What it help you to point out that Kw = Ka ? Kb, and so Ka = Kw/Kb ?
Do you know what Ka and Kb? They are the acid dissociation constant and base hydrolysis constant.
Ka = [H+] [A-]
[HA]
Kb = [BH+][OH-]
Where HA is the concentration of the un-dissociated acid and B is the concentration of the base. That is what you have in your cases, the concentration of the original solutions (HA and/or B).
You know the HA -> H+ and A-. There for Ka = [X][X]
[HA – X]
Simple solving for x, once you know x you know the concentration of the H+ and therefore the pH (once you "-log it) .
Using the one with Kb (base hydrolysis) you get the concentration of the OH-, so use the Kw=Ka*Kb and convert to Ka to H+ concentration. Or pH + pOH = 14 (you have to be careful with this one though, because very dilute or very concentrated solutions can throw this off)
Note: The calculations will change some if it is a polyprotic acid.