hw 9 - rhodes(ajr2283 Homework#9 Holcombe(52460 This...

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rhodes (ajr2283) – Homework #9 – Holcombe – (52460) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find 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. 13 2. 7 3. 5 4. 9 correct 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. 0.012 M 2. 0.006 M 3. 0.0075 M correct 4. 0.050 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. pH = 7 correct 2. pH < 7 3. pH > 7 4. The p K a of HCl needs to be provided to answer this question.
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