Exam 2-solutions - Version 409 Exam 2 Laude (52390) 1 This...

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Unformatted text preview: Version 409 Exam 2 Laude (52390) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. K a = [ H + ] [ A ] [ HA ] pK a =- log K a pH =- log [H + ] pOH =- log [OH ] [ OH ] = ( K b C b ) 1 / 2 [ OH ] = K b parenleftbigg C b C a parenrightbigg 10 14 = K w [H + ] [OH ] = K w K a K b = K w pK a + pK b = pK w 14 = pK w ( K a C a ) 1 / 2 = [ H + ] K a parenleftbigg C a C b parenrightbigg = [ H + ] [ H + ] = ( K ax K ay ) 1 / 2 pH = 0 . 5( pK ax + pK ay ) 0 = [ H + ] 2- C a [ H + ]- K w K sp = [ C ] c [ A ] a E o cell = E o cathode- E o anode E cell = E o cell- parenleftbigg . 05916 n parenrightbigg log Q Q = [ C ] c [ D ] d [ A ] a [ B ] b G o =- n F E o cell =- R T ln K E o cell = parenleftbigg R T ln K n F parenrightbigg parenleftbigg I t n F parenrightbigg = moles of product Half reaction E o Au 3+ + 3 e - Au +1 . 50 Cl 2 + 2 e - 2 Cl +1 . 36 Ag + + e - Ag +0 . 80 O 2 + 2 e + 2 H+- H 2 O 2 +0 . 68 I 2 + 2 e - 2 I +0 . 53 Pb 2+ + 2 e - Pb- . 13 Ni 2+ + 2 e - Ni- . 25 Mn 2+ + 2 e - Mn- 1 . 12 Na + + e - Na- 2 . 71 Sr 2+ + 2 e - Sr- 2 . 89 F = 96 , 485 . 3 C per mole of e 1 Ampere = 1 C s 1 R = 8 . 314 J mol 1 K 1 N = 6 . 022 10 23 001 6.0 points Far a battery, the cathode is the (posi- tive/negative) terminal and the electrons flow through the external circuit from (anode to cathode/cathode to anode). 1. negative, anode to cathode 2. positive, anode to cathode correct 3. negative, cathode to anode 4. positive, cathode to anode Explanation: For a discharging battery, the cathode is the positive terminal. The cathode always has an inward flow of electrons, because it is by definition the site of reduction. 002 6.0 points Consider a tetraprotic acid of the form H 4 A. If a buffer is formed by placing Li 2 H 2 A and Li 3 HA in solution, which K a is used to solve the buffer equation? 1. K a 4 2. not enough information 3. K a 4. K a 3 correct 5. K a 2 6. K a 1 Explanation: The two salts when dissolved in water will yield H 2 A 2 and HA 3 . These two species are related by K a 3 . Version 409 Exam 2 Laude (52390) 2 003 6.0 points Which of the following would be equal to K a 3 for orthocarbonic acid, H 4 CO 4 ? 1. [H 4 CO 4 ] [H + ] 3 [HCO 3 4 ] 2. [H + ] 3 [HCO 3 4 ] [H 4 CO 4 ] 3. [H 2 CO 2 4 ] [H + ] [HCO 3 4 ] 4. [H + ] 3 [HCO 3 4 ] [H 2 CO 2 4 ] 5. [H + ] [HCO 3 4 ] [H 2 CO 2 4 ] correct Explanation: K a 3 for orthocarbonic acid, H 4 CO 4 , describes the 3 rd deprotonation event....
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This note was uploaded on 05/04/2009 for the course CH 52390 taught by Professor Laude during the Spring '09 term at Abilene Christian University.

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Exam 2-solutions - Version 409 Exam 2 Laude (52390) 1 This...

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