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 0.029 M in NaClO at 25 °C? pH=

Answer 1

Look at the hydrolysis of NaClO...

NaClO + H2O ==> HClO + NaOH

Since NaClO is acting as a base, we need to use the Kb which is calculated as follows:

KaKb = 1x10^-14

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

Kb = 2.5x10^-7

Now, write the Kb expression:

Kb = 2.5x10^-7 = [HClO][NaOH] / [NaClO]

Solve for [NaOH]:

2.5x10^-7 = (x)(x) / 0.029 - x (since Kb is so small, we can ignore x in demoninator)

2.5x10^-7 = (x)(x) / 0.029

x² = 7.25x10^-9

x = 8.51x10^-5 M (note: this is ~0.3% of 0.029 so above assumption was valid)

pOH = -log OH-

pOH = -log 8.51x10^-5

pOH = 4.07

pH = 14 - 4.07

pH = 9.93