Chemistry by Whitten, Atwood, Morrison Chapter 11 solutions

Chemistry by Whitten, Atwood, Morrison Chapter 11 solutions...

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97 Reactions in Aqueous Solutions II: Calculations 11 11-1 Molarity is a convenient method for expressing the concentration of a solution because a specific volume of the solution will contain a fixed number of moles of solute dissolved in the solvent. 11-3 (a) ? mol H 3 AsO 4 L = 27.8 g H 3 AsO 4 0.400 L x 1 mol H 3 AsO 4 141.95 g H 3 AsO 4 = 0.490 M H 3 AsO 4 (b) ? mol (COOH) 2 L = 6.33 g (COOH) 2 0.500 L x 1 mol (COOH) 2 90.0 g (COOH) 2 = 0.141 M (COOH) 2 (c) ? mol (COOH) 2 ·2H 2 O L = 6.33 (COOH) 2 ·2H 2 O 0.500 L x 1 mol (COOH) 2 ·2H 2 O 126 g (COOH) 2 ·2H 2 O x 1 mol (COOH) 2 1 mol (COOH) 2 ·2H 2 O = 0.100 M (COOH) 2 11-5 ? mol Fe NO 3 ( ) 2 L = 47.5 g Fe NO 3 ( ) 2 0.750 L x 1 mol Fe NO 3 ( ) 2 179.87 g Fe NO 3 ( ) 2 = 0.352 M Fe NO 3 ( ) 2 11-7 ? mol HNO 3 L = 1110 g soln L x 19.0 g HNO 3 100 g soln x 1 mol HNO 3 63.0 g HNO 3 = 3.35 M HNO 3 11-9 HCl + LiOH LiCl + H 2 O, LiCl is lithium chloride. Final volume of solution = 500 mL + 500 mL = 1000 mL or 1.000 L ? mol LiCl L 0.500 L x 3.00 mol HCl 1.00 L x 1 mol LiCl 1 mol HCl x 1 1.000 L soln = 1.50 M LiCl 11-11 2HCl + Ba(OH) 2 BaCl 2 + 2H 2 O, BaCl 2 is barium chloride. Final volume of solution = 16.00 mL + 9.00 mL = 25.00 mL ? mol BaCl 2 L = 0.01600 L x 4.50 M HCl x 1 mol BaCl 2 2 mol HCl x 1 0.02500 L soln = 1.44 M BaCl 2 Note: Both volumes above should be in the same unit. Also, 4.50 M stands for 4.50 mol HCl 1 L soln . You could use 0.01600 L of 4.50 M HCl or 0.00900 L of 4.00 M Ba(OH) 2 to arrive at the same 1.44 M BaCl 2 . ? mol BaCl 2 L = 0.00900 L x 4.00 M Ba(OH) 2 x 1 mol BaCl 2 1 mol Ba OH ( ) 2 x 1 0.02500 L soln = 1.44 M BaCl 2

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98 11-13 NH 3 + HCl NH 4 Cl, 0.0175 L x 8.00 M NH 3 = 0.140 moles of NH 3 0.0210 L x 12.00 M HCl = 0.252 moles of HCl NH 3 is the limiting reactant. Final volume of solution = 21.0 mL + 17.5 mL = 38.5 mL = 0.0385 L ? mol NH 4 Cl L = 0.140 moles of NH 3 x 1 mol NH 4 Cl 1 mol NH 3 x 1 0.0385 mL soln = 3.64 M NH 4 Cl 11-15 Final volume of solution = 150. mL + 200. mL = 350. mL ? mmol HCl added = 150. mL x 0.200 mmol HCl mL = 30.0 mmol HCl ? mmol Ba(OH) 2 added = 200. mL x 0.0400 mmol Ba(OH) 2 mL = 8.0 mmol Ba(OH) 2 2HCl + Ba(OH) 2 BaCl 2 + 2H 2 O rxn ratio: 2 mol 1 mol 1 mol start: 30.0 mmol 8.0 mmol 0 mmol change: -16.0 mmol -8.0 mmol +8.0 mmol final: 14.0 mmol 0 mmol 8.0 mmol Molarity BaCl 2 = 8.0 mmol BaCl 2 350. mL = 0.0229 M BaCl 2 Molarity HCl = 14.0 mmol HCl 350. mL = 0.0400 M HCl 11-17 Consider 1 L (1000. mL) of solution: ? mol NH 3 = 1000 mL/L x 0.979 g soln 1.00 mL soln x 5.03 g NH 3 100 g soln x 1 mol 17.0 g NH 3 = 2.90 mol ? M = 2.90 mol 1.00 L = 2.90 M 11-19 (a) standard solution: a solution of accurately known concentration (b) titration : the process by which the volume of a standard solution required to react with a specific amount of a substance is determined. (c) primary standard : a substance of known high degree of purity that undergoes one invariable reaction with the other reactant of interest. (d) secondary standard : a solution that has been titrated against a primary standard. A standard solution is a secondary standard. 11-21 The net ionic equation is the equation that results from canceling spectator ions and eliminating brackets from a total ionic equation.
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