17. What is the % ionization for a 0.400M solution of
HN
3
(hydrozoic acid)?
K
a
= 1.9 x 10
-5
Balanced equation:
HN
3
(aq)
H
+
(aq) + N
3
-
(aq)
I.C.E. grid:
HN
3
(aq)
H
+
(aq)
N
3
-
(aq)
INITIAL
0.400M
0M
0M
CHANGE
-xM
+xM
+xM
EQUILIBRIUM
(0.400M-x)M
xM
xM
K
a
= 1.9 x
10
-5
= {x
2
} / {(0.400-x)}
Now what must we assume about
x
?
a)
x
<< 1.9 x
10
-5
b)
x
<<
x
2
c)
x
>> 0.400
d)
x
<< 0.400
