16. 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

Now we substitute the equilibrium concentrations into the Ka expression. Which of the following represents the correct substitution?

a) 1.9 x 10-5 = {(0.400-x)} / {x2}
b) (0.400-x) = {x2} / {1.9 x 10-5}
c) 1.9 x 10-5 = {x2} / {(0.400-x)}
d) x2 = {(0.400-x)} / {1.9 x 10-5}