what is the pH of 2.50M HF (aq)? Ka=7.2X10^-4
...............HF ==> H^+ + F^-
initial......2.50......0......0
change.......-x.........x......x
equil.......2.5-x......x.......x
Substitute into Ka expression and solve for x = (H^+), then pH = -log(H^+)