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**Unformatted text preview: **Version 291 – Exam 2 – Holcombe – (52460) 1 This print-out should have 33 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Some constants: Kw=1e-14; R=8.314 J/K- mol 001 10.0 points When dissolved in water, bases produce 1. salts. 2. H + ions. 3. OH − ions. correct 4. hydronium ions. 5. H 3 O + ions. Explanation: Bases produce hydroxide ions (OH − ) in wa- ter; acids produce hydronium ions (H 3 O + ). 002 10.0 points A diprotic acid H 2 A has values of K a1 = 1 . × 10 − 6 and K a2 = 1 . × 10 − 10 . What is the [A 2 − ] in a 0.10 M solution of H 2 A? 1. 1 . × 10 − 10 M correct 2. 3 . 2 × 10 − 6 M 3. 0.10 M 4. 0.20 M 5. 3 . 2 × 10 − 4 M Explanation: 003 10.0 points How many grams of HNO 3 are needed to make 315 mL of a solution with a pH of 2.60? The molar mass of HNO 3 is 63.02 g/mol. 1. 19.9 g 2. 0.158 g 3. 1 . 25 × 10 − 5 g 4. 0.0498 g correct 5. 0.502 g Explanation: pH = 2 . 6 V = 315 mL = 0 . 315 L MM = 63 . 02 g/mol [H + ] = 10 − pH = 10 − 2 . 6 = 0 . 00251189 Since HNO 3 is a strong acid, HNO 3 (aq) → H + (aq) + NO − 3 (aq) . One HNO 3 → one H + , so [H + ] = [HNO 3 ] = 0 . 00251189 mol and (0 . 315 L) × . 00251189 mol HNO 3 1 L = 0 . 000791244 mol HNO 3 . Thus (0 . 000791244 mol HNO 3 ) × 63 . 02 g HNO 3 1 mol HNO 3 = 0 . 0498642 g HNO 3 004 10.0 points What would be the pH of a 1 M solution of Na 2 HPO 4 ? Assume H 3 PO 4 has a p K a 1 of 2 and a p K a 2 of 7 and a p K a 3 of 13. 1. 10 correct 2. 13 3. 7.5 4. 4.5 5. 2 6. 7 Explanation: For a solution composed of a single ampho- teric species (HPO 4 2 − ), pH = 0 . 5(p K a 2 + p K a 3 ) = 0 . 5(7 + 13) = 10 Version 291 – Exam 2 – Holcombe – (52460) 2 005 10.0 points A solution containing an unspecified amount of HCl and ammonia (NH 3 ) has a pH of 8 . 81 What is the ratio of NH 3 to NH + 4 in the solution? (For NH 3 , K b = 1 . 8 × 10 − 5 ) 1. . 233 2. . 359 correct 3. 2 . 06 4. cannot be determined 5. . 485 6. 2 . 79 7. 5 . 6 × 10 − 10 Explanation: Consider this equilibria... NH + 4 ⇀ ↽ NH 3 + H + Rearranging the equilibrium expression gives... K a [H + ] = [NH3] [NH4+] = ratio K a = K w / 1 . 8 × 10 − 5 = 5 . 556 × 10 − 10 [H+] = 10 − 8 . 81 = 1 . 54882 × 10 − 9 M ratio = 5 . 556 × 10 − 10 1 . 54882 × 10 − 9 M = 0 . 359 Note that the same answer could be ob- tained by using the base equilibria and K b 006 10.0 points You have a weak molecular base with K b = 6 . 6 × 10 − 9 . What is the pH of a 0.0500 M solution of this weak base? 1. None of these 2. pH = 7.12 3. pH = 3.63 4. pH = 9.26 correct 5. pH = 4.74 Explanation: [base] = 0.05 M As mentioned, this is a weak base, so use the equation to calculate weak base [OH − ] concentration: [OH − ] = radicalbig K b C b = radicalBig (6 . 6 × 10 − 9 ) (0 . 05) = 1 . 81659 × 10 − 5 After finding [OH − ], you can find pH using either method below: A) pOH =- log ( 1 . 81659 × 10 − 5 ) = 4 . 74074...

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