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=
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=
Chemistry by OpenStax (2015-05-04)
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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 =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fab9c6c64-11a9-4770-afb5-aa519577362f%2Fe72bb4ba-e700-40b4-a1f0-6843ae8bf252%2Fwu7l9rh_processed.png&w=3840&q=75)
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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