Kaplan Equilibrium Question

[El-Rami]

I had the following problem from Kaplan (simplified somewhat):

H3A(aq) + H2O(l) ⥨ H3O+(aq) + H2A-(aq)

The answer key says the reaction should shift LEFT if you remove water. However, I thought that removing solvent or solids in equilibrium has no effect on the thermodynamics? In the thermodynamic expression, solvents and solids have a value of 1. Therefore, Q should not be different from Kc, meaning that removing water has no effect on the equilibrium.

Am I correct? If not, why? Thanks very much.

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[rtmsf95]

So this might be a bit trippy, but Keq is the quotient of products over reactants. And although water is not included in the equilibrium expression, it definitely contributes to the concentration of aqueous substances. So if you remove water, the concentrations of the aqueous products will be greater (Q>K), and so there will be a shift to the left to decrease their concentrations.

Now if we were looking at another scenario with gases and solids, and the question asked us what shift would take place if we increased the solid concentration, then there would be no change in K's value, since increasing or decreasing the solid concentration will not have any affect on the concentration of given gases.

I guess just know that adding or losing water will affect the equilibrium expression if there are aqueous substances present, since it will further concentrate or dilute those substances, respectively.

 

[El-Rami]

That makes sense. Thank you.

EDIT: Now that you mentioned that, let me ask you this quick question. Considering what you said (removing water increases solute concentration), wouldn't the RATIO of products over reactants still be the same, thus Q = Kc still?

 

[desertrat12]

Shifting to the left or to the right, in your presented situation, doesn't involved a change in equilibrium. A change in equilibrium would occurs if you removed water and the reaction didn't shift in order to maintain the current equilibrium. The reaction shifts to maintain the current equilibrium. So, because the equilibrium remains the same (aka, ratio of products over reactants at equilibrium stays the same) you are right to say the ratio stays the same

 

[rtmsf95]

The thing to keep in mind is that Keq will usually only change when under the effect of temperature. Concentrations will change if you add or take out substances, but the shift will occur such that the equilibrium constant will stay the same! So individual concentrations/pressures may change, but they'll change in a way such that the ratio of them will remain the same.

 

[sazerac]

If you removed water and the concentration of everything doubled, there are two aqueous products and only one aqueous reactant. The Q would therefore double. The reaction would have to shift to the left to restore it to K.

 

[EParker37]

Just going to add that this is the concept behind Le Chatelier's principle. Which would be good to study up on for the MCAT.