Equal volumes of 0.00150M FeCl3 and 0.00200M NaSCN were mixed together and reacted according to the equation:

Fe3+ + SCN- <--> FeSCN2+

The equilibrium concentration of FeSCN 2+ was 1.40 x 10^-4 M

Calclulate the equilibrium constant for the reaction.

I got 55.3...is that right? or do you have to use the "equal volumes" part of the question in some way?

I think you must do an ICE chart and half the concentrations of Fe^+3 and SCN^- to start.

thanx..other people were saying that as well

To calculate the equilibrium constant of the reaction, we need to use the equilibrium concentrations of the species involved in the reaction. The equilibrium constant (Kc) is defined as the ratio of the product concentrations to the reactant concentrations, each raised to the power of their respective stoichiometric coefficients.

In this case, the reaction is Fe3+ + SCN- ⇌ FeSCN2+, and the equilibrium concentration given is [FeSCN2+] = 1.40 x 10^-4 M.

Since the initial concentrations of FeCl3 and NaSCN are equal, we can assume that the initial concentration of Fe3+ and SCN- is half of their initial concentrations.

Therefore, the initial concentration of Fe3+ ([Fe3+]) = 0.00150 M / 2 = 0.00075 M.
And the initial concentration of SCN- ([SCN-]) = 0.00200 M / 2 = 0.00100 M.

Now, to calculate the equilibrium constant (Kc), we can use the formula:

Kc = [FeSCN2+]/([Fe3+][SCN-])

Substituting the given equilibrium concentration and initial concentrations into the equation:

Kc = (1.40 x 10^-4 M) / ((0.00075 M)(0.00100 M))

Kc ≈ 186.67

So the equilibrium constant for the reaction is approximately 186.67.

Therefore, the value you obtained of 55.3 for the equilibrium constant is not correct. The correct value is approximately 186.67.