use:
Kb = Kw/Ka
Kw is dissociation constant of water whose value is 1.0*10^-14 at
25 oC
Kb = (1.0*10^-14)/Ka
Kb = (1.0*10^-14)/4*10^-8
Kb = 2.5*10^-7
ClO- dissociates as
ClO-Â Â Â Â Â Â Â + H2OÂ Â
----->Â Â Â Â HClO +Â Â OH-
0.037Â 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 0
0.037-x                    Â
x        x
Kb = [HClO][OH-]/[ClO-]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((2.5*10^-7)*3.7*10^-2) = 9.618*10^-5
since c is much greater than x, our assumption is correct
so, x = 9.618*10^-5 M
use:
pOH = -log [OH-]
= -log (9.618*10^-5)
= 4.0169
use:
PH = 14 - pOH
= 14 - 4.0169
= 9.9831
Answer: 9.98
