posted by Anonymous on .
Calculate the pH at the equivalence point of 25.0 mL of a 0.100 M acetic acid, CH3COOH, titrated with a 0.100 M sodium hydroxide, NaOH. Ka of acetic acid is 1x10^-5
You must recognize what you have at the equivalence point.
CH3COOH + NaOH ==> CH3COONa + H2O.
At the equivalence point, we have converted all of the CH3COOH to CH3COONa (sodium acetate) so we have M x L = 0.1 x 0.025 = 0.0025 moles sodium acetate. That is in a volume of 50 mL (the other 25 came from 25 mL of 0.1 M NaOH). The pH will be determined by the hydrolysis of water. If we call acetate, Ac^-, and acetic acid HAc, then
Ac^- + HOH ==> HAc + OH^-
Kb for acetate = (Kw/Ka) = (HAc)(OH^-)/(Ac^-).
Set up an ICE chart and solve.
Post your work if you get stuck.