HW8-ex3 - dsj294 hw08 Holcombe (53570) 1 This print-out...

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Unformatted text preview: dsj294 hw08 Holcombe (53570) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points At the stoichiometric point in the titra- tion of 0.130 M HCOOH(aq) with 0.130 M KOH(aq), 1. the pH is 7.0. 2. [HCOOH] = 0.0650 M. 3. the pH is less than 7. 4. the pH is greater than 7. correct 5. [HCO- 2 ] = 0.130 M. Explanation: 002 10.0 points What is the pH at the half-stoichiometric point for the titration of 0.22 M HNO 2 (aq) with 0.01 M KOH(aq)? For HNO 2 , K a = 4 . 3 10- 4 . 1. 7.00 2. 2.16 3. 2.31 4. 3.37 correct 5. 2.01 Explanation: 003 10.0 points For the titration of 50.0 mL of 0.020 M aque- ous salicylic acid with 0.020 M KOH(aq), cal- culate the pH after the addition of 55.0 mL of KOH(aq). For salycylic acid, p K a = 2.97. 1. 12.02 2. 12.30 3. 11.26 4. 7.00 5. 10.98 correct Explanation: 004 10.0 points Consider the titration of 50.0 mL of 0.0200 M HClO(aq) with 0.100 M NaOH(aq). What is the formula of the main species in the solution after the addition of 10.0 mL of base? 1. NaOH 2. ClOH 3. ClO- correct 4. HClO 2 5. ClO 2 Explanation: 005 (part 1 of 2) 10.0 points 10 20 30 40 50 60 1 2 3 4 5 6 7 8 9 10 11 12 Titration Curve mL of NaOH pH What is the pH at the equivalence point of this titration? 1. 5 . 13 2. 8 . 94 correct dsj294 hw08 Holcombe (53570) 2 3. 5 . 89 4. 7 . 04 5. 3 . 06 6. 10 . 47 7. 4 . 09 Explanation: The inflection points are shown below. 10 20 30 40 50 60 1 2 3 4 5 6 7 8 9 10 11 12 Titration Curve mL of NaOH pH (17 . 5, 5 . 13) (35, 8 . 94) 006 (part 2 of 2) 10.0 points What is the p K a of this acid? 1. 3 . 06 2. 5 . 89 3. 7 . 04 4. 8 . 94 5. 4 . 09 6. 5 . 13 correct 7. 10 . 47 Explanation: 007 10.0 points You have a solution that is buffered at pH = 2.0 using H 3 PO 4 and H 2 PO- 4 (p K a1 = 2 . 12; p K a2 = 7 . 21; p K a3 = 12 . 68). You decide to titrate this buffer with a strong base. 15.0 mL are needed to reach the first equivalence point. What is the total volume of base that will have been added when the second equivalence point is reached? 1. > 30 mL correct 2. A second equivalence point in the titra- tion will never be observed. 3. 30 mL 4. < 30 mL Explanation: 008 10.0 points 50.0 mL of 0.0018 M aniline (a weak base) is titrated with 0.0048 M HNO 3 . How many mL of the acid are required to reach the equiva- lence point? 1. 133 mL 2. Bad titration since HNO 3 is not a strong acid. 3. Need to know the K b of aniline. 4. 18.8 mL correct 5. 4.21 mL Explanation: V aniline = 50 mL [Aniline] = 0.0018 M [HNO 3 ] = 0.0048 M Aniline is a monobasic base ( i.e. , it pro- duces one OH- in solution). Thus you can expect that aniline and HNO 3 will react in a one-to-one fashion....
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This note was uploaded on 02/26/2009 for the course CH 58734 taught by Professor Bronzewook during the Spring '07 term at UT Chattanooga.

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HW8-ex3 - dsj294 hw08 Holcombe (53570) 1 This print-out...

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