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# solution_pdf - Version 288 Make Up Exam 2 VANDEN BOUT(53585...

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Version 288 – Make Up Exam 2 – VANDEN BOUT – (53585) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points How long will it take for a current of 3.00 amperes to plate 6 . 355 grams of Cu out of a solution of Cu(NO 3 ) 2 ? (1 Faraday = 96,485 coulombs, Cu = 63 . 55 g/mol) 1. 53.6 min 2. None of these 3. 2.68 hr 4. 12.9 sec 5. 26.8 min 6. 1.79 hr correct Explanation: I = 3.0 A m Cu = 6 . 355 g The reaction that is occuring is Cu 2+ + 2 e Cu(s) Number of moles of Cu(s) plated out 6 . 355 g 63 . 55 g / mol = 0 . 1 mol . Number of moles of e used 0 . 1 mol × 2 = 0 . 2 mol e . Number of Coulombs = Faraday × moles of e = (96485 C / mol) × (0 . 2 mol e ) = 19297 C . current = charge time 3 A = 19297 C time time = 19297 C 3 A = 6432 . 33 seconds 6432 seconds 6432 seconds = 6432 seconds 3600 seconds / hour = 1 . 78676 hrs 1 . 79 hrs 002 10.0 points A buffer (of pH 3.74) was prepared by mixing 1.00 mole of formic acid and 1.00 mole of sodium formate to form an aqueous solution with a total volume of 1.00 liter. To 250 mL of this solution was added 10.0 mL of 1.00 M NaOH. What is the pH of this solution? 1. 4.29 2. 3.51 3. 3.78 correct 4. 4.08 5. 3.32 Explanation: [NaOH] = 1 M [HF] = 1 M [F ] = 1 M pH ini = 3.74 Initial condition (ini): n NaOH = 10 . 0 × 1 . 0 = 10 mmol n HF = 200 × 1 . 0 = 200 mmol n Na + = 250 × 1 . 0 = 250 mmol n F - = 250 × 1 . 0 = 250 mmol NaOH + HF Na + + F + H 2 O ini 10 250 250 250 Δ - 10 - 10 10 10 fin 0 240 260 260 Na + is a spectator ion. HF / F is a buffer system. Since [HF] = [F ] in the original buffer p K a = pH ini = 3 . 74. pH fin = p K a + log parenleftBigg bracketleftbig F bracketrightbig [HF] parenrightBigg = 3 . 74 + log parenleftbigg 260 240 parenrightbigg = 3 . 77476 003 10.0 points A buffer is formed by mixing 100 mL of 0.2 M

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Version 288 – Make Up Exam 2 – VANDEN BOUT – (53585) 2 HClO 2 and 200 mL of 0.7 M KClO 2 . What volume of 0.2 M KOH can be added before the buffer capacity is reached? 1. 300 mL 2. 100 mL correct 3. 10 mL 4. 700 mL 5. 150 mL Explanation: 004 10.0 points All components are present in 0.10 M concen- trations. I) HCN and NaCN II) NH 3 and NH 4 Cl III) HNO 3 and NH 4 NO 3 IV) HClO 3 and NaClO 3 Which will give buffer solutions? 1. I and II only correct 2. I, III and IV only 3. I and III only 4. II, III and IV only 5. III and IV only Explanation: Buffers are formed in one of two ways, by combining a weak acid and its conjugate base or by combining a weak base and its conjugate acid. HNO 3 and HClO 3 are both strong acids and cannot be used to make effective buffer solutions. HCN is a weak acid and NaCN is the salt of its conjugate base, CN . NH 3 is a weak base and NH 4 Cl is the salt of its conjugate acid, NH + 4 . Therefore 1 and 2 can be used to make effective buffer solutions. 005 10.0 points Arrange the agents I) Fe 3+ + e Fe 2+ E red = +0 . 77 II) Cu 2+ + e Fe + E red = +0 . 15 III) S + 2 e S 2 E red = - 0 . 48 IV) Mn 3+ + e Mn 2+ E red = +1 . 51 V) Ca 2+ + 2 e Fe E red = - 2 . 87 in increasing order of oxidizing agent strength.
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solution_pdf - Version 288 Make Up Exam 2 VANDEN BOUT(53585...

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