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When doing a titration with $\ce{HCl}$ and $\ce{NaOH}$, both being strong and dissociating completely, if you add a volume of $\ce{NaOH}$ that is less than needed to reach the equivalence point, you say all the $\ce{OH-}$ reacted with an equal amount of $\ce{H+}$ from $\ce{HCl}$.
It would be amount of substance $\ce{HCl}$ - amount of substance of $\ce{OH-}$ = $[\ce{H3O+}]$, and all of the $\ce{OH-}$ is gone, so $[\ce{OH-}] = 0$. But if you take the pH found using the $[\ce{H3O+}]$ value, use that to find pOH and then use that to find $[\ce{OH-}]$ you get a very small value, but not zero. So if I need to report a $[\ce{OH-}]$ value, do I just put 0 or the one gotten from pOH?

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The problem that you are facing comes from the fact that technically, not all of the $\ce {OH-}$ reacts with the $\ce {H+}$. It is only because the amount of OH- that doesn't react is so negligible compared to the amount that does, that the $\ce {[H_3O+]}$ can be taken to be $\ce{[HCl]-[$\ce{OH-}$]}$.

So, yes, to find the remaining (negligible) amount of $\ce{OH-}$, you must use the "very small value" given by either subtracting $\ce {pH}$ from $14 =\ce{pOH}$ or doing $\ce K_w/[\ce{H+}]=[\ce{OH-}]$ (which amounts to the same thing).

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