# Le Chatelier Principle

1. Mar 3, 2006

### PPonte

http://img15.imgspot.com/u/06/61/13/out1141410754.gif [Broken]

Someone dissolved $$Fe(SCN)_3$$ in water, in the recipient A.

This ion dissociates. The chemical equation is:

$$Fe(SCN)^{2+}$$ -->* $$Fe^{3+}$$ + $$SCN^-$$
$$Fe(SCN)^{2+}$$ is red.
$$SCN^-$$ is yellow.

* The reaction is incomplete.

Select the recipient that has the resulting solution of the adition of NaSCN to the initial solution.

To add NaSCN is the same to add SCN mononegative ions to the solution. The reaction will produce more reactants, in this case, $$Fe(SCN)^{2+}$$. Therefore, the resulting solution would be more red. Recipient B. I can understand but I can't agree. The resulting solution when reaches the equilibrium should have the same colour since the proportion of the concentrations of products and the concentration of reactants does not change because the equilibrium constant is the same although the rate of the direct reaction, when I perturbate the system, is lower than the rate of the inverse reaction. Am I missing something?

Last edited by a moderator: May 2, 2017
2. Mar 3, 2006

### Staff: Mentor

$$Fe(SCN)^{2+}$$ One has to put the 2+ in {} -> {2+} in order to group the exponent.

And $$Fe^{3+}$$

Solution A would have a mix of $Fe^{2+}$ and $Fe^{3+}$.

Anyway, the solution with $Fe^{2+}$ is red, which occurs if the pH is more acidic.

I presume C is mostly $Fe^{3+}$ and adding Na would make it more basic, IIRC.

Of course, I am stretching my memory back 30+ years.

Last edited: Mar 3, 2006
3. Mar 3, 2006

### PPonte

Sorry, you know I appreciate very much your help, but that is not the point. Please, forget acids and bases. Remember Le Chatelier Principle:

4. Mar 3, 2006

### Staff: Mentor

What I was trying to say is that Solution A contains a mix of FeSCN2+ (red) and Fe3+ (colorless) ions, and SCN-(yellow).

Solution A is orange or reddish-yellow because one 'sees' the color of both FeSCN2+ (red) and SCN-(yellow), because there sufficient Fe3+ (colorless) ions to allow for sufficient SCN- to show yellow.

Solution B is mostly FeSCN2+ (red) with equilibrium SCN-. The red dominates the yellow.

Solution C is mostly Fe3+ (colorless) ions, and SCN-(yellow). The Na drives the reaction from FeSCN2+ (red) to Fe3+ (colorless), and there is SCN-(yellow).

See http://www.polaris.nova.edu/~pomeroy/LE_CHAT.html particularly 3. A Complex Ion Equilibrium (Fe thiocyanide).

5. Mar 3, 2006

### PPonte

How? To add NaSCN is to increase the concentration of SCN-. And if I increase the concentration of the products, the system will react in order to decrease the concentration of products consuming them and forming more reactants: FeSCN2+(red). Therefore, the solution is reddish.

6. Mar 3, 2006

### Staff: Mentor

Possibly because Na promotes OH-, which would tend to favor Fe3+ - IIRC, it has to do with oxidation potential.

See - [URL [Broken] of the Equilibrium Constant
for the Formation of FeSCN2+[/url]

One could add thiocyanic acid to the solution, but that would tend to promote FeSCN2+

Last edited by a moderator: May 2, 2017
7. Mar 4, 2006

### PPonte

I didn't study yet oxidation power. But even if Na promotes OH- it does not react, since the chemical equation is:

$$Fe(SCN)^{2+}$$ --> $$Fe^{3+}$$ + $$SCN^-$$

But if you could explain how... Thanks

It requires some advanced knowledges. Wikipedia gave a hand.

8. Mar 5, 2006

### GCT

You can think of the iron adduct having a "saturation" point, as you may have learned with some relatively insoluble solids. If you add any of the ion components, there's going to be more precipitation. It's pretty much the same case here, at least, pertaining to what the problem is trying to illustrate.

9. Mar 7, 2006

### Staff: Mentor

I am trying to find the explanation of why pH or the presence of Na vs H would affect the ionized state of Fe. I think it is somewhat like chromate/dichromate equilibrium, which is pH dependent.