chapter17_part1

chapter17_part1 - Chapter 17: Additional Aqueous Equilibria...

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Unformatted text preview: Chapter 17: Additional Aqueous Equilibria 772 End-of-Chapter Solutions for Chapter 17 Summary Problem 1. Answer: (a) Mix 0.100 mol HCOOH and 0.090 mol NaHCOO and add water to make 1.00 L of solution (b) 3.65 (c) 4.00 g Strategy and Explanation: Given a pH and an acid/base conjugate pair, describe how you prepare a buffer. Given a quantity of strong acid added, determine the new pH of the solution prepared. Determine the mass of a base needed to exceed the capacity of the buffer made. (a) Use Problem-Solving Example 17.3 (page 828-9) as a guide. Get K a for CHOOH from Table 16.2: K a = 1.8 10 –4 , then calculate pK a . pK a = – log(1.8 10 –4 ) = 3.7447 3.74 (report to 2 decimal places) Use the Henderson-Hasselbalch equation (from Section 17.1, page 827) to find the concentration ratio: pH = pK a + log conj. base [ ] conj. acid [ ] 3.70 = 3.74 + log HCOO HCOOH [ ] – 0.04 = log HCOO HCOOH [ ] (report to 2 decimal places) 0.90 = HCOO HCOOH [ ] (report to 2 sig figs) Any solution with this concentration ratio would work. At this point, we will arbitrarily select the concentration of HCOOH to be 0.100 M, so we can calculate the HCOO – concentration for the proper ratio: 0.90 0.100 M = [HCOO – ] 0.090 M = [HCOO – ] (report to 2 sig figs) To prepare a buffer with a pH of 3.70, combine 0.100 mol HCOOH and 0.090 mol NaHCOO, then add enough water to make a solution of 1.00 L. (b) The details of this question, and its answer, depend on the choice of concentrations in the preparation of the buffer. Other answers will result, if the selection of HCOOH concentration had been different in (a). The addition of 0.0050 mol of strong acid, HCl, into the solution results in an equivalent amount of H 3 O + being formed. The base in the buffer, HCOO – , reacts with the acid in a product-favored neutralization reaction: HCOO – (aq) + H 3 O + (aq) HCOOH(aq) + H 2 O( l ) This reaction is product-favored, so we will first make as many products as possible. Using the method of limiting reactants, run the reaction towards products until one of the reactants runs out. HCOO – (aq) H 3 O + (aq) HCOOH(aq) initial conc. (M) 0.090 0.005 0.100 change as reaction occurs (M) – 0.005 – 0.005 + 0.005 final conc. (M) 0.085 0 0.105 Chapter 17: Additional Aqueous Equilibria 773 The solution is still a buffer solution, containing an acid-base conjugate pair, so we can find the pH using the Henderson-Hasselbalch equation: pH = 3.74 + log 0.085 0.105 = 3.65 (c) The details of this question, and its answer, also depend on the choice of concentrations in the preparation of the buffer. Other answers will result, if the selection of HCOOH concentration had been different in (a)....
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This note was uploaded on 07/15/2009 for the course CHE 131 taught by Professor Kerber during the Fall '08 term at SUNY Stony Brook.

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chapter17_part1 - Chapter 17: Additional Aqueous Equilibria...

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