After the addition of HF, but before any dissociation can occur, the concentration of HF, #["HF"] = 0.750" M"#. #["F"^-]=0# and #["H"_3"O"^+]# is negligible.

We need to know the amount of HF dissociated. Since the volume of the solution, #V#, is assumed to be constant, denote #x# to be the amount of HF dissociated divided by #V#. This is to facilitate calculations later since the acid dissociation constant is expressed in terms of the product of the concentration of the species involved.

After dissociation,

#["HF"] = 0.750" M"-x#

#["F"^-]=["H"_3"O"^+]=0+x=x#

At equilibrium,

#frac{["F"^-]*["H"_3"O"^+]}{["HF"]}=K_a#.

Therefore,

#frac{x^2}{0.750-x}=6.8xx10^{-4}#.

Solving the quadratic gives

#x=0.0222#

Since #["H"_3"O"^+]=x#, the pH of the solution is given by

#-"lg"(0.0222)=1.7#