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 0.037 M in NaClO at 25 °C?

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)/4*10^-8
Kb = 2.5*10^-7
ClO- dissociates as

ClO-        + H2O   ----->     HClO +   OH-
0.037                        0         0
0.037-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)*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