... ethanoic acid to sodium hydroxide is 1:1 the moles for both are equal. NaOH = 0.02185 moles 4. Concentration = Moles Volume 0.1moldm-3= 0.02185 V Volume of NaOH = 0.02185 0.1 = 0.21854dm3 = 218.54cm3 5. Because we diluted the vinegar by the factor of 10 we will need to take this in consideration in our NaOH therefore: 218.54 10 = 0.021854dm3 = 21.854cm3 Therefore I predict that the volume of NaOH needed to titrate the ethanoic acid will be 0.021854dm3, which is the same as 21.854cm3. By calculating the mount of NaOH needed for the titration we will be able to foresee whether the titration will be feasible in a school laboratory. Looking at the volume of NaOH needed from our calculations we can see that such a titration can be carried out. Justification: The need to carry out a titration between sodium hydroxide and hydrochloric acid was because we ...

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