# Find the pH during the titration of 20.00 mL of 0.2380 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54 10-5), with 0.2380 M NaOH solution after the following additions of titrant.

(a) 0 mL
Ive got the answer of 2.72
(b) 10.00 mL
Ive got the answer of 4.81
(c) 15.00 mL

(d) 19.00 mL

(e) 19.95 mL

(f) 20.00 mL

(g) 20.05 mL

(h) 25.00 mL

c,d,e are all done essentially the same way. I will go through c.
For simplicity in typing, I will call butanoic acid a simple HB. Then
butanoic acid + KOH = potassium salt of butanoic acid + H2O OR in simple terms:
HB + KOH ==> KB + H2O
mols HB at beginning of titration are L x M = 0.020 x 0.2380 = ??
mols KOH added = (for c) 0.015 x 0.2380 = ??
Subtract from HB to see how much HB is left. Add to KB to see how much B^- is there. Plug into the Henderson-Hasselbalch equation and solve for pH.
d and e are done essentially the same way.
For must recognize that 20.00 mL KOH is the EXACT equivalence point; therefore, the pH is determine SOLELY by the amount of salt (KB) present and that is determined by the hydrolysis of the salt. Using the Bronsted-Lowry theory of acids/bases, write the hydrolysis of the anion, B^-.
B^- + HOH ==> HB + OH^-
Post back if you don't know what to do with that.

For the others AFTER the equivalence point, these are just excess KOH so you need to calculate how much excess KOH is present, the total volume of the solution, and the molarity of the KOH present at those points. Then use pH + pOH = pKw
I shall be happy to check your work if you post it.