Exam 2-solutions-1 - Version 049 Exam 2 laude (51635) 1...

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Unformatted text preview: Version 049 Exam 2 laude (51635) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 6.0 points Consider the titration of equal volumes of 0.1 M HCl and 0.1 M HC 2 H 3 O 2 with 0.1 M NaOH. Which of the following would be the same for both titrations? 1. Two of the other answers are correct. 2. the initial pH 3. the pH at the half equivalence point 4. the volume of NaOH added to reach the equivalence point correct Explanation: Both HCl and CH 2 C 3 O 2 H 2 are monoprotic acids and so if the volume concentrations are identical, the amounts of replaceable H + in both solutions are identical. 002 6.0 points What is the [H + ] of 2 10- 8 M HNO 3 ? 1. 1 . 10499 10- 7 correct 2. 2 . 00998 10- 7 3. 2 10- 8 M 4. 4 . 04 10- 14 Explanation: HNO 3 is a strong acid, so it completely dissociates. This adds H + ions to solution, which initially will violate the equilibrium ex- pression for the dissociation of water. The some of the H + will be neutralized with the OH- until the expression is obeyed again. K w = [H + ][OH- ] 10- 14 = (10- 7 + 2 10- 8- x )(10- 7- x ) = [H + ]([H + ]- 2 10- 8 ) [H + ] 1 . 10499 10- 7 003 6.0 points What is K sp for M 3 X, if its molar solubility is 2 10- 6 mol / L? 1. 4 . 32 10- 22 correct 2. 1 . 6 10- 23 3. 7 . 2 10- 17 4. 1 . 296 10- 21 5. 1 . 44 10- 22 6. 4 . 8 10- 23 Explanation: S = 2 10- 6 mol / L The solubility equilibrium is M 3 X(s) 3 M + (aq) + X 3- (aq) [M + ] = 3 S = 6 10- 6 mol / L [X 3- ] = S = 2 10- 6 mol / L K sp = [M + ] 3 [X 3- ] = ( 6 10- 6 ) 3 (2 10- 6 ) = 4 . 32 10- 22 004 6.0 points What is the pH at the half-stoichiometric point for the titration of 0.22 M HNO 2 (aq) with 0.01 M KOH(aq)? For HNO 2 , K a = 4 . 3 10- 4 . 1. 2.31 2. 7.00 3. 3.37 correct 4. 2.01 5. 2.16 Explanation: Version 049 Exam 2 laude (51635) 2 005 6.0 points Which of the following solutions will produce a buffer? I) 20 mL of 0.5 M (CH 3 ) 3 NHCl + 50 mL of 0.1 M (CH 3 ) 3 N II) 20 mL of 0.5 M HNO 2 + 50 mL of 0.1 M NaOH III) 20 mL of 0.5 M HCl + 50 mL of 0.1 M NH 3 IV) 20 mL of 0.5 M HClO 2 + 50 mL of 0.1 M CH 3 COOH V) 20 mL of 0.5 M NH 4 Cl + 50 mL of 0.1 M NaOH 1. I, II, and V only correct 2. II only 3. I, II, IV, and V only 4. II and IV only...
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This note was uploaded on 06/14/2011 for the course MATH 305G taught by Professor Cathy during the Spring '11 term at University of Texas at Austin.

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Exam 2-solutions-1 - Version 049 Exam 2 laude (51635) 1...

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