Does the number of H+ ions in solution go up on dilution of a weak acid? In my textbook, a footnote says: 


  In case of weak acids, on dilution the total number of $\ce{H^{+}}$ ions in solution increases because dissociation of the weak acid increases


This didn't make sense to me however. Consider the general dissociation reaction of a weak acid:

$$\ce{HA (aq)  + H_2O \rightleftharpoons H_3O^{+} (aq) + A^{-}(aq)}$$

Now, the Chatelier's principle tells us that if we add more reactants ($\ce{H_2O}$), the reaction will proceed more to the right, and so the dissociation of the acid will increase, but the part about the total number of $\ce{H^{+}}$ ions in water being higher was still unclear to me. 

Delving into the math a bit, we can see that

$$K_a = \frac{[\ce{H_3O^{+}}][\ce{A^{-}}]}{[\ce{HA}]}$$

Since the concentrations of the two products are equal, 

$$[\ce{H_3O^{+}}] = \sqrt{[\ce{HA}]\cdot K_a}$$

Now, let us assume that we have a $1 \ M$ solution of the acid in question dissolved in 1 litre of water, plugging this into the equation, we have a $\sqrt{K_a}$ molar solution of $H^{+}$ ions. Since only 1 litre of solvent is there, we have $\sqrt{K_a}$ moles of ions. 

Now, say we dump 1 litre of water into this solution. The concentration of our acid will be halved. Plugging this into the equation, we get the concentration of $\ce{H_3O^{+}}$ ions as approx. $0.7 \sqrt{K_a} \ M$. We have two litres of solvent, so we have $1.4 \sqrt{K_a}$ moles of ions, which is considerably higher. So, the numbers all work out, but what's the intuition behind it?
Let's say we have got a solution of an acid with a pK$_a=5$. The resulting pH dependent on the concentration of that acid would be like:

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Examples: $0.1~\ce{M}$: pH$=3.0$; $0.001~\ce{M}$: pH$=4.0$; $0.00001~\ce{M}$: pH$=5.2$; $0.0000001~\ce{M}$: pH$=6.8$

Depending on the pH resulting from these example concentrations we can have a look at percentage composition of the whole acid base system. (Degree of association $\alpha'$ in blue, degree of dissociation $\alpha$ in light orange.)

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Examples: pH$=3.0$: $\alpha\approx 1.0~\%$; pH$=4.0$: $\alpha\approx 9.5~\%$; pH$=5.2$: $\alpha\approx 61.8~\%$; pH$=6.8$: $\alpha\approx 98.4~\%$

One can see that low acid concentrations lead to higher pH values than higher acid concentrations which automatically result in an increasing degree of dissociation.

So for me this is no big intuition thing but rather a small sequence of actions.



Lets say it in math:

The proton concentration for your system can be described by $$c(\ce{H3O+}) = \sqrt{c_0(\ce{HA})~k_a+k_W}$$ (adding $k_W$ for very low concentrations) and the equation for the degree of dissociation is $$\alpha = \frac{k_a}{k_a+c(\ce{H3O+})}$$ which can be combined to give $$\alpha\left(c_0(\ce{HA})\right)=\frac{k_a}{k_a+\sqrt{c_0(\ce{HA})~k_a+k_W}}$$



One could expand this system a little more (using the exact pH equation including the self-ionization of water) and come up with a thorough scheme like the following.

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