We interrogate the equilibrium...........
HF(aq) + H_2O(l) rightleftharpoons H_3O^+ + F^-
Unlike the other hydrogen halides, HF is NOT a strong Bronsted acid, and I believe this is an entropy effect. Anyway, we need a K_a value for the above equilibrium.
This site reports that K_a for HF = 6.3xx10^-4. (This really should have been quoted in the question!)
So we set up the equilibrium expression as usual:
K_a=([H_3O^+][F^-])/([HF])=6.3xx10^-4.
Now initially, [HF]=0.200*mol*L^-1; we posit that "x"*mol*L^-1 dissociates, and we substitute these values into the equilibrium expression.
K_a=("x"xx"x")/(0.200-"x")=x^2/(0.200-x)=6.3xx10^-4.
This is a quadratic in x, which we could solve EXACTLY. Because we are lazy we ASSUME that 0.200">>"x, and that (0.200-x)~=0.200. We must justify this approximation later.
So x_1=sqrt(6.3xx10^-4xx0.200)=0.0112*mol*L^-1.
Now that we have an approx. value, we can resubstitute this value into the equilibrium expression, and see how our approx. evolves.
x_2=sqrt(6.3xx10^-4xx(0.200-0.0112))=0.0109*mol*L^-1
x_3=sqrt(6.3xx10^-4xx(0.200-0.0109))=0.0109*mol*L^-1
Given that our approximations have CONVERGED, we accept x=0.0109*mol*L^-1.
Now x=[H_3O^+], and thus [H_3O^+]=0.0109*mol*L^-1, and pH=1.96.
