Calculating pH and [H3O+] after dilution

[Bohrok]

A solution contains 0.200 mol HC₂H₃O₂, 0.200 mol NaC₂H₃O₂, and enough water to make 1.00 L of solution.
a) Calculate the hydronium ion concentration.
b) Calculate the new hydronium ion concentration after the solution is diluted to 2.00 L

For a), I know the pH is just the same as the pK_a of the acetic acid using the Henderson-Hasselbalch equation since log(^[0.200]/_[0.200) = 0, and the hydronium concentration is easy to find next. For b), I think the new concentration is just half of a) since adding more water shouldn't affect anything else other than the volume, correct?

However, if the 0.200 mol HC₂H₃O₂ and 0.200 mol NaC₂H₃O₂ were added to enough water to make 2.00 L of solution instead of being diluted from 1.00 L to 2.00 L later, the pH would be the same as in part a) since log(^[0.100]/_[0.100]) = 0.

I was helping a student with this problem; I can't see an error in my thinking, but I can't figure out how to explain the apparent difference between starting with 1.00 L and diluting to 2.00 L as opposed to starting with 2.00 L.

 

[Borek]

Buffer pH is - to a large extent - independent of dilution, so answer to b is identical to a. See Henderson-Hasselbalch equation - note that in the ratio of concentrations volume cancels out and you are left with ratio of number of moles of acid and conjugate base.