Posted by Audrey on Tuesday, November 23, 2010 at 11:44am.
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.
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