# Can you create a buffer with a strong acid?

So I was wading through several textbooks yesterday reviewing the concepts of buffers. I found them lot more challenging than I used to. One thing in particular disturbed me: The books and internet all said "Yeah, buffers work pretty great in as long as the $\rm pH \approx pK_a\pm 1$." But I am not one to leave these things alone. I finally found the Quantitative explanation I was looking for over at buffer capacity on Wikipedia.

However, the graph inside the link says the buffering capacity becomes great when you solution becomes very acidic or very basic, as well. And, of course, looking at titration curves there seem to be a few "buffer" zones e.g. where the solution is highly acidic or highly basic, the $\rm pH$ doesn't change as much after a certain point:

So, obviously there is something wrong with my reasoning, since nobody calls those areas buffer zones. Is it because it is not useful to have such a highly acid/basic solution? Or is it just because technically, there is no buffering, just small $\rm pH$ changes? Or maybe something else entirely...

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I totally agree with trb456. I'd like to add that what you're seeing in the figure you provided is actually a titration of water (that starts out at acidic pH). Remember that water is both an acid and a base. When you add NaOH as in your figure, that titration curve is what you see. That's why the inflection point occurs at pH = 7.0. You add enough NaOH to neutralize the nitric acid and get to pH 7, then as you add more NaOH you increase the concentration of hydroxide ion - increasing the pH. Also, remember that pH is calculated on a log scale, which helps explain the shape of these plots. –  Phillip Mar 4 '13 at 23:33

$\ce{HA <-> H+ + A-}$
where $\ce{HA}$ is a weak acid, and there is an excess of $\ce{A-}$, the conjugate base. Because $\ce{HA}$ is weak, it is a relative poor proton donor, but $\ce{A-}$ is a relatively good proton acceptor. So if you now add a strong acid, the $\ce{H+}$ it donates reacts with the $\ce{A-}$ to produce $\ce{HA}$, shifting the equilibrium to the left, rather than to the right as one might expect. Or perhaps, preventing as much shifting to the right as might be expected. This gets to your issue of buffering capacity.