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Exam 2-solutions(2)

# Exam 2-solutions(2) - Version 312 Exam 2 Laude(53590 This...

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Version 312 – Exam 2 – Laude – (53590) 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 6.0 points Arrange the agents I) Na + + e Na E red = +2 . 71 II) Ba 2+ + 2 e Ba E red = - 2 . 91 III) I 2 + 2 e 2 I E red = +0 . 54 IV) Au 3+ + 3 e Au E red = +1 . 40 V) Ti 3+ + e Ti 2+ E red = - 0 . 37 in increasing order of reducing agent strength. 1. I, II, III, IV, V 2. IV, I, V, II, III 3. V, IV, III, II, I 4. II, V, III, IV, I 5. I, IV, III, V, II correct Explanation: 002 6.0 points What is the concentration of Pb 2+ and F in a saturated solution of PbF 2 ? ( K sp = 3 . 7 × 10 8 ) [Pb 2+ ] [F ] 1. 1 . 9 × 10 4 M 3 . 8 × 10 4 M 2. 2 . 1 × 10 3 M 4 . 2 × 10 3 M correct 3. 1 . 05 × 10 3 M 2 . 1 × 10 3 M 4. 2 . 1 × 10 3 M 2 . 1 × 10 3 M 5. 3 . 3 × 10 3 M 6 . 6 × 10 3 M 6. 9 . 6 × 10 5 M 1 . 9 × 10 4 M Explanation: 003 6.0 points The pH of 0.010 M H 3 PO 4 (aq) is 2.24. Esti- mate the concentration of HPO 2 4 in the solu- tion. For H 3 PO 4 , the values of K a1 , K a2 , and K a3 are 7 . 6 × 10 3 , 6 . 2 × 10 8 , and 2 . 1 × 10 13 , respectively. 1. 2 . 1 × 10 13 M 2. 0.010 M 3. 7 . 6 × 10 3 M 4. 6 . 2 × 10 8 M correct 5. 5 . 8 × 10 3 M Explanation: 004 6.0 points A buffer (pH 4.74) was prepared by mixing 1.00 mole of acetic acid and 1.00 mole of sodium acetate to form a 1.0 liter aqueous solution. To 100 mL of this solution, 10.0 mL of 2.00 molar NaOH was added. What is the new pH? 1. 5.31 2. 5.20 3. 4.47 4. 4.92 correct 5. 4.72 Explanation: K a = 1 . 8 × 10 5 pH = 4.74 Acetic acid is CH 3 COOH and acetate is CH 3 COO (Na + is a spectator so we just leave it out). OR for simplicity, we can just write acetic acid as HA and acetate as A . 100 mL of the buffer solution will contain 0.1 moles of HA and 0.1 moles of A . 10.0 mL of 2.0M NaOH contains 0.02 moles OH (again, Na + is just a spectator). HA + OH A + H 2 O Initial 0.1 0.01 Change 0 . 1 - 0 . 02 - 0 . 02 0 . 1+0 . 02 Final 0 . 08 0 . 12

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Version 312 – Exam 2 – Laude – (53590) 2 Use the Weak Acid Buffer Equation: [H + ] = K a parenleftbigg [HA] [A ] parenrightbigg = 1 . 8 × 10 5 × parenleftbigg 0 . 08 0 . 12 parenrightbigg = 1 . 2 × 10 5 M pH = - log ( 1 . 2 × 10 5 ) = 4 . 92 005 6.0 points What is the concentration of SO 2 4 in 0.100 M H 2 SO 4 ? For H 2 SO 4 , K a1 is very large and K a2 = 1 . 2 × 10 2 . 1. 0.0098 M correct 2. 0.012 M 3. 0.0086 M 4. 0.029 M 5. 0.017 M 6. 0.10 M 7. 0.035 M Explanation: [H 2 SO 4 ] = 0.10 M K a2 = 1 . 2 × 10 2 Because H 2 SO 4 is a strong acid and has such a high K a1 , we know that it will disso- ciate completely into 0.1 M H + and 0.1 M HSO 4 . Then, the 0.1 M HSO 4 will dissociate only partially according to K a2 . So we need to figure out how much SO 2 4 will be present at equilibrium: HSO 4 H + SO 2 4 Initial 0.1 0.1 0 Change - x + x + x Equilibrium 0 . 1 - x 0 . 1 + x x 0 . 012 = x (0 . 1 + x ) (0 . 1 - x ) Solving for x , x = 0 . 00984831 (Yes, you MUST use the quadratic for- mula.) 006 6.0 points The overall reaction for the discharge of a nickel/cadmium cell is Cd + NiO 2 + 2 H 2 O Cd(OH) 2 + Ni(OH) 2 Which of the following statements is true?
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Exam 2-solutions(2) - Version 312 Exam 2 Laude(53590 This...

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