exam 3 - Version 231 Exam 3 Holcombe (52460) 1 This...

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Unformatted text preview: Version 231 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 Chlorine, bromine, and iodine are good 1. bases. 2. reducing agents. 3. oxidizing agents. correct Explanation: 002 10.0 points What molar ratio of sodium acetate to acetic acid (NaAc/HAc) should be used in preparing a buffer having a pH of 4.35? ( K a = 1 . 8 10 5 for acetic acid.) 1. 1 . 0 : 0 . 40 2. 1 . 0 : 1 . 3. 2 . 0 : 1 . 4. . 59 : 1 . 5. . 40 : 1 . correct Explanation: pH = 4.35 K a = 1 . 8 10 5 pH = p K a + log parenleftbigg [NaAc] [HAc] parenrightbigg log parenleftbigg [NaAc] [HAc] parenrightbigg = pH- p K a = 4 . 35- [- log(1 . 8 10 5 )] =- . 394727 [NaAc] [HAc] = 10 . 394727 = 0 . 40297 003 10.0 points What is the molar solubility of CaF 2 ? ( K sp = 3 . 9 10 11 .) 1. 3 . 9 10 11 2. 4 . 4 10 6 3. 2 . 1 10 4 correct 4. 3 . 4 10 4 5. 6 . 2 10 6 Explanation: CaF 2 Ca 2+ + 2 F K sp = [Ca 2+ ] [F ] 2 3 . 9 10 11 = ( x ) (2 x ) 2 = 4 x 3 x = 2 . 1 10 4 004 10.0 points Consider the cell Ag(s) | Ag + (aq , . 100 M) || Ag + (aq , . 100 M) | Ag(s) . What is the voltage of this cell? 1. +0.80 V 2. +0.0296 V 3. +0.0592 V 4. correct Explanation: 005 10.0 points A saturated solution of MX 4 (of MW 135 g/mol) contains 0 . 0235 grams per liter. What is the solubility product constant of MX 4 ? 1. 1 . 6 10 19 2. 2 . 56 10 18 3. 3 . 03 10 8 4. 2 . 11 10 11 Version 231 Exam 3 Holcombe (52460) 2 5. 2 . 48 10 14 6. 8 . 18 10 17 7. 4 . 09 10 17 correct Explanation: 006 10.0 points If E is positive, G must be (posi- tive/negative) and K must be (greater/less) than 1. 1. positive, less 2. negative, less 3. positive, greater 4. negative, greater correct Explanation: G =- nFE =- RT ln K 007 10.0 points What is the pH of a solution which is 0.600 M in dimethylamine ((CH 3 ) 2 NH) and 0.400 M in dimethylamine hydrochloride ( (CH 3 ) 2 NH + 2 Cl ) ? K b for dimethylamine = 7 . 4 10 4 . 1. 2.95 2. 10.78 3. 3.31 4. 10.69 5. 10.87 6. 11.05 correct 7. 11.21 Explanation: K w = 1 10 14 K a = 0 . 00074 [(CH 3 ) 2 NH] = 0.6 M [(CH 3 ) 2 NH + 2 ] = 0.4 M K a , (CH 3 ) 2 NH + 2 = K w K b , (CH 3 ) 2 NH Applying the Henderson-Hasselbalch equa- tion, pH = p K a + log parenleftbigg [(CH 3 ) 2 NH] [(CH 3 ) 2 NH + 2 ] parenrightbigg =- log parenleftbigg K w K a parenrightbigg + log parenleftbigg [(CH 3 ) 2 NH] [(CH 3 ) 2 NH + 2 ] parenrightbigg =- log parenleftbigg 1 10 14 . 00074 parenrightbigg + log parenleftbigg . 6 . 4 parenrightbigg = 11 . 0453 008 10.0 points Oxidation occurs 1. at either, depending on whether the cell is electrochemical or electrolytic....
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exam 3 - Version 231 Exam 3 Holcombe (52460) 1 This...

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