$$\ce{Fe^3+(aq) + SCN-(aq) <=> [Fe(SCN)]^2+(aq)}$$

$\pu{12.0 mL}$ of $\pu{0.00110 mol L-1}$ $\ce{Fe^3+}$ was added to $\pu{6.00 mL}$ of $\pu{0.00140 mol L-1}$ $\ce{SCN-}$. The absorbance of the resulting solution was $0.625$. If the constant $(\varepsilon b) = \pu{3800 (mol L-1)-1}$, determine the value of $K_c$ that should be reported for the equilibrium.

a. $\pu{1.64e-4}$
b. $\pu{3.03e-4}$
c. $\pu{4.67e-4}$
d. $\pu{5.69e-4}$
e. $957$

Answer given: e.

Given what I know about $K_c$ it should be:

$$K_c = \frac{\left[\ce{[Fe(SCN)]^2+}\right]}{[\ce{Fe^3+}][\ce{SCN-}]}$$

which gives me the answer a. I tried calculating $C$ by manipulating Beer's law, but I still don't end up with $957$ no matter what I try.

I would appreciate it if someone could walk me through the steps of solving this problem.

  • $\begingroup$ answer choice e is just 957? Without even going through the question, it seems like it would be noticeable that this option is several orders of magnitude larger than any of the other options and is also the only option greater than 1. $\endgroup$ – Tyberius Apr 1 '19 at 21:45
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    $\begingroup$ I noticed that and on the test I would pick e, I just don't understand how to actually arrive at that answer through calculations. I would like to understand it because I feel like the question won't be so straight forward on the real exam. $\endgroup$ – Codeblockz Apr 1 '19 at 22:04

To start with, its often helpful to make an ICE table when dealing with problems of chemical equilibrium. $$\begin{array}{cccc} \begin{array}{c|ccc} \hline & \ce{Fe^3+} & \ce{SCN-} & \ce{[Fe(SCN)]^2+} \\ \hline \mathrm{initial} & 1.32\times10^{-5} & 8.40\times10^{-6} & 0\\ \mathrm{change} & -x & -x &+x \\ \mathrm{equil} & 1.32\times10^{-5}-x & 8.40\times10^{-6}-x & x\\ \hline\\ \mathrm{conc.} & \frac{1.32\times10^{-5}-x}{V}& \frac{8.40\times10^{-6}-x}{V} & \frac{x}{V}\end{array} \end{array}$$

Then we can use Beer's Law to find x. $$\frac{x}{V}=\frac{A}{\epsilon b}=\pu{1.645\times10^{-3}mol/L} \to x= (1.645\times10^{-3})(.018)=\pu{2.961\times10^{-6}mol}$$

With x, we can find all the concentrations:




Finally, we can then compute $K_c$: $$K_c=\frac{[\ce{[Fe(SCN)]^2+}]}{[\ce{Fe^3+}][\ce{SCN-}]}=956.77$$

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  • $\begingroup$ Thank you very much. I wasn't finding the concentrations properly and instead I was using the initial concentrations instead of the equilibrium ones. Very informative answer. $\endgroup$ – Codeblockz Apr 1 '19 at 22:49
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    $\begingroup$ @Codeblockz definitely look at MaxW's answer as well. It can help with understanding a problem to see slightly different ways of getting to the solution. $\endgroup$ – Tyberius Apr 1 '19 at 22:53
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    $\begingroup$ Wonder when ICE tables became the rage. I don't ever remember seeing one before I came onto this site. -- When I was in chemistry classes there were just four elements, earth, air, water and fire. ;-) $\endgroup$ – MaxW Apr 1 '19 at 23:08
  • $\begingroup$ @MaxW Thats an interesting point. It was what I was taught back in high school, so its at least been around for the last decade or so. However, I couldn't find the original source of the idea. Might be a worthwhile question here or History of Science and Math depending on if they allow pedagogical history. $\endgroup$ – Tyberius Apr 1 '19 at 23:15
  • $\begingroup$ I left high school in 1969. $\endgroup$ – MaxW Apr 1 '19 at 23:16

The problem is a bit strange since the details of the absorbance measurement aren't detailed very well. I'll assume the book defines the relationship as: $$A = \epsilon bc$$


  • A=absorbance
  • $\epsilon$ = molar attenuation coefficient
  • $b$ = path length
  • $c$ = concentration

So the concentration, c, of the $\ce{[Fe(SCN)]^2+}$ is:

$$ c = \dfrac{A}{\epsilon b} = \dfrac{0.625}{3800} = 1.64\times10^{-4}$$

Considering the dilutions...

$\ce{[Fe^{+3}] + [Fe(SCN)]^2+} = 0.00110 \times \dfrac{12}{18} = 7.33\times10^{-4}$

$\ce{[SCN^-] + [Fe(SCN)]^2+} = 0.00140 \times \dfrac{6}{18} = 4.67\times10^{-4}$

But there is $1.64\times10^{-4}$ of $\ce{[Fe(SCN)]^2+}$ so:

$\ce{[Fe^{+3}]} = 7.33\times10^{-4} - 1.64\times10^{-4} = 5.69\times10^{-4}$

$\ce{[SCN^-]} = 4.67\times10^{-4} - 1.64\times10^{-4} = 3.03\times10^{-4}$


$K_c = \dfrac{1.64\times10^{-4}}{(5.69\times10^{-4})(3.03\times10^{-4})} = 951$

I'd assume the small difference is just due to a difference in rounding somewhere...

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  • $\begingroup$ @MathewMahindaratne - Made the correction. Thanks for pointing out the error. $\endgroup$ – MaxW Apr 1 '19 at 23:59
  • $\begingroup$ Thanks, cleared a lot up. $\endgroup$ – Codeblockz Apr 3 '19 at 17:32

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