When you add ethanoic acid to water, pH changes because you increase H . Am I right to think that pOH changes because of the equilibrium H20 <----> H OH-. As H increase, equilibrium shifts to the left and amount of OH- decreases?

The ethanoate also reacts with water to form ethanoic acid once more and OH-. In the solution there are H which are coming from the ethanoic acid dissociation and thus in truth OH- and H ions react to form water.

Thus, it seams that the equilibrium of ethanoic acid and ethanoate occurs in two steps? First the dissocaiation of ethanoic acid to form ethanoate ions and H . H remain in solution and ethanoate and water react to form ethanol and OH- ions. The OH- ions react with H to form water while ethanoic acid remains in solution.

1. Yes. Also, you should observe that pH + pOH = 14 (this is always true due to Kw)

2. You can say it like that, because it's actually an equilibrium reaction, meaning that the reaction never really stops and there are always reactants and products reacting. That is why it is called dynamic equilibrium