Find the pH during the titration of a 20.00mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka=1.54 x10^-5), with the addition of 10 mL of 0.1000 M NaOH solution??

Answer 1

Butanoic acid we'll call HB and you watch the significant figures.
millimols HB = 20 mL x 0.1 M = 2
mmols NaOH added = 10 mL x 0.1M = 1

..........HB + OH^-==> H2O + B^-
initial...2......0......0
add.............1...........
change....-1....-1......1.....1
equil.....1......0.......1....1.

So you have formed 1 mmol base and have left 1 mmols HB (this is a buffer) so use the Henderson-Hasselbalch equation to solve for pH.
pH = pKa + log(base)/(acid)
base = 1
acid = 1
pKa = -log Ka
Solve for pH. You should get 4.81