Posted by Anonymous on Wednesday, June 30, 2010 at 2:28am.
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.
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?
• Chem - DrBob222, Tuesday, June 29, 2010 at 4:30pm
• 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 = ??
Hi, I am still confuse on how to work this problem out
So I have 0.2 NaOH
Can I convert it to CH3COOH?
CH3COOH + OH^- --> H2O + CH3COO^-
0.2 OH^- = 0.2 CH3COO^-
Then, 2.7 mmol would be how much CH3COO produced?
Total Volume: 52 mL
[CH3COO]= 2.7/52= ....M
[CH3COOH] = 0.2/52=....M
I would like to know if I am going in the right direction in this problem, or if I'm doing it completely wrong. Thanks.
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