Finding the average concentrations given average Keq

[DRC12]

Homework Statement

How would I find the average concentrations of Fe(SCN)⁺², Fe³⁺, and SCN^- where the equation is Fe³⁺+ SCN^-→Fe(SCN)⁺² and the K_eq was found to be 394.

Homework Equations

I know the equation for K_eq=[Fe(SCN)⁺²]/([Fe³⁺]*[SCN^-])
but I have no idea how what other equations to use.

The Attempt at a Solution

and I haven't had any attempts at solving it

 

[Borek]

No such thing as "average" here - you are looking just for concentrations using known value of K.

At the moment there is not enough information to solve the problem. Final concentrations depend on amounts of substances present - for example, if you add more Fe³⁺, equilibrium shifts slightly to the right and some more FeSCN²⁺ is made.

What is the whole questions?

 

[DRC12]

The only reason its average is because I averaged four different solutions Keq. I already have the concentrations of the Fe(SCN)+2, Fe3+, and SCN- for four individual tests but the question asks to use the Keq average to find the average concentrations. I just realized that the questions says for solution #4 for which has 3mL of Fe3+ at ,002M, 6mL SCN- also .002M and 1.0mL water so do i find the molarity of each using M1V1=M2V2 then find the average Fe(SCN)2+ using the equation for Keq and the known average Keq
then 395=Fe(SCN)2+/(.00040+*.001)

 

[Borek]

It would be better to call your K "experimental" than "average" then. At least that would be my preference.

In general in such cases you have to combine stoichiometry with equilibrium. You know initial concentrations (from dilution), you know from the reaction stoichiometry how the concentrations change during the reaction (for example - for each mole of FeSCN²⁺ created one mole of Fe³⁺ is consumed), combine this information with K (for example using ICE table) and calculate concentrations for each sample separately. If all samples were prepared in the identical way you can average these concentrations afterwards, although I still don't see what for.