Question

A solution containing 0.5 M of iron(III) ions and 0.5 M of thiocyanate ions is allowed to reach equilibrium. If the equilibrium constant of the formation of the FeSCN2+ complex ion is 3.0 x 10^3, calculate the concentration of FeSCN2+ complex ion at equilibrium when the initial concentration of thiocyanate ions is doubled.

Answer 1

To solve this problem, we will use the ICE  Initial, Change, Equilibrium  table and the equilibrium constant expression. The balanced chemical equation for the formation of the FeSCN2+ complex ion is:Fe3+  aq  + SCN-  aq   FeSCN2+  aq Let's first find the equilibrium concentrations of Fe3+, SCN-, and FeSCN2+ in the original solution.Initial concentrations:[Fe3+] = 0.5 M[SCN-] = 0.5 M[FeSCN2+] = 0 MChange in concentrations:[Fe3+] = -x[SCN-] = -x[FeSCN2+] = +xEquilibrium concentrations:[Fe3+] = 0.5 - x[SCN-] = 0.5 - x[FeSCN2+] = xNow we can write the equilibrium constant expression:K = [FeSCN2+] /  [Fe3+] * [SCN-]  = 3.0 x 10^3Substitute the equilibrium concentrations:3.0 x 10^3 = x /   0.5 - x  *  0.5 - x  Now we need to solve for x, which represents the equilibrium concentration of FeSCN2+.Since K is large, we can assume that x is small compared to 0.5, so we can simplify the equation:3.0 x 10^3  x /  0.5 * 0.5 x  3.0 x 10^3 * 0.25x  0.75 MNow let's find the equilibrium concentration of FeSCN2+ when the initial concentration of thiocyanate ions is doubled:Initial concentrations:[Fe3+] = 0.5 M[SCN-] = 1.0 M  doubled [FeSCN2+] = 0 MChange in concentrations:[Fe3+] = -x[SCN-] = -x[FeSCN2+] = +xEquilibrium concentrations:[Fe3+] = 0.5 - x[SCN-] = 1.0 - x[FeSCN2+] = xSubstitute the equilibrium concentrations into the equilibrium constant expression:3.0 x 10^3 = x /   0.5 - x  *  1.0 - x  Again, since K is large, we can assume that x is small compared to 0.5 and 1.0:3.0 x 10^3  x /  0.5 * 1.0 x  3.0 x 10^3 * 0.5x  1.5 MSo, the concentration of FeSCN2+ complex ion at equilibrium when the initial concentration of thiocyanate ions is doubled is 1.5 M.