In a certain trial, the initial concentrations of Fe3+ and SCN- are both 1.00x10^-3 M and the initial concentration of FeSCN2+ is zero. Suppose that after the reation reaches equilibrium, the concentration of FeSCN2+ was found to be 1.19x10^-4 M. Use this information to calculate the value of Kc for the reaction, Fe3+ + SCN- <-> FeSCN2+. Show work.

Remember how to set up an ICE chart.

.............Fe^3+ + SCN^- ==> FeSCN^2+
initial.....0.001..0.001........0
change........-x.....-x.........+x
equil.....0.001-x...0.001-x...0.000119
Therefore, x must be 0.000119
Now you can calculate 0.001-x and plug into the Kc expression and solve for Kc.

To calculate the value of Kc for the given reaction, we need to use the concentrations of the reactants and the product at equilibrium.

Let's denote the concentration of Fe3+ as [Fe3+], the concentration of SCN- as [SCN-], and the concentration of FeSCN2+ as [FeSCN2+].

Given information:
[Fe3+] = 1.00x10^-3 M
[SCN-] = 1.00x10^-3 M
[FeSCN2+] = 1.19x10^-4 M

The balanced equation for the reaction is:
Fe3+ + SCN- <-> FeSCN2+

At equilibrium, the concentrations of Fe3+, SCN-, and FeSCN2+ can be represented as:
[Fe3+]equilibrium = [Fe3+]initial - x
[SCN-]equilibrium = [SCN-]initial - x
[FeSCN2+]equilibrium = [FeSCN2+]initial + x

Where 'x' represents the change in the concentrations of Fe3+, SCN-, and FeSCN2+ at equilibrium.

Since the initial concentration of FeSCN2+ is zero, [FeSCN2+]initial = 0. Therefore, [FeSCN2+]equilibrium = x.

Now, substitute the given values into the equation:
1.19x10^-4 M = x

To calculate x, we need to use an equilibrium equation that relates the concentrations of Fe3+, SCN-, and FeSCN2+. The equation is given by the equation:
Kc = [FeSCN2+]/([Fe3+][SCN-])

Now plug the equilibrium concentrations into the equation:
Kc = (1.19x10^-4 M) / [(1.00x10^-3 M) * (1.00x10^-3 M)]

Kc = (1.19x10^-4) / (1.00x10^-6)

Kc = 119.

Therefore, the value of Kc for the given reaction is 119.