Exam2_HW4-7

# Exam2_HW4-7 - Version 026 – Exam 2 – Mccord –(52445 1...

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Unformatted text preview: Version 026 – Exam 2 – Mccord – (52445) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. McCord CH302 10am This exam is only for McCord’s MWF 10am CH302 class. 001 10.0 points What is the pH of a 0.020 M solution of hydrosulfuric acid, a diprotic acid? K a1 = 1 . 1 × 10 − 7 K a2 = 1 . × 10 − 14 1. 5.22 2. 3.65 3. 7.00 4. 9.67 5. 4.33 correct 6. 4.69 7. 7.84 Explanation: Solve using ONLY the 1st ionization. So this works like any other monoprotic acid where the assumption [H + ] = radicalbig (Conc)( K a1 ) is valid. 002 10.0 points The figure represents a reaction at 298 K. b b b b b G A B C D E rxn progress Which statement is FALSE? 1. The position of equilibrium will change if the temperature is changed. 2. The standard free energy of reaction is negative. 3. K is less than 1. correct 4. Pure reactants are represented by point A. Explanation: K is greater than 1 because Δ G ◦ is negative (point E has lower free energy than point A). 003 10.0 points Consider the following reaction: PCl 5 (g) → PCl 3 (g) + Cl 2 (g) . At equilibrium, [PCl 5 ] = 2.00 M and [PCl 3 ] = [Cl 2 ] = 1.00 M. If suddenly 1.00 M PCl 5 (g), PCl 3 (g), and Cl 2 (g) are each added, calculate the equilibrium concentration of PCl 3 (g). 1. 1.35 M correct 2. 2.32 M 3. 2.18 M 4. 1.43 M 5. 3.35 M 6. 1.95 M Explanation: K = (1)(1) 2 = 0 . 5 after the addition each concentration in- creases by 1 M and then the reaction must shift to the left to reestablish equilibrium. K = (2 − x ) 2 (3 + x ) = 0 . 5 which converts to 0 = x 2 − 4 . 5 x + 2 . 5 Version 026 – Exam 2 – Mccord – (52445) 2 The quadratic formula will yield two an- swers for x . x = 0 . 64929 or 3 . 85078 The 3.85078 answer is not physically possible and therefore x = 0 . 64929. [PCl 3 ] = 2 − x = 1 . 35078 M 004 10.0 points The conjugate base of H 2 SO 4 is: 1. H 3 O + 2. H 2 SO 3 3. H 3 SO + 4 4. SO 2 − 4 5. HSO − 4 correct 6. OH − 7. H 2 O Explanation: Remove one H + ion from a species to find its conjugate base. 005 10.0 points Which of the following would be equal to K a 4 for orthocarbonic acid, H 4 CO 4 ? 1. [HCO 3 − 4 ] [H + ] · [CO 4 − 4 ] 2. [H + ] 4 · [CO 4 − 4 ] [HCO 3 − 4 ] 3. [H + ] 4 · [CO 4 − 4 ] [H 4 CO 4 ] 4. [H 4 CO 4 ] [H + ] 4 · [CO 4 − 4 ] 5. [H + ] · [CO 4 − 4 ] [HCO 3 − 4 ] correct Explanation: K a 4 for orthocarbonic acid, H 4 CO 4 , describes the 4 th deprotonation event. HCO 3 − 4 ↔ H + + CO 4 − 4 006 10.0 points A solution of HClO is mixed and found to have a pH of 3 . 99. Find what the initial concentration of HClO was for this solution. K a = 3 . 00 × 10 − 8 for HClO....
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Exam2_HW4-7 - Version 026 – Exam 2 – Mccord –(52445 1...

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