Thank you SO much for clearing that up!
One last point, the concentrations of acid and alkali here are equal, but if they weren't, i.e. if the concentration of NaOH was 2 mol dm
-3 while 100 cm
3 of a 1 mol dm
-3 ethanoic acid was used... would this be the volume of alkali then:
One-third of the acid would still react with the alkali, since the acid-alkali mol ratio is still 1:1, thus, one-third of 100 cm
3 = 33.33 cm
3.
Thus, the mol of acid that reacts with the alkali in 33.33 cm
3 of acid is:
33.33 x 1 / 1000 = 0.03333 mol
The same mol of alkali reacts. Thus, in 2 mol dm
-3 of alkali, the volume that reacts with the acid is:
0.03333 x 2 / 1000 = 16.665 cm
3?