Chapter16ISM

Chapter16ISM - CHAPTER 16 ACID-BASE EQUILIBRIA AND...

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CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA Problem Categories Biological : 16.16, 16.103, 16.121, 16.126, 16.131. Conceptual : 16.19, 16.20, 16.21, 16.22, 16.35, 16.36, 16.67, 16.68, 16.102, 16.117, 16.123, 16.127, 16.129. Descriptive : 16.36, 16.79, 16.80, 16.83, 16.85, 16.86, 16.93, 16.95, 16.111, 16.113, 16.114, 16.116, 16.118. Environmental : 16.124. Difficulty Level Easy : 16.5, 16.6, 16.9, 16.10, 16.11, 16.12, 16.19, 16.20, 16.39, 16.49, 16.51, 16.52, 16.54, 16.91, 16.95, 16.114. Medium : 16.13, 16.14, 16.15, 16.16, 16.21, 16.22, 16.25, 16.26, 16.27, 16.28, 16.41, 16.42, 16.50, 16.53, 16.55, 16.56, 16.57, 16.63, 16.64, 16.65, 16.66, 16.67, 16.68, 16.69, 16.70, 16.71, 16.75, 16.79, 16.80, 16.83, 16.84, 16.85, 16.86, 16.87, 16.88, 16.89, 16.93, 16.98, 16.101, 16.102, 16.103, 16.104, 16.106, 16.111, 16.115, 16.116, 16.117, 16.118, 16.120, 16.122, 16.123, 16.125, 16.130, 16.133, 16.134. Difficult : 16.17, 16.18, 16.29, 16.30, 16.31, 16.32, 16.33, 16.34, 16.35, 16.36, 16.40, 16.58, 16.59, 16.60, 16.72, 16.76, 16.77, 16.78, 16.90, 16.92, 16.94, 16.96, 16.97, 16.99, 16.100, 16.105, 16.107, 16.108, 16.109, 16.110, 16.112, 16.113, 16.119, 16.121, 16.124, 16.126, 16.127, 16.128, 16.129, 16.131, 16.132. 16.5 (a) This is a weak acid problem. Setting up the standard equilibrium table: CH 3 COOH( aq ) U H + ( aq ) + CH 3 COO ( aq ) Initial ( M ): 0.40 0.00 0.00 Change ( M ): x + x + x Equilibrium ( M ): (0.40 x ) x x 3 a 3 [H ][CH COO ] [CH COOH] +− = K 22 5 1.8 10 (0.40 ) 0.40 ×= x x x x = [H + ] = 2.7 × 10 3 M pH = 2.57 (b) In addition to the acetate ion formed from the ionization of acetic acid, we also have acetate ion formed from the sodium acetate dissolving. CH 3 COONa( aq ) CH 3 COO ( aq ) + Na + ( aq ) Dissolving 0.20 M sodium acetate initially produces 0.20 M CH 3 COO and 0.20 M Na + . The sodium ions are not involved in any further equilibrium (why?), but the acetate ions must be added to the equilibrium in part (a). CH 3 COOH( aq ) U H + ( aq ) + CH 3 COO ( aq ) Initial ( M ): 0.40 0.00 0.20 Change ( M ): x + x + x Equilibrium ( M ): (0.40 x ) x (0.20 + x )
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CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 467 3 a 3 [H ][CH COO ] [CH COOH] +− = K 5 () ( 0 . 2 0 ) ( 0 . 2 0 ) 1.8 10 (0.40 ) 0.40 + ×= xx x x x = [H + ] = 3.6 × 10 5 M pH = 4.44 Could you have predicted whether the pH should have increased or decreased after the addition of the sodium acetate to the pure 0.40 M acetic acid in part (a)? An alternate way to work part (b) of this problem is to use the Henderson-Hasselbalch equation. a [conjugatebase] pH p log [acid] =+ K 5 0.20 log(1.8 10 ) log 4.74 0.30 0.40 =− × + = = pH 4.44 M M 16.6 (a) This is a weak base calculation. NH 3 ( aq ) + H 2 O( l ) U NH 4 + ( aq ) + OH ( aq ) Initial ( M ): 0.20 0 0 Change ( M ): x + x + x Equilibrium ( M ): 0.20 x x x 4 b 3 [NH ][OH ] [NH ] = K 2 5 1.8 10 0.20 0.20 x x x = 1.9 × 10 3 M = [OH ] pOH = 2.72 pH = 11.28 (b) The initial concentration of NH 4 + is 0.30 M from the salt NH 4 Cl. We set up a table as in part (a). NH 3 ( aq ) + H 2 O( l ) U NH 4 + ( aq ) + OH ( aq ) Initial ( M ): 0.20 0.30 0 Change ( M ): x + x + x Equilibrium ( M ): 0.20 x 0.30 + x x 4 b 3 [NH ][OH ] = K 5 ( )(0.30 ) (0.30) 1.8 10 0.20 0.20 + x x x = 1.2 × 10 5 M = [OH ] pOH = 4.92 pH = 9.08
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CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 468 Alternatively, we could use the Henderson-Hasselbalch equation to solve this problem. Table 15.4 gives the value of K a for the ammonium ion. Substituting into the Henderson-Hasselbalch equation gives: 10 a [conjugate base] (0.20) pH p log log(5.6 10 ) log acid (0.30) =+ = × + K pH = 9.25 0.18 = 9.07 Is there any difference in the Henderson-Hasselbalch equation in the cases of a weak acid and its conjugate base and a weak base and its conjugate acid? 16.9 (a) HCl (hydrochloric acid) is a strong acid. A buffer is a solution containing both a weak acid and a weak base. Therefore, this is not a buffer system.
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Chapter16ISM - CHAPTER 16 ACID-BASE EQUILIBRIA AND...

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