Question

Calculate the pH at 25C of a 0.92 M solution of sodium hypochlorite (NaClO). Note that hypochlorous acid (HClO) is a weak acid with a pKa of 7.50 . Round your answer to 1 decimal place.

Answer 1

use:

pKa = -log Ka

7.5 = -log Ka

Ka = 3.162*10^-8

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)/3.162*10^-8

Kb = 3.162*10^-7

ClO- dissociates as

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

0.92 0 0

0.92-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 ((3.162*10^-7)*0.92) = 5.394*10^-4

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

so, x = 5.394*10^-4 M

use:

pOH = -log [OH-]

= -log (5.394*10^-4)

= 3.2681

use:

PH = 14 - pOH

= 14 - 3.2681

= 10.7319

Answer: 10.7