chapter16_part1

# chapter16_part1 - 722 Chapter 16 Acids and Bases...

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Chapter 16: Acids and Bases 722 End-of-Chapter Solutions for Chapter 16 Summary Problem Answer: (a) L form (b) pK a = 3.91 (c) see equation below (d) L form: 2.40; D form: 2.41 (e) see equation below (f) 2.41 (g) 43.0 mL (h) 8.94, 8.93, 8.94 Strategy and Explanation: Given the K a values for two forms of lactic acid, determine which form is a stronger acid and find the pH of a solution containing a mixture of these forms. Write equations for the ionization of lactic acid in water. Given initial concentrations of each form in separate solutions, determine the resulting pH. Identify the reaction in water that determines the pH in the absence of any acid. Given mass of lactic acid and volume of solvent, calculate the pH of a solution made. Given the molarity of a base and a mass of lactic acid, determine the volume of the base needed to neutralize it, the pH if the lactic acid were D form, L form, or a mixture. (a) Section 16.5 introduces K a , page 785. Acids strengths are related to their K a . The larger K a , the stronger the acid. Because the L form of the acid has a slightly larger K a (1.6 × 10 –4 ) than the D form (1.5 × 10 –4 ), the L form is stronger. (b) When a solution contains a mixture of two acids, the observed value of the K a must be a combination of the two K a ’s. If the mixture is a 50:50 ratio of D:L, then half the reaction have pH dictated by K a,D and half are dictated by K a,L . We might imagine that some sort of average is needed. But we want to be careful. K is logarithmic, so let’s look more closely at the combined reactions here. When reactions are added, K’s are multiplied: HLac D (aq) + H 2 O( l ) Lac D (aq) + H 3 O + (aq) K a,D HLac L (aq) + H 2 O( l ) Lac L (aq) + H 3 O + (aq) K a,L HLac D (aq) + HLac L (aq) + 2 H 2 O( l ) Lac D (aq) + Lac L (aq) + 2 H 3 O + (aq) K = K a,D K a,L So, when 50:50 ratio of the two forms of lactic acid are present, the K can be identified this way: 1 2 HLac D (aq) + 1 2 HLac L (aq) + H 2 O( l ) 1 2 Lac D (aq) + 1 2 Lac L (aq) + H 3 O + (aq) K a = (K a,D K a,L ) 1/2 pK a = – log K a = –log[(K a,D K a,L ) 1/2 ] Using the fact that: log(x n ) = n logx and log(yz) = logy + logz: pK a = 1 2 (–log[(K a,D K a,L )]) = 1 2 [( –log(K a,D ) – log(K a,L )] pK a = 1 2 (pK a,D + pK a,L ) Therefore, it is the average of the pK a values that will provide us the observed pK a : pK a,L = – log K a,L = – log (1.6 × 10 –4 ) = 3.80 pK a,D = – log K a,D = – log (1.5 × 10 –4 ) = 3.82 pK a = 1 2 (pK a,D + pK a,L ) = 1 2 (3.80 + 3.82) = 3.81 (c) Ionization causes the transfer of the carboxyl hydrogen ion from lactic acid to water: CH 3 CH(OH)COOH(aq) + H 2 O( l ) CH 3 CH(OH)COO (aq) + H 3 O + (aq) (d) The equation for the equilibrium constant expression is: K a = [C 3 CH(OH)COO " ][H 3 O + ] [C 3 CH(OH)COOH]

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Chapter 16: Acids and Bases 723 As the reactants decompose, the concentrations of the products increase stoichiometrically, until they reach
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chapter16_part1 - 722 Chapter 16 Acids and Bases...

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