# percent ionization explanation [duplicate]

Using a weak acid as an example: $$\ce{CH_3COOH <=> CH_3COO^- + H^+}$$

The position of this equilibrium shifts to the right with increasing dilution.

I have found textbooks that say this, but do not provide an explanation. Why is there increasing ionization with increasing dilution?

I was wondering if it has to do with the free energy available; since ethanoic acid is a weak acid, energy has to be provided to promote the proton to a higher free energy state (transfer from $\ce{CH_3COOH}$ to $\ce{H_2O}$), therefore, "more" free energy for promotion and less contact with other $\ce{CH_3COOH}$ molecules?

## 1 Answer

For $\ce{HA <=> A- + H+}$ with a particular $K_a$ value at a particular temperature (Also, remember it usually applies to weak acid as $K_a$ of strong acid is rather meaningless),
Consider its equilibrium expression: $$K_a=\frac{\ce{[A^{-}][H+]}}{\ce{[HA]}}$$ when water (of the same temperature) is added, all three component concentration will decrease with the same extent.

Think about whether the numerator and denominator has a bigger decrease and you will find the answer (temperature is relatively constant). And for your free energy concern, being out of equilibrium necessarily means there is a favorable driving force pushing the reaction in one direction to reach back eqm.

Acquire a physical picture about the initial and equilibrium state and the dynamic state of a reaction in terms of collision theory. Think equilibrium in the form of competing forward and backward reaction and equilibrium constant is actually derived from rate constant.

Another reminder: Temperature and concentration/pressure all affect free energy. At the initial state (when you just add water and no reaction has taken place yet), there is only change in concentration but not temperature. So even if the temperature is changed upon reaction, the percentage of ionization will also be higher in all circumstances.