Exam 2-solutions(2) - Version 312 Exam 2 Laude (53590) 1...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 =- . 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 . 1- . 02- . 02 . 1+0 . 02 Final . 08 . 12 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 . 08 . 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...
View Full Document

This note was uploaded on 04/06/2008 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

Page1 / 9

Exam 2-solutions(2) - Version 312 Exam 2 Laude (53590) 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online