AgBr m o Ag Br Initial Change Equilibrium 18 x 10 8 X Ag Br K sp 18 x 10 8 X 50

Agbr m o ag br initial change equilibrium 18 x 10 8 x

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AgBr m o Ag + + Br - Initial Change Equilibrium 1.8 x 10 -8 X [Ag + ][Br - ] = K sp [1.8 x 10 -8 ][X] = 5.0 x 10 -13 X = 2.78 x 10 -5 = [Br - ] in solution when AgCl just about to precipitate.
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68 CHEM 162-2007 EXAM II + ANSWERS CHAPTER 16A – EQUILIBRIA SOLUBILITY PRODUCT 11. For CaF 2 , K sp = 5.3 x 10 -9 . What amount of CaF 2 would dissolve in 1.00 L of a solution which contains 0.60 mol of dissolved NaF? (a) 1.5 x 10 -8 mol (b) 7.4 x 10 -5 mol (c) 1.1 x 10 -3 mol (d) 2.2 x 10 -4 mol (e) 6.4 x 10 -6 mol CaF 2 m o Ca 2+ + 2F - CaF 2 m o Ca 2+ + 2F - Initial Y 0 0.60 Change -X +X +2X Equilibrium Y - X +X 0.60 + 2X [Ca 2+ ][F - ] 2 = K sp [X][0.60 + 2X] 2 = 5.3 x 10 -9 Use the small K rule to drop the “+2X” [X][0.60] 2 = 5.3 x 10 -9 ((([X] x ([0.60] 2 )) = (5.3 x 10 -9 )),X) X = 1.47 x 10 -8 M CHEM 162-2007 EXAM II + ANSWERS CHAPTER 16A – EQUILIBRIA SOLUBILITY PRODUCT 14. A saturated solution of calcium hydroxide, Ca(OH) 2 has pH = 12.40. Calculate K sp for calcium hydroxide. (a) 7.9 x 10 -6 (b) 4.1 x 10 -5 (c) 2.2 x 10 -6 (d) 3.8 x 10 -4 (e) 1.7 x 10 -8 pH = 12.40 pH + pOH = 14.00 pOH = 1.60
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69 [OH - ] = 10 -pOH = 10 -1.60 = 2.51 x 10 -2 Ca(OH) 2 m o Ca 2+ + 2OH - Ca(OH) 2 (s) m o Ca 2+ (aq) + 2OH - (aq) Initial Y 0 0 Change -X +X +2X Equilibrium Y - X +X 2.51 x 10 -2 0 + 2X = 2.51 x 10 -2 X = 1.255 x 10 -2 Ca(OH) 2 (s) m o Ca 2+ (aq) + 2OH - (aq) Initial Change Equilibrium Y – (1.255 x 10 -2 ) +1.255 x 10 -2 2.51 x 10 -2 [Ca 2+ ][OH - ] 2 = K sp [1.255 x 10 -2 ][2.51 x 10 -2 ] 2 = K sp ([1.255 x 10 -2 ]) x ([2.51 x 10 -2 ] 2 ) = 7.91 x 10 -6
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70 COMPLEX ION EQUILIBRIA 39 Chem 162-2011 Final exam Chapter 16 Equilibria Complex ion equilibria A solution with an initial [Zn 2+ ] = 0.20M and initial [CN - ] = 1.0M reacts to form a complex ion [Zn(CN) 4 2- ] (aq). What is the [Zn 2+ ](aq) at equilibrium? K f (Zn(CN) 4 2- ) = 4.7x10 19 A. 5.3 x 10 -21 M B. 1.0 x 10 -20 M C. 2.7 x 10 -18 M D. 4.0 x 10 -17 M E. 4.7 x 10 -19 M Zn 2+ (aq) + 4CN - (aq) m o (Zn(CN) 4 2- )(aq) Zn 2+ (aq) + 4CN - (aq) m o (Zn(CN) 4 2- )(aq) Initial 0.20 1.0 0 Change -X -4X +X Equilibrium 0.20 -X 1.0 -4X X [Zn(CN) 4 2- ]/([Zn 2+ ][CN - ] 4 ) = 4.7 x 10 19 Quadratic equation, possibly actually a cubic equation. Use large K rule. Zn 2+ is the limiting reactant. Zn 2+ (aq) + 4CN - (aq) m o (Zn(CN) 4 2- )(aq) Initial 0.20 1.0 0 Change -0.20 -0.80 +0.20 Equilibrium 0 0.20 0.20 To find the [Zn 2+ ], use the Right-to-Left Rule. Zn 2+ (aq) + 4CN - (aq) m o (Zn(CN) 4 2- )(aq) New initial 0 0.20 0.20 Change +X -4X -X Equilibrium +X 0.20-4X 0.20-X [Zn(CN) 4 2- ]/([Zn 2+ ][CN - ] 4 ) = 4.7 x 10 19 [0.20-X]/([X][0.20+4X] 4 ) = 4.7 x 10 19 Use small K rule to avoid quadratic equation (actually higher powered than quadratic). [0.20]/([X][0.20] 4 = 4.7 x 10 19 X = 2.66 x 10 -18 M = [Zn 2+ ]
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71 Chem 162-2011 Exam III + Answers Chapter 16 - Equilibria Complex Ion Equilibria 17 . Given AgBr(s) ĺ Ag + (aq) + Br - (aq) K sp = 7.7 x 10 -13 Ag + (aq) + 2CN - (aq) ĺ Ag(CN) 2 - (aq) K f = 5.6 x 10 8 Calculate the solubility of AgBr(s) in 0.10 M NaCN(aq) solution. A. 9.8 x 10 -10 mol/L B. 7.7 x 10 -12 mol/L C. 6.8 x 10 -4 mol/L D. 5.2 x 10 -3 mol/L E. 2.0 x 10 -3 mol/L AgBr(s) ĺ Ag + (aq) + Br - (aq) K sp = 7.7 x 10 -13 Ag + (aq) + 2CN - ĺ Ag(CN) 2 K = 5.6 x 10 8 AgBr(s) + 2CN - ҡ Ag(CN) 2 - + Br - K=(7.7 x 10 -13 ) x (5.6 x 10 8 ) = 4.3 x 10 -4 AgBr(s) + 2CN - ҡ Ag(CN) 2 - + Br - Initial Y 0.10 0 0 Change -X -2X +X +X Equilibrium Y-X 0.10-2X +X +X [Ag(CN) 2 - ][Br - ]/([CN - ] 2 ) = 4.3 x 10 -4 [X][X]/([0.10-2X] 2 ) = 4.3 x 10 -4 Use small K rule to avoid quadratic equation. [X][X]/([0.10] 2 ) = 4.3 x 10 -4 X = 2.07 x 10 -3 M
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72 Chem 162-2011 Exam III + Answers Chapter 16 - Equilibria Complex Ion Equilibria 19 , C 2 O 4 2- , forms a complex ion with manganese ion, Mn 2+ , as shown in the following equation: Mn 2+ + 2 C 2 O 4 2- ҡ [Mn(C 2 O 4 ) 2 ] 2- K f = 6.3 x ҏ 10 5 Suppose that 500.0 mL of 0.020 M Mn(NO 3 ) 2 is mixed with 1.0 L of 1.0 M Na 2 C 2 O 4 , forming the complex ion. What is [Mn 2+ ] at equilibrium.
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