construct the ICE table
   NaOCl (aq) + H2O (l) <====> HOCl (aq) +
NaOH (aq)
IÂ Â Â
0.25Â 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
0
CÂ Â Â
-x                                              Â
+x             Â
+x
E
0.25-x                                           Â
+x            Â
+x
Kb = [HOCl] [NaOH] / [NaOCl]
3.4 x 10-7 = [x] [x] / [0.25-x]
x2 + x 3.4 * 10-7 - 8.5 x 10-8
= 0
solve the quadratic equation
x = 2.91 x 10-4 M = [NaOH]
pOH = -log(OH-) = -log(2.91 x 10-4) = 3.53
pH + pOH = 14
pH = 14-pOH = 14-3.53 = 10.46M
