The ionization constant (Ka) of HF is 6.6 x 10-4. Which of the
following is true in a 0.1 M solution of this acid?
a. [HF] is greater than [H+] x [F-]
b. [HF] is less than [H+] x [F-]
c. [HF] is equal to [H+] x [F-]
d. [HF] is equal to [H+] + [F-]

For a weak acid, Ka=[H+][F-]/[HF].  Therefore, [HF]=[H+][F-]/Ka. 
1/Ka=1/6.6*10^-4=1515.15.  Therefore, [HF]=1515.15[H+][F-], so [HF] is
greater than [H+]X [F-].  We also know that Co=0.1 mol/L, that
Co=[HF]+[F-], and that [H+]=[F-].

Using the above equations, 6.6*10^-4=[H+]^2/(0.1-[H+]). 
[H+]=7.79*10^-3=[F-].  [HF]=0.1-7.79*10^-3=9.22*10^-2.  [HF] <>
[H+]+[F-] or [H+] X [F-].  Therefore, the correct answer is A.

Sincerely,

Wonko