Given :
Ka of HClO = 4.0 E-8
[NaClO] = 0.017 M
We know
[NaClO]= [ClO-] ….(salt dissociates completely)
Reaction of ClO- with water
          Â
ClO- (aq)Â Â + H2O (l)Â Â Â ---- > HClO
(aq)Â Â Â + OH- (aq)
I Â Â Â Â Â Â Â Â Â 0.017
                                    Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0
CÂ Â Â Â Â Â Â Â
-x                               Â
  +x
                          Â
+x
EÂ Â Â Â Â (0.017-x)
         Â
                      x                      Â
x
kb = 1.0 E-14 / ka
= 1.0 E-14 / 4.0 E-8
= 2.5 E -7
Les find x
2.5 E-7 = x2 / (0.017-x)
2.5 E-7 x 0.017 =
x2Â Â Â Â Â Â Â Â (By
using 5 % approximation)
x = 6.52 E-5
x = [OH-] = 6.52 E-5
pOH = - log ([OH-]) = -log ( 6.52 E-5)
= 4.19
pH = 14 – pOH = 14- 4.19 = 9.81
pH = 9.81
