Chapter 16 Lecture 3

Chapter 16 Lecture 3 - One case: Complex ion equilibrium...

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The Last of Chapter 16 Complex Ion Formation: Lewis Acid: An electron pair acceptor For us these will be metal ions Lewis Base: An electron pair donor We will have three of these H 2 0 NH 3 Cl Examples: [Fe(H 2 O) 6 ] 3+ [Fe(CN) 6 ] 4 - [Cu(NH 3 ) 4 ] 2+
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Cu 2+ + 4NH 3 Cu(NH 3 ) 4 2+ In this case: 13 2 4 3 2 4 3 10 1 . 1 ] [ ] [ ] ) ( [ X Cu NH NH Cu K f = = + + K f is called the formation constant Just like any equilibrium problem we have worked before: What must be the ammonia concentration in a solution that is 0.015 M in [Ag(NH 3 ) 2 ] + if [Ag + ] is to be kept at 1.0 X 10 - 8 M? Write the reaction: Look up Kf Write equilibrium expression Invent “x” if necessary Solve by algebra
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Your book has several complicated “x” type problems that we will not bother with.
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Unformatted text preview: One case: Complex ion equilibrium where a precipitate might form. Consider a solution with a complex ion: Ag + + 2NH 3 [Ag(NH 3 ) 2 ] + Kf = 1.6 X 10 7 Now we add to the beaker KBr AgBr Ag + + Br-Ksp = 5.0 X 10-13 If the Ag + concentration is set by the complex equilibrium, At what concentration of Br-would a precipitate start to form? Work the same problem again. Only ask the question like one out of the book. How many mg of KBr can be added before AgBr(s) precipitates from 1.42 L of an aqueous solution that is 0.220 M in [Ag(NH 3 ) 2 ] + and has [NH 3 ] equal to 0.805 M?...
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This note was uploaded on 04/16/2008 for the course CHEM 1123 taught by Professor Paul during the Fall '06 term at Arkansas.

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Chapter 16 Lecture 3 - One case: Complex ion equilibrium...

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