Monoprotic Titration Curve Question?

I think you might have combined a couple problems here. If you are titrating HCl with NaOH, this is a strong acid/strong base titration. HCl does not have a pKa = 4.75. That would indicate a weak acid. Acetic acid is a weak acid and does have a pKa of 4.75, and I'll assume that.

Volume 1 (0 mL)

You have to use the expression for Ka for the weak acid and its ionization.

HAc <--> H+ + Ac-

Ka = [H+][Ac-]/[HAc] = 10^-4.75 = 1.8X10^-5

Let [H+] = [Ac-], and [HAc] = 0.01585 M. Then,

1.8X10^-5 = x^2 / 0.01585

x = [H+] = 5.3X10^-4

pH = -log (5.3X10^-4) = 3.27

Volume 2 (1/2 Ve):

At this point, the acetic acid is half neutralized, and [HAc] = [Ac-], Then,

Ka = [H+] = 1.8X10^-5

pH = 4.75

Volume 3 (Ve)

At the equivalence point, you have a solution of just Ac- ions. These act as a base by the equation:

Ac- + H2O <--> HAc + OH-

Kb = [HAc][OH-] / [Ac-] = Kw / Ka = 1X10^-14 / 1.8X10^-5 = 5.6X10^-10

Let [OH-] = [HAc] and [Ac-] = 0.01585. Then

5.6X10^-10 = x^2 / 0.01585

x = [OH-] = 2.98X10^-6

pOH = 5.52

pH = 14 - 5.52 = 8.44

Volume 4: 2 Ve

You have added an additional 0.01585 moles of OH- in (ignoring dilutions) 100 mL

[OH-] = 0.01585 mol / 0.1 L = 0.1585 M

pOH = 0.80

pH = 13.20

Hope that helps....