chemistry
posted by Audrey on .
what is the pH of a solution that is .15 M in HOCl and .25 M NaOCl after .05 mol HCl/L has been bubbled into the solution?
this is what I did:
HOCl + H2O > H3O+ + OCl
.15 .25
.05 +.05

.1 .30
3.5E8=[H3O][.3]/[.1]
H3O= 1.167E8
pH= log(1.67E8)
pH=7.78
But the answer was 7.46. What did I do wrong?

When you add 0.05M H^+, that decreases NaOCl by 0.05 (to make it 0.20M), and you increase HOCl by 0.05 (to make it 0.20M) so you have
3.5E8=(H^+)(0.20)/(0.20)
3.5E8 = (H^+) and pH = 7.46.
Do you use the HendersonHasselbalch equation. That makes buffer problems a lot faster to work.
pH = pKa + log[(base)/(acid)]
pH = 7.46 + log (0.20/0.20)
pH = 7.46.