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
