first from the Ka value find the Kb
Ka x Kb = 1.0 x 10^-14
Kb = 1.0 x 10^-14 / Ka = 1.0 x 10^-14 / 3.0 x 10^-8Â Â
= 3.3 x 10^-7
now construct ICE table
    NaOCl (aq) + H2O (l) <----> HOCl
(aq) + NaOH (aq)
IÂ Â Â
0.05Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0
CÂ Â
-x                                                   Â
+x               Â
+x
E 0.05 -
x                                             Â
+x              Â
+x
Kb = [HOCl] [NaOH] / [NaOCl]
3.3 x 10^-7 = [x] [x] / [0.05-x]
x^2 + x 3.3 * 10^-7 - 1.65 * 10^-8 = 0
solve the quadratic rquation
x = 0.0001283
from ICE table equilibrium concentration of NaOH = x =
0.0001283
[OH-] = 0.0001283 or 1.28 x 10^-3 M
