Can you help me with Chemistry?

If the Ka is < 10^-5 you need an algebraic solution. write the equation for ionization and use x for the hydronium ion concnetration:
                                          HF(aq)   +   H2O  <==>  H3O+   +   F-
Start                                     0.25M                                          0.10M
amt. H3O+                                                                   X
at equilibrium                         0.25 -x                             x          0.10 + x
equil. calculation                      Ka = 0.00068  = [H3O+]*[F-] / [HF]
                                                                    = [ x ]*[0.10 + x] / [0.25 - x ]
                                                x  = 0.00166  pH = -log10 [0.00166]  =  2.56
Using the HH equation you are ignoring the decrease in the concentration of the HF that ionizes and the increase of the fluoride ion due to the ionization of the HF.
Using the HH equation you find the pH = 3.56 which is significantly different.