Solution Found!
In the presence of CN, Fe3 forms the complex ion Fe(CN)6
Chapter 16, Problem 61(choose chapter or problem)
In the presence of \(\mathrm{CN}^-,\mathrm{\ Fe}^{3+}\) forms the complex ion \(\mathrm{Fe}(\mathrm{CN})_6^{\ 3-}\). The equilibrium concentrations of \(\mathrm{Fe}^{3+}\) and \(\mathrm{Fe}(\mathrm{CN})_6^{\ 3-}\) are \(8.5\times10^{-40}\ M\) and \(1.5\times10^{-3}\ M\), respectively, in a 0.11 M KCN solution. Calculate the value for the overall formation constant of \(\mathrm{Fe}(\mathrm{CN})_6^{\ 3-}\).
\(\mathrm{Fe}^{3+}(aq)+6\mathrm{CN}^-(aq)\leftrightharpoons\mathrm{Fe}(\mathrm{CN})_6^{\ 3-}\quad\ \ \ K_{\text{overall }}=?\)
Questions & Answers
QUESTION:
In the presence of \(\mathrm{CN}^-,\mathrm{\ Fe}^{3+}\) forms the complex ion \(\mathrm{Fe}(\mathrm{CN})_6^{\ 3-}\). The equilibrium concentrations of \(\mathrm{Fe}^{3+}\) and \(\mathrm{Fe}(\mathrm{CN})_6^{\ 3-}\) are \(8.5\times10^{-40}\ M\) and \(1.5\times10^{-3}\ M\), respectively, in a 0.11 M KCN solution. Calculate the value for the overall formation constant of \(\mathrm{Fe}(\mathrm{CN})_6^{\ 3-}\).
\(\mathrm{Fe}^{3+}(aq)+6\mathrm{CN}^-(aq)\leftrightharpoons\mathrm{Fe}(\mathrm{CN})_6^{\ 3-}\quad\ \ \ K_{\text{overall }}=?\)
ANSWER:Step 1 of 2
From the given,
Molarity of KCN = 0.11 m
The concentration of \(F e^{3+}=1.5 \times 10^{-3}\)
The concentration of \(F e^{2+}=8.5 \times 10^{-40}\)
The reaction between \(\mathrm{Fe}^{2+}\) and \(\mathrm{CN}