Well, it's been many many years since i've even attempted a gen chem problem so this might be completely wrong. Here's what i did.
You know you want the pH = 2 AFTER you add the HCl. Therefore, using the definition pH you get pH = -log [H+] --> 2 = -log [H+]. 10^-2 = .01 = [H+]. So you want the molarity of H+ to equal .01. Since HCl and NaOH are in a one to one ratio it will take 20mL of HCl before the solution is neutral so anything you add after 20ml will contribute to the H+ concentration (adding it before will just neutralize into water).
Now you take the definition of molarity
Number of moles (H+) / total volume of solution = .01
Let x = amount of HCl added.
(x - .020) / (x + .020) = .01
(exess moles of H+ in solution) / (total volume in solution) = .01
Solving that for x you get x = .0204L or 20.4mL. So 0.4mL of H+ needs to be in excess.
You can test the answer by doing -log ( .0004 moles of H+/ (.0404L) = 2.00.
Hope this Helps and hopefully if i did something wrong (it's been forever since i had chem) someone will catch it. GL!
