Exam 3 - Version 170 Exam 3 Holcombe (52460) 1 This...

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Unformatted text preview: Version 170 Exam 3 Holcombe (52460) 1 This print-out should have 33 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 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 2. < 30 mL 3. > 30 mL correct 4. A second equivalence point in the titra- tion will never be observed. Explanation: 002 10.0 points Which of the following acid-base indicators should be used for the titration of CH 3 COOH with KOH? For acetic acid, K a = 1 . 8 10 5 . 1. None of these is suitable. 2. Neutral red, color change red/yellow 6 . 8 < pH < 8 . 3. Methyl red, color change red/yellow at 4 . 4 < pH < 6 . 2 4. Any of these is suitable. 5. Phenolphthalein, color change colorless/red-violet 8 . < pH < 10 . correct Explanation: The endpoint of the weak acid (CH 3 COOH)- strong base (KOH) titration is above 7. An appropriate indicator has color change range pH = p K a 1. 003 10.0 points If 100 mL of 0.040 M NaOH solution is added to 100 mL of solution which is 0.10 M in CH 3 COOH and 0.10 M in NaCH 3 COO, what will the pH of the new solution be? ( K a = 1 . 8 10 5 ) 1. 5.00 2. 5.11 correct 3. 4.89 4. 4.81 5. 4.74 Explanation: [CH 3 COOH] = 0.10 M [NaOH] = 0.040 M [CH 3 COO ] = 0.10 M K a = 1 . 8 10 5 Initial condition (ini): n NaOH = 100 . 04 = 4 mmol n CH 3 COOH = 100 . 10 = 10 mmol n Na + = 100 . 10 = 10 mmol n CH 3 COO- = 100 . 10 = 10 mmol NaOH +CH 3 COOH Na + +CH 3 COO + H 2 O ini 4 . 10 . 10 . 10 . - 4 .- 4 . 4 . 4 . fin 6 . 14 . 14 . Na + is a spectator ion. CH 3 COOH / CH 3 COO is a buffer system. pH = p K a + log parenleftBigg bracketleftbig CH 3 COO bracketrightbig [CH 3 COOH] parenrightBigg =- log ( 1 . 8 10 5 ) + log parenleftbigg 14 . 6 . parenrightbigg = 5 . 1127 004 10.0 points Ferrous hydroxide is a slightly soluble base. In which of the following would Fe(OH) 2 be most soluble? It is not necessary to know K sp for Fe(OH) 2 . 1. 0.1 M Ca(OH) 2 Version 170 Exam 3 Holcombe (52460) 2 2. 0.1 M KOH 3. 0.1 M HCl correct 4. 0.1 M FeCl 2 Explanation: 005 10.0 points A battery formed from the two half re- actions below dies (reaches equilibrium). If [Fe 2+ ] was 0 . 24 M in the dead battery, what would [Cd 2+ ] be in the dead battery? Half reaction E Fe 2+- Fe- . 44 Cd 2+- Cd- . 40 1. 5 . 4 M 2. 120 . 3 M 3. . . 0005 M 4. . 01 M correct Explanation: E = +0 . 04 E cell = E cell- . 05916 N e log Q 0 = 0 . 04- . 05916 2 log . 24 [Cd 2+ ] log . 24 [Cd 2+ ] = 1 . 35 . 24 [Cd 2+ ] = 10 1 . 35 [Cd 2+ ] = . 24 10 1 . 35 = 0 . 0107 M 006 10.0 points Fe(OH) 3 (s) is very insoluble in water ( K sp...
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This note was uploaded on 04/22/2011 for the course CHEM 302 taught by Professor Mccord during the Spring '10 term at University of Texas at Austin.

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Exam 3 - Version 170 Exam 3 Holcombe (52460) 1 This...

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