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homework8 - savage(jws775 Homework#8 Holcombe(52460 This...

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savage (jws775) – Homework #8 – Holcombe – (52460) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Which of the following is NOT true for a solution buffered at a selected pH? 1. The solution pH will increase a lesser amount than a non-buffered solution, when base is added. 2. The buffered solution contains an acid and its conjugate base. 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 10.0 points What is the pH of a solution containing 0.3 M NH 4 Cl and 0.6 M NH 3 ? The p K a of the ammonium ion is 9.25. 1. 9.55 correct 2. 12.25 3. 5.05 4. 4.45 5. 8.95 Explanation: For NH + 4 , p K a = 9.25 p K b = p K w - p K a = 14 - 9 . 25 = 4 . 75 p K b = - log K b K b = 10 - p K b = 10 - 4 . 75 = 1 . 77828 × 10 - 5 Building an ICE using molarities, NH 3 (aq)+H 2 O( ) NH + 4 (aq)+OH - (aq) 0.6 - x + x + x 0 . 6 - x 0 . 3 + x + x K b = [NH + 4 ] [OH - ] [NH 3 ] = (0 . 3 + x ) x 0 . 6 - x 0 . 3 x 0 . 6 x = 0 . 6 K b 0 . 3 = 0 . 6 (1 . 77828 × 10 - 5 ) 0 . 3 = 3 . 55656 × 10 - 5 = [OH - ] [H + ] = K w [OH - ] = 1 × 10 - 14 3 . 55656 × 10 - 5 = 2 . 81171 × 10 - 10 Thus pH = - log(2 . 81171 × 10 - 10 ) = 9 . 55103 003 10.0 points What is the pH of an aqueous solution that is 0.10 M HCOOH ( K a = 1 . 8 × 10 - 4 ) and 0.10 M NaHCO 2 ? 1. 3.74 correct 2. 5.74 3. 10.26 4. 5.62 5. 2.38 Explanation: 004 10.0 points
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savage (jws775) – Homework #8 – Holcombe – (52460) 2 Which of the following combinations CAN- NOT produce a buffer solution? All compo- nents are present in 0.10 M concentration.
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