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University of Texas CH 302 McCord H09 - pH curves Indicators

University of Texas CH 302 McCord H09 - pH curves...

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byrne (cmb3744) – 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 – fnd all choices beFore answering. 001 10.0 points At the stoichiometric point in the titration oF 0.130 M HCOOH(aq) with 0.130 M KOH(aq), 1. the pH is greater than 7. correct 2. [HCO - 2 ] = 0.130 M. 3. the pH is 7.0. 4. the pH is less than 7. 5. [HCOOH] = 0.0650 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)? ±or HNO 2 , K a = 4 . 3 × 10 - 4 . 1. 2.01 2. 7.00 3. 2.16 4. 3.37 correct 5. 2.31 Explanation: 003 10.0 points ±or 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). ±or salycylic acid, p K a = 2.97. 1. 10.98 correct 2. 12.30 3. 12.02 4. 11.26 5. 7.00 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. ClO - correct 2. ClOH 3. NaOH 4. ClO 2 5. HClO 2 Explanation: 005 10.0 points You have a solution that is bu²ered 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 bu²er with a strong base. 15.0 mL are needed to reach the frst equivalence point. What is the total volume oF base that will have been added when the second equivalence point is reached? 1. A second equivalence point in the titra- tion will never be observed. 2. 30 mL 3. > 30 mL correct 4. < 30 mL Explanation: 006 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?

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byrne (cmb3744) – H09: pH curves Indicators – mccord – (51620) 2 1. 4.21 mL 2. Need to know the K b of aniline. 3. 18.8 mL correct 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
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University of Texas CH 302 McCord H09 - pH curves...

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