8 - Elbel Brittany – Homework 8 – Due noon – Inst J A...

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Unformatted text preview: Elbel, Brittany – Homework 8 – Due: Mar 29 2006, noon – Inst: J A Holcombe 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Which of the following is NOT true for a solution buffered at a selected pH? 1. The buffered solution contains an acid and its conjugate base. 2. The solution pH will increase a lesser amount than a non-buffered solution, when base is added. 3. The buffered solution could be made to be either acidic or basic. 4. The solution pH will not change with the addition of an acid. correct Explanation: Buffers are solutions which contain compar- ative amounts of a conjugate weak acid/base part and allow for only small changes in the pH of a chemical system, compared to non- buffered solutions, when an acid or a base is introduced. 002 (part 1 of 1) 10 points The diprotic acid H 2 CO 3 can produce two buffers depending on the pH of the solution. Identify the conjugate acid/base pair from carbonic acid that would form the most basic buffer. 1. CO 2- 3 and OH- 2. HCO- 3 and CO 2- 3 correct 3. H 2 CO 3 and CO 2- 3 4. H 2 CO 3 and HCO- 3 5. H + and H 2 CO 3 Explanation: H + and H 2 CO 3 cannot accept protons and are not a conjugate acid/base pair. CO 2- 3 can accept 2 protons and OH- can accept 1 pro- ton, but they are not a conjugate acid/base pair either. H 2 CO 3 cannot accept any protons and HCO- 3 can accept only 1 proton. H 2 CO 3 can- not accept any protons and CO 2- 3 can accept 2 protons. HCO- 3 can accept 1 proton and CO 2- 3 can accept 2 protons, which is more than the other conjugate acid/base pairs can accept. 003 (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 . 15 moles of HCl into 743 mL of solution. What is the pH of this buffer? K b = 3 . 1 × 10- 6 for R NH 2 . Your answer must be within ± 0.35%. Correct answer: 8 . 87157 . Explanation: n R NH 2 = 0 . 51 mol n HCl = 0 . 15 mol V soln = 743 mL = 0 . 743 L K b = 3 . 1 × 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 . 15- . 15- . 15 . 15 . 15 . 36 . 15 . 15 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 ] = . 36 mol . 743 L [R NH + 3 ] = . 15 mol . 743 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 ] Elbel, Brittany – Homework 8 – Due: Mar 29 2006, noon – Inst: J A Holcombe 2 =- log µ 1 . × 10- 14 3 . 1 × 10- 6 ¶ + log µ . 36 / . 743 . 15 / . 743 ¶ = 8 . 87157 004 (part 1 of 1) 10 points Which of the following combinations CAN- NOT produce a buffer solution?...
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8 - Elbel Brittany – Homework 8 – Due noon – Inst J A...

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