Solved: In a dilute nitric acid solution. Fe3+ reacts with

Chapter 23, Problem 46P

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In a dilute nitric acid solution, \(\mathrm{Fe}^{3+}\) reacts with thiocyanate ion \(\left(\mathrm{SCN}^{-}\right)\) to form a dark-red complex:

\(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}+\mathrm{SCN}^{-} \rightleftharpoons  \mathrm{H}_{2} \mathrm{O}+\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+}\)  

The equilibrium concentration of \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+}\) may be determined by how darkly colored the solution is (measured by a spectrometer). In one such experiment, 1.0 mL of 0.20 M \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) was mixed with 1.0 mL of \(1.0 \times 10^{-3}\) M KSCN and 8.0 mL of dilute \(\mathrm{HNO}_{3}\). The color of the solution quantitatively indicated that the \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+}\) concentration was \(7.3 \times 10^{-5}\) M. Calculate the formation constant for \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+}\).