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Name: Anna
Status: educator
Grade:  n/a
Country: Canada
Date: Spring 2011

Is it possible that sometimes adding a pure liquid (like water) or a solid to a system in equilibrium will produce a shift in the equilibrium? (adding or removing solids or liquids is not supposed to change equilibrium according to theory...)

Yes, it can - I am afraid you may have an incomplete understanding of the theory. A good source to consult is Physical Chemistry by Atkins and DePaula (any edition).

Equilibrium constant expressions do not usually include solvents because the standard states are chosen to have unit activity for pure solids and liquids. So changing the amount of a pure solid or pure liquid does not enter into the equilibrium constant expression... *unless* the liquid is a *solvent* in the reaction. In that situation, it matters whether the solution is dilute or concentrated. If the solution is very dilute we can treat the solvent as a "pure liquid" and neglect it from the equilibrium constant expression. However, if the solution is very concentrated then the activity of the solvent is no longer equal to one, and then it needs to be explicitly included in the equilibrium constant expression.

Here is an example. The reaction of acetic acid with water is HAc + H2O <-> Ac- + H3O+. When the acid is dilute we can get away with writing the equilibrium constant as Ka = [Ac-][H3O+]/[Hac]. However, when the acid is very concentrated, reaction of the acid with the water it is dissolved in measurably changes the concentration of water in the solution, and so the full equilibrium constant formula K = [Ac-][H3O+]/[Hac][H2O] must be used.

Hope this helps!

best, dr. topper


Remember that the equilibrium constant is equal to the concentrations of the substances raised to their stoichiometric coefficients. Since pure liquids and solids do not have concentrations, they do not enter into the equation. So, as reactants and products, pure liquids and solids do not affect the chemical equilibrium.

However, water can act as a solvent. So if there are more moles of reactant in aqueous solution form tan there are of the products, than adding water will preferentially dilute more moles of reactants, this would preferentially slow down the forward reaction, and more reactants are produced than products.

I cannot think of a special case for solids.

Greg (Roberto Gregorius)
Canisius College

I am not sure whose theory you are citing, that," adding or removing solids or liquids is not supposed to change equilibrium according to theory..." That simply is not true. There are many counter-examples, but here are just a few.

(1) Add acetone, or most any other pure water soluble organic compound to an aqueous solution of a water soluble ionic salt greatly reduces the water solubility of the salt.

(2) Conversely, adding a pure water soluble ionic salt to an aqueous solution of an organic compound reduces its water solubility. In the "chemical" jargon, this is called the>"salting out effect". Starting with a false premise will lead to incorrect consequences. You need to correct your original premise.

Vince Calder


You can definitely change the equilibrium of a reaction by adding water or a solid. The equilibrium is based on the ratio of concentrations for each component in the reaction vessel. This is the Keq value for the system. So if the addition of water or solid changes the concentration ratio the equilibrium will adjust. This is shown by using Le Chatelier's principle in chemistry classes. When the concentration is adjusted the equilibrium will shift in predictable a manner to react to the change. The times when this addition would not change the equilibrium would be when the solid/liquid does not interact with the solution itself or does not change the ratios between concentrations. If the addition of water simply dilutes all of the concentrations the same then the overall equilibrium will not shift as the overall ration does not change.

Brad Sieve

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