Exam 3 Final Version-solutions

Exam 3 Final Version-solutions - Version 065 Exam 3 Final...

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Version 065 – Exam 3 Final Version – Lyon – (53565) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. PLEASE CHECK TO BE SURE YOUR NAME, EID AND EXAM VERSION NUM- BER IS CORRECTLY BUBBLED BEFORE YOU TURN IN YOUR EXAM ANSWER SHEET. 001 10.0 points The pH o± 0.50 M HNO 2 (aq) is 1.8. There±ore, the pH o± a solution that is 0.50 M HNO 2 (aq) and 0.10 M KNO 2 (aq) is greater than 1.8. 1. False 2. True correct Explanation: 002 10.0 points The un-ionized ±orm o± an acid indicator is yellow and its anion is blue. The K a o± this indicator is 10 - 5 . What will be the color o± the indicator in a solution o± pH 3? 1. orange 2. red 3. yellow correct 4. blue 5. green Explanation: K a = 10 - 5 pH = 3 HIn yellow + H 2 O H 3 O + + In - blue K a = 10 - 5 p K a = - log ( 10 - 5 ) = 5 The color change range is pH = p K a ± 1. At pH values above 6 the indicator will be ionized and at pH values below 4 the indicator will be un-ionized. 003 10.0 points Consider the titration o± 15.0 mL o± 0.200 M H 3 PO 4 (aq) with 0.200 M NaOH(aq). What is/are the major species in solution a±ter the addition o± 15.0 mL o± base? 1. H 2 PO - 4 (aq) and HPO 2 - 4 (aq) 2. H 3 PO 4 (aq) and H 2 PO - 4 (aq) 3. H 2 PO - 4 (aq) correct 4. PO 3 - 4 (aq) Explanation: 004 10.0 points Fe(OH) 3 (s) is very insoluble in water ( K sp = 1 . 6 × 10 - 39 ; however, Fe 3+ ±orms a strong complex with EDTA (FeEDTA - , K f = 1 . 3 × 10 25 ). For a solution which is at equi- librium and contains Fe(OH) 3 (s) precipitate, which o± the ±ollowing occurs i± EDTA is added to the solution? 1. Nothing happens because K sp is much smaller than K f . 2. More Fe(OH) 3 (s) precipitates because [Fe 3+ ] decreases. 3. No more Fe(OH) 3 (s) dissolves, but Fe 3+ complexes with EDTA. 4. More Fe(OH) 3 (s) dissolves. correct Explanation: 005 10.0 points What is K sp ±or Ag 3 PO 4 , i± its molar solubility is 2 . 7 × 10 - 6 mol / L? 1. 2 . 0 × 10 - 17 2. 5 . 3 × 10 - 16 3. 1 . 4 × 10 - 21 correct 4. 7 . 3 × 10 - 12 5. 4 . 8 × 10 - 22
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Version 065 – Exam 3 Final Version – Lyon – (53565) 2 6. 1 . 7 × 10 - 14 7. 5 . 3 × 10 - 23 Explanation: S = 2 . 7 × 10 - 6 mol / L The solubility equilibrium is Ag 3 PO 4 (s) 3 Ag + (aq) + PO 3 - 4 (aq) [Ag + ] = 3 S = 8 . 1 × 10 - 6 mol / L [PO 3 - 4 ] = S = 2 . 7 × 10 - 6 mol / L K sp = [Ag + ] 3 [PO 3 - 4 ] = ( 8 . 1 × 10 - 6 ) 3 (2 . 7 × 10 - 6 ) = 1 . 43489 × 10 - 21 006 10.0 points A 0.50 M solution of dimethylamine (a weak base with K b = 7 . 4 × 10 - 4 ) would be resistant to large changes of pH Z1) when a strong acid is added, but not if a strong base is added. Z2) when a strong base is added, but not if a strong acid is added. Z3) when either a strong acid or a strong base
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This note was uploaded on 04/27/2008 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

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Exam 3 Final Version-solutions - Version 065 Exam 3 Final...

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