HW 8 - ) Homework #8 holcombe This print-out should have 26...

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) – Homework #8 – holcombe – 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A 1 M solution oF NaOH is used to titrate a 1 M solution oF C 6 H 5 COOH (benzoic acid). IF the K a oF C 6 H 5 COOH is 5 × 10 - 5 , what is the pH oF the solution at the equivalence point? 1. 5 2. 9 correct 3. 13 4. 7 5. not enough inFormation Explanation: Equal volumes oF the titrant and analyte will be used to reach the equivalance point. Regardless oF the starting volume, at the equivalance point the volume will be double the starting value and all oF the benzoic acid will have been converted to benzoate. The solution will be 0 . 5 M C 6 H 5 COO - . K b = K w /K a = 10 - 14 / (5 × 10 - 5 ) = 2 × 10 - 10 [OH - ] = ( K b C b ) 1 / 2 = (2 × 10 - 10 · 0 . 5) 1 / 2 = (10 - 10 ) 1 / 2 = 10 - 5 pOH = 5 pH = 9 002 10.0 points It was Found that 25 mL oF 0.012 M HCl neutralized 40 mL oF NaOH solution. What was the molarity oF the base solution? 1. 0.050 M 2. 0.006 M 3. 0.0075 M correct 4. 0.012 M Explanation: V HCl = 25 mL M HCl = 0 . 012 M V NaOH = 40 mL = 0 . 04 L The base is NaOH. To neutralize, mol H + = mol OH - . n H + = 0 . 012 mol L (25 mL HCl) × 1 L 1000 mL × 1 mol H + 1 mol HCl = 0 . 0003 mol H + = n OH - = n NaOH M NaOH = mol L = 0 . 0003 mol NaOH 0 . 04 L = 0 . 0075 M NaOH 003 10.0 points What is the pH when 100 mL oF 0.1 M HCl is titrated with 50 mL oF 0.2 M NaOH? 1. The p K a oF NAOH needs to be provided to answer this question. 2. The p K a oF HCl needs to be provided to answer this question. 3. pH > 7 4. pH < 7 5. pH = 7 correct Explanation: 004 (part 1 of 5) 10.0 points Consider the Following titration curve.
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– Homework #8 – holcombe – 2 0 2 4 6 8 10 12 14 0 20 40 60 80 100 120 140 Volume of base (mL) pH This is a plot of a (weak, strong) acid and a (weak, strong) base titration. 1. strong; strong 2. weak; strong correct 3. strong; weak Explanation: The curve shape follows that for a weak acid titrated with a strong base, identiFed by the concave down-to-concave up shape of the curve before the equivalence point. 005 (part 2 of 5) 10.0 points What is the approximate p K a of the acid? 1. 12.5 2. 7.5 correct 3. 10 4. There is no p K a because it is a strong acid. 5. 13 6. 5 Explanation: The p K a of the strong acid can be estimated by inspection of the buffer region, or by the pH at half the equivalence point volume. The equivalence point is at 70 mL added base, so the pKa is at 35 mL, about pH=7.5. 006 (part 3 of 5) 10.0 points What is the pH at the equivalence (stoichio- metric) point? 1. about 10 correct 2. about 5 3. exactly 7 4. about 7.5 Explanation: The equivalence point is the vertical part of the curve, which is for this curve the second in±ection point (changes from concave up to concave down) at pH = 10. 007 (part 4 of 5) 10.0 points
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This note was uploaded on 11/12/2010 for the course CH 302 taught by Professor Holcombe during the Fall '07 term at University of Texas.

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HW 8 - ) Homework #8 holcombe This print-out should have 26...

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