Chapter09 - Solutions for Chapter 9 End-of-Chapter Problems...

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March 2005 ACS Chemistry Cahpter 9 suggested solutions 1 Solutions for Chapter 9 End-of-Chapter Problems Problem 9.1. (a) Balanced chemical equations for the formation of pale blue Cu(OH) 2 (s) and deep blue-violet [Cu(NH 3 ) 4 ] 2+ (aq) when NH 3 (aq) is slowly added to a light blue solution of Cu 2+ (aq) are: 2NH 3 (aq) + 2H 2 O (l) + Cu 2+ (aq) Cu(OH) 2 (s) + 2NH 4 + (aq) light blue pale blue Cu(OH) 2 (s) + 4NH 3 (aq) [Cu(NH 3 ) 4 ] 2+ (aq) + 2OH (aq) pale blue deep blue-violet (b) To understand how the formation of Cu(OH) 2 (s) from Cu 2+ (aq) will be affected by basic conditions, we need to consider the reactions that make up the net reaction written in part (a): 2NH 3 (aq) + 2H 2 O (l) 2NH 4 + (aq) + 2OH (aq) Cu 2+ (aq) + 2OH (aq) Cu(OH) 2 (s) Additional OH (aq) is a disturbance to these equilibria (Le Chatelier’s principle) and the system will respond by using up some of the additional OH (aq) . Thus, the first reaction will shift toward reactants, because OH (aq) is a product, and the second reaction will shift toward products because OH (aq) is a reactant. The net result will be to favor formation of Cu(OH) 2 (s) from Cu 2+ (aq) . (c) The equation written in part (a) suggests that the formation of [Cu(NH 3 ) 4 ] 2+ (aq) from Cu(OH) 2 (s) will be favored under mildly acidic conditions, because H 3 O + (aq) will react with OH (aq) ions to form water. This reaction is a disturbance to the equilibrium, because it removes a product of the reaction, The system should respond by trying to form more OH (aq) , which also would mean the formation of more [Cu(NH 3 ) 4 ] 2+ (aq) . This analysis is flawed because it neglects the basicity of NH 3 (aq) , which will also react with added H 3 O + (aq) : NH 3 (aq) + H 3 O + (aq) NH 4 + (aq) This reaction removes a reactant, NH 3 (aq) , necessary for formation of [Cu(NH 3 ) 4 ] 2+ (aq) , so the presence of H 3 O + (aq) in the solution will not favor formation of [Cu(NH 3 ) 4 ] 2+ (aq) . Note that the reaction of H 3 O + (aq) with OH (aq) also removes a reactant in the second reaction in part (b), so Cu(OH) 2 (s) will tend to dissolve or not form under acidic conditions. (d) None of the reactions involves oxidation or reduction. Copper has an oxidation number of +2 in all cases. Problem 9.2. If Fe 3+ (aq) and SCN (aq) formed a one-to-two metal ion complex, the complex in Figure 9.1 would have to show the new ratio of two SCN (aq) ions to every Fe 3+ (aq) ion. Twice as many SCN (aq) ions would be used in forming the complex and there would be four more unreacted Fe 3+ (aq) ions in the right-hand part of the figure, as shown in this alternative figure:
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Chemical Equilibria Chapter 9 2 ACS Chemistry Chapter 9 suggested solutions Problem 9.3. (a) Addition of CN (aq) to the equilibrium reaction, Cd 2+ (aq) + 6CN (aq) Cd(CN) 6 2– (aq) , will cause the reaction to shift toward formation of more of the product, Cd(CN) 6 2– (aq) . Addition of more of a reactant is a disturbance to the equilibrium system (Le Chatelier’s principle) and the system responds by minimizing the disturbance, using up some of the added CN (aq) to form more of the product.
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Chapter09 - Solutions for Chapter 9 End-of-Chapter Problems...

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