Question

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.

Answer 1

Six problems in here. What do you know about any of these?

Answer 2

nothing

Answer 3

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
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