What is the pH of a 0298 M aqueous solution of potassium hyp

What is the pH of a 0.298 M aqueous solution of potassium hypochlorite, KCIO at 25 °C? (Ka for HCIO 3.5x103) pH =

Solution

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)/3.5*10^-8
Kb = 2.857*10^-7
ClO- dissociates as

ClO-        + H2O   ----->     HClO +   OH-
0.298                        0         0
0.298-x                      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 ((2.857*10^-7)*0.298) = 2.918*10^-4

since c is much greater than x, our assumption is correct
so, x = 2.918*10^-4 M



use:
pOH = -log [OH-]
= -log (2.918*10^-4)
= 3.53


use:
PH = 14 - pOH
= 14 - 3.53
= 10.47
Answer: 10.47