$$\ce{HA(aq) +H2O(l) <=>H3O+(aq) +A- (aq)}$$

Say the acid solution is currently in equilibrium. How would diluting the acid by adding water change the equilibrium position? Le Chatelier's Principle says pure liquids like water do not affect the equilibrium position. Using the equilibrium constant, it is easy to show the reaction should shift towards the right, but how can we show that with Le Chatelier's Principle?


Le Chatelier's Prinicple also says that the system will adjust itself based on a change in concentration.

Concentration of an acid is (normally) expressed in moles/liter. If you were to increase the amount of solvent (water) which is on the reactant side of the equation, you would be diluting the acid solution that is in equilibrium and causing the system to adjust accordingly.

As far as the last portion of your request is concerned I am slightly confused. Le Chatelier's Prinicple is built into the equilibrium expression. If I wanted to show how the system were to react based on the addition of more reactants/products I would just use the equilibrium expression for the system.
Hope this helps, D

  • $\begingroup$ So when the water is added, the concentrations of HA, H+, and A-, will all decrease, so we can't really use Le Chatelier's Principle right? But using the equilibrium constant will always work. $\endgroup$ – carbenoid Oct 6 '15 at 2:10
  • $\begingroup$ Le Chatelier's Principle can still be used to get an idea of how the reaction is going to adjust according to what is added. The equilibrium constant is a mathematical way of determining exactly how it will adjust. $\endgroup$ – Dupin Oct 6 '15 at 8:01

As concentration of weak acid decreases, a dilution is occurring, thus an increase in water. So equilibrium shifts right to make up more products to make up for increase in h20. Products/reactants. More products = bigger numerator = increase percent dissociation.


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