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I'm a little confused on something: A buffer has a certain capacity and then runs out.
But, if the buffer is given by: $$\ce{A- + H2O <=> HA + OH-},$$ when the $\ce{A-}$ reacts with the water, the reaction shifts to the left. So how can it "run out"? If $Q$ just keeps becoming $0$, shouldn't the reaction keep shifting to maintain $K$? Does the right side, perhaps, approach $0$ as it keeps decreasing in concentration to maintain $K$, but it can never become $0$ molar, because that wouldn't maintain $K$?
I guess if $\ce{H2O}$ was "aqueous"/had a concentration and was therefore used in the $K$ expression, this the $K$ would always be maintained.
But how does the buffer "run out" of $\ce{A-}$ to react when $\ce{H2O}$ is liquid (not aqueous)? It doesn't make sense as wouldn't the reaction always adjust to make more $\ce{A-}$ to maintain $K$?

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If strong acid is added to a buffer system, conjugate base removes it. Eventually, the concentration of conjugate base falls to such a low level that its effect on pH can be neglected. At this point, the solution behaves more like a weak acid solution. If even more acid is added, the system behaves like a strong acid system in terms of calculating pH.

If strong base is added to a buffer system, conjugate acid removes it. Eventually the concentration of conjugate acid falls to such a low level that its effect on pH can be neglected. At this point. the solution behaves more like a weak base solution. If even more base is added, the system behaves like a strong base system in terms of calculating pH.

In the equation you wrote, $$\ce{A- + H2O <=> HA + OH-},$$ think about the effect of adding acid. Acid reacts with the $\ce{OH-}$, producing water and removing $\ce{A-}$ from solution. The $\ce{OH-}$ and $\ce{A-}$ cannot be removed from solution but their concentrations can be driven to negligible levels by the addition of sufficient quantities of acid.

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