# 10_new - Homework#8 holcombe(51160 This print-out should...

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Homework #8 – holcombe – (51160) 1 This print-out should have 20 questions. Multiple-choice questions may continue on 2. 0.006 M the next column or page find all choices before answering. 3. 0.012 M 1 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. 7 2. 5 3. 13 4. not enough information 5. 9 correct 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. 0.0075 M correct 4. 0.050 M Explanation: V HCL = 25 mL M HCL = 0.012 M V N A OH = 40 mL = 0.04 L The base is NaOH. To neutralize, mol H + = mol OH . NH + = 0.012 mol (25 mL HCl) L 1 L 1 mol H + × 1000 mL × 1 mol HCl = 0.0003 mol H + = N OH = N N A OH mol 0.0003 mol NaOH M N A OH = L = 0.04 L = 0.0075 M NaOH 3 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 pK A of NAOH needs to be provided to answer this question. 2. The pK 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 – (51160) 2 14 12 10 8 pH 6 4 2 0 0 20 40 60 80 100 120140 Volume of base (mL) This is a plot of a (weak, strong) acid and a (weak, strong) base titration. 1. weak; strong correct 2. strong; weak 3. strong; strong Explanation: The curve shape follows that for a weak acid titrated with a strong base, identified by the concave down-to-concave up shape of the curve before the equivalence point. 005 (part 2 of 5) 10.0 points
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10_new - Homework#8 holcombe(51160 This print-out should...

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