Posted by Anonymous on Tuesday, June 29, 2010 at 4:02pm.
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 = ??
Related Questions
Chemistry - Calculate the pH at the equivalence point of 25.0 mL of a 0.100 M ...
Chem - What is the pH of the resulting solution if 30.00 mL of 0.100M acetic ...
Chemistry - 25.0 mL of 0.100 M acetic acid (Ka= 1.8 x 10^-5) is titrated with 0....
Chemistry - Consider a weak acid-strong base titration in which 25.0 mL of 0.100...
chem-please help!!!!! - when a 25.0 mL sample of an unknown acid was titrated ...
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--please help me!! - when a 25.0 mL sample of an unknown acid was titrated ...
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...
For Further Reading