Chemistry Forum

Have you already been presented to the theory of "third medium" of a burette on a titration? I'm not sure but I don't think this makes sense. Some teachers say the volume of solution from the substance must be at the third medium of a burette after a titration (complete neutralizarion, complete oxirreduction etc). I am not sure but it doesnt make sense to me.

What should be this third Medium. The titration solution in the burette is dissolved in water normaly, like NaOH, HCl, KMnO₄, etc. And the substance to determine in the Erlenmeyer flask or beaker glas as well.

I think you will understand better now what I meant

http://i.ibb.co/55s0w7r/A₁₂₁₃-A₆₄-343-A ... 226536.jpg

Consider the burette is divided into equal parts.
See that I demarcated the third medium. The teachers say that after a titration has reached the end point, the volume should be at the third medium of the burette. Some say this occurs because of the liquid adhesion to glass, but I don't think it matters. Also all the burette have margin error to consider, so it matters even less to me.

Remove the spaces between the link of image

Simple No. I can't figure out what your drawing should be.

Hmm I'll try to explain just with words then.
Consider a burette of 50 mL. Divide it into 3:
50/3 = 16,67

So we have 3 parts of 16,67 mL

We have part 1 at the bottom of the burette, part 2 at the medium and part 3 at the top.

If you use 20 mL of reagent, you "fall" into the third medium (30 mL) of the burette which ranges from 16,67 and 33,33.

So on the standardization of NaOH, we have NaOH in the burette and HCl on the erlenmeyer flask
The unknown (actual) concentration of NaOH must be determined.
mol OH- = mol H+
x mol/L.20,00mL = 0,1 mol/L.25mL hcl aliq

So the actual concentration of NaOH would be 0,125 mol/L and with more precision. But I don't think it makes sense.

So on the standardization of NaOH, we have NaOH in the burette and HCl on the erlenmeyer flask
The unknown (actual) concentration of NaOH must be determined.
mol OH- = mol H+
x mol/L.20,00mL = 0,1 mol/L.25mL hcl aliq

In this case its the opposit. The known concentration HCl is in the burette the unknown NaOH in the beaker.

The Medium discussion is nonsense, because the concentration is in the whole burette from top over medium to bottom the same.