Question

Ka for hypochlorous acid, HClO, is 3.0x10-8.Calculate the pH after 10.0, 20.0, 30.0, and 40.0 mL of 0.100MNaOH have been added to 40.0 mL of 0.100 M HClO. I am mostly confused on when 40.0 mL is added, because that will be the equivalence point I believe.

Answer 1

we know that

moles = molarity x volume (L)

so

at 40 ml

moles of HCl0 = 0.1 x 40 x 10-3 = 4 x 10-3

moles of NaoH = 0.1 x 40 x 10-3 = 4 x 10-3

now

the reaction is

HCl0 + NaOH ---> NaCl0 + H20

now

from the above reaction

moles of NaCl0 formed = moles of HCl0 reacted = moles of NaOH reacted

moles of NaCl0 formed = 4 x 10-3

at this point

all the acid and base are reacted

only NaCl0 an H20 are present

NaCl0 undergoes hydrolysis

Cl0- + H20 ---> HCl0 + OH-

so

NaCl0 is a weak base

for weak bases

[OH-] = sqrt ( Kb x C)

also

Kb = Kw / ka

so

Kb = 10-14 / 3 x 10-8

Kb = 3.33 x 10-7

now

final volume = 40 + 40 = 80 ml

now

conc of NaCl0 (C) = moles / volume (L)

= 4 x 10-3 / 80 x 10-3

= 0.05

conc (C) = 0.05 M

so

[OH-] = sqrt ( kb x C)

[OH-] = sqrt ( 3.33 x 10-7 x 0.05 )

[OH-] = 1.29 x 10-4

now

pOH = -log [OH-]

pOH = -log 1.29 x 10-4

pOH = 3.89

now

pH = 14 - pOH

pH = 14 - 3.89

pH = 10.11

so

the pH is 10.11