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
3.5E-8=(H^+)(0.20)/(0.20)
3.5E-8 = (H^+) and pH = 7.46.
Do you use the Henderson-Hasselbalch 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.
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
3.5E-8=[H3O][.3]/[.1]
H3O= 1.167E-8
pH= -log(1.67E-8)
pH=7.78
But the answer was 7.46. What did I do wrong?
1 answer