Le Chatelier's principle with Fe3+

alingy1
"The Beer-Lambert relationship must be solved using a FeSCN2+ standard solution. This standard solution can be prepared by mixing a SCN- solution that has a very low concentration with a Fe3+ solution that has avery high concentration. In this situation, the reaction is driven to completion instead of equilibrium, according to Le Châtelier [...]. "

I do not see how Le Châtelier applies here. Q=[FeSCN2+]/([SCN-][Fe3+]) How does having a low and a very high concentration help...

Mentor
The equilibrium still applies, but, since the SCN is mixed with an overabundance of Fe3, virtually all the SCN gets consumed. So the number of moles of FeSCN2 in the final solution is virtually identical to the number of moles of SCN that was present in the original SCN solution.

Chet

alingy1
I am not able to picture myself this with calculations. In my head, when we calculate Q, the bottom part becomes a normal value (product of big value with small value)...

Mentor
I am not able to picture myself this with calculations. In my head, when we calculate Q, the bottom part becomes a normal value (product of big value with small value)...
Q is not calculated. Q is the equilibrium constant.

Let SCN0 be the initial concentration of SCN in the mixture and Fe30 be the initial concentration of Fe3 in the mixture. Let x be the number of moles per liter of SCN that react with Fe3 to produce x moles /liter of FeSCN2. Then:

$$\frac{x}{(SCN_0-x)(Fe3_0-x)}=Q$$
Since Fe30 is going to be much greater than SCN_0, we can neglect x compared to Fe30. So the above equation becomes:
$$\frac{x}{(SCN_0-x)}=Q(Fe3_0)$$
So,
$$\frac{x}{(SCN_0)}=\frac{Q(Fe3_0)}{Q(Fe3_0)+1}$$
So, if Fe3_0 is high enough such that Q(Fe30)>>1,

x -->SCN0

This is the final concentration of FeSCN2 in the solution.

alingy1
Wow, this is really clear! Thank you. One small question though. What is the algebra behind your third latex code and your second latex code. I do not understand how you got rid of the -x and divided the left side by Q(Fe30)+1.
*gulp* Sorry, my math is really weak. I admit it. EDIT: IT'S FINE. I COULD FIND IT! :)