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:


$\hskip2in$ ![dependence of the pH of an acid solution with pK_a 5](https://i.stack.imgur.com/HrDec.png)


**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.)


$\hskip2in$ ![composition of an acid base system depending on the given pH](https://i.stack.imgur.com/K4hwb.png)


**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 $$\alpha\left(c\_0(\ce{HA})\right)=\frac{k\_a}{k\_a+\sqrt{c\_0(\ce{HA})~k\_a+k\_W}}$$




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*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.*


$\hskip2in$ ![enter image description here](https://i.stack.imgur.com/cF16V.png)


 12

Intuition behind this concept is thermodynamically based.  
  
 If you think of $K\_a$ as a thermodynamically favorable constant that describes the analyte concentrations associated with acid dissociation then it is not too difficult to reconcile the increase in $\ce{H+}$ ions in solution as a necessary means to maintain a thermodynamically favorable state when water is added to solution. 


 4

This is one of the common problems of ionic equilibrium. Generally questions are asked to calculate the $\mathrm{pH}$ of a diluted weak acidic solution and if the $\mathrm{pH}$ calculation turns out to be $7$ or more than that, we have to add the contribution of water also. I am giving you an example that will resolve your matter completely:


**Calculate the $\mathrm{pH}$ of $\pu{1.0E-8 M}$ solution of $\ce{HCl}$.**


For the reaction
$$\ce{HCl + H2O <=> H3O+ + Cl-}\tag1\label{hcl-dissociation},$$


if you calculate the pH as 
$$\text{pH} = -\log(10^{-8}),\tag2$$ 
then $\text{pH}=8$.
Now you can see that the $\mathrm{pH}$ of $\ce{HCl}$ can not be $8$, it has to be less than $7$, therefore **you have to include the contribution of water**.


Consider
\begin{align}
\ce{2H2O &<=> H3O+ + OH-}\tag3\label{selfionisation}\\
K\_\mathrm{w} &=\ce{[OH-][H3O+]}\\
K\_\mathrm{w} &=10^{-14}
\end{align}


Therefore the $\ce{H3O+}$ or $\ce{H+}$ concentration increases from:


1. Ionization of $\ce{HCl}$ dissolved \eqref{hcl-dissociation},
2. Self ionization of water \eqref{selfionisation}.


So in very dilute solutions like this solution we have to consider the contribution of both sources of $\ce{H3O+}$; this also explains why the concentration of hydronium ions increases upon dilution of a weak acidic solution. 


now going back to problem


\begin{align}
\text{let } x = [\ce{OH-}] &= [\ce{H3O+}]\\
[\ce{H3O+}] &= 10^{-8} + x\\
K\_\mathrm{w} &= (10^{-8} + x)(x) = 10^{-14}\\
0 &= x^2 +10^{-8}x -10^{-14}\\
[\ce{OH-}] &= x = \pu{9.5E-8}\\
\mathrm{pOH}&=7.02\\
\mathrm{pH}&=6.98
\end{align}


Now you can see that we get $\mathrm{pH} = 6.98$ after adding the contribution of water, otherwise we would have gotten the absurd answer of $\mathrm{pH}=8$, which can not be true.


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