Determining Kc for the equilibrium involving iron thiocyanate using spectrophotometric data

Question

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

$\pu{12.0 mL}$ 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.

Answer 1

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 $K_c$: $$K_c=\frac{[\ce{[Fe(SCN)]^2+}]}{[\ce{Fe^3+}][\ce{SCN-}]}=956.77$$

Answer 2

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...