Posted by Kevin on Tuesday, October 25, 2011 at 1:22am.
100 mL HAc x 0.1M = 10 mmoles.
50 mL NaOH x 0.1M = 5 mmoles.
.........HAc + OH^- => Ac^- + H2O
begin....10....0........0.......0
add............5.0.................
change..-5.0...-5.0.....+5.0....+5.0
equil....5.0.....0.......5.0
Then I would use the Henderson-Hasselbalch equation to solve for pH.
Since (base) = (acid), then log base/acid = log 1 = 0 so pH = pKa or about 4.74
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