EXAM3SOL - Moore, Robert – Exam 3 – Due: Apr 12 2007,...

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Unformatted text preview: Moore, Robert – Exam 3 – Due: Apr 12 2007, noon – Inst: McCord 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. SnS = 150.78 g/mol 001 (part 1 of 1) 10 points When a moderate amount of acid is added to a buffer, the pH should 1. decrease greatly. 2. increase greatly. 3. decrease slightly. correct 4. increase slightly. 5. stay the same. Explanation: A buffer contains comparable amounts of a weak acid or a weak base and its conjugate. HA + H 2 O * ) A- + H 3 O + B + H 2 O * ) BH + + OH- When H 3 O + is added, it will react with the base in both systems: B + H 3 O + → BH + + H 2 O A- + H 3 O + → HA + H 2 O However, the ratio of the conjugate DOES change (more acid is made) and therefore the pH lowers. 002 (part 1 of 1) 10 points A buffer solution is made by dissolving 0 . 51 moles of a primary amine (R NH 2 ) and 0 . 21 moles of HCl into 624 mL of solution. What is the pH of this buffer? K b = 3 × 10- 6 for R NH 2 . Your answer must be within ± 0.35%. Correct answer: 8 . 63202 . Explanation: n R NH 2 = 0 . 51 mol n HCl = 0 . 21 mol V soln = 624 mL = 0 . 624 L K b = 3 × 10- 6 for R NH 2 K a , NH + 3 = K w K b , R NH 2 R NH 2 + HCl → R NH + 3 + Cl- . 51 . 21- . 21- . 21 . 21 . 21 . 3 . 21 . 21 R NH + 3 is the conjugate acid of the weak base R NH 2 . Together they form a buffer system. At equilibrium [R NH 2 ] = . 3 mol . 624 L [R NH + 3 ] = . 21 mol . 624 L Cl- is a spectator ion. pH = p K a + log [R NH 2 ] [R NH + 3 ] =- log µ K w K b ¶ + log [R NH 2 ] [R NH + 3 ] =- log µ 1 . × 10- 14 3 × 10- 6 ¶ + log µ . 3 / . 624 . 21 / . 624 ¶ = 8 . 63202 003 (part 1 of 1) 10 points A buffer solution contains 0.0200 M acetic acid and 0.0200 M sodium acetate. What is the pH after 2.0 millimoles of HCl are added to 1.00 L of this buffer? p K a = 4.75 for acetic acid. 1. 4.84 2. 4.66 correct 3. 4.80 4. 4.75 Moore, Robert – Exam 3 – Due: Apr 12 2007, noon – Inst: McCord 2 5. 4.70 Explanation: 004 (part 1 of 1) 10 points The solubility product constant of PbCl 2 is 1 . 7 × 10- 5 . What is the maximum concentra- tion of Pb 2+ that can be in ocean water that contains 0.0500 M NaCl? 1. 3 . 4 × 10- 3 M 2. 6 . 8 × 10- 3 M correct 3. 4 . 2 × 10- 8 M 4. 1 . 7 × 10- 3 M 5. 8 . 5 × 10- 7 M Explanation: K sp of PbCl 2 = 1.7 × 10- 5 PbCl 2 → Pb 2+ + 2FCl- K sp = [Pb 2+ ][Cl- ] 2 1 . 7 × 10- 5 = [Pb 2+ ](0 . 05) 2 [Pb 2+ ] = 0 . 0068 005 (part 1 of 1) 10 points What is the concentration of CN- ion in a 0.400 molar solution of K 4 Fe(CN) 6 ? K d for (Fe(CN) 6 ) 4- is 1 . 3 × 10- 37 . 1. 1 . 02 × 10- 6 M 2. 5 . 00 × 10- 6 M 3. 8 . 33 × 10- 7 M 4. 6 . 09 × 10- 6 M correct 5. 2 . 83 × 10- 6 M Explanation: 006 (part 1 of 1) 10 points Calculate the ratio of the molarities of HPO 2- 4 and H 2 PO- 4 ions required to achieve buffering at pH = 7.00. For H 3 PO 4 , p K a1 = 2.12, p K a2 = 7.21, and p K a3 = 12.68....
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This note was uploaded on 04/12/2008 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

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EXAM3SOL - Moore, Robert – Exam 3 – Due: Apr 12 2007,...

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