Hi

Please help
Initial [Fe3+] = 0.00050 M
Initial [SCN-] = 0.0010 M
Equilibrium [Fe(SCN)2+] = 5.7 x 10-5

Find [Fe3+]E and [SCN-]E (in mol/L)
Find K.

Balanced equation is:
Fe3+ + SCN- -> Fe(SCN)2+ (reversible)

I know I can get K if I have the equilibrium concentrations, but not sure how to calculate these with so little info.

Thanks.

...........Fe^+3 + SCN^- ==> FeSCN^+2

begin......5E-4....1E-3......0
change
equilib......................5.7E-5
Wouldn't you think that if the equilibrium concn FeSCN^+2 = 5.7E-5 and it was zero initially, then 5.7E-5 M must have been used from the initial SCN^- of 1E-3 and 5.7E-5 must have been used from the initial 5E-4 of Fe^+3.
So equilibrium concn Fe^+3 is 5E-4 - 5.7E-5 and equilibrium concn SCN^- is 1E-3 - 5.7E-5. Check for typos

To calculate the equilibrium concentrations [Fe3+]E and [SCN-]E, we need to use the given information and the equilibrium expression for the reaction. Let's break down the steps:

Step 1: Write the equilibrium expression.
The equilibrium expression for the reaction is:
K = ([Fe(SCN)2+]/[Fe3+]E[SCN-]E)

Step 2: Substitute the given values.
Given: [Fe(SCN)2+] = 5.7 x 10-5 M, [Fe3+] = 0.00050 M, and [SCN-] = 0.0010 M.

Step 3: Solve for the unknowns.
We can rearrange the equilibrium expression to solve for [Fe3+]E and [SCN-]E:
K = ([Fe(SCN)2+]/[Fe3+]E[SCN-]E)
K = (5.7 x 10-5 M)/([Fe3+]E[SCN-]E)

Now, substitute the numerical values:
K = (5.7 x 10-5 M)/(0.00050 M * 0.0010 M)

Step 4: Calculate K.
Let's perform the calculation:
K = (5.7 x 10-5)/(0.00000050 * 0.0010)
K ≈ 0.114

Therefore, the value of K is approximately 0.114.

However, based on the given information, it seems that the equilibrium concentration of [Fe3+]E and [SCN-]E were not directly provided. We would need additional information, such as the volume of the solution or any changes in the concentration over time, in order to calculate the equilibrium concentrations.