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 0.065 M in NaClO at 25 °C?

Answer 1

First calculate the value of Kb as follows:

Kb= Kw/ Ka

Kb= 1 x 10^-14 / Ka

Here Ka = 4.0 × 10–8
Kb = 1 x 10^-14 / 4.0 x 10^-8
Kb = 2.5 x 10^-7

NaClO= Na+ ClO-

ClO- + H2O = HClO+ OH-

Therefore the Kb for this reaction:
Kb = [HOCl][OH-] / [OCl-]

Here HOCl = OH-, and NaClO = ClO-= 0.065
Kb = [OH-]^2 / [OCl-]

2.5 x 10^-7 = [OH-]^2 / 0.065

[OH-]^2 = 1.625*10^-8

[OH-]=1.275 *10^-4

Now pOH = -log [OH-]= - log 1.275 *10^-4

=3.89

pH+pOH =14

pH 14- pOH=14-3.89

pH= 10.11