2
$\begingroup$

When given the equation: $$\ce{Fe^3+_{(aq)} + SCN^-_{(aq)} <=> FeSCN^2+_{(aq)}}$$

How do you calculate the equilibrium constant when given the slope of the absorbance vs concentration graph ($\pu{4317 M-1}$) and the absorbance of $\ce{FeSCN^{2+}}$ (0.276)

The following information is also given: $2.000\ \mathrm{mL}$ of a $0.00200\ \mathrm{M}$ solution of $\mathrm{KSCN}$ with $5.00\ \mathrm{mL}$ of $0.00200\ \mathrm{M}$ solution of $\ce{Fe(NO3)3}$ and $3.00\ \mathrm{mL}$ of water is combined.

So from this, I assumed that the concentrations of $\ce{SCN-}$ and $\ce{Fe^{3+}}$ are both $0.002\ \mathrm{M}$.

I found the concentration of $\ce{FeSCN^2+_{(aq)}}$ to be $\pu{6.39e-5 M}$ using this equation:

$$\mathrm{Absorbance} = \mathrm{slope}\cdot \mathrm{conc.}$$

$$\mathrm{conc.} = \frac{\mathrm{Absorbance}}{\mathrm{slope}}$$

$$\mathrm{conc.} = \frac{0.276}{\pu{4317 M-1}}$$

$$\mathrm{conc.} = \pu{6.39e-5 M}$$

Using this value, I used the equation for the K constant of an equilibrium:

$$\mathrm{K} = \frac{[\ce{FeSCN^2+}]}{[\ce{Fe^3+}][\ce{SCN^-}]}$$

$$\mathrm{K} = \frac{\pu{6.39e-5}}{0.002^2}$$

$$\mathrm{K} = 15.975$$

This did not turn out to be the correct answer, but I'm not sure why. I suspect the concentrations for the two reactions are not correct since the volumes are also given.

$\endgroup$
  • 1
    $\begingroup$ The thing you did wrong is to assume that the concentration of the substance you have at the beginning is the same in the 'endmix'. Because you see, when you add all these things together the volume is bigger thus changing the concentration of the substances you added previously. Also besides that you should then correct in the denominator for concentratioms Fe3+ and SCN- that have reacted by substracting with concentration of formed FeSCN2. $\endgroup$ – user21398 Oct 27 '16 at 10:11
1
$\begingroup$

I'm following the outline from the comment by user21398.

Initial concentrations after mixing

The volume of the mixture is $V_\text{mix} = \pu{10 mL}$. Using the dilution law, I get the following concentrations:

$$\ce{[Fe^3+]_\text{initial}} = \pu{2.00 mM} \cdot \frac{\pu{5 mL}}{\pu{10 mL}} = \pu{1.00 mM}$$

$$\ce{[SCN-]_\text{initial}} = \pu{2.00 mM} \cdot \frac{\pu{2 mL}}{\pu{10 mL}} = \pu{0.400 mM}$$

Equilibrium concentrations

The equilibrium concentration of the complex is already calculated, $\ce{[FeSCN^2+]_\text{equil}}=\pu{6.39e−5 M}.$

To get the equilibrium concentrations of the reactants, we have to consider that some reacted:

$$\ce{[Fe^3+]_\text{equil}} = \ce{[Fe^3+]_\text{initial}} - \ce{[FeSCN^2+]_\text{equil}} $$ $$ = \pu{1.00e-3 M} - \pu{6.39e−5 M} = \pu{0.94e-3 M}$$

And similar for thiocyanide:

$$\ce{[SCN-]_\text{equil}} = \ce{[SCN-]_\text{initial}} - \ce{[FeSCN^2+]_\text{equil}} $$ $$ =\pu{0.400e-3 M} - \pu{6.39e−5 M} = \pu{0.336e-3 M}$$

Equilibrium constant

This is copied from the question, using the values calculated in this answer above and dropping the unit M because the standard state is 1 M:

$$\mathrm{K} = \frac{[\ce{FeSCN^2+}]}{[\ce{Fe^3+}][\ce{SCN^-}]}$$

$$\mathrm{K} = \frac{\pu{6.39e-5}}{\pu{} \pu{0.94e-3}\cdot \pu{0.336e-3}}$$

$$\mathrm{K} = 202$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.