Why does pH increase as a weak acid becomes more dissociated? From Wade's Organic Chemistry [1, p. 71]: 


  In most cases, the $\mathrm{p}K_\mathrm{a}$ of an acid corresponds to the $\mathrm{pH}$ where the acid is about half dissociated. At a lower (more acidic) $\mathrm{pH}$, the acid is mostly undissociated; at a higher (more basic) $\mathrm{pH}$, the acid is mostly dissociated.


I'm having a hard time intuiting this concept. Consider placing $\pu{1 mol}$ of $\ce{HF}$ in $\pu{1 L}$ of water. As the $\ce{HF}$ dissociation reaches equilibrium, I'm assuming the concentration of $\ce{H3O+}$ ions in solution increases, until the maximum concentration of $\ce{H3O+}$ ions are reached at equilibrium. If the concentration of $\ce{H3O+}$ ions in water increases as the dissociation of $\ce{HF}$ proceeds, then wouldn't the $\mathrm{pH}$ of the solution decrease (become more acidic) as the acid dissociates over time?

If any one has any thoughts on how to best interpret the quoted statement, that would be very much appreciated! Any and all new perspectives are welcomed. Thanks in advance for your generous help.

References


Wade, L. G.; Simek, J. W. Organic Chemistry, 9th ed.; Pearson: Glenview, IL, 2017. ISBN 978-0-321-97137-1.
A weak acid (HA) in water: $\ce{HA + H2O &lt;=&gt; H3O+ + A-}$, and thus,

$$K_\mathrm{a} = \frac{\ce{[H3O+][A-]}}{\ce{[HA]}}$$

By Le Chatelier's principle, if you add some $\ce{H3O+}$ from an external source (see Mithoron's comment above) to increase $\mathrm{pH}$, the equilibrium favors the backward reaction in order to reduce some $\ce{A-}$. Consequently, the amount of the dissociation of acid decreases and $K_\mathrm{a}$ will remain constant. 

In similar manor, if you decrease $\mathrm{pH}$ of the solution by adding some $\ce{OH-}$, that will react with equal amount of $\ce{H3O+}$. Therefore, the equilibrium favors the forward reaction in order to produce some $\ce{H3O+}$ to keep $K_\mathrm{a}$ constant. Accordingly, $\ce{[A-]}$ in the solution increases while $\ce{[HA]}$ decreases, meaning the amount of the dissociation of acid increases. At one point, $\ce{[A-]} = \ce{[HA]}$, and hence, $K_\mathrm{a} = \ce{[H3O+]}$. At that point, therefore, $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$.

Applying this to your example ($\ce{HA} = \ce{HF}$), $\mathrm{p}K_\mathrm{a}$ of $\ce{HF}$ is $3.14$ ($K_\mathrm{a} = 7.2 \times 10^{-4}$).

When $\pu{1.0 mol}$ of $\ce{HF}$ is dissolve in $\pu{1.0 L}$ of deionized water (no added external acid to adjust $\mathrm{pH}$), $\ce{[HF]} = \pu{1.0 mol\cdot L^{-1}}$ and $\ce{[F-]} = \ce{[H3O+]} = \alpha \gg 1.0$. Thus,
$$\ce{[H3O+][A-]} = K_\mathrm{a} \cdot \ce{[HA]}$$
$$\alpha^2 = K_\mathrm{a} \cdot 1.0~~~~\text{or}~~~~\alpha = \sqrt{K_\mathrm{a}}= \sqrt{7.2 \times 10^{-4}}= 2.7 \times 10^{-2}$$

Therefore $\mathrm{pH}$ of the solution is as follows:

$$\mathrm{pH} = -\log \ce{[H3O+]} = -\log \alpha = -\log 2.7 \times 10^{-2}= 1.57$$

Thus, at $\mathrm{pH} = 1.57$, the ionization of the solution is: $\ce{[F-]} = \ce{[H3O+]}= \pu{2.7 \times 10^{-2} mol \cdot L^{-1}}$.

Now, if you added an external base to the solution to increase its $\mathrm{pH}$ until it becomes $\mathrm{pH} = 3.14$, then, at $\mathrm{pH} = 3.14$, $\ce{[F-]} = \ce{[HA]}= \pu{0.5 mol \cdot L^{-1}}$.