Question

HClO is a weak acid (Ka = 4.0 × 10–8) and so the salt NaClO acts as a weak base. What is the pH of a solution that is 0.011 M in NaClO at 25 °C?

Answer 1

calculate Kb for ClO-
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-        + H2O   ----->     HClO +   OH-
0.011                        0         0
0.011-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 ((2.5*10^-7)*1.1*10^-2) = 5.244*10^-5

since c is much greater than x, our assumption is correct
so, x = 5.244*10^-5 M



use:
pOH = -log [OH-]
= -log (5.244*10^-5)
= 4.28


use:
PH = 14 - pOH
= 14 - 4.28
= 9.72
Answer: 9.72