Just wondering, with the titration curve above, why is the graph so steep/vertical around the equivalence point (ph 8-11ish)?
Because the end point (the point at which the inducator changes colour, and theoretically, this occurs at the same time as the equivalence point, where reactants are in stoichiometrically equal proportions) is sharp. That is, there is a very precise and quick colour change, represented by the steepness.
When looking at back titrations, you notice that they're essentially done because the end point is broad; that is, there is no real quick & precise colour change of the indicator. In the circumstance of a broad end point, there is a nore diagonal slope which indicates the end point, and so to obtain accurate results (this is obtained with a sharp end point), you carry out a back titration. Hope this helped