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CHAPTER 15 2 nd HALF APPLIC. OF ACID & BASE EQUILIBRIA (BUFFERS & TITRATIONS) Problems to prepare students for hourly exam II. Buffer concepts Buffer calculations Titrations and Indicators E. Tavss 4/06 1 BUFFER CONCEPTS 11 Chem 162-2008 Final Exam + Answers Chapter 15B- Acids, Bases and Acid-Base Equilibria Buffer concepts Which of the following mixtures does not result in a buffer? A. 0.100mol HCl with 0.300mol NH 3 B. 0.200mol NH 4 Cl with 0.400mol NH 3 C. 0.100mol NaOH with 0.200mol HBrO D. 0.100mol NaBrO with 0.200mol HBrO E . 0.100mol NaOH with 0.200mol NH 3 Assume everything is in a 1 L solution. A. NH 3 + H 3 O + NH 4 + + H 2 O NH 3 + H 3 O + NH 4 + + H 2 O Initial 0.3M 0.1M Change-0.1-0.1 +0.1 +0.1 Equilibrium 0.2M 0.1M 0.2 M of a base & 0.1M of its conjugate acid is a buffer. B. NH 4 + + NH 3 NR 0.2M of a conjugate acid & 0.4M of its base is a buffer. C. HBrO + OH- BrO- + H 2 O HBrO + OH- BrO- + H 2 O Initial 0.2M 0.1M Change-0.1-0.1 +0.1 +0.1 Equilibrium 0.1M 0M 0.1M 0.1M 0.1 M weak acid & 0.1M conjugate base is a buffer. D. HBrO + BrO- NR 0.2M of a weak acid & 0.1M of its conjugate base is a buffer E. NH 3 + OH- NR A weak base & a strong base is not a buffer. E 2 Chem 162-2008 Exam II + Answers Chapter 15B Applic. of Acid & Base Equilibria (Buffers & Titrations) Buffer concepts Definition of buffers; relationship between pH and pK a 20. Phosphoric acid, H 3 PO 4 , is a triprotic acid with: K a1 = 7.5 x 10-3 K a2 = 6.3 x 10-8 K a3 = 4.3 x 10-13 Which of the following combinations would be best for preparing a pH 7 buffer? (a) H 3 PO 4 and Na 3 PO 4 (b) H 3 PO 4 and NaH 2 PO 4 (c) Na 2 HPO 4 and Na 3 PO 4 (d) H 3 PO 4 and HCl (e) NaH 2 PO 4 and Na 2 HPO 4 Henderson-Hasselbalch equation: pH = pK a + log([A- ]/[HA]) According to the Henderson-Hasselbalch equation, a good buffer is one which has approximately equal molar quantities of the weak acid and its conjugate base, resulting in pH = pK a + log(1) pH = pKa. H 3 PO 4 + H 2 O H 3 O + + H 2 PO 4- K a1 = 7.5 x 10-3 pK a1 = -log(7.5 x 10-3 ) = 2.12 H 2 PO 4- + H 2 O H 3 O + + HPO 4 2- K a2 = 6.3 x 10-8 pK a2 = -log(6.3 x 10-8 ) = 7.20 HPO 4 2- + H 2 O H 3 O + + PO 4 3- K a3 = 4.3 x 10-13 pK a3 = -log(4.3 x 10-13 ) = 12.4 (a) False. A buffer is defined as a combination of a weak acid and its conjugate base; the conjugate base of H 3 PO 4 is NaH 2 PO 4 , not Na 3 PO 4 . (b) False. Since the pKa of this combination is 2.12, it would make a good buffer at pH approximately 2. (c) False. Since the pKa of this combination is 12.4, it would make a good buffer at pH approximately 12. (d) False. A buffer is defined as a combination of a weak acid and its conjugate base; this is a combination of a weak acid and a strong acid.... View Full Document

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