Question

What is the K_a of a 2.41 M aqueous acid solution that has a % dissociation of 2.4 %?

Answer 1

CONSIDER dissociation of an acid HA IN WATER

HA + H₂O   H₃O⁺ + A^-

INITIAL CONCENTRATION 2.41M 0 0

AT EQUILIBRIUM 2.41-2.4% of 2.41M 2.4%of 2.41M 2.4%of 2.41M

Acid dissociation constant K_a = [H₃O⁺][ A^- ] / [HA]

now  [H₃O⁺] = 2.4% of 2.41M

= (2.4/100) X2.41

=(2.4X2.41)/100

=5.784/100

[H₃O⁺] =5.784X10^-2M SO    [A^- ] ALSO 5.784X10^-2M  

CONCENTRATION OF UNDISSOCIATED ACID [HA] = 2.41M-5.784X10^-2M  

= 2.41M-0.05784M

=2.35216M

NOW K_a = [ (5.784X10^-2 )x (5.784X10^-2 )]/2.35216

=(33.454656 X 10^-4)/2.35216

=14.22295 X10^-4

^~ 14.223 X10^-4