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51011 and the inverse of waters dissociation k 1001014

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Unformatted text preview: e dissociation reaction for F– (Kb = 1.5!10–11), and the inverse of water’s dissociation (K = 1.00!10+14). The equilibrium constant for the reaction, therefore, is 11, which suggests a favorable reaction. For the reaction H3PO4(aq) + 2F–(aq) D 2HF(aq) + HPO4–(aq) we add together the acid dissociation reaction for H3PO4 (Ka = 7.11!10–3), the acid dissociation reaction for H2PO4– (Ka = 6.32!10–8), the base dissociation reaction for F– (Kb = 1.5!10–11) taken twice, and the inverse of water’s dissociation (K = 1.00!10+14) taken twice. The equilibrium constant for the reaction, therefore, is 0.0010, which suggests an unfavorable reaction. 4. Construct a single ladder diagram for the metal ligand complexes: Ag(CN)2–, Ni(CN)42–, and Fe(CN)63–. What range of concentrations for the cyanide ion will ensure that all three metal ions are present as cyanide complexes? Are there conditions where only one of the three metal ions forms a complex with cyanide? If yes, what are those conditions and which metal ion is present as a cyanide complex? The ladder diagram for this set of metal-ligand complexes is shown to the right. Note that for βn, the step is at (1/n)logβn. The metal-ligand complex is the predominate...
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This homework help was uploaded on 04/01/2014 for the course CHEM 352 taught by Professor Davidharvey during the Fall '13 term at DePauw.

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