What about that ratio in the Henderson-Hasselbalch equation?

A few people have asked about some buffer questions that use the Henderson-Hasselbalch equation. In the proper Henderson-Hasselbalch equation derived from Ka, the pH and pKa are related by "log of the concentration of conjugate base over concentration of conjugate acid". Great, that works. When we're actually using the H-H equation, we can get away with a little shortcut in the calculation BECAUSE we're talking about a single buffer solution that has some conjugate acid and some conjugate base. Here's an example from an old exam:

The set-up calls for concentration {implied by the square brackets}, but when I worked through the calculation, I used moles. THAT'S CHEATING!!!! Well, not really... If we wanted to convert both of those moles to molarity, we'd divide each of them by 0.5000L, and you can go ahead and do that if you like, BUT since we're dividing the numerator and the denominator by the same volume, those volumes cancel out (mathematically) and give the same ratio as when we use moles directly. Let's look at a simpler example of this using simple whole numbers so I don't have to find a calculator... Let's say we have 2 liters of buffer solution that contains 4 moles of conjugate base and 2 moles of conjugate acid. The concentration of conjugate base is (4moles/2L = 2M) and the concentration of conjugate acid is (2moles/2L = 1M), so the ratio of conjugate base concentration to conjugate acid concentration is (2M/1M = 2). If I just use moles, the ratio is (4moles/2moles = 2). Same ratio.
So when we're using the Henderson-Hasselbalch equation correctly for buffer calculations, we can often just use the ratio of moles rather than the ratio of concentrations because it gives the same answer.