exam 3 ch 302

exam 3 ch 302 - Version 119 Exam 3 Holcombe (52460) 1 This...

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Unformatted text preview: Version 119 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 Consider the addition of 0.130 moles of solid NaOH to 1.0 liter of 0.230 M HCl. What will be the pH of the resulting solution? As- sume no volume change due to the addition of NaOH. 1. pH = 1.00 correct 2. pH = 1.50 3. pH = 0.50 4. pH = 2.00 5. pH = 0.75 Explanation: 002 10.0 points Chlorine, bromine, and iodine are good 1. oxidizing agents. correct 2. bases. 3. reducing agents. Explanation: 003 10.0 points For the reaction Cu(s) + Ag + (aq) Cu + (aq) + Ag(s) 1. Cu is reduced at the cathode and Ag + is oxidized at the anode. 2. Cu is oxidized at the cathode and Ag + is reduced at the anode. 3. Cu is oxidized at the anode and Ag + is reduced at the cathode. correct 4. Cu is reduced at the anode and Ag + is oxidized at the cathode. Explanation: Oxidation (loss of electrons) occurs at the anode. Reduction (gain of electrons) occurs at the cathode. 004 (part 1 of 2) 10.0 points Citric acid is a triprotic acid (p K a1 = 3 . 13, p K a2 = 4 . 76, p K a3 = 6 . 40). 50 mL of 0.020 M citric acid is titrated with 0.050 M NaOH. How many mL of base solution are needed to reach the first equivalence point? 1. 50.0 mL 2. 2.0 mL 3. 5.0 mL 4. 20.0 mL correct 5. 60.0 mL Explanation: 50 mL (0.02 M) = 1.0 mmol of acid This means you need 1 mmol of base to reach the first equivalence point. 1 mmol of NaOH / 0.05 M = 20.0 mL of base solution. 005 (part 2 of 2) 10.0 points Which answer best approximates the pH of the solution half way to the first equivalence point of this titration? 1. 8.6 2. 4.8 3. 7.0 4. 4.0 5. 3.1 correct 6. 6.4 Version 119 Exam 3 Holcombe (52460) 2 7. 2.2 Explanation: At the half-way point to the first equiv- alence point, the solution pH should equal p K a1 which is 3.13. The choice of 3.1 is a good approximation. 006 10.0 points Oxidation occurs 1. at both anode and cathode. 2. at either, depending on whether the cell is electrochemical or electrolytic. 3. at the anode. correct 4. in the electrolyte. 5. at the cathode. Explanation: 007 10.0 points Consider the following voltaic cell at room temperature: Pt | Sn 4+ (1.0 M), Sn 2+ (1.0 M) || Ag + (1.0 M) | Ag Sn 4+ + 2 e Sn 2+ E = +0 . 15 V Ag + + 1 e Ag E = +0 . 80 V What is the equilibrium constant for the cell reaction? 1. 3 . 4 10 9 2. 21.9 3. 8 . 6 10 19 4. 9 . 7 10 21 correct 5. 19.9 Explanation: 008 10.0 points The solubility product constant of CaCO 3 is K sp = 5 10 9 . How many moles of CaCO 3 dissolve in one liter of water? 1. 2 . 5 10 9 mole 2. 1 10 8 mole 3. 7 10 5 mole correct 4. 5 10 9 mole Explanation: K sp = 5 10 9 CaCO 3 Ca 2+ + CO 2 3 Let x = molar solubility....
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exam 3 ch 302 - Version 119 Exam 3 Holcombe (52460) 1 This...

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