Look at the two equations and the values of their equilibrium constants Ag Cl

Look at the two equations and the values of their

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Look at the two equations and the values of their equilibrium constants: Ag + + Cl AgCl(s) K = 1/K sp = 5.6 × 10 9 Ag + + 2 Cl [AgCl 2 ] K f = 2.5 × 10 5 The precipitation reaction is more product-favored, but less susceptible to changes in the concentration of the chloride ion. In the dilute chloride solution, the silver ion reacts with just one chloride ion to make the solid, and we see precipitation. In the presence of much higher chloride concentration, each silver ion must react with two chloride ions, forming a complex ion, which is why we see no precipitation. Complex Ion Formation 72 . Answer: see equations and expressions below Strategy and Explanation: The charge of the reactant metal ion is determined by subtracting the Lewis base’s charge(s), if any, from the complex ion charge. (a) Ag + + 2 CN [Ag(CN) 2 ] K f = [[Ag(CN) 2 ] " ] [Ag + ][CN " ] 2 (b) Cd 2+ + 4 NH 3 [Cd(NH 3 ) 4 ] 2+ K f = [[Cd(NH 3 ) 4 ] 2 + ] [Cd 2 + ][NH 3 ] 4 73. Answer: see equations and expressions below Strategy and Explanation: The charge of the reactant metal ion is determined by subtracting the Lewis base’s charge(s), if any, from the complex ion charge. (a) Co 3+ + 6 Cl [CoCl 6 ] 3– K f = [[CoCl 6 ] 3 " ] [Co 3 + ][Cl " ] 6 (b) Zn 2+ + 4 OH [Zn(OH) 4 ] 2– K f = [Zn(OH) 4 ] 2 " ] [Zn 2 + ][OH " ] 4 74 . Answer: 7.8 × 10 –3 mol or more Strategy and Explanation: Adapt the method described in the solution to Question 70, and Problem-Solving Example 17.13. The Na 2 S 2 O 3 salt provides a source of S 2 O 3 2– , a Lewis base capable of forming a complex ion with the silver ion. If all the solid is dissolved it must be have enough S 2 O 3 2– to complex enough the silver ions to prevent the precipitation of AgBr in the resulting Br solution. First, we get the balanced equation for the reaction from Problem-Solving Example 17.13 and determine the value of its equilibrium constant using Tables 17.2 and 17.3: AgBr(s) + 2 S 2 O 3 2– (aq) [Ag(S 2 O 3 ) 2 ] 3– (aq) + Br (aq) K = [[Ag(S 2 O 3 ) 2 ] 3 " ][Br " ] [S 2 O 3 2 " ] 2 K = K sp × K f = [Ag + ][Br " ] # [[Ag(S 2 O 3 ) 2 ] 3 " ] [Ag + ][S 2 O 3 2 " ] 2 = [[Ag(S 2 O 3 ) 2 ] 3 " ][Br " ] [S 2 O 3 2 " ] 2 K = K sp × K f = (3.3 × 10 –13 ) × (2.0 × 10 13 ) = 6.6 Now, calculate the [Br ] and [[Ag(S 2 O 3 ) 2 ] 3– ], once all the AgBr has dissolved:
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Chapter 17: Additional Aqueous Equilibria 805 [Br ] = 0.020 mol AgBr 1.0 L " 1 mol Br # 1 mol AgBr = 0.020 M, Similarly, [[Ag(S 2 O 3 ) 2 ] 3– ] = 0.020 M Now, we can calculate the necessary [S 2 O 3 2– ]: 6.6 = (0.020)(0.020) [S 2 O 3 2 " ] 2 [S 2 O 3 2– ] = 7.8 × 10 –3 M One mole of S 2 O 3 2– is found in each mole of Na 2 S 2 O 3 , so we must add 7.8 × 10 –3 mol of Na 2 S 2 O 3 to this 1.0 L solution. 75. Answer: 1.2 × 10 –7 M Strategy and Explanation: Take the product-favored reaction completely to products, then return it to equilibrium using one variable, x. Ag + (aq) + 2 NH 3 (aq) [Ag(NH 3 ) 2 ] + (aq) initial conc. (M) 0.063 c 0 change as forward reaction occurs (M) – 0.063 – 0.126 + 0.063 final conc. (M) 0 c – 0.126 0.063 change as reverse reaction occurs (M) + x + 2x – x equilibrium conc. (M) x 0.18 M = c – 0.126 + 2x 0.063 – x K f = [[Ag(NH 3 ) 2 ] + ] [Ag + ][NH 3 ] 2 = 1.6 × 10 7 1.6 × 10 7 = (0.063 " x) (x)(0.18) 2 Solve for x: (1.6 × 10 7 ) × (x)(0.18) 2 = 0.063 – x (5.2 × 10 5 )x = 0.063 – x [1 + (5.2 × 10 5 )]x = 0.063 x = 1.2 × 10 –7 [[Ag(NH 3 ) 2 ] + ] = 0.063 M – 1.2 × 10 –7 = 0.063 M [Ag + ] = 1.2 × 10 –7 M 76 . Answer: see equations below
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