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Concentration of $\ce{Fe(SCN)^{2+}}$ in calibration solutions
Recall that the concentration of $\ce{Fe(SCN)^{2+}}$ in the calibration solutions is essentially the concentration of the $\ce{SCN-}$ initially in the solution. So, determine the concentration of $\ce{SCN-}$ initially in each of the $\pu{10.0 mL}$ calibration solutions. This is a simple dilution!
The lab checker is saying that this is wrong. Am I even going about doing this problem correctly?
So, to find the concentration of $Fe(SCN)^{2+}$, the problem says "the calibration solutions is essentially the concentration of the $[SCN^−]$ in the solution." So, you simply take the volume of $K(SCN)$ solution divided by the total volume, then multiply by $[K(SCN)]$:
$[Fe(SCN)^{2+}]=[SCN^−]=[K(SCN)]=0.002M*\frac{0.5mL}{10.0mL}=0.0001M$ (repeat for 2, 3, 4)
The reason we do not involve $K_c$ is because the large amount of $[Fe^{3+}]$ drives the reaction to the right. The absorbance measurements are going to be useful for determine the amount of $[Fe(SCN)^{2+}]$ in a system where $[Fe^{3+}]$ is not high. In these cases, the concentrations $[Fe^{3+}]$ and $[SCN^-]$ are known, and you use the absorbances you collected to draw a line relating concentration $[Fe(SCN)^{2+}]$ to absorbance. When you know the concentration of your three parts of $K_c$, you can simply use arithmetic to find $K_c$.
