[jennielynn_1980]

I am doing a lab about the chromate/dichromate equilibrium represented by the following equation:

H+(aq) + 2CrO₄^2- (aq) <--> Cr₂O₇^2- (aq) + OH- (aq)

I am trying to figure out why
1) NH₃ acts as a base when added to chromate and dichromate solution
2) why Ca(OH)₂ only has a small effect on the equilibrium
3) why C₂H₅OH has on effect on the equilibrium


For 1)  I know it has something to do with the two unpaired electrons in the ammonia but other than that, I don't know specifically what is happening

For 2) I think it has to do with the low solubility of Ca(OH)₂.  If it has low solubiility, it is unable to react with the other species in an aqueous solution

For 3) I am pretty stumped.  The only thing I really know about ethanol is that it react strangely with water.


Thanks

[DevaDevil]

First of all the equilibrium is generally noted as 2H⁺(aq) + 2CrO₄^2- (aq) <--> Cr₂O₇^2- (aq) + H₂O (l)
As you can never have both OH^- and H⁺ in abundance in solution; think of the water equilibrium.

1) Are protons (hydronium ions) in abundance in the solution? So what does that tell you about the acid / base equilibrium of ammonia-ammonium?

2) true low solubility means the acidity [H+] doesn't change much, hence according to Le Chatelier's principle the equilibrium will not change much.

3) Dichromate combined with an acidic solution is used as an oxidiser. It will Oxidise Ethanol to Ethanal; so you can write down the formula?

[jennielynn_1980]

So 1) for the ammonia - ammonium because there is lots of H+, the ammonia reacts with water to give hydroxide ions and ammonium.  So because one of the products is hydroxide ions, the ammonia added will act as a base.  The equation being:
NH₃ + H₂O -->  NH₄⁺ + OH-

For 3) I don't know what would happen here.  So if ethanol is oxidized it loses electrons meaning the hydrogen?  My only guess is that it dissociates kind of so that there is both OH- and H+ and they cancel each other out.   I haven't studied oxidation reactions yet.

[Borek]

So 1) for the ammonia - ammonium because there is lots of H+, the ammonia reacts with water to give hydroxide ions and ammonium.

I don't get this 'because there is lots of H⁺' part.

So because one of the products is hydroxide ions, the ammonia added will act as a base.  The equation being:
NH₃ + H₂O -->  NH₄⁺ + OH-

That's nothing else but Bronsted-Lowry base definition at work

[jennielynn_1980]

I meant that "there is lots of H+" in the equilibrium equation.  The H+ would react with the NH3.

But the equation is correct?

[DevaDevil]

1) like I said in the first reply; if [H⁺] is big, [OH^-] will be minimal because of the water equilibrium. You might want to change your equilibrium with ammonia to include H⁺, not OH^-, else it is fine. Like borek said; acid-base chemistry

3) can you write a reaction with Cr₂O₇^2-, H⁺ and ethanol, with final products (a.o.) Cr³⁺ and ethanal (which is the final product of ethanol, it will not dissociate)?
Also, what will this do to the original chromate/dichromate equilibrium?

[Borek]

I meant that "there is lots of H+" in the equilibrium equation.  The H+ would react with the NH3.

But the equation is correct?

If there is 'lots of H⁺' ammonia doesn't have to react with water - it can get protonated directly

[jennielynn_1980]

So for the NH₃, the equation would be:

NH₃ + H+ --> NH₄+

And there would be H+ ions in the solution already from the chromate/dichromate equilibrium.

For the ethanol, the equation would be:

Cr₂O₇^2- + 8H+ + 3C₂H₅OH --> 2Cr³⁺ + 3C₂H₄O + H₂O

That would give us ethanal that will not further dissociate.    I don't know what it will do the to the equilbrium though.  According to the experiment, it has no effect whatsoever.  Why, I still don't know.

[DevaDevil]

For the ethanol, the equation would be:

Cr₂O₇^2- + 8H+ + 3C₂H₅OH --> 2Cr³⁺ + 3C₂H₄O + H₂O

That would give us ethanal that will not further dissociate.    I don't know what it will do the to the equilbrium though.  According to the experiment, it has no effect whatsoever.  Why, I still don't know.

If you'd balance it completely (H₂O) you have it right

Well, the effect should be that you lower the Cr₂O₇^2- content, thereby pushing the equilibrium a little to this side (Le Chatelier's principle). Meaning the concentration of CrO₄^2- will be reduced as well to counterbalance the loss in dichromate concentration form the reaction.

[jennielynn_1980]

So, because in effect the concentration of both chromate and dichromate are lowered, there APPEARS to be no change when the ethanol is added?  Or there should be a somewhat visible change (in the colour of the solution I mean)?  I am asking because the experiment says there should be no change in colour.  I didn't actually do the experiment by the way.  It was a simulation in which I was given the results and asked to analyze the data.

Thanks for all your help

[DevaDevil]

well; any reaction in equilibrium has an equilibruim constand as you know.

K = conc products / conc starting
in this case K = [OH-]*[dichromate] / [H+]*[chromate], so K = Kw*[dichromate] / [H+]²*[chromate]

In this case you remove more H+ than dichromate. (8:1) so what should happen to the equilibrium? I must recall my words before, the conc. chromate may even increase a little. It mainly depends on the change of [H+] as compared to the change of [dichromate].

But in amounts; I don't know If your acid was very strong to start with (high [H+], aka low change in conc) and your alcohol volume low, the change may be negligable, if your acid was weak, or your dichromate concentration very low, the change will be more pronounced.

Since you influence both sides of the equilibrium, it is not a clear thing to say unless you know in what proportions the products are there