What is the actual concentration of the molecular form of HF in a 1.00 M HF solution given that K_a of HF is 6.8 $\times$ 10^-⁴?

Question

Quizplus · Accepted Answer

The actual concentration of the molecular form of HF can be calculated using the equation for the ionization constant of HF: 
K = [H+][F-]/[HF]
Assuming that x is the concentration of the molecular form of HF that ionizes, then the concentration of H+ and F- are both x, since HF is a strong acid. 
Therefore, 
K = x^2 / (1 - x) = 6.8 x 10^-4
Solving for x gives x = 0.03103 M, which is the concentration of the molecular form of HF. 
However, this represents only the ionization of a small fraction of the HF in solution, and so the actual concentration of HF in the 1.00 M solution is (1.00 - 0.03103) M = 0.96897 M. Therefore the best choice is B.

What Is the Actual Concentration of the Molecular Form of HF