CHM 201 Problems Solution for Chap 8, Fall 2013

270 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA Buffers 15. Only the third beaker represents a buffer solution. A weak acid and its conjugate base must both be present in large quantities in order to have a buffer solution. This is only the case in the third beaker. The first beaker respresents a beaker full of strong acid which is 100% dissociated. The second beaker represents a weak acid solution. In a weak acid solution, only a small fraction of the acid is dissociated. In this representation, 1/10 of the weak acid has dissociated. The only B present in this beaker is from the dissociation of the weak acid. A buffer solution has B added from another source. 16. A buffer solution is a solution containing a weak acid plus its conjugate base or a weak base plus its conjugate acid. Solution c contains a weak acid (HOCl) plus its conjugate base (OCl), so it is a buffer. Solution e is also a buffer solution. It contains a weak base (H2NNH2) plus its conjugate acid (H2NNH3 +). Solution a contains a strong acid (HBr) and a weak acid (HOBr). Solution b contains a strong acid (HClO4) and a strong base (RbOH). Solution d contains a strong base (KOH) and a weak base (HONH2). 17. pH = pKa + log acid] [ [base] ; when [acid] > [base], then acid] [ [base] < 1 and log       acid] [ [base] < 0. From the Henderson-Hasselbalch equation, if the log term is negative, then pH < pKa. When one has more acid than base in a buffer, the pH will be on the acidic side of the pKa value; that is, the pH is at a value lower than the pKa value. When one has more base than acid in a buffer ([conjugate base] > [weak acid]), then the log term in the Henderson-Hasselbalch equation is positive, resulting in pH > pKa. When one has more base than acid in a buffer, the pH is on the basic side of the pKa value; that is, the pH is at a value greater than the pKa value. The other scenario you can run across in a buffer is when [acid] = [base]. Here, the log term is equal to zero, and pH = pKa. 18. NH3 + H2O ⇌ NH4 + + OH Kb = NH ] [ [] OH ] NH [ 3 4   ; taking the log of the Kb expression: log Kb = log[OH]  NH ] [ ] [NH log 3 4  , log[OH] = log Kb + NH ] [ ] [NH log 3 4  pOH = pKb + log NH ] [ ] NH [ 3 4  or pOH = pKb + base] [ log[acid] CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 271 19. A buffer solution is one that resists a change in its pH when either hydroxide ions or protons (H+) are added. Any solution that contains a weak acid and its conjugate base or a weak base and its conjugate acid is classified as a buffer. The pH of a buffer depends on the [base]/[acid] ratio. When H+ is added to a buffer, the weak base component of the buffer reacts with the H+ and forms the acid component of the buffer. Even though the concentrations of the acid and base components of the buffer change some, the ratio of [base]/[acid] does not change that much. This translates into a pH that doesn’t change much. When OH is added to a buffer, the weak acid component is converted into the base com- ponent of the buffer. Again, the [base]/[acid] ratio does not change a lot (unless a large quantity of OH is added), so the pH does not change much. H+(aq) + CO3 2(aq)  HCO3 (aq); OH(aq) + HCO3 (aq)  H2O(l) + CO3 2(aq) 20. When [HA] = [A] (or [BH+] = [B]) for a buffer, the pH of the solution is equal to the pKa value for the acid component of the buffer (pH = pKa because [H+] = Ka). A best buffer has equal concentrations of the acid and base components so it is equally efficient at absorbing H+ and OH. For a pH = 4.00 buffer, we would choose the acid component having a Ka close to 10 .4 00 = 1.0 × 4 10 (pH = pKa for a best buffer). For a pH = 10.00 buffer, we would want the acid component of the buffer to have a Ka close to 1010.00 = 1.0 × 1010 . Of course, we can have a buffer solution made from a weak base and its conjugate acid. For a pH = 10.00 buffer, our conjugate acid should have Ka  1.0 × 1010 , which translates into a Kb value of the base close to 1.0 × 4 10 (Kb = Kw/Ka for conjugate acid-base pairs). The capacity of a buffer is a measure of how much strong acid or strong base the buffer can neutralize. All the buffers listed have the same pH (= pKa = 4.74) because they all have a 1 : 1 concentration ratio between the weak acid and the conjugate base. The 1.0 M buffer has the greatest capacity; the 0.01 M buffer the least capacity. In general, the larger the concen- trations of weak acid and conjugate base, the greater is the buffer capacity, that is, the more strong acid or strong base that can be neutralized with little pH change. 21. a. This is a weak acid problem. Let HC3H5O2 = HOPr and C3H5O2  = OPr. HOPr(aq) ⇌ H+(aq) + OPr(aq) Ka = 1.3 × 105 Initial 0.100 M ~0 0 x mol/L HOPr dissociates to reach equilibrium Change x  +x +x Equil. 0.100  x x x Ka = 1.3 × 105 = .0 100 .0 100 [HOPr] [H ][OPr ] 2 2 x x x      x = [H+] = 1.1 × 103 M; pH = 2.96; assumptions good by the 5% rule. 272 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA b. This is a weak base problem. OPr(aq) + H2O(l) ⇌ HOPr(aq) + OH(aq) Kb = a w K K = 7.7 × 1010 Initial 0.100 M 0 ~0 x mol/L OPr reacts with H2O to reach equilibrium Change x  +x +x Equil. 0.100  x x x Kb = 7.7 × 1010 = .0 100 .0 100 ] [OPr [HOPr][OH ] 2 2 x x x      x = [OH] = 8.8 × 106 M; pOH = 5.06; pH = 8.94; assumptions good. c. Pure H2O, [H+] = [OH] = 1.0 × 107 M; pH = 7.00 d. This solution contains a weak acid and its conjugate base. This is a buffer solution. We will solve for the pH through the weak acid equilibrium reaction. HOPr(aq) ⇌ H+(aq) + OPr(aq) Ka = 1.3 × 105 Initial 0.100 M ~0 0.100 M x mol/L HOPr dissociates to reach equilibrium Change x  +x +x Equil. 0.100  x x 0.100 + x 1.3 × 105 = 100 .0 ) .0 100 () ( 100 .0 ) () ( .0 100 x x x x    = x = [H+] [H+] = 1.3 × 105 M; pH = 4.89; assumptions good. Alternately, we can use the Henderson-Hasselbalch equation to calculate the pH of buffer solutions. pH = pKa + log acid] [ [base] = pKa + log       .0 100 .0 100 = pKa = log(1.3 × 105) = 4.89 The Henderson-Hasselbalch equation will be valid when an assumption of the type 0.1 + x  0.1 that we just made in this problem is valid. From a practical standpoint, this will almost always be true for useful buffer solutions. If the assumption is not valid, the solu- tion will have such a low buffering capacity it will not be of any use to control the pH. Note: The Henderson-Hasselbalch equation can only be used to solve for the pH of buffer solutions. 22. a. We have a weak acid (HOPr = HC3H5O2) and a strong acid (HCl) present. The amount of H+ donated by the weak acid will be negligible. To prove it lets consider the weak acid equilibrium reaction: CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 273 HOPr ⇌ H+ + OPr Ka = 1.3 × 105 Initial 0.100 M 0.020 M 0 x mol/L HOPr dissociates to reach equilibrium Change x  +x +x Equil. 0.100  x 0.020 + x x [H+] = 0.020 + x ≈ 0.020 M; pH = 1.70; assumption good (x = 6.5 × 105 is << 0.020). Note: The H+ contribution from the weak acid HOPr was negligible. The pH of the solution can be determined by only considering the amount of strong acid present. b. Added H+ reacts completely with the best base present, OPr. OPr + H+  HOPr Before 0.100 M 0.020 M 0 Change 0.020 0.020  +0.020 Reacts completely After 0.080 0 0.020 M After reaction, a weak acid, HOPr , and its conjugate base, OPr, are present. This is a buffer solution. Using the Henderson-Hasselbalch equation where pKa = log (1.3 × 105) = 4.89: pH = pKa + log acid] [ [base] = 4.89 + log .0 020) ( ( .0 080) = 5.49; assumptions good. c. This is a strong acid problem. [H+] = 0.020 M; pH = 1.70 d. Added H+ reacts completely with the best base present, OPr. OPr + H+  HOPr Before 0.100 M 0.020 M 0.100 M Change 0.020 0.020  +0.020 Reacts completely After 0.080 0 0.120 A buffer solution results (weak acid + conjugate base). Using the Henderson- Hasselbalch equation: pH = pKa + log acid] [ [base] = 4.89 + log .0 120) ( ( .0 080) = 4.71; assumptions good. 23. a. OH will react completely with the best acid present, HOPr. HOPr + OH  OPr- + H2O Before 0.100 M 0.020 M 0 Change 0.020 0.020  +0.020 Reacts completely After 0.080 0 0.020 274 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA A buffer solution results after the reaction. Using the Henderson-Hasselbalch equation: pH = pKa + log acid] [ [base] = 4.89 + log .0 080) ( ( .0 020) = 4.29; assumptions good. b. We have a weak base and a strong base present at the same time. The amount of OH- added by the weak base will be negligible. To prove it, let’s consider the weak base equilibrium: OPr + H2O ⇌ HOPr + OH Kb = 7.7 × 1010 Initial 0.100 M 0 0.020 M x mol/L OPr reacts with H2O to reach equilibrium Change x  +x +x Equil. 0.100  x x 0.020 + x [OH] = 0.020 + x  0.020 M; pOH = 1.70; pH = 12.30; assumption good. Note: The OH contribution from the weak base OPr was negligible (x = 3.9 × 109 M as compared to 0.020 M OH- from the strong base). The pH can be determined by only considering the amount of strong base present. c. This is a strong base in water. [OH] = 0.020 M; pOH = 1.70; pH = 12.30 d. OH will react completely with HOPr, the best acid present. HOPr + OH  OPr + H2O Before 0.100 M 0.020 M 0.100 M Change 0.020 0.020  +0.020 Reacts completely After 0.080 0 0.120 Using the Henderson-Hasselbalch equation to solve for the pH of the resulting buffer solution: pH = pKa + log acid] [ [base] = 4.89 + log .0 080) ( ( .0 120) = 5.07; assumptions good. 24. Consider all the results to Exercises 21, 22, and 23: Solution Initial pH After Added H+ After Added OH a 2.96 1.70 4.29 b 8.94 5.49 12.30 c 7.00 1.70 12.30 d 4.89 4.71 5.07 The solution in Exercise 21d is a buffer; it contains both a weak acid (HC3H5O2) and a weak base (C3H5O2 ). Solution d shows the greatest resistance to changes in pH when either a strong acid or a strong base is added, which is the primary property of buffers. CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 275 25. Major species: HF, F, K+, and H2O. K+ has no acidic or basic properties. This is a solution containing a weak acid and its conjugate base. This is a buffer solution. One appropriate equilibrium reaction you can use is the Ka reaction of HF, which contains both HF and F. However, you could also use the Kb reaction for F and come up with the same answer. Alternately, you could use the Henderson-Hasselblach equation to solve for the pH. For this problem, we will use the Ka reaction and set up an ICE table to solve for the pH. HF ⇌ F + H+ Initial 0.60 M 1.00 M ~0 x mol/L HF dissociates to reach equilibrium Change x  +x +x Equil. 0.60  x 1.00 + x x Ka = 7.2 × 4 10 = HF] [ [] H ] [F   = x x x   60 .0 )( ) .1( 00  60 .0 .1( 00)( ) x (assuming x << 0.60) x = [H+] = 0.60  (7.2 × 104) = 4.3 × 4 10 M; assumptions good. pH = log(4.3 × 104) = 3.37 26. Major species: HONH2 (Kb = 1.1 × 8 10 ), HONH3 +, Cl, and H2O; Cl has no acidic/basic properties. We have a weak base and its conjugate acid present at the same time in solution. We have a buffer solution. To solve for the pH of a buffer, one can set up an ICE table using the Ka reaction for HONH3 +, or set up an ICE table using the Kb reaction for HONH2, or use the Henderson-Hasselbalch equation. Using the Henderson-Hasselbalch equation: pH = pKa + log ] HONH [ [HONH ] log 10 1.1 10 0.1 log acid] [ base] [ 3 2 8 14               pH = log(9.1 × 107) + log       .0 100 .0 100 = 6.04 + 0.00 = 6.04 Note that pH = pKa for a buffer solution when [weak base] = [conjugate acid]. 27. Major species after NaOH added: HF, F, K+, Na+, OH, and H2O. The OH from the strong base will react with the best acid present (HF). Any reaction involving a strong base is assumed to go to completion. Because all species present are in the same volume of solution, we can use molarity units to do the stoichiometry part of the problem (instead of moles). The stoichiometry problem is: OH + HF  F + H2O Before 0.10 mol/1.00 L 0.60 M 1.00 M Change 0.10 M 0.10 M  +0.10 M Reacts completely After 0 0.50 1.10 After all the OH reacts, we are left with a solution containing a weak acid (HF) and its conjugate base (F). This is what we call a buffer problem. We will solve this buffer problem 276 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA using the Ka equilibrium reaction. One could also use the Kb equilibrium reaction or use the Henderson-Hasselbalch equation to solve for the pH. HF ⇌ F + H+ Initial 0.50 M 1.10 M ~0 x mol/L HF dissociates to reach equilibrium Change x  +x +x Equil. 0.50  x 1.10 + x x Ka = 7.2 × 4 10 = x x x   50 .0 )( ) .1( 10  50 .0 .1( 10)( ) x , x = [H+] = 3.3 × 4 10 M; pH = 3.48; assumptions good. Note: The added NaOH to this buffer solution changes the pH only from 3.37 to 3.48. If the NaOH were added to 1.0 L of pure water, the pH would change from 7.00 to 13.00. Major species after HCl added: HF, F, H+, K+, Cl, and H2O; the added H+ from the strong acid will react completely with the best base present (F). H+ + F  HF Before 00 L .1 .0 20 mol 1.00 M 0.60 M Change 0.20 M 0.20 M  +0.20 M Reacts completely After 0 0.80 0.80 After all the H+ has reacted, we have a buffer solution (a solution containing a weak acid and its conjugate base). Solving the buffer problem: HF ⇌ F + H+ Initial 0.80 M 0.80 M 0 Equil. 0.80  x 0.80 + x x Ka = 7.2 × 4 10 = x x x   80 .0 )( ) ( .0 80  80 .0 ( .0 80)( ) x , x = [H+] = 7.2 × 4 10 M; pH = 3.14; assumptions good. Note: The added HCl to this buffer solution changes the pH only from 3.37 to 3.14. If the HCl were added to 1.0 L of pure water, the pH would change from 7.00 to 0.70. 28. Major species: H2O, Cl, Na+, HONH2, HONH3 +, and OH; the added strong base dominates the initial reaction mixture. Let the OH react completely with the best acid present (HONH3 +). HONH3 + + OH  HONH2 + H2O Before 0.100 M 0.020 M 0.100 M Change 0.020 0.020  +0.020 Reacts completely After 0.080 0 0.120 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 277 A buffer solution results. Using the Henderson-Hasselbalch equation: pH = pKa + log                     8 14 HONH ,b w a 10 1.1 10 0.1 log K K log pK acid ;] [ base ] [ 2 = 6.04 pH = 6.04 + log ] HONH [ HONH ] [ 3 2  = 6.04 + log .0 080) ( ( .0 120) = 6.04 + 0.18 = 6.22 Major species: H2O, Cl, HONH2, HONH3 +, and H+; the added strong acid domi- nates the initial reaction mixture. Let the H+ react completely with HONH2, the best base present. HONH2 + H+  HONH3 + Before 0.100 M 0.020 M 0.100 M Change 0.020 0.020  +0.020 Reacts completely After 0.080 0 0.120 A buffer solution results after reaction. Using the Henderson-Hasselbalch equation: pH = 6.04 + log ] HONH [ HONH ] [ 3 2  = 6.04 + log .0 120) ( ( .0 080) = 6.04  0.18 = 5.86 29. C5H5NH+ ⇌ H+ + C5H5N Ka = 9 14 b w 10 7.1 10 0.1 K K      = 5.9 × 106 pKa = log(5.9 × 106) = 5.23 We will use the Henderson-Hasselbalch equation to calculate the concentration ratio necessary for each buffer. pH = pKa + log acid] [ [base] , pH = 5.23 + log C H NH ] [ C H N] [ 5 5 5 5  a. 4.50 = 5.23 + log C H NH ] [ C H N] [ 5 5 5 5  b. 5.00 = 5.23 + log C H NH ] [ C H N] [ 5 5 5 5  log C H NH ] [ C H N] [ 5 5 5 5  = 0.73 log C H NH ] [ C H N] [ 5 5 5 5  = 0.23 C H NH ] [ C H N] [ 5 5 5 5  = 100.73 = 0.19 C H NH ] [ C H N] [ 5 5 5 5  = 100.23 = 0.59 278 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA c. 5.23 = 5.23 + log C H NH ] [ C H N] [ 5 5 5 5  d. 5.50 = 5.23 + log C H NH ] [ C H N] [ 5 5 5 5  C H NH ] [ C H N] [ 5 5 5 5  = 100.0 = 1.0 C H NH ] [ C H N] [ 5 5 5 5  = 100.27 = 1.9 30. pH = pKa + log HNO ] [ ] NO [ 2 2  , 3.55 = log(4.0 × 104) + log HNO ] [ ] NO [ 2 2  3.55 = 3.40 + log HNO ] [ ] NO [ 2 2  , HNO ] [ ] NO [ 2 2  = 100.15 = 1.4 Let x = volume (L) HNO2 solution needed, then 1.00  x = volume of NaNO2 solution needed to form this buffer solution. HNO ] [ ] NO [ 2 2  = 1.4 = L 50 molHNO .0 L .0 50 molNaNO ) 00 .1( 2 2    x x = x x .050) ( ( .0 50) .0 50  (0.70)x = 0.50  (0.50)x , (1.20)x = 0.50, x = 0.42 L We need 0.42 L of 0.50 M HNO2 and 1.00  0.42 = 0.58 L of 0.50 M NaNO2 to form a pH = 3.55 buffer solution. 31. Ka for H2NNH3 + = Kw/Kb, H2NNH2 = 1.0 × 1014/3.0 × 106 = 3.3 × 109 pH = pKa + ] H NNH [ [H NNH ] log 3 2 2 2  = log(3.3 × 109) +       .0 80 .0 40 log = 8.48 + (0.30) = 8.18 pH = pKa for a buffer when [acid] = [base]. Here, the acid (H2NNH3 +) concentration needs to decrease, while the base (H2NNH2) concentration needs to increase in order for [H2NNH3 +] = [H2NNH2]. Both of these changes are accomplished by adding a strong base (like NaOH) to the original buffer. The added OH from the strong base converts the acid component of the buffer into the conjugate base. Here, the reaction is H2NNH3 + + OH  H2NNH2 + H2O. Because a strong base is reacting, the reaction is assumed to go to completion. The following set-up determines the number of moles of OH(x) that must be added so that mol H2NNH3 + = mol H2NNH2 . When mol acid = mol base in a buffer, then [acid] = [base] and pH = pKa. H2NNH3 + + OH  H2NNH2 + H2O Before 1.0 L × 0.80 mol/L x 1.0 L × 0.40 mol/L Change x x  +x Reacts completely After 0.80  x 0 0.40 + x We want mol H2NNH3 + = mol H2NNH2. So: 0.80  x = 0.40 + x, 2x = 0.40, x = 0.20 mol OH CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 279 When 0.20 mol OH is added to the initial buffer, mol H2NNH3 + is decreased to 0.60 mol, while mol H2NNH2 is increased to 0.60 mol. Therefore, 0.20 mol of NaOH must be added to the initial buffer solution in order to produce a solution where pH = pKa. 32. [H+] added = 2500L .0 .0 010mol = 0.040 M; the added H+ reacts completely with NH3 to form NH4 +. a. NH3 + H+  NH4 + Before 0.050 M 0.040 M 0.15 M Change 0.040 0.040  +0.040 Reacts completely After 0.010 0 0.19 A buffer solution still exists after H+ reacts completely. Using the Henderson- Hasselbalch equation: pH = pKa + log ] NH [ NH ] [ 4 3  = log(5.6 × 1010) + log       .0 19 .0 010 = 9.25 + (1.28) = 7.97 b. NH3 + H+  NH4 + Before 0.50 M 0.040 M 1.50 M Change 0.040 0.040  +0.040 Reacts completely After 0.46 0 1.54 A buffer solution still exists. pH = pKa + log ] NH [ NH ] [ 4 3  , 9.25 + log       .1 54 .0 46 = 8.73 Note: The two buffers differ in their capacity and not their initial pH (both buffers had an initial pH = 8.77). Solution b has the greatest capacity since it has the largest concentrations of weak acid and conjugate base. Buffers with greater capacities will be able to absorb more added H+ or OH. 33. [HC7H5O2] = 2000L .0 .12g 122 1molHC H O 21 5. g HC H O 2 5 7 2 5 7  = 0.880 M [C7H5O2 ] = 2000L .0 NaC H O mol molC H O 1 .10g 144 1molNaC H O 7. g NaC H O 37 2 5 7 2 5 7 2 5 7 2 5 7    = 1.31 M We have a buffer solution since we have both a weak acid and its conjugate base present at the same time. One can use the Ka reaction or the Kb reaction to solve. We will use the Ka reaction for the acid component of the buffer. HC7H5O2 ⇌ H+ + C7H5O2  Initial 0.880 M ~0 1.31 M x mol/L of HC7H5O2 dissociates to reach equilibrium Change x  +x +x Equil. 0.880 – x x 1.31 + x 280 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA Ka = 6.4 × 5 10 = x x x   880 .0 ) .1( 31  880 .0 31) .1(x , x = [H+] = 4.3 × 5 10 M pH = log(4.3 × 5 10 ) = 4.37; assumptions good. Alternatively, we can use the Henderson-Hasselbalch equation to calculate the pH of buffer solutions. pH = pKa + log acid] [ [base] = pKa + log HC H O ] [ ] C H O [ 2 5 7 2 5 7  pH = log(6.4 × 5 10 ) + log       .0 880 .1 31 = 4.19 + 0.173 = 4.36 Within round-off error, this is the same answer we calculated solving the equilibrium problem using the Ka reaction. The Henderson-Hasselbalch equation will be valid when an assumption of the type 1.31 + x ≈ 1.31 that we just made in this problem is valid. From a practical standpoint, this will almost always be true for useful buffer solutions. If the assumption is not valid, the solution will have such a low buffering capacity that it will be of no use to control the pH. Note: The Henderson-Hasselbalch equation can only be used to solve for the pH of buffer solutions. 34. pH = pKa + HOCl] [ [OCl ] log  = log(3.5 × 108) +       .0 20 .0 90 log = 7.46 + 0.65 = 8.11 pH = pKa when [HOCl] = [OCl] (or when mol HOCl = mol OCl). Here, the moles of the base component of the buffer must decrease, while the moles of the acid component of the buffer must increase in order to achieve a solution where pH = pKa. Both of these changes occur when a strong acid (like HCl) is added. Let x = mol H+ added from the strong acid HCl. H+ + OCl  HOCl Before x 1.0 L × 0.90 mol/L 1.0 L × 0.20 mol/L Change x x  +x Reacts completely After 0 0.90  x 0.20 + x We want mol HOCl = mol OCl. Therefore: 0.90  x = 0.20 + x, 2x = 0.70, x = 0.35 mol H+ When 0.35 mol H+ is added, mol OCl is decreased to 0.55 mol, while the mol HOCl is increased to 0.55 mol Therefore, 0.35 mol of HCl must be added to the original buffer solution in order to produce a solution where pH = pKa. 35. pH = pKa + log HC H O ] [ ] C H O [ 2 3 2 2 3 2  ; pKa = log(1.8 × 105) = 4.74 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 281 Because the buffer components, C2H3O2  and HC2H3O2, are both in the same volume of water, the concentration ratio of [C2H3O2 ]/[HC2H3O2] will equal the mole ratio of mol C2H3O2 /mol HC2H3O2. 5.00 = 4.74 + log 2 3 2 2 3 2 HC H O mol molC H O  ; mol HC2H3O2 = 0.5000 L × L .0 200mol = 0.100 mol 0.26 = log 100mol .0 molC H O , 100mol .0 molC H O 2 3 2 2 3 2   = 100.26 = 1.8, mol C2H3O2  = 0.18 mol Mass NaC2H3O2 = 0.18 mol NaC2H3O2 × mol 82.03g = 15 g NaC2H3O2 36. Added OH converts HC2H3O2 into C2H3O2 : HC2H3O2 + OH  C2H3O2  + H2O From this reaction, the moles of C2H3O2  produced equal the moles of OH added. Also, the total concentration of acetic acid plus acetate ion must equal 2.0 M (assuming no volume change on addition of NaOH). Summarizing for each solution: [C2H3O2 ] + [HC2H3O] = 2.0 M and [C2H3O2 ] produced = [OH] added a. pH = pKa + log HC H O ] [ ] C H O [ 2 3 2 2 3 2  ; for pH = pKa, log HC H O ] [ ] C H O [ 2 3 2 2 3 2  = 0 Therefore, HC H O ] [ ] C H O [ 2 3 2 2 3 2  = 1.0 and [C2H3O2 ] = [HC2H3O2]. Because [C2H3O2 ] + [HC2H3O2] = 2.0 M: [C2H3O2 ] = [HC2H3O2] = 1.0 M = [OH] added To produce a 1.0 M C2H3O2  solution, we need to add 1.0 mol of NaOH to 1.0 L of the 2.0 M HC2H3O2 solution. The resulting solution will have pH = pKa = 4.74. b. 4.00 = 4.74 + log HC H O ] [ ] C H O [ 2 3 2 2 3 2  , HC H O ] [ ] C H O [ 2 3 2 2 3 2  = 100.74 = 0.18 [C2H3O2 ] = 0.18[HC2H3O2] or [HC2H3O2] = 5.6[C2H3O2 ] Because [C2H3O2 ] + [HC2H3O2] = 2.0 M: [C2H3O2 ] + 5.6[C2H3O2 ] = 2.0 M, [C2H3O2 ] = 6.6 0.2 = 0.30 M = [OH] added We need to add 0.30 mol of NaOH to 1.0 L of 2.0 M HC2H3O2 solution to produce 0.30 M C2H3O2 . The resulting solution will have pH = 4.00. c. 5.00 = 4.74 + log HC H O ] [ ] C H O [ 2 3 2 2 3 2  , HC H O ] [ ] C H O [ 2 3 2 2 3 2  = 100.26 = 1.8 282 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 1.8[HC2H3O2] = [C2H3O2 ] or [HC2H3O2] = 0.56[C2H3O2 ] 1.56[C2H3O2 ] = 2.0 M, [C2H3O2 ] = 1.3 M = [OH] added We need to add 1.3 mol of NaOH to 1.0 L of 2.0 M HC2H3O2 to produce a solution with pH = 5.00. 37. When H+ is added, it converts C2H3O2  into HC2H3O2: C2H3O2  + H+ → HC2H3O2. From this reaction, the moles of HC2H3O2 produced must equal the moles of H+ added and the total concentration of acetate ion + acetic acid must equal 1.0 M (assuming no volume change). Summarizing for each solution: [C2H3O2 ] + [HC2H3O2] = 1.0 M and [HC2H3O2] = [H+] added a. pH = pKa + log HC H O ] [ ] C H O [ 2 3 2 2 3 2  ; for pH = pKa, [C2H3O2 ] = [HC2H3O2]. For this to be true, [C2H3O2 ] = [HC2H3O2] = 0.50 M = [H+] added, which means that 0.50 mol of HCl must be added to 1.0 L of the initial solution to produce a solution with pH = pKa. b. 4.20 = 4.74 + log HC H O ] [ ] [C H O ,] HC H O [ ] C H O [ 2 3 2 2 3 2 2 3 2 2 3 2   = 10 .0 54 = 0.29 [C2H3O2 ] = 0.29[HC2H3O2]; 0.29[HC2H3O2] + [HC2H3O2] = 1.0 M [HC2H3O2] = 0.78 M = [H+] added 0.78 mol of HCl must be added to produce a solution with pH = 4.20. c. 5.00 = 4.74 + log HC H O ] [ ] [C H O ,] HC H O [ ] C H O [ 2 3 2 2 3 2 2 3 2 2 3 2   = 100.26 = 1.8 [C2H3O2 ] = 1.8[HC2H3O2]; 1.8[HC2H3O2] + [HC2H3O2] = 1.0 M [HC2H3O2] = 0.36 M = [H+] added 0.36 mol of HCl must be added to produce a solution with pH = 5.00. 38. 50.0 g NH4Cl × .49g NH Cl 53 molNH Cl 1 4 4 = 0.935 mol NH4Cl added to 1.00 L; [NH4 +] = 0.935 M Using the Henderson Hasselbalch equation to solve for the pH of this buffer solution: pH = pKa + log ] NH [ NH ] [ 4 3  = log(5.6 × 1010) + log       .0 935 .0 75 = 9.25  0.096 = 9.15 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 283 39. a. pKb for C6H5NH2 = log(3.8 × 1010) = 9.42; pKa for C6H5NH3 + = 14.00  9.42 = 4.58 pH = pKa + log ] C H NH [ C H NH ] [ 3 5 6 2 5 6  , 4.20 = 4.58 + log ] C H NH [ 50 .0 3 5 6  M 0.38 = log ] C H NH [ 50 .0 3 5 6  M , [C6H5NH3 +] = [C6H5NH3Cl] = 1.2 M b. 4.0 g NaOH × NaOH mol molOH 1 .00g 40 1molNaOH   = 0.10 mol OH; [OH] = L 0.1 .0 10 mol = 0.10 M C6H5NH3 + + OH  C6H5NH2 + H2O Before 1.2 M 0.10 M 0.50 M Change 0.10 0.10  +0.10 After 1.1 0 0.60 A buffer solution exists. pH = 4.58 + log       1.1 .0 60 = 4.32 40. pH = pKa + log HC H O ] [ ] C H O [ 2 3 2 2 3 2  , 4.00 = log(1.8 × 5 10 ) + log HC H O ] [ ] C H O [ 2 3 2 2 3 2  HC H O ] [ ] C H O [ 2 3 2 2 3 2  = 0.18; this is also equal to the mole ratio between C2H3O2  and HC2H3O2. Let x = volume of 1.00 M HC2H3O2 and y = volume of 1.00 M NaC2H3O2 x + y = 1.00 L, x = 1.00 – y x(1.00 mol/L) = mol HC2H3O2; y(1.00 mol/L) = mol NaC2H3O2 = mol C2H3O2  Thus: x y = 0.18 or y y .1 00  = 0.18; solving: y = 0.15 L, so x = 1.00  0.15 = 0.85 L. We need 850 mL of 1.00 M HC2H3O2 and 150 mL of 1.00 M NaC2H3O2 to produce a buffer solution at pH = 4.00. 41. a. pH = pKa + log acid] [ [base] , 7.15 = log(6.2 × 8 10 ) + log ] H PO [ ] HPO [ 4 2 2 4   7.15 = 7.21 + log ] H PO [ ] HPO [ 4 2 2 4   , ] H PO [ ] HPO [ 4 2 2 4   = 10 .0 06 = 0.9, ] HPO [ ] H PO [ 2 4 4 2   9.0 1  = 1.1 ≈ 1 b. A best buffer has approximately equal concentrations of weak acid and conjugate base, so pH ≈ pKa for a best buffer. The pKa value for a H3PO4/H2PO4  buffer is log(7.5 × 10 3)  = 2.12. A pH of 7.15 is too high for a H3PO4/H2PO4  buffer to be effective. At this high of pH, there would be so little H3PO4 present that we could hardly consider it a buffer; this solution would not be effective in resisting pH changes, especially when a strong base is added. 284 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 42. pH = pKa + H CO ] [ ] [HCO log 3 2 3  , 7.40 = log(4.3 × 107) + 0012 .0 ] [HCO log 3  0012 .0 ] [HCO log 3  = 7.40 – 6.37 = 1.03, 0012 .0 ] [HCO 3  = 101.03, [HCO3 ] = 1.3 × 102 M 43. At pH = 7.40: 7.40 = log(4.3 × 107) + H CO ] [ ] [HCO log 3 2 3  H CO ] [ ] [HCO log 3 2 3  = 7.40 – 6.37 = 1.03, H CO ] [ ] HCO [ 3 2 3  = 101.03, ] HCO [ H CO ] [ 3 3 2  = 101.03 = 0.093 At pH = 7.35: H CO ] [ ] [HCO log 3 2 3  = 7.35 – 6.37 = 0.98, H CO ] [ ] HCO [ 3 2 3  = 100.98 ] HCO [ H CO ] [ 3 3 2  = 100.98 = 0.10 The [H2CO3] : [HCO3 ] concentration ratio must increase from 0.093 to 0.10 in order for the onset of acidosis to occur. 44. Because we have added two solutions together, the concentration of each reagent has changed. What hasn’t changed is the moles or millimoles of each reagent. Let’s determine the millimoles of each reagent present by multiplying the volume in milliters by the molarity in units of mmol/mL. 100.0 mL × 0.100 M = 10.0 mmol NaF; 100.0 mL × 0.025 M = 2.5 mmol HCl H+ + F  HF; 2.5 mmol H+ converts 2.5 mmol F into 2.5 mmol HF. After the reaction, a buffer solution results containing 2.5 mmol HF and (10.0  2.5 =) 7.5 mmol F in 200.0 mL of solution. pH = pKa + log HF] [ [F ]  = 3.14 + log       mmol/ 2000. mL 5.2 mmol/ 2000. mL 5.7 = 3.62; assumptions good. 45. A best buffer has large and equal quantities of weak acid and conjugate base. Because [acid] = [base] for a best buffer, pH = pKa + log acid] [ [base] = pKa + 0 = pKa (pH  pKa for a best buffer). The best acid choice for a pH = 7.00 buffer would be the weak acid with a pKa close to 7.0 or Ka ≈ 1 × 107. HOCl is the best choice in Table 7.2 (Ka = 3.5 × 108; pKa = 7.46). To make this buffer, we need to calculate the [base]/[acid] ratio. 7.00 = 7.46 + log HOCl] [ [OCl ] acid ,] [ [base]  = 100.46 = 0.35 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 285 Any OCl/HOCl buffer in a concentration ratio of 0.35 : 1 will have a pH = 7.00. One possibility is [NaOCl] = 0.35 M and [HOCl] = 1.0 M. 46. For a pH = 5.00 buffer, we want an acid with a pKa close to 5.00. For a conjugate acid-base pair, 14.00 = pKa + pKb. So for a pH = 5.00 buffer, we want the base to have a pKb close to (14.0  5.0 =) 9.0 or a Kb close to 1 × 109. The best choice in Table 7.3 is pyridine (C5H5N) with Kb = 1.7 × 109. pH = pKa + log 9 14 b w a 10 7.1 10 0.1 K K K acid ;] [ base] [       = 5.9 × 106 5.00 = log(5.9 × 10-6) + log C H NH ] [ [C H N] acid ,] [ base] [ 5 5 5 5  = 100.23 = 0.59 There are many possibilities to make this buffer. One possibility is a solution of [C5H5N] = 0.59 M and [C5H5NHCl] = 1.0 M. The pH of this solution will be 5.00 because the base to acid concentration ratio is 0.59 : 1. 47. a. No; a solution of a strong acid (HNO3) and its conjugate base (NO3 ) is not ge