Sample Final Solution - Solutions to Practice Problems for...

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Solutions to Practice Problems for Final Exam CHEM 102, Section E2 PART A: True-False/Short Answer For the True-False questions, circle the correct answer. For short answer problems, write your answer in the space provided. A.1. (4 marks) When CuCl (s) dissolves in 2.0 M NH 3 (aq) solution, the complex formed binds only 4 ammonia molecules to the metal [Cu(NH 3 ) 4 ] + (aq). (a) What is the formula and name of the metal salt in solution after reaction? ANSWER: [Cu(NH 3 ) 4 ]Cl, Tetraamminecopper(I) chloride (b) What is the shape of this complex? ANSWER: Tetrahedral. CN = 4 and it has 10 d-electrons (Only d 8 complexes with CN = 4 are square planar). (c) Is this complex diamagnetic or paramagnetic? ANSWER: This is a d 10 complex, so it has no unpaired electrons, and is diamagnetic. A.2. The activation energy of a reaction depends on the reaction temperature. True False A.3. The activation energy of a reaction depends on the fraction of reactants with the proper orientation. True False A.4. The higher the activation energy, the slower the reaction rate. True False A.5. A bimolecular reaction is generally twice as fast as a unimolecular reaction. True False PART B: Long Answer B.1. (8 marks) Quinoline (C 9 H 7 N) is a weak base (pK b = 9.5) used to preserved anatomical specimens. Its solubility in water at 25 o C is 0.6 g/100 mL. (a) What is the pH of a saturated solution of quinoline in water at this temperature? ANSWER: K b = 10 -9.5 = 3.2 ! 10 -10 Convert solubility to M: s = (0.6 g/100 mL)(1000 mL/L)(1 mol/129.16 g) = 0.046 M Build ice table for the base hydrolysis reaction in water: C 9 H 7 N (aq) + H 2 O (l) " C 9 H 8 N + (aq) + OH - (aq) i. 0.046 --- 0 0 c. -x --- +x +x e. 0.046-x --- x x
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3.2 ! 10 -10 = x 2 / 0.046-x ~ x 2 / 0.046 x = 3.8 x 10 -6 (error = 0.008 % OK) x = [OH - ] = 3.8 ! 10 -6 pOH = 5.4 and pH = 14 - pOH = 8.6 (b) When 100 mL of this saturated solution of quinoline is converted to the conjugate weak-acid formquinoline hydrochloride (C 9 H 8 NCl), what is the pH of this solution? (Assume no dilution) ANSWER: If there is no dilution, then when the weak base quinoline is completely converted to the weak acid form, [C 9 H 8 N + ] = 0.046 M Now build an ice table for the acid dissociation in water: C 9 H 8 N + (aq) + H 2 O (l) " C 9 H 7 N (aq) + H 3 O + (aq) i. 0.046 --- 0 0 c. -x --- +x +x e. 0.046-x --- x x The above reaction is governed by K a , so find K a and solve for x: K w /3.2 ! 10 -10 = 3.125 ! 10 -5 = x 2 / 0.046-x ~ x 2 / 0.046 x = 1.2 ! 10 -3 (error = 2.6 % OK) x = [H 3 O + ] = 1.2 x ! 10 -3 pH = 2.9 B.2. (8 marks) The following questions refer to the plot of reaction progress given below. (a)
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This note was uploaded on 04/22/2010 for the course CHEM 102 taught by Professor Brown during the Winter '10 term at University of Alabama at Birmingham.

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Sample Final Solution - Solutions to Practice Problems for...

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