Answer:
pH = 7.49
Step-by-step explanation:
This is a buffer solution. We can apply Henderson Hasselbach equation:
pH = pKa + log ((total mmoles - mmoles HCl) / mmoles HCl)
We have the same amount of mmoles.
35 mL . 0.400 M = 14 mmoles HCl
35 mL . 0.400 M = 14 mmoles NaClO
Total moles are 28 mmoles.
We replace data:
pH = pKa + log ((28 mmoles -14 mmoles) / 14 mmoles)
Notice that the relation in log = 1. So pH = pKa
pKa for HClO is 7.49
When pH = pKa we have the same amount of base and acid. Notice, that we have the same mmoles of HCl and NaClO.
Remember that a buffer can be prepared with:
a) A weak acid and its conjugate base (acetic /acetate)
A weak base and its conjugate acid (ammonia/ammonium)
b) A weak acid with a strong base, where the acid can be in excess.
c) A conjugate base (salt from weak acid) with a strong acid, where the base can be in excess. (This case but, the base is not in excess, that's why ph = pKa)