG (rrrrrrrrrrrrrrrrrr) rr GG = ff GG (aaaa) ff GG CCCC 3 CCCC 2 CCCC 2 CCCC 2(aaaa) ff GG CCCC 3 CCCC 2 CCCC 2 CCCC 2 (aaaa) ff GG CCCC 3 CCCC 2 CCCC 2 CCCC 2(aaaa) = rr GG ff GG (aaaa) ff GG CCCC 3 CCCC 2 CCCC 2 CCCC 2 (aaaa) ff GG CCCC 3 CCCC 2 CCCC 2 CCCC 2(aaaa) = 28.0 kkkk kkkk kkkk 0 399.57 mmmmmm mmmmmm mmmmmm ff GG CCCC 3 CCCC 2 CCCC 2 CCCC 2(aaaa) = 371.6 kkkk mmmmmm Step 6: Check your work 12. When acid is added to NN 3, NNNN 4 is generated. The pppp of the solution is determined by the relative concentrations of these two species and by the KK aa value of the conjugate acid (NNNN 4 ) since KK aa = aa aa cccccccc.bbbbbbbb. aa cccccccc.aaaaaaaa Taking the logarithm of both sides of this equation, multiplying by -1 and rearranging to isolate pppp gives = pppp aa llllll aa cccccccc.bbbbbbbb. This is known as the enderson-asselbalch equation. aa cccccccc.aaaaaaaa This question can be answered using either of the two equations above. In either case, start by using the pppp bb of NNNN 3 to calculate the KK aa or pppp aa of its conjugate acid, NNNN 4. Then calculate the desired ratio of NNNN 3 to NNNN 4. Finally, use stoichiometry to calculate the amount of Cl necessary to generate that ratio. Step 1: Calculate KK aa or pppp aa of NNNN 44 = KK aa (cccccccccccccccccc aaaaaaaa) KK bb (cccccccccccccccccc bbbbbbbb) KK aa (cccccccccccccccccc aaaaaaaa) = KK aa (NNNN 4 ) = KK bb (NNNN 3 ) KK bb (cccccccccccccccccc bbbbssee) @ 25, = 1 10 14 KK bb (NNNN 3 ) = 10 pppp bb (NNNN 3 ) = 10 4.74 = 1.8 10 5 KK aa (NNNN 4 ) = 1 1014 1.8 10 5 KK aa (NNNN 4 ) = 5.5 10 10 and pppp aa (NNNN 4 ) = llllll[kk aa (NNNN 4 )] = llllll(5.5 10 10 ) = 9.26 At 25, you can use a shortcut that pppp aa (cccccccccccccccccc aaaaaaaa) and pppp bb (cccccccccccccccccc bbbbbbbb) must add up to 14. This does not work at any other temperature. They always add up to pppp ww, but pppp ww varies with temperature. Step 2: Calculate desired ratio of NNNN 44 to NNNN 33 to give p 9.26 Note that the desired pppp is equal to the pppp aa of the conjugate acid. We know that pppp = pppp aa when there is a 1 : 1 ratio of conjugate acid and conjugate base. So, in this case, we must have equal amounts of NNNN 3 and NNNN 4.
If you forgot that (or if you were doing a different question where the pppp and pppp aa values didn t match) the following approach also works: pppp = llllll(aa ) aa = 10 pppp = 10 9.26 = 5.5 10 10 KK aa = aa aa cccccccc.bbbbbbbb aa cccccccc.aaaaaaaa = aa aa NNNN 3 aa NNNN4 aa NNNN 3 aa NNNN4 aa NNNN 3 aa NNNN4 = KK aa aa = 5.5 1010 5.5 1010 = 1.0 Step 3: Use stoichiometry to calculate the amount of Cl required to convert enough NNNN 33 into NNNN 44 to give a 1 : 1 ratio Initially, nn NNNN3 = 0.016 mmmmmm LL 0.5000LL = 0.0080 mmmmmm Since all of the NNNN 4 must have been made from NNNN 3, that means that the final amounts of NNNN 4 and NNNN 3 must add up to 0.0080 mol. Combine this with the fact that the two amounts must be equal and you find that you must have 0.0040 mol of each. NNNN 3(aaaa) (aaaa) ninitial 8.0 10 3 4.0 10 3 0 NNNN 4(aaaa) nchange 44. 00 1111 33 44. 00 1111 33 44. 00 1111 33 nfinal 44. 00 1111 33 00 44. 00 1111 33 Step 4: Calculate the volume of 0.025 M Cl required to provide 44. 00 1111 33 mmmmmm VV = 4.0 103 mmmmmm = 0.16 LL 0.025 mmmmmm LL Step 5: Check your work The concentrations of Cl and NNNN 3(aaaa) are similar, so it makes sense that it would take approximately half as much Cl to convert half the NNNN 3 into NNNN 4. 13. (a) (b) Since the reaction is favoured in the forward direction, 2 2 must be a stronger acid than water (and a stronger base than 2 ). This implies that 2 (the conjugate base of 2 2 ) must be more stable than (the conjugate base of 2 ). This is likely because of the electronegative oxygen atom bonded to the oxygen bearing the negative charge in 2. The second oxygen atom will pull electron density away from the, stabilizing it by lessening the charge density on that negative oxygen atom. This is an example of an inductive effect.
(c) (d) Step 1: Add the equation for deprotonation of 22 22 to the reverse of the equation for dissociation of water into and Those two equations add up to give the equation in the question. Therefore, the equilibrium constants for those two equations multiply to give the equilibrium constant for the equation in the question. Recall that, if you reverse the direction in which the reaction is written, the equilibrium constant is inverted. So, if the equilibrium constant for water dissociating is then the equilibrium constant for it reforming is 1 2 2(aaaa) (aaaa) (aaaa) (aaaa) 2 2(aaaa) (aaaa). 2(aaaa) KK 1 = KK aa ( 2 2 ) 2 (ll) KK 2 = 1 2 (ll) 2(aaaa) KK oooooooooooooo = KK aa ( 2 2 ) We were given a value for K, and is a constant at 25, so we can solve for KK aa ( 2 2 ). KK oooooooooooooo = KK 1 KK 2 = KK aa ( 2 2 ) 1 = KK aa ( 2 2 ) KK aa ( 2 2 ) = KK oooooooooooooo = (200)(1 10 14 ) = 2 10 12 Step 2: Use KK aa to calculate pppp aa ppkk aa = llllllkk aa ppkk aa = llllll(2 10 12 ) = 11.7 Step 3: Check your work The pppp aa of water is 14. It seems reasonable that the pppp aa of 2 2 would be similar to the pppp aa of water. Step 1: Calculate KK bb of conjugate base from KK aa of conjugate acid = KK aa (cccccccccccccccccc aaaaaaaa) KK bb (ccccccccuugggggggg bbbbbbbb) @ 25, = 1 10 14 KK bb (cccccccccccccccccc bbbbbbbb) = KK bb ( 2 ) = KK aa ( 2 2 ) KK aa (cccccccccccccccccc aaaaaaaa) KK bb ( 2 ) = 1 1014 2 10 12 KK bb ( 2 ) = 5 10 3 Step 2: Use KK bb to calculate pppp bb ppkk bb = llllllkk bb ppkk bb = llllll(5 10 3 ) = 2.3 Step 3: Check your work At 25, we expect the pppp aa of an acid and the pppp bb of its conjugate base to add up to 14. (This could have been used as an alternative approach to calculating pppp bb. Then calculate KK bb from pppp bb.)
14. The following answers were accepted. This list may not be exhaustive. The forward reaction is favoured by entropy since more moles of gas are produced than consumed. The forward reaction is favoured by enthalpy because a very strong bond is formed. The bond is so strong because is a very small atom so the internuclear distance is short. is a more stable anion than because is more electronegative than and therefore better able to sustain a negative charge. is a more stable anion than because is larger than therefore the charge density of is lower than that of. is a stronger base than the conjugate base of water ( ) so it cannot exist in water. We know this because water is a stronger acid than the conjugate acid of ( 2 ).