Question

A 35.0 mL sample of 0.150 M acetic acid is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added:
a) 0mL, b) 17.5 mL, c) 34.5 mL, d) 35 mL

I've figured out a,b, and c. But at 35 mL the moles become equal so when I plug the numbers into Henderson Hasselbach equation I get log to be 1 which makes the pH be 4.74 which is obviously not right. What am I doing wrong?

Answer 1

I don't think you can use the HH equation for the equivalence point. You have the base (acetate concn) but there is no acid and a number/0 is undefined.
CH3COOH + NaOH ==> CH3COONa + H2O

What you do is the pH at the equivalence point is due to the hydrolysis of th salt.
CH3COO^- + HOH ==> CH3COOH + OH^-

Kb = (Kw/Ka) = (CH3COOH)(OH^-)/(CH3COO^-)
You let x = CH3COOH and x = OH^-, you know Kw, Ka, and acetate, solve for x and convert OH^- to pH.

Answer 2

35.0 mL sample of 0.150 M acetic acid (HC2H3O2) (Ka = 1.8 x 10-5) is titrated with 0.150 M NaOH solution. Calculate the pH after 17.5 mL volumes of base have been added

Answer 3

47 mL