#Chem 162-2008 Exam II + answers

#Chem 162-2008 Exam II + answers - Exam II Spring 2008...

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Exam II Spring 2008 White version Chem 162-2008 Exam II + Answers Chapter 16 Equilibria Solubility product Given K sp and common ion concentration of salt, find solubility of salt. 1. Calcium carbonate, CaCO 3 has K sp = 2.8 x 10 -9 What amount of calcium carbonate will dissolve in 250 mL of water which contains 0.10 moles of Na 2 CO 3 already dissolved in it? (a) 1.3 x 10 -5 moles (b) 3.3 x 10 -6 moles (c) 2.8 x 10 -8 moles (d) 1.8 x 10 -9 moles (e) 1.4 x 10 -10 moles 0.10 mol/0.250L = 0.40M CaCO 3 Ca 2+ + CO 3 2- CaCO 3 (s) Ca 2+ (aq) + CO 3 2- (aq) Initial Y 0 0.40 Change -X +X +X Equilibrium Y-X +X 0.40 + X [Ca 2+ ][CO 3 2- ] = 2.8 x 10 -9 = K sp [X][0.40 + X] = 2.8 x 10 -9 Avoid a quadratic equation by using the small K rule. [X][0.40] = 2.8 x 10 -9 X = 7.0 x 10 -9 M 7.0 x 10 -9 mol/L of CaCO 3 will dissolve/L; therefore 1.75 x 10 -9 mol of CaCO 3 will dissolve in 250 mL of water. Chem 162-2008 Exam II + Answers Acid and base concepts Strong acids vs. Lewis acids 2. Which of the following ions is the strongest Lewis acid? (a) Na + (b) Cl - (c) CH 3 COO - (d) Ba 2+ (e) Fe 3+
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Na + is a Lewis acid. However, since it is the conjugate acid of a strong base it is such a weak acid that it is not even considered to be an acid. It is considered to be a spectator ion. (b) False. Cl - is a Lewis base. (c) False. CH 3 COO - is a Lewis base. (d) False. Ba 2+ is a Lewis acid. However, since it is the conjugate acid of a strong base it is such a weak acid that it is not even considered to be an acid. It is considered to be a spectator ion. (e) True. Fe 3+ is a Lewis acid in that it reacts with an electron pair, such as OH - (to form Fe(OH) 3 ). Chem 162-2008 Exam II + Answers Chapter 14 Non Acid/Base Chemical Equilibrium Non acid-base equilibrium concepts Adding equilibrium equations and multiplying K’s 3. Consider the following two reactions and their equilibrium constants. Reaction 1: NO 2 (g) NO(g) + O(g) K 1 = 6.8 x 10 -49 Reaction 2: NO 2 (g) + O 2 (g) O 3 (g) + NO(g) K 2 = 1.7 x 10 33 Calculate K for the reaction: O 2 (g) + O(g) O 3 (g) (a) 1.2 x 10 -15 (b) 4.0 x 10 -82 (c) 8.7 x 10 14 (d) 2.5 x 10 81 (e) 4.3 x 10 41 This is Hess’ law. Reaction 1 gets reversed; therefore, K 1 gets inverted. Reaction 2 doesn’t change. The sum of reversed reaction 1 and reaction 2 equals the product of K 2 and inverted K 1 . NO(g) + O(g) NO 2 (g) 1/K 1 = 1/(6.8 x 10 -49 )= 1.47 x 10 48 NO 2 (g) + O 2 (g) O 3 (g) + NO(g) K 2 = 1.7 x 10 33 O 2 (g) + O(g) O 3 (g) (1/K 1 ) x (K 2 ) = (1.47 x 10 48 ) x (1.7 x 10 33 ) = 2.50 x 10 81 Chem 162-2008 Exam II + Answers Chapter 14 Non Acid/Base Chemical Equilibrium Non acid-base equilibrium calculations Given initial value and K, find equilibrium values 4. 2NOCl(g) 2NO(g) + Cl 2 (g) K p = 7.2 x 10 -6 NOCl(g) is added to a container, exerting a pressure of 0.80 atm before any reaction. After equilibrium is established, what is the pressure of Cl 2 in the container? (a)
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#Chem 162-2008 Exam II + answers - Exam II Spring 2008...

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