13 - Chem204 Lecture13 Recapping some points from last...

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Chem 204 Lecture 13 Recapping some points from last lecture. With large K sp values, few ions are in solution. Precipitates can be dissolved by using acids with insoluble carbonates and hydroxides. This works with the anions (primarily). An alternative to removing anions for dissolving precipitates is to sequester the cation! Transition metals can form strong coordination complexes with a variety of ligands.
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Lecture 13 –Page 2 The number of ligands can vary 6: Co(H 2 O) 6 2+ , Ni(NH 3 ) 6 2+ 4: CoCl 4 2- , Cu(NH 3 ) 4 2+ 2: Ag(NH 3 ) 2 + Let’s consider what happens with Ag + Ag + (aq)+NH 3 (aq) AgNH 3 + (aq); K 1 =2.1x10 3 AgNH 3 + (aq)+NH 3 (aq) Ag(NH 3 ) 2 + ; K 2 =8.2x10 3 These equilibria lie far to the right. Will this impact the solubility of AgCl? AgCl(s)+H 2 O(l) Ag + (aq)+Cl - (aq); K sp =1.6x10
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Lecture 13 –Page 3 Before we get started, are there other relevant reactions to consider? What about NH 3 ? NH 3 (aq)+H 2 O(l) NH 4 + (aq)+OH - (aq); K b =1.8x10 -5 This equilibrium lies far to the left, so there is plenty of NH 3 (aq) to complex with Ag + . Ag + (aq)+NH 3 (aq) AgNH 3 + (aq); K 1 =2.1x10 3 AgNH 3 + (aq)+NH 3 (aq) Ag(NH 3 ) 2 + ; K 2 =8.2x10 3 Since K 1 and K 2 are large, the net reaction is Ag + (aq)+2NH 3 (aq) Ag(NH 3 ) 2 + (aq); K=K 1 •K 2 K = 1.7 x 10 7
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13 - Chem204 Lecture13 Recapping some points from last...

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