Question

the ionization constant of HF is 6.7 * 10^-4. which of the following is true in a 0.1 M soultion of this acid?

the answer is HF is greater than [H][F]. HOW COME??

Answer 1

HF ==> H^+ + F^-
Ka = (H^+)(F^-)/(HF) = 6.7 x 10^-4
You can do this one of two ways: (1) mathematical logic or (2)work it out. I'll do both starting with #2.

Before ionization:
(HF) = 0.1M
(H^+) = 0
(F^-) = 0

change at ionization:
(HF) = -y (I use y to avoid x looking like the times sign).
(H^+) = +y
(F^-) = +y

At equilibrium: just add the two numbers from above to obtain equilibrium.
(HF) = 0.1 - y = 0.1 - y
(H^+) = 0 + y = y
(F^-) = 0 + y = y

Now plug the equilibrium values into the Ka expression to obtain
(H^+)(F^-)/(HF) = 6.7 x 10^-4
(y)*(y)/(0.1 - y) = 6.7 x 10^-4
solve for y = 7.86 x 10^-3 (It's a quadratic); therefore,
(H^+)= 0.00786 = 0.0079
(F^-) = 0.00786 = 0.0079
(HF) = 0.1 - 0.00786 = 0.09214 = 0.092
Now you can multiply (H^+)(F^-) and see that it is less than (HF).

(1)math logic and this is far easier to do because no calculations are necessary. I did the first one to show you with numbers that the statement in the problem is true. This is the logic we could have gone through.
(H^+)(F^-)/(HF) = 6.7 x 10^-4
For Ka to be a very small number (6.7 x 10^-4), it means that the numerator (which is [H][F]) is smaller than the denominator (which is [HF]). The only way you can get a number less than 1 for Ka (and 10^-4 is <1) is for the numerator to be smaller than the denominator.
I hope this helps.