[EdwinD]

First off, I guess I should introduce myself. My name is Edwin and I live in DC. I came here because I need help with some homework and my friends weren't able to explain it to me.

The question I have is the following:
I have a 1 Molar solution of benzoic acid (pKa=4.19) dissolved .1L of water. Titration is with .1 Molar NaOH. The excercise is to calculate the pH of the solution when 0ml, 5 ml, 10ml and 20ml of NaOH is added.

I was able to calculate the initial pH with the equation (sorry, I don't know the name of it) pH=[pKa -log(c₀(HA)]/2 , which is 2.095 .
I wasen't sure about using the Henderson-Hasselbalch eq. to calculate the others because I wasn't sure what c(A^-) was in these cases. Since benzoic acid is a weak acid and NaOH is a strong base, does this mean that the mols of base added to the solution will be equal to the mols of deprotonized acid? Also, does c(HA) in the HH eq. mean the initial concentration of acid in the solution (versus c₀(HA)?)?

Thanks in advance!

[Borek]

First of all - take a look at pH calculation lectures.

Since benzoic acid is a weak acid and NaOH is a strong base, does this mean that the mols of base added to the solution will be equal to the mols of deprotonized acid?

Yep.

Also, does c(HA) in the HH eq. mean the initial concentration of acid in the solution (versus c₀(HA)?)?

No, it means equilibrium concentration. Hint: total amount of benzoic acid doesn't change, and you already know [A^-].