18. What is the % ionization for a 0.400M solution of HN3 (hydrozoic acid)? Ka = 1.9 x 10-5

Balanced equation: HN3 (aq) H+ (aq) + N3- (aq)

I.C.E. grid:

HN3 (aq) H+ (aq) N3- (aq)
INITIAL
0.400M
0M
0M
CHANGE
-xM
+xM
+xM
EQUILIBRIUM
(0.400M-x)M
xM
xM

Ka = 1.9 x 10-5 = {x2} / {(0.400-x)}
After solving for x, which is found to be 2.8 x 10-3M, how do we calculate the % ionization from the [H+]?
a) % ionization = {[H+]EQUILIBRIUM} / {[HN3]INITIAL} x 100
b) % ionization = {[HN3]INITIAL} / {[HN3] + [H+]EQUILIBRIUM} x 100
c) % ionization = {[HN3] + [H+]EQUILIBRIUM} / {[HN3]INITIAL} x 100
d) % ionization = {[H+]EQUILIBRIUM} / {[HN3] + [H+]EQUILIBRIUM} x 100