Understanding the Le Chatelier Principle: Impact of Adding NaSCN

[PPonte]

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

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?

 

[Astronuc]

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.

 

[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:

If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

Follow my thinking and comment, please.

To add NaSCN is the same to add SCN mononegative ions to the solution. The reaction will produce more reactants, in this case, LaTeX graphic is being generated. Reload this page in a moment.. 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?

 

[Astronuc]

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).

 

[PPonte]

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

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.

 

[Astronuc]

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.

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

See - 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⁺

 

[PPonte]

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

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

See - Determination of the Equilibrium Constant
for the Formation of FeSCN2+

It requires some advanced knowledges. Wikipedia gave a hand.

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

Yes. Because to add HSCN is to add SCN-. Correct?

 

[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.

 

[Astronuc]

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.