[Discord120]

I titrated vinegar with 0.1M NaOH to find the equivalence point. I graphically determined the equivalence point to occur 36.39mL, which means that at 18.195mL, I find the pKa of my vinegar sample, which was 4.59, which I don't matters in the question I'm about to ask.

From this, I'm supposed to find the concentration of acetic acid in the original solution. In the procedure, I had 5mL of the acetic acid and diluted it with 20mL CO2 free water for 25mL total.

To find the concentration of acetic acid in the original vinegar solution, I believe that I can just use M1V1 (acetic acid)= M2V2 (NaOH).

My question is should I be using (x)(25mL) = (0.1M)(18.195mL) to find the original concentration of acetic acid in the original vinegar solution, or should i replace 25mL with the 5mL?

[Borek]

5 mL for the concentration in the original sample.

It is all about mass conservation. Amount of acid didn't change, its concentration did when you diluted. M₂V₂ is a number of moles of NaOH - and it is also a number of moles of acetic acid, as they react 1:1. You have number of moles - just use concentration definition to calculate concentration in the initial volume.