Determining Kc for the equilibrium involving iron thiocyanate using spectrophotometric data


> 
> $$\ce{Fe^3+(aq) + SCN-(aq) <=> [Fe(SCN)]^2+(aq)}$$
> 
> 
> $\pu{12.0 mL}$ 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.




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:


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


$[\ce{SCN-}]=3.02\times10^{-4}$


$[\ce{[Fe(SCN)]^2+}]=1.64\times10^{-4}$


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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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:
$$A = \epsilon bc$$


where:


* 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:


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


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


Therefore:


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


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


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