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?

marked as duplicate by bon, M.A.R. ಠ_ಠ, Klaus-Dieter Warzecha, Jannis Andreska, user15489 Jul 28 '15 at 20:29

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.