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, \(1.0 \mathrm{~mL}\) of \(0.20 \mathrm{M} \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) was mixed with \(1.0 \mathrm{~mL}\) of \(1.0 \times 10^{-3} M \mathrm{KSCN}\) and \(8.0 \mathrm{~mL}\) 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 \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+}\)

Step by step solution

01

Calculation of Initial Concentrations

The initial concentrations of the reactants and products need to be calculated before the reaction occurs. Since \(\mathrm{Fe^{3+}}\) is derived from \(\mathrm{Fe(NO_3)_3}\) and thiocyanate (\(\mathrm{SCN^{-}}\)) comes from \(\mathrm{KSCN}\), we get the following initial concentrations: Concentration \(\mathrm{Fe^{3+}} = C1 = \frac{0.20 mol/L * 1.0 mL}{10 mL} = 0.02 M\), Concentration \(\mathrm{SCN^-} = C2 = \frac{1.0 * 10^{-3} mol/L * 1.0 mL}{10 mL} = 1.0 * 10^{-4} M\), The concentration of the complex \([\mathrm{Fe(H_2O)_5NCS}]^{2+}\) is zero initially.

02

Calculation of Final Concentrations

Next, calculate the final concentrations after the reaction has reached equilibrium. The final concentration of \([\mathrm{Fe(H_2O)_5NCS}]^{2+}\) is given as \(7.3 * 10^{-5} M\). Because the reaction is 1:1, for every M of \([\mathrm{Fe(H_2O)_5NCS}]^{2+}\) formed, the same amount of \(\mathrm{SCN^-}\) and \(\mathrm{Fe^{3+}}\) will have reacted. This means that Concentration \(\mathrm{Fe^{3+}} = C1' = 0.02 M - 7.3 * 10^{-5} M\) and Concentration \(\mathrm{SCN^-} = C2' = 1.0 * 10^{-4} M - 7.3 * 10^{-5} M\)

03

Calculation of the Formation Constant

Finally, with the equilibrium concentrations calculated, the formation constant \(K_c\) could be calculated with the corresponding equation: \(K_c = \frac{[Fe(H_2O)_5NCS]^{2+}}{[Fe^{3+}][SCN^-]}\). By substituting the equilibrium concentrations: \(K_c = \frac{7.3 * 10^{-5}}{(C1' * C2')}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!