Chapter 22 Transition Metals and Coordination Chemistry Complex Ion Equilibria

Chapter 22 transition metals and coordination

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Chapter 22 - Transition Metals and Coordination Chemistry Complex Ion Equilibria Given K f and initial concentrations, find equilibrium concentration. 28. 0.40 moles of AgNO 3 and 2.5 moles of Na 2 S 2 O 3 (sodium thiosulfate) are dissolved in 1.00L of aqueous solution. A coordination complex forms. Ag + (aq) + 2S 2 O 3 2- (aq) [Ag(S 2 O 3 ) 2 ] 3- (aq) K f = 1.7 x 10 13 Calculate [Ag + ] in this solution. A . 8.1 x 10 -15 M B. 5.3 x 10 -15 M C. 1.7 x 10 -13 M D. 2.0 x 10 -12 M E. 1.7 x 10 -13 M Ag + (aq) + 2S 2 O 3 2- (aq) [Ag(S 2 O 3 ) 2 ] 3- (aq) K f = 1.7 x 10 13 Ag + (aq) + 2S 2 O 3 2- (aq) [Ag(S 2 O 3 ) 2 ] 3- (aq) Initial 0.40 2.5 0 Change -X -2X +X Equilibrium 0.40-X 2.5-2X +X [Ag(S 2 O 3 ) 2 ] 3- /([Ag + ][S 2 O 3 2- ] 2 )= K f = 1.7 x 10 13 ([X])/([0.40-X] x ([2.5-2X] 2 )) = K f = 1.7 x 10 13 Quadratic equation. Use the large K rule. Bring the reaction to completion. Ag + is the limiting reactant. Ag + (aq) + 2S 2 O 3 2- (aq) [Ag(S 2 O 3 ) 2 ] 3- (aq) Initial 0.40 2.5 0 Change -0.40 -0.80 +0.40 Equilibrium 0 1.70 +0.40 [Ag + ] = 0, but we need a finite value for the [Ag + ]. Use the Right-to-Left Rule. Ag + (aq) + 2S 2 O 3 2- (aq) [Ag(S 2 O 3 ) 2 ] 3- (aq) New initial 0 1.70 0.40 Change +X +2X -X Equilibrium +X 1.70 + 2X 0.40-X [Ag(S 2 O 3 ) 2 ] 3- /([Ag + ][S 2 O 3 2- ] 2 ) = K f = 1.7 x 10 13 [0.40-X]/([X] x ([1.70 + 2X] 2 )) = 1.7 x 10 13 Quadratic equation. Get around this by recognizing that due to a large K, there is much substance on the right side and very little on the left side. Therefore, the 2X is very small relative to the 1.70, and the –X is very small relative to the 0.40. That is, use the right-to-left rule. [0.40]/([X][1.70] 2 ) = 1.7 x 10 13 X = 8.142 x 10 -15 = [Ag + ]
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Chem 162-2016 Chapter 18 Acid-base & solubility equilibria lecture notes 67 Complex Ion Formation Effect on Salt Solubility Ligands can compete with counteranions for bonding with positive ions. Adding a potential ligand to the solution might thus affect solubility. That is, if the ligand is strong enough, it can attract a positive ion away from its electrostatic bond with a negatively charged counterion, thus forming a new bond, and dissolving a precipitate. Consider silver chloride, AgCl, as an example. AgCl(s) Ag + (aq) + Cl - (aq) K sp = 1.6 x 10 -10 If NH 3 is added to a solution containing silver ions, a complex ion, Ag(NH 3 ) 2 + , will form: Ag + + 2NH 3 Ag(NH 3 ) 2 + K f = 1.5 x 10 7 According to LeChatelier’s principle, as Ag + is removed (“ consumed ” by combining with NH 3 ), more AgCl will dissolve.
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Chem 162-2016 Chapter 18 Acid-base & solubility equilibria lecture notes 68 Chem 162-2003 Quiz 3 Version A1 ET: Given K sp and K f , find K c . Then, using K c and initial concentration, find solubility. 2mod. What is the solubility of AgI in water? What is the solubility of AgI in a 3.00M KSCN solution? K sp (AgI) = 1.5 x 10 -16 ; K f ([Ag(SCN) 4 ] 3- ) = 1.2 x 10 10 Solubility of AgI in H 2 O: AgI(s) Ag + (aq) + I - (aq) Initial Y 0 0 Change -X +X +X Equilibrium Y-X X X [Ag + ][I - ] = K sp [X][X] = 1.5 x 10 -16 [X] = 1.2 x 10 -8 M = AgI solubility in H 2 O Solubility of AgI in KSCN: AgI(s) Ag + (aq) + I - (aq) K sp = 1.5 x 10 -16 Ag + (aq) + 4SCN - (aq) Ag(SCN) 4 3- (aq) K f = 1.2 x 10 10 AgI(s) + 4SCN - I - + Ag(SCN) 4 3- K = 1.8 x 10 -6 AgI(s) + 4SCN - (aq) I - (aq) + Ag(SCN) 4 3- (aq) Initial Y 3.00 0 0 Change -X -4X +X +X Equilibrium Y-X 3.00 - 4X +X +X ([I - ][Ag(SCN) 4 3- ])/[SCN - ] 4 = 1.8 x 10 -6 ([X][X])/[3.00 - 4X] 4 = 1.8 x 10 -6 Quadratic equation; therefore, use small K rule.
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