Finding the average concentrations given average Keq

  • Thread starter DRC12
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  • #1
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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 Keq was found to be 394.

Homework Equations


I know the equation for Keq=[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
 

Answers and Replies

  • #2
Borek
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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?
 
  • #3
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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)
 
  • #4
Borek
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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.
 

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