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 0.090 M in NaClO at 25 °C? pH=

Answer 1

Looking at the hydrolysis of the conjugate base (ClO^-), we have...

ClO^- + H₂O ==> HClO + OH^-

Since ClO^- is acting as a base, we can write the Kb expression as follows:

Kb = [HClO][OH^-] / [ClO^-].

To find the Kb, we use KaKb = Kw = 1x10^-14

Kb = 1x10^-14 / 4.0x10^-8

Kb = 2.5x10^-7

Substituting in the Kb expression, we have...

2.5x10^-7 = (x)(x) / 0.090 - x (and assume x is small relative to 0.09 and ignore it)...

2.5x10-7 = x² / 0.09

x = 1.5x10^-4 M = [HClO] = [OH^-] and this is less than 1% of 0.09 so above assumption was valid.

Now we have the [OH^-], so we can find the pH...

pOH = -log [OH-] = -log 1.5x10^-4

pOH = 3.82

pH = 14 - pOH

pH = 10.18