H+ concentration decreasing over time reasoning in weak acid strong base titration

We know that, when $\ce{CH3COOH}$ is just present in water, it is in equilibirum with water and having some $\ce{H+}$ dissociated to give some $\mathrm{pH}$ to it. Now when we start adding $\ce{NaOH}$ to it, we say as $\ce{OH-}$ is having tendency to react with $\ce{H+}$ formed intially, so now onwards $\ce{H+}$ starts decreasing, but by Le Chatlier this means weak acid should increase its dissociation. The $\mathrm{pH}$ is given by $\mathrm pK\_\mathrm a + \log\frac{\ce{[salt]}}{\ce{[acid]}} $. But how to show that net $\ce{H+}$ concentration will decrease at any time? So that $\mathrm{pH}$ will keep on increasing, as its not easy to see that $\ce{H+}$ conc will decrease over time from just the two chemical equations given by -
$$ \ce{CH3COOH <<=> CH3COO- + H+}$$
$$ \ce{H+ + OH- <<=> H2O}$$
Because forming rate of $\ce{H+}$ from $\ce{CH3COOH}$ is different from reacting rate with $\ce{OH-}$ isn't?




If starting generally with [Le Chatelier's principle](https://en.wikipedia.org/wiki/Le_Chatelier%27s_principle), the reaction of the system against the external change compensates it just *partially*. If we increase pressure of a gaseous reaction mixture in equilibrium two times, it will react to partially decrease the pressure, but will not return to the original pressure.


Back to the reaction: Note that the net reaction during the titration is:


$$\ce{CH3COOH(aq) + OH-(aq) -> CH3COO-(aq) + H2O(l)} \tag{1}$$


As you have written yourself, the equation for $\mathrm{pH}$ is:


$$\mathrm{pH} = \mathrm{p}K\_\mathrm{a} + \log{\frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]}}\tag{2}$$


which can be directly derived from the dissociation equilibrium reaction:


$$K\_\mathrm{a} = \frac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}\tag{3}$$


belonging to the equilibrium reaction:


$$\ce{CH3COOH(aq) <=> CH3COO-(aq) + H+(aq)} \tag{4}$$


which partially resupplies H+(aq) ions consumed by the neutralization reaction


$$\ce{H+(aq) + OH-(aq) -> H2O}\tag{5}$$


which has its own equilibrium equation


$$K\_\mathrm{w} = [\ce{H+}][\ce{OH-}]\tag{6}$$


The reaction (4) resupplies H+ just partially, because as the titration is progressing, the value of the logarithm in (2) increases and therefore $\mathrm{pH}$ increases.


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