Is it true that a pH higher than the pKa of an acid make the H automatically disassociate?
I propose the following  answer:

Let's consider the following equilibrium:

$$ \ce{H}A+\ce{H_2O} \leftrightarrows A^-+\ce{H_3O^+}$$
The equilibrium constant $\ce{K}_A=\dfrac{[A^-][\ce{H_3O^+}]}{[\ce{H}A]}$. So if we take ''$-\log$'' of this equation, we can define the $\ce{pK}_A$ of an acid as: $\ce{pK}_A=\ce{pH}- \log\dfrac{[\ce{A^-}]}{[\ce{H}A]}$.

we rewrite this equation as follows:

$$
\ce{pH}=\ce{pK}_A+\log\dfrac{[A^-]}{[\ce{H}A]}
$$
From this equation, we can deduce that when $\ce{pH}&gt;\ce{pK}_A$, $[A^-]&gt;[\ce{H}A]$,  i.e. the concentration of the conjugated base is bigger than the concentration of the acid. But this doesn't mean that all the acid molecules are dissociated. The bigger the difference between the pH and the $\ce{pK}_A$, the bigger is the percent of dissociated acid molecules.