J.R. S. answered 03/25/23
Ph.D. University Professor with 10+ years Tutoring Experience
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