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.
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?
__________________________
• 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
2.7-2.5=0.2
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.
No one has answered this question yet.
Answer this Question
Related Questions
Chem - 25.0 mL of 0.100 M acetic acid (Ka= 1.8 x 10^-5) is titrated with 0.100 M...
Chemistry - Calculate the pH at the equivalence point of 25.0 mL of a 0.100 M ...
Chemistry - Consider a weak acid-strong base titration in which 25.0 mL of 0.100...
Chemistry - 40.0 ml of an acetic acid of unknown concentration is titrated with ...
Chemistry - 40.0 ml of an acetic acid of unknown concentration is titrated with...
Chem - What is the pH of the resulting solution if 30.00 mL of 0.100M acetic ...
Chemistry - Greetings: I was wondering how to calculate the weight/volume % of ...
chemistry - A volume of 25.0 mL of 0.100 M CH3CO2H is titrated with 0.100 M NaOH...
Chemistry - A volume of 25.0 ml of 0.100 M CH3C02H is titrated with 0.100 M NaOH...
chem-please help!!!!! - when a 25.0 mL sample of an unknown acid was titrated ...
For Further Reading
