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#.
