what is the OH of a 0.025 M HF solution?
...........HF ==> H^+ + F^-
I.........0.025...0.....0
C..........-x.....x.....x
E.......0.025-x...x.....x
Substitute the E line into the Ka expression for HF and solve for x = (H^+), then
(H^+)(OH^-) = Kw = 1E-14
Substitute (H^+) from above and you know Kw, solve for (OH^-).