Chem 162 2011 Hourly Exam II Answers Chapter 15B Applic Of Acid Base Equilibria

Chem 162 2011 hourly exam ii answers chapter 15b

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24A. 5.8 x 10-4 B. 7.9 x 10-5C. 2.3 x 10-6D. 3.0 x 10-4 E. 1.6 x 10-5 HA + OH-ĺH2O + A-mol = g/MW; 1.50g/178gmol-1= 8.427 x 10-3mol pH = 4.00; therefore pOH = 10.00; therefore [OH-] = 10-pOH= 10-10.00= 1 x 10-10HA + OH-ĺH2O + Initial (8.427x10-3mol/ (0.050+0.0125) =0.1348M (0.300x0.0125)/ (0.0125+0.050) =0.06M 0 Change Equilibrium 1 x 10-10Large K rule. HA + OH-ĺH2O + Initial (8.427x10-3mol/ (0.050+0.0125) =0.1348M (0.300x0.0125)/ (0.0125+0.050) =0.06M 0 Change -0.06 -0.06 +0.06 Equilibrium 0.0748 0 +0.06 But the “0” for the [OH-] is only approximately zero. It is really 1 x 10-10, and the [H+] is therefore 1 x 10-4, and the pH is therefore 4.00. pH = pKa+ log(base/acid) 4.00 = pKa+ log(0.06/0.0748) pKa= 4.096 Ka= 10-pKa= 10-4.096= 8.02 x 10-5 Chem 162-2011 Hourly Exam II + Answers Chapter 15B - Applic. Of Acid & Base Equilibria (Buffers & Titrations) Titrations and Indicators 9. 1.50 g of a weak acid (molar mass 176) is dissolved in 50.0 mL of water, and the resultant solution is titrated with 0.300 M NaOH. When 12.5 mL of 0.300 M NaOH is added, the pH of the resultant solution is 4.00. Calculate Kafor this acid. A - A -
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25 Chem 162-2011 Hourly Exam II + Answers Chapter 15B - Applic. Of Acid & Base Equilibria (Buffers & Titrations) Titrations and Indicators 21. For which of the following acid-base titrations in aqueous solution will the pH at the equivalence point equal to 7.0? A. titration of HF with NaOH B. titration of CH 3 COOH with KOH C. titration of NH 3 with HBr D. titration of KOH with HNO 3 E. titration of Ba(OH) 2 with HNO 2 A. False. Titration of a weak acid with a strong base should provide an equivalent point pH of ~ 9. B. False. Titration of a weak acid with a strong base should provide an equivalent point pH of ~ 9. C. False. Titration of a weak base with a strong acid should provide an equivalent point pH of ~ 5. D. True. Titration of a strong base with a strong acid should provide an equivalent point pH of ~ 7. E. False. Titration of a weak acid with a strong base should provide an equivalent point pH of ~ 9. Chem 162-2011 Hourly Exam II + Answers Chapter 15B - Applic. Of Acid & Base Equilibria (Buffers & Titrations) Titrations and Indicators 23. Pyridine, C5H5N, is a weak base with Kb= 1.5 x 10-9. What is the pH at the equivalence point when 0.50 M pyridine is titrated with 0.50 M HCl? 2 0
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26 2 0 0 O + 0 +X +X CHEM 162-2010 Final exam Chapter 15B - Applic. of Acid & Base Equilibria (Buffers & Titrations) Titrations and Indicators
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