A 30.0 mL sample of 0.165 molL−1 propanoic acid (Ka=1.3×10−5) is titrated with 0.300 molL−1KOH.

Calculate the pH at 0, 5,10, at equivalence point, 1/2 of the equivalence point, 20, 25 mL of added base.

3 answers

Six problems in here. What do you know about any of these?
nothing
at 0mL base pH = 2.83
at 5mL base pH = 4.52
at 10mL base pH = 5.07
at equivalance point pH = 8.96
at half the equivalance point pH = 4.88
at 20mL base pH = 12.32
at 25mL base pH = 12.67
your welcome.