Chem 162 2008 Exam II Answers Chapter 15B Applic of Acid Base Equilibria

Chem 162 2008 exam ii answers chapter 15b applic of

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38(a) 1 x 10-4 (b)1 x 10-6 (c) 1 x 10-8 (e) 1 x 10-10 (e) 1 x 10-12 Henderson-Hasselbalch equation: pH = pKa+ log([A-]/[HA]) At the half-way point [A-] = [HA]. Therefore, log([A-]/[HA]) = 0. Therefore, pH = pKaThe equivalence point is at 50 mL of NaOH. Therefore, the half-equivalence point is at 25 mL NaOH. The pH corresponding to the addition of 25 mL of NaOH = 6.0. Therefore, pKa = 6.0. Ka= 1 x 10 Chem 162-2008 Exam II + Answers Chapter 15B Applic. of Acid & Base Equilibria (Buffers & Titrations) Titrations and Indicators Given the pH at the half-way point in a titration, what is the Ka? 24. In the above graph, NaOH is added to a solution of a weak acid. Based on the graph, what is the approximate Kavalue of the acid? . -6
39 31 Chem 162-2007 Final exam + answers Chapter 15B – Applications of Acid & Base Equilibria Titrations and Indicators The graph below represents the titration of 50 mL of a 0.10 M acid with 0.10 M NaOH 024681012140102030405060V base added (mL)pHWhich of the following statements are consistent with this titration curve? X. The equivalence point occurs when approximately 25 mL of NaOH are added Y. The curve represents the titration of a weak acid. Z. The pKaof the acid is approximately 9.4.
44 Chem 162-2007 Final exam + answers Chapter 15B – Applications of Acid & Base Equilibria Titrations and indicators Calculate the pH after 10.0 mL of 0.400 M NaOH is added to 20.0 mL of 0.50 M CH3COOH. (Ka CH3COOH = 1.8 × 10í5 ). - A 0
41 CHEM 162-2007 EXAM II + ANSWERS CHAPTER 15B – APPLICATIONS OF ACID & BASE EQUILIBRIA (BUFFERS & TITRATIONS) TITRATIONS AND INDICATORS

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