Asked by Anonymous
25.0 mL of 0.100 M acetic acid (Ka= 1.8 x 10^-5) is titrated with 0.100 M NaOH. Calculate the pH after the addtion of 27.00 mL of 0.100M NaOH.
my work
CH3COOH + H2O <-> H3O^+ + CH3COO^-
25mL x 0.100 mmol/ml = 2.5mmol CH3COOH
27mL x 0.100 mmol/ml = 2.7mmol NaOH
There are more base than acid.... so what do I do?
my work
CH3COOH + H2O <-> H3O^+ + CH3COO^-
25mL x 0.100 mmol/ml = 2.5mmol CH3COOH
27mL x 0.100 mmol/ml = 2.7mmol NaOH
There are more base than acid.... so what do I do?
Answers
Answered by
DrBob222
So the base neutralizes all of the acetic acid and one is left with 2.5 mmoles of the salt (sodium acetate) + an excess of (2.7-2.5 = 0.2 mmoles NaOH. The volume is 25 + 27 mL = ??
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