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 0.020 M in NaClO at 25 °C? pH=

Answer 1

Since NaClO is acting as a weak base, we need to look at the hydrolysis of this weak base, and then use Kb to find [OH^-], and from that, we can calculate the pH.

NaClO + H2O ==> NaOH + HClO .. hydrolysis reaction

ClO^- + H2O ==> OH^- + HClO .. hydrolysis reaction (Na+ as a spectator ion)

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

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

Kb = 2.5x10^-7

2.5x10^-7 = (x)(x) / 0.020 - x (assume x is small relative to 0.02 and ignore it in the denominator)

2.5x10^-7 = x² / 0.020

x² = 5x10^-9

x = 7.07x10^-5 (note: this is insignificant relative to 0.02 so above assumption was valid)

This is the [OH^-] and so we can now find pOH and then pH:

pOH = -log [OH^-] = -log 7.07x10^-5

pOH = 4.15

pH = 14 - pOH = 14 - 4.15

pH = 9.85