Question

what is the molarity of the acetic acid solution if 29.7 mL of a 0.205 M KOH solution is required to titrate 25.0 mL of a solution of HC2H3O2?

HC₂H₃O₂(aq)+KOH(aq)--->H₂O(l)+KC₂H₃O₂(aq)

Answer 1

: firstly have to convert mL to cm3, but since 1 ml = 1 cm3, its the same.
concentration also has to be in mol/dm3 not M, but again 1 M = 1 mol/dm3 so its the same.

29.7 ml = 29.7 cm3

Now we have to find the number of moles of KOH using:

moles (KOH) = volume x concentration/ 1000 = 29.7 x 0.205/ 1000 = 6.09 x10^-3

since 1 mole KOH reacted with 1 mole HC2H3O2; moles HC2H3O2 = 6.09 x 10^-3

concentration HC2H3O2 = moles/ volume x1000 = 6.09 x 10^-3/ 25.0 x1000 = 0.244 mol/dm3

0.244 mol/dm3 = 0.244 M