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#
