solut08 - CHAPTER EIGHT APPLICAT IONS OF AQUEO US EQU...

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194 CHAPTER EIGHT APPLICATIONS OF AQUEOUS EQUILIBRIA Buffers 15. A buffered solution must contain both a weak acid and a weak base. Most buffered solutions are prepared using a weak acid plus the conjugate base of the weak acid (which is a weak base). Buffered solutions are useful for controlling the pH of a solution since they resist pH change. 16. When strong acid or strong base is added to a sodium bicarbonate/sodium carbonate buffer mixture, the strong acid/base is neutralized. The reaction goes to completion resulting in the strong acid/base being replaced with a weak acid/base. This results in a new buffer solution. The reactions are: H + (aq) + CO 3 2- (aq) ÷ HCO 3 - (aq); OH - (aq) + HCO 3 - (aq) ÷ CO 3 2- (aq) + H 2 O(l) 17. 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 (= pK a = 4.74) since 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 concentrations of weak acid and conjugate base, the greater the buffer capacity, i.e., the greater the ability to neutralize added strong acid or strong base. 18. NH 3 + H 2 O NH 4 + + OH - K b = ; Taking the -log of the K b expression: - log K b = - log [OH - ] - log , - log [OH - ] = - log K b + log pOH = pK b + log or pOH = pK b + log 19. a. This is a weak acid problem. Let HC 3 H 5 O 2 = HOPr and C 3 H 5 O 2 - = OPr - . HOPr(aq) H + (aq) + OPr - (aq) K a = 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
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CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 195 K a = 1.3 × 10 -5 = x = [H + ] = 1.1 × 10 -3 M ; pH = 2.96 Assumptions good by the 5% rule. b. This is a weak base problem. OPr - (aq) + H 2 O(l) HOPr(aq) + OH - (aq) K b = = 7.7 × 10 -10 Initial 0.100 M 0 ~0 x mol/L OPr - reacts with H 2 O to reach equilibrium Change - x + x + x Equil. 0.100 - x x x K b = 7.7 × 10 -10 = x = [OH - ] = 8.8 × 10 -6 M ; pOH = 5.06; pH = 8.94 Assumptions good. c. pure H 2 O, [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) K a = 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 = = 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 = pK a + log = pK a + log = pK a = -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 solution 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.
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196 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 20. a. We have a weak acid (HOPr = HC 3 H 5 O 2 ) and a strong acid (HCl) present. The amount of H + donated by the weak acid will be negligible.
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