Question

A student mixes 5.0 mL of 0 00200 M Fe(NO_3)_3 with 5.0 mL 0.00200 KSCN She finds that the concentration of FeSCN^2+ in the equilibrium mixture is 0.000125 M. Follow these steps to determine the corresponding experimental value of K_e for the reaction of Fe63+ and 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. Calculate the molarity of Fe^3+, SCN^-. and FeSCN^2+ initially present after mixing the two solutions, but prior to any reaction taking place (M_1 V_1 = M_2 V_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. Fill in the equilibrium value for the molarity of FeSCN^2+. From this, you can determine the value of x. Given the value of x, determine the equilibrium molarities of Fe^3+ and SCN^-. Give the correct expression for K_e for this equation Then calculate the value of K_c for the reaction from the equilibrium concentrations. Use correct significant figures. On the reverse side, complete an 'ICE' table using this same procedure, but using this same procedure, but using a different reaction stoichiometry: Fe^3+ + 2 SCN^- FeSCN^2+. Assume that the equilibrium concentration of FeSCN^2+ is 0.0000625 M, or one- half its previous value. Remember how the reaction stoichiometry affects the expression for K_c.

Answer 1