Calculate the pH during the titration of 20.00 mL of 0.1000 M CH3CH2CH2COOH with 0.1000 M NaOH after 26.5 mL of the base have been added. Ka of butanoic acid = 1.54 x 10-5.

Answer 1

The secret to these titration problems is to know where you are on the titration curve.
Let's call the acid HBu.
HBu + NaOH ==> NaBu + H2O

mmoles HBu = M x L = 20.00 mL x 0.1 = 2 mmoles.
mmoles NaOH = M x L = 26.5 mL x 0.1 = 2.65.
It should be obvious that all of the HBu acid has been neutralized, you have passed the equivalence point and the pH is determined by the excess OH^- in the solution.
So how much past? 2.65 mmoles - 2.00 mmoles = 0.65 mmoles and the concn OH^- = 0.65 mmoles/(20 + 26.5)mL = ??
Then convert to pOH and to pH.