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pH Curves Indicators H09

# pH Curves Indicators H09 - rote(cmr2495 H09 pH curves...

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rote (cmr2495) – H09: pH curves Indicators – mccord – (51620) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points At the stoichiometric point in the titration of 0.130 M HCOOH(aq) with 0.130 M KOH(aq), 1. the pH is less than 7. 2. [HCO - 2 ] = 0.130 M. 3. the pH is greater than 7. correct 4. [HCOOH] = 0.0650 M. 5. the pH is 7.0. Explanation: 002 10.0points 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. 3.37 correct 2. 2.01 3. 2.16 4. 2.31 5. 7.00 Explanation: 003 10.0points 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. 7.00 2. 11.26 3. 12.02 4. 12.30 5. 10.98 correct Explanation: 004 10.0points 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. ClO - correct 2. NaOH 3. ClOH 4. ClO 2 5. HClO 2 Explanation: 005 10.0points 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 2. > 30 mL correct 3. A second equivalence point in the titra- tion will never be observed. 4. < 30 mL Explanation: 006 10.0points 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?

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rote (cmr2495) – H09: pH curves Indicators – mccord – (51620) 2 1. 18.8 mL correct 2. Need to know the K b of aniline. 3. 4.21 mL 4. Bad titration since HNO 3 is not a strong acid. 5. 133 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. With this ratio, we can determine how much HNO 3 will be required to react with all of the aniline. First, convert 50.0 mL aniline into L of aniline: 50.0 mL aniline parenleftBig 1 L 1000 mL parenrightBig = 0.0500 L aniline Then use the ratio to determine the volume of HNO 3 needed: (0.0500 L aniline) parenleftBig 0 . 0018 mol aniline 1 L aniline parenrightBig ×
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pH Curves Indicators H09 - rote(cmr2495 H09 pH curves...

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