Question

A. The dissociation constant of the weak acid HF is K₂ = [H][F]. If at equilibrium

Ka
[HF]
the concentrations of H* and F are each 2.6 x 10-3 M and the concentration of
HF is 0.010 M, what is K₂ for HF? (3 points)

Answer 1

6.76*10^-4 , or 6.8*10^-4 if you consider sig figs

Step-by-step explanation:

This is a bit of a confusing question, since it incorrectly states that the acid disassociation constant is simply [H+][F-].
Ka = [H+][F-]/[HF] , so plugging in the values, we get

`K=([2.6*10^(-3)][2.6*10^(-3)])/([0.01])\\K=([2.6*10^(-3)]^2)/([0.01])\\K=0.000676=6.76*10^(-4)`

If we were to use the equation in the question, Ka=[H+][F-], we would get the answer 6.76 * 10^-6 , which is not the Ka for HF