Exam 2 - Version 033 Exam 2 Holcombe (52460) This print-out...

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Version 033 – Exam 2 – Holcombe – (52460) 1 This print-out should have 33 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. Some constants: Kw=1e-14; R=8.314 J/K- mol 001 10.0 points Consider the Fractional composition diagram For phosphoric acid. 0.2 0.4 0.6 0.8 1.0 2 4 6 8 10 12 14 pH ±ractional Composition What is the approximate pH oF a solu- tion prepared by adding several grams oF Na 2 HPO 4 to a Few hundred mL oF water? 1. 5 2. 7 3. 12 4. 10 correct 5. 2 Explanation: Na 2 HPO 4 is the salt oF the monoprotic species which is the dominate species between pH’s oF 7.2 and 12.7 which is the orange or shortest dashed curve. The best estimate oF the pH is the dead center oF that range which is a pH oF 10 which is the average oF p K 2 and p K 3 For the acid. 002 10.0 points A diprotic acid H 2 A has values oF K a1 = 1 . 0 × 10 6 and K a2 = 1 . 0 × 10 10 . What is the [A 2 ] in a 0.10 M solution oF H 2 A? 1. 0.20 M 2. 0.10 M 3. 1 . 0 × 10 10 M correct 4. 3 . 2 × 10 4 M 5. 3 . 2 × 10 6 M Explanation: 003 10.0 points The pH oF 0 . 1 M (C 3 H 7 NH 2 ) propylamine aqueous solution was measured to be 11 . 86. What is the value oF p K b oF propylamine? 1. 0.11 2. 3.35 3. 2.14 4. 0.96 5. 3.25 correct 6. 11.86 Explanation: M = 0 . 1 M pH = 11 . 86 Analyzing the reaction with molarities, C 3 H 7 NH 2 + H 2 O C 3 H 7 NH + 3 + OH 0 . 1 - 0 0 - x - x x 0 . 1 - x - x x pOH = 14 - 11 . 86 = 2 . 14 . [C 3 H 7 NH + 3 ] = [OH ] = 10 pOH = 10 2 . 14 = 0 . 00724436 mol / L The K b is K b = [C 3 H 7 NH + 3 ][OH ] [C 3 H 7 NH 2 ] = (0 . 00724436) 2 0 . 1 - 0 . 00724436 = 0 . 000565796 .
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Version 033 – Exam 2 – Holcombe – (52460) 2 and the p K b is p K b = - log(0 . 000565796) = 3 . 24734 . 004 10.0 points A solution has [H + ] = 1 . 5 × 10 7 M. What is the [OH ] in this solution? 1. 1 . 5 × 10 7 M 2. None of these 3. 6 . 7 × 10 6 M 4. 6 . 7 × 10 7 M 5. 6 . 7 × 10 8 M correct Explanation: [H + ] = 1 . 5 × 10 7 M K w = [H + ][OH ] = 1 . 0 × 10 14 [OH ] = K w [H + ] = 1 . 0 × 10 14 1 . 5 × 10 7 = 6 . 7 × 10 8 M 005 10.0 points The conjugate acid of HCO 3 is ? and the conjugate base of HCO 3 is ? . 1. CO 2 3 ; H 2 CO 3 2. CO 2 3 ; CO 2 3 3. CO 2 3 ; OH 4. H 2 CO 3 ; H 2 O 5. H + ; OH 6. H + ; CO 2 3 7. H 2 CO 3 ; CO 2 3 correct 8. H 2 CO 3 ; H 2 CO 3 Explanation: To produce a conjugate acid, HCO 3 must be acting as a base. Therefore it is a proton acceptor forming H 2 CO 3 . To produce a conjugate base, it must be acting as an acid. Therefore it is a proton donator forming CO 2 3 . 006
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Exam 2 - Version 033 Exam 2 Holcombe (52460) This print-out...

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