Question

Calculate the equilibrium constant (K) for the formation of the complex ion [Fe(SCN)₂⁺] when 3.0 x 10⁻³ moles of Fe(NO₃)₃ and 1.5 x 10⁻³ moles of NaSCN are mixed in a 250 mL solution. The solution shows that the concentration of Fe(SCN)₂⁺ is 1.2 x 10⁻³ M at equilibrium.

Answer 1

To calculate the equilibrium constant  K  for the formation of the complex ion [Fe SCN ], we first need to write the balanced chemical equation for the reaction:Fe aq  + 2SCN aq   [Fe SCN ] aq Next, we need to determine the initial concentrations of the reactants and products. Since we know the moles and volume of the solution, we can calculate the initial concentrations:Initial concentration of Fe =  3.0 x 10 moles  /  0.250 L  = 0.012 MInitial concentration of SCN =  1.5 x 10 moles  /  0.250 L  = 0.006 MInitial concentration of [Fe SCN ] = 0 M  since it has not formed yet Now, we can set up an ICE  Initial, Change, Equilibrium  table to determine the concentrations at equilibrium:                 Fe aq  + 2SCN aq   [Fe SCN ] aq Initial  M :        0.012        0.006             0Change  M :         -x          -2x               +xEquilibrium  M :  0.012-x      0.006-2x         1.2 x 10We are given the equilibrium concentration of [Fe SCN ] as 1.2 x 10 M, so we can substitute that value for x:1.2 x 10 = xNow, we can find the equilibrium concentrations of Fe and SCN:[Fe] = 0.012 - x = 0.012 - 1.2 x 10 = 0.0108 M[SCN] = 0.006 - 2x = 0.006 - 2 1.2 x 10  = 0.0036 MFinally, we can calculate the equilibrium constant  K  using the equilibrium concentrations:K = [Fe SCN ] /  [Fe][SCN] K =  1.2 x 10  /   0.0108  0.0036  K =  1.2 x 10  /  0.0108 x 0.00001296 K  9.2So, the equilibrium constant  K  for the formation of the complex ion [Fe SCN ] is approximately 9.2.