A student mixes 5.0 mL of 0.00200 M \(\ce{Fe(NO3)3}\) with 5.0 mL 0.00200 M \(\ce{KSCN}\). She finds that the concentration of \(\ce{FeSCN^{2+}}\) in the equilibrium mixture is 0.000125 M. Follow these steps to determine the corresponding experimental value of \(K_{c}\) for the reaction of \(\ce{Fe^{3+}}\) and \(\ce{SCN^{-}}\) to produce this complex ion. Show your calculations for each step below and then place the appropriate value(s) in the equilibrium (or 'ICE') table near the bottom of the page. Step 1. Calculate the molarity of \(\ce{Fe^{3+}}\), \(\ce{SCN^{-}}\), and \(\ce{FeSCN^{2+}}\) initially present after mixing the two solutions, but prior to any reaction taking place. (\(M_{1}V_{1} = M_{2}V_{2}\)) Step 2. Determine the expression and initial value for \(Q_{c}\). Then give the appropriate signs of the concentration changes for each species in terms of the reaction's shift, or \(x\), into the 'ICE' table. Step 3. Fill in the equilibrium value for the molarity of \(\ce{FeSCN^{2+}}\). From this, you can determine the value of \(x\). Step 4. Given the value of \(x\), determine the equilibrium molarities of \(\ce{Fe^{3+}}\) and \(\ce{SCN^{-}}\). Unnamed: 0 \(\ce{Fe^{3+}}\) (aq) \(+ \quad \ce{SCN^{-} (aq)}\) \(\ce{ <=>\quad FeSCN^{2+} (aq)}\) I NaN NaN NaN C NaN NaN NaN E NaN NaN NaN Step 5. Give the correct expression for \(K_{c}\) for this equation. Then calculate the value of \(K_{c}\) for the reaction from the equilibrium concentrations. Use correct significant figures. Step 6. On the reverse side, complete an 'ICE' table using this same procedure, but using a different reaction stoichiometry: \(\ce{Fe^{3+} + 2 SCN^{-} <=> Fe(SCN)2^{2+}}\) (ignore that fact that this is incorrect since the reaction is not balanced with respect to charge). Assume that the equilibrium concentration of \(\ce{FeSCN^{2+}}\) is 0.0000625 M, or one-half its previous value. Remember how the reaction stoichiometry affects the expression for \(K_{c}\).