Solutions Chapter 16 2009

Solutions Chapter 16 2009 - CHAPTER 16 ACID-BASE EQUILIBRIA...

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CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 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 ) 3 a 3 [H ][CH COO ] [CH COOH] = K 5 ( )(0.20 ) (0.20) 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)?
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CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 455 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 xx 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 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
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CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 456 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. (b) H 2 SO 4 (sulfuric 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. (c) This solution contains both a weak acid, H 2 PO 4 and its conjugate base, HPO 4 2 . Therefore, this is a buffer system. (d) HNO 2 (nitrous acid) is a weak acid, and its conjugate base, NO 2 (nitrite ion, the anion of the salt KNO 2 ), is a weak base. Therefore, this is a buffer system.
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This note was uploaded on 05/18/2011 for the course CHM 2046 taught by Professor Veige/martin during the Spring '07 term at University of Florida.

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Solutions Chapter 16 2009 - CHAPTER 16 ACID-BASE EQUILIBRIA...

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