Calculate the percent ionization of hydrofluoric acid at 0.60 M:?

Answer 1

pKa=3.15

.. .. .. .. .. .. HF <=> H+ +F-
Initial .. .. .. C
Dissociate x
Produce .. .. .. .. .. .. x.. ..x
At Equil. .. C-x .. .. .. x .. ..x

Ka= [H+][F-]/[HF] =x^2/(C-x) (1)

The percent ionization is equal to
a=mole dissociated / mole initial = x/C => x=aC (2)

Substitute x in (1) using (2)
Ka = a^2C^2 /(C-aC) =a^2C/(1-a) =>
C*a^2=Ka-aKa=>
C*a^2+ (Ka)a- Ka=0
C=0.60
Ka=10^-pKa =10^-3.15=7.08*10^-4
so we have

0.6 a^2 + (7.08*10^-4) a -7.08*10^-4=0
solving the quadratic
a= 0.033766 = 3.4%

Answer 2

no, you calculate it