34.09 ml of 0.251 M KOH is required to tritate a 10.2951 g sample of acetic acid solution. The density of the solution had been previously found to be 1.007g/ml. what is the molarity of the aetic acid solution?.

Answer 1

MolarityAcid*VolumeAcid=MolarityBase*Volumebase

MolarityAcit=.251* 34,08/volumeacid

but volume acid=massacid/density

And, there you have it.

molarityAcit=.251* 34.08/(10.2951/1.007)