HW06-solutions

HW06-solutions - kamalska (mk23835) HW06 Quigley (104001) 1...

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Unformatted text preview: kamalska (mk23835) HW06 Quigley (104001) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Calculate the value of the equilibrium con- stant for the reaction AgCl(s) + 2 NH 3 (aq) Ag(NH 3 ) + 2 (aq) + Cl- (aq) given K sp = 1 . 6 10- 10 for silver chloride and K f = 1 . 6 10 7 for the ammonia complex Ag(NH 3 ) + 2 of Ag + ions. 1. 6 . 3 10 9 2. 6 . 3 10- 8 3. 1 . 10 17 4. 1 . 10- 17 5. 2 . 6 10- 3 correct Explanation: 002 10.0 points A buffer is formed by mixing 100 mL of 0.1 M NH 4 Cl with 50 mL of 0.3 M NH 3 . To this solution 19 mL of 0 . 1 M NaOH is added. What is the approximate pH of the new so- lution after neutralization? Assume a K b of 1 . 8 10- 5 for the weak base. 1. 9 . 49 2. 9 . 57 correct 3. 4 . 43 4. 4 . 58 5. 9 . 42 Explanation: V NH 4 Cl = 100 mL M NH 4 Cl = 0.1 M V NH 3 = 50 mL M NH 3 = 0.3 M V NaOH = 19 mL M NaOH = 0 . 1 M K b = 1 . 8 10- 5 Neutralization Step: 19 mL of 0 . 1 M NaOH (strong base, 1 . 9 mmol) reacts with NH + 4 (weak acid, 10 mmol) to give 8 . 1 mmol NH- 4 and 16 . 9 mmol NH 3 : NH 3 +H 2 O NH + 4 + OH- 50 mL 100 mL . 3 M . 1 M 15 mmol 10 mmol starting buffer +1 . 9 mmol- 1 . 9 mmol neutraliz . reaction 16 . 9 mmol 8 . 1 mmol neutraliz . products 16 . 9 mmol 169 mL 8 . 1 mmol 169 mL x = 0 . 1 M = 0 . 047929 M K b = [OH- ] [NH + 4 ] [NH 3 ] [OH- ] = K b [NH 3 ] [NH + 4 ] = ( 1 . 8 10- 5 ) (0 . 1 M) . 047929 M = 3 . 75556 10- 5 pOH =- log[OH- ] =- log(3 . 75556 10- 5 ) = 4 . 42533 pH = 14- pOH = 9 . 57467 003 10.0 points A ligand is best described as which of the following? 1. an Arrhenius acid 2. a Lowry-Bronsted acid 3. an Arrhenius base 4. a Lewis base correct 5. a Lewis acid 6. a Lowry-Bronsted base Explanation: Ligands must have lone pair electrons to donate to the transition metal to make a com- plex ion. That is essentially the definition of a Lewis base. kamalska (mk23835) HW06 Quigley (104001) 2 004 10.0 points What is the pH of a solution containing 0.3 M NH 4 Cl and 0.6 M NH 3 ? The p K a of the ammonium ion is 9.25. 1. 12.25 2. 5.05 3. 4.45 4. 9.55 correct 5. 8.95 Explanation: For NH + 4 , p K a = 9.25 p K b = p K w- p K a = 14- 9 . 25 = 4 . 75 p K b =- log K b K b = 10- p K b = 10- 4 . 75 = 1 . 77828 10- 5 Building an ICE using molarities, NH 3 (aq)+H 2 O( ) NH + 4 (aq)+OH- (aq) 0.6- x + x + x . 6- x . 3 + x + x K b = [NH + 4 ] [OH- ] [NH 3 ] = (0 . 3 + x ) x . 6- x . 3 x . 6 x = . 6 K b . 3 = . 6 (1 . 77828 10- 5 ) . 3 = 3 . 55656 10- 5 = [OH- ] [H + ] = K w [OH- ] = 1 10- 14 3 . 55656 10- 5 = 2 . 81171 10- 10 Thus pH =- log(2 . 81171 10- 10 ) = 9 . 55103 005 10.0 points Choose the effective pH range of a HF- NaF buffer. For HF, K a = 3 . 5 10- 4 ....
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HW06-solutions - kamalska (mk23835) HW06 Quigley (104001) 1...

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