Question

What is the pH of a 0.10 M solution of KClO, Ka (HClO) = 3.0x10^8

Answer 1

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*10^-8

Kb = 3.333*10^-7

ClO- dissociates as

ClO- + H2O -----> HClO + OH-

0.1 0 0

0.1-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 ((3.333*10^-7)*0.1) = 1.826*10^-4

since c is much greater than x, our assumption is correct

so, x = 1.826*10^-4 M

use:

pOH = -log [OH-]

= -log (1.826*10^-4)

= 3.7386

use:

PH = 14 - pOH

= 14 - 3.7386

= 10.2614

Answer: 10.26