Titanium in HF vs HCl So we tried the reaction between Titanium metal with hydrochloric acid and hydrofluoric acid at a similar concentration. While boiling temperatures and constant heat over a long period of time are required to etch $\ce{Ti}$ using $\ce{HCl}$ even cold $\ce{HF}$ is enough to immediately break the protective oxide layer and the metal vigorousely reacts in the solution afterwards. 

My question is now why $\ce{HF}$ is so much better at etching and dissolving $\ce{Ti}$ compared to aqua regia or $\ce{HCl}$?

Of course there is a protective oxide layer on the metal but boiling it in $\ce{HCl}$ will also form a purple solution after a while and the color intensifies if heat is applied over a long period of time. So even after the layer is destroyed the metal itself continues to react with the $\ce{HCl}$. For $\ce{HF}$ it reacts way faster so I assume it's not an issue with the protective layer itself but rather the reaction with the pure metal. 

I know for glass and $\ce{HF}$ it's a combination where $\ce{H2F+}$ reacts with $\ce{Si-OH}$ endgroups and increases the electron density on the bridging oxygen which makes them more basic and the $\ce{Si-O-Si}$ bond breaks to form $\ce{Si-OH}$ bonds. But this is all for the oxides and I assume the titanium dioxide will be dissolved in a similar manner. 

For metals I couldn't find much information on the topic. I know there are some similar elements, Niobium, Silicon, etc. they all require $\ce{HF}$ to be etched or dissolved. 

Maybe it's the case because a $\ce{Ti-F}$ bond, much like a $\ce{Ti-O}$ "bond" is way stronger than a $\ce{Ti-Cl}$ bond? Which would explain why $\ce{HF}$ starts the reaction faster as this unique $\ce{HF/H2F+}$ mechanism seems to work pretty well for tough oxides. 

Maybe it's also just the fact that more energy is produced when a $\ce{Ti-F}$ bond forms but the fast reaction starts immediately so it shouldn't just be the heat that is produced enhancing the reaction. 

The process itself is a redox-reaction but the essential part here is the $\ce{H+}$ not the counter-ion. So from a redox perspective it should be pretty much the same, unless the redox potential and the overall redox reaction is influenced by the $\ce{Ti-F}$ bond formation, like the hydration enthalphy adds to the oxidative potential of oxygen and suddenly $\ce{Cu(I)}$ turns into $\ce{Cu(II)}$ once the insoluble $\ce{CuCl}$ is added to water. 

(Also I'm sorry, but I cannot create a suitable tag with my score at the moment)
TL;DR. Are you asking a question about energy (thermodynamics) or a question about rate? What possible relation is there between boiling HCl(aq) and HF at room temperature as far as rate? I see no value in comparing those two reactions, but I'm probably missing something. So, if I understand you correctly, you are claiming that once you see some "purple" color, that therefore the Ti° is "bare", right? That, of course, doesn't (necessarily) follow. I'm not familiar with the reaction of Ti° + H+ → Ti(IV) + H2. (I'm assuming the +4 oxidation state) but almost certainly (I'd guess) it is not a single step reaction. Where the intermediate steps occur and how the presence of F- or Cl- ions impact one or more of those steps is the question. They could be absorbed on the metal surface, form complexes with intermediates, compete with oxide or hydroxide or oxyhalide species. It seems to me that the slow reaction rate of HCl is telling you that there continues to be something (some barrier) on the surface of the Ti° (by "continues" I don't mean to imply that it was there prior to the oxide films dissolution). About the only thing the energy can tell you about a reaction is that if its exothermic then you'd expect hotter and if hotter then faster (but there are (fairly rare) exceptions to this where hotter means slower). Since the value of comparing reaction A at 25°C to reaction B at 100°C isn't particularly informative (except if B is slower at 100° than A at RT, then we'd predict A will be even faster that B at 100° and B be even slower at RT) - especially if the reaction conditions are so different that we can't assume the mixing is the same (viscosity, acid dissociation, diffusion (turbulence?, H2 behavior?) all will be substantially different). Again, I'm not familiar with this reaction, but since it is a redox reaction, you can't necessarily even assume that the reduction is occurring near (on the molecular scale) to the oxidation, electrochemical cell formation is another possible complication... Could be complicated, IDK.