Question

If the solution in #3 (1.54x10^-4) was prepared by mixing 1.000 mL of 0.0020 M Fe3+ and 9.000 mL of 0.0020 M SCN- (for a final total volume of 10.000 mL), what is the equilibrium concentration of Fe3+?

M1V1=M2V2
[Fe(SCN)2+]𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 = x
[Fe3+]𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 = [Fe3+]𝑝𝑟𝑒−𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 - x

Answer 1

I'm not sure I understand the problem, especially the 1.54E-4 from #3 and I don't know what #3 is all about.
Anyway, the reaction is
................Fe^3+ + SCN^- ==> FeSCN^2+
Initially, (Fe^3+) = (0.002 x 1/10) = ?
Initially, (SCN^-) = (0.002 x 9/10) = ?
Kf for FeSCN^2+ = (FeSCN)^2+/(Fe^3+)(SCN^-)
................Fe^3+ + SCN^- ==> FeSCN^-
I.............0.0002.....0.0018............0
C............-x............-x......................x
E..........0.0002-x.....0.0018-x..........x
Plug the E line into the Kf expression and solve for x, then evaluate 0.0002-x.
Post your work if you get stuck or let me know if I have interpreted the problem incorrectly.