In the presence of CN", Fe3+ forms the complex ion Fe(CN)63-. The equilibrium concentrations of Fe3+ and Fe(CN)63- are 8.5 x 10-40 M and 1.4 x 10-3 M, respectively, in a 0.12 M KCN solution. Calculate the value for the overall formation constant of Fe(CN)63-. Fe3*(aq) + 6CN (aq) → Fe(CN)63- K=

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Chapter15: Equilibria Of Other Reaction Classes
Section: Chapter Questions
Problem 65E: Sometimes equilibria fur complex ions are described in terms of dissociation constants. Kd . For the...
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In the presence of CN", Fe3+ forms the complex ion Fe(CN)63- The
equilibrium concentrations of Fe3+ and Fe(CN)63- are 8.5 × 10¬40 M and
1.4 x 10-3 M, respectively, in a 0.12 M KCN solution. Calculate the value
for the overall formation constant of Fe(CN)63-.
Fe"(aq) + 6CN (aq) → Fe(CN)63-
K =
Transcribed Image Text:In the presence of CN", Fe3+ forms the complex ion Fe(CN)63- The equilibrium concentrations of Fe3+ and Fe(CN)63- are 8.5 × 10¬40 M and 1.4 x 10-3 M, respectively, in a 0.12 M KCN solution. Calculate the value for the overall formation constant of Fe(CN)63-. Fe"(aq) + 6CN (aq) → Fe(CN)63- K =
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