Hello.
I done an experiment where I titrated NaOH into CH3COOH.
I started by putting 0.025L of 0.1M CH3COOH into a beaker, I then added 0.025L of deionized water to this, the initial reading of pH was 3.27. I added NaOH until the pH was 12(0.046L NaOH ) in total.
I want to work out what the pH is at the equivalence point.
Can you tell me how to do this please?
is CH3COOH M0.05 now as 0.025L of water diluted it?
I have this done so far....
n(CH3COOH)=0.025 x 0.05/1=1.25x10-3mol
and
n(OH-)= 0.046 x 0.1/1= 4.6 x 10-3 mol
How can I calculate the pH at the equivalence point please?
Thanks